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MECHANICS: ILLUSTRATIONS 1.
A particle moves along a path ABCD. Then the magnitude of net displacement of the particle from position A to D is: 10 m
(B)
5 2 m
(C)
9m
(D)
7 2 m
As can be seen 2
FD 7 2 m 2
A truck travelling due north at 50 km/hr turns west and travels at the
je
2.
AF
e.
The displacement is
in
Sol.
(A)
50 km/hr north-west
(B)
50 2 km/hr north-west
(C)
50 km/hr south-west
cr
ac k
(A)
(D) Sol.
iit
same speed. What is the in velocity.
50 2 km / hr south-west
ˆ Vi 50j Vf 50ˆi
V Vf Vi i 50 ˆj = 50 ˆ
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V
50
2
50
2
= 50 2
V 50 2 SW 3.
A boy start towards east with uniform speed 5m/s. After t = 2 second he turns right and travels 40 m with same speed. Again he turns right and travels for 8 second with same speed. Find out the displacement;
e.
The particle starts from point A & reaches point D passing through B & C
ac k
Displacement = AD
iit
DE = 30 m
je
Now, AE = 40 m
=
AE2 DE2
=
402 302
cr
Sol.
in
average speed, average velocity and total distance travelled.
= 5 cm
Time taken in the motion = tAB + tBC + tCD =2
40 8 8
= 18 sec
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Total distance travelled = AB + BC + CD = 10 + 40 + 40 = 90 m
25 m/s 9
=
90 18
cr
= 5 m/s
Distance time
ac k
Average speed =
e.
=
je
50 18
iit
=
Displacement time
in
Average velocity =
A radius vector of point A relative to the origin varies with time t as
r at ˆi bt2 ˆj where a and b are constants. Find the equation of point's trajectory.
r at ˆi bt2 ˆj X = at y = – bt2
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x
t= a
x y b a
b 2 x. a2
y 4.
2
A man moves on his motor bike with 54 km/h and then takes a u turn
in
(180°) and continues to move with e speed. The time of u turn is 10s.
e.
Find the magnitude of average acceleration during u turn. (B)
(C)
1.5 2 ms–2
je
0
(D)
3 ms–2
none of these
Vi 15 ˆi Vf 15ˆi
cr
Sol.
ac k
iit
(A)
V 15ˆi 15 ˆi 54 km/h = 54
5 18
= 15 m/s =
15 15 10
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= 3 m/s2 A particle whose speed is 50 m/s moves along the line from A(2, 1) to B(9, 25). Find its velocity vector in the from of aiˆ bˆj.
14ˆi 48ˆj i ˆj Position vector of point A = 2ˆ
e.
AB 9ˆi 25ˆj 2ˆi ˆj
in
Position vector of point B = 9ˆ i 25ˆj
je
= 7ˆ i 24ˆj
AB
7ˆi 24ˆj 25
cr
=
AB
ac k
AB
iit
Unit vector in the direction of AB
vector vector unit vector
V 50 AB i 48ˆj m / s = 14ˆ
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11.
ˆ What should be A particle is moving with initial velocity u ˆ i ˆj k. its acceleration so that it can remain moving in the same straight line without change of direction?
(B)
ˆ a 2ˆi 2ˆj 2k
(C)
ˆ a 3ˆi 3ˆj 2k
(D)
a 1ˆi 1ˆj
in
ˆ a 2ˆi 2ˆj 2k
e.
Sol.
(A)
To move in a straight line
iit
je
a || u
a 2u In B,
cr
a 2u
ac k
In A,
After sometime direction will get reversed of same straight. 12.
Two balls are moving on the same smooth horizontal plane. Their velocity components along one edge of the square plane are 10 3 & 20 m/s. Their velocity components along a perpendicular edge are 30 & 20 m/s. Find the angle between their directions of motion.
Sol.
VA 10
3 ˆi 30ˆj
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VB 10 2ˆi 2ˆj VA.VB
VA VB
100 2 3 6
10 3 9 10 2 2 2
3 3 2 6
=
3 2 2
cr
= 15°
je
=
3 2 2 33
e.
2
iit
cos
13.
ac k
=
in
cos
An insect starts from rest from point (3, 4) and moves with an acceleration 2 2 m/s2 in x – y plane along a line, equally inclined to both the axis. After 3 sec insect turns towards right in perpendicular direction without wasting any time and keeping speed same at the movement of turning. For the further motion acceleration is 2 2 m/sec2 in the direction of motion. Find the position of insect after 5 seconds from the starting.
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In 1st three seconds
ˆi ˆj 1 2 ˆ ˆ S 3i 4i 2 2 3 2 2
= 12ˆ i 13ˆj
V= 2
ˆi ˆj 2 3 2
in
i 6ˆj = 6ˆ
je
e.
v ' 6ˆi 6ˆj 2 2 ˆi ˆj a' =
ac k
iit
2
i 2ˆj = 2ˆ
In next 2 seconds
cr
Sol.
1 S 6ˆi 6 ˆj 2 2 ˆi ˆj 22 2 i 12ˆj 4ˆi 4ˆj = 12ˆ
i 16ˆj = 16ˆ
S' S S i 13ˆj 16ˆi 16ˆj = 12ˆ
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S' 28ˆi 3ˆj – 3ˆ i 4ˆj (A)
It speed of a body is varying, its velocity must be varying and it must have zero acceleration It velocity of a body is varying, its speed must be varying
(C)
A body moving with varying velocity may have constant speed
(D)
A body moving with varying speed may have constant velocity if its
If speed of a particle changes, the velocity of the particle definitely
je
Sol.
e.
direction of motion remains constant.
in
(B)
iit
changes and hence the acceleration of the particle is nonzero. Velocity of a particle change without change in speed. (In uniform
ac k
circular motion)
When speed of a particle varies, its velocity cannot be constant.
The position vector of a particle is given as r = (t2 – 4t + 6) ˆ i t2 ˆj.
cr
8.
The time after which the velocity vector and acceleration vector becomes perpendicular to each other is equal to
Sol.
(A)
1 sec
(B)
(C)
1.5 sec (D)
not possible
2 sec
r t2 4t 6 ˆi t2 ˆj; r-
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V
dr dt
= 2t 4 ˆ i 2t ˆj, a =
dv 2ˆi 2ˆj dt
If a and v are perpendicular
in
a.v 0
e.
2 2t 4 2 2t 0
t = 1 sec.
9.
ac k
t = 1 sec.
iit
je
4t 8 4t 0
The velocity time relation of an electron starting from rest is given by v
(A) (C) Sol.
a=
cr
= K t, where K = 2 m/s2. The distance traversed in 3 sec is : 9m
(B)
16m
27 m
(D)
36m
dv dt
= k = 2 m/s2 a = constant S= 0
1 2 2 3 2
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= 9 m. 10.
i 4ˆj (m/s) and an A particle has an initial velocity (i.e., at t = 0) of 3ˆ acceleration of 0.3ˆi + 0.4ˆj (m/s2). The speed of particle at t = 10 sec is :
7 m/s 2
(C)
7 2 m/s
(D)
14 2 m / s
in
(B)
je
v u at
ac k
when a is constant
iit
Sol.
7 m/s
e.
(A)
velocity at t = 10 sec.
cr
i 4ˆj 10 0.4ˆi 0.3ˆj is v 3ˆ
i 7ˆj = 7ˆ
v 7 2 m/s 14.
A particle P is moving with a constant speed or 6 m/s in a direction
ˆ When t = 0, P is at a point whose position vector is 2iˆ ˆj 2k. ˆ Find the position vector of the particle P after 4 ˆ 7k. 3iˆ 4j seconds.
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Sol.
V6
ˆ 2iˆ ˆj 2k 3
ˆ = 2 2iˆ ˆj 2k
ˆ is The equation of unit vector along 2iˆ ˆj 2k =
ˆ 2iˆ ˆj 2k 22 12 22
r r0 vt
je
e.
in
ˆ 2iˆ ˆj 2k = 3
ac k
iit
ˆ 4 2 2iˆ ˆj 2k ˆ ˆ 7k = 3iˆ 4j ˆ ˆ 23k r 19iˆ 4j
The velocity of a car moving on a straight road increases linearly
cr
15.
according to equation, v = a + bx, where a & b are positive constants. The acceleration in the course of such motion : (x is the distance travelled) (A)
increases
(B)
decreases
(C)
stay constant
(D)
becomes zero
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Sol.
V = a + bx (V increases as x increases)
dV dx b bV dt dt hence acceleration increases as V increases with x. 16.
The displacement of a body from a reference point is given by,
x 2t 3, where 'x' is in meters and t in seconds. This shows that
e.
3 2
rest at t =
(B)
is accelerated
(C)
is decelerated
(D)
is in uniform motion
ac k
iit
je
(A)
x = (2t – 3)
cr
Sol.
in
the body:
x = (2t – 3)2 V=
dx 2 2t 3 2 dt
= 4(2t – 3) = 86 – 12 V = 4(2t – 3) = 0 2t – 3 = 0
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t
3 sec 2
rest at t = 8 a = 8 m/s a= 17.
dV 8 m / s2 dt
The position of a particle moving rectilinearly is given by X = t3 – 3t2 –
in
10. Find the distance travelled by the particle in the first 4 seconds
= 3t2 – 6t
iit
dx dt
ac k
v=
je
X = t3 – 3t2 – 10
v = 0; gives t=0
cr
Sol.
e.
starting from t = 0.
& t = 2 sec.
Velocity will become zero at t = 2 sec., so particle will change direction after t = 2 sec. At t = 0 x(0 sec) = – 10 At t = 2 sec.
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X(2 sec) = 23 – 3(2)2 – 10 = 8 – 12 – 10 = – 14 At t = 4 sec. x(4 sec) = 43 – 3(4)2 – 10 = 64 – 48 – 10
in
=6
iit
= 4 + 20 = 24
je
= |– 14 – (– 10)| + |6 – (–14)|
e.
Distance travelled = x1 + x2
18.
ac k
Distance Travelled = 24 units.
A particle moving with uniform acceleration along x - axis has speed v
cr
m/s at a position x metre given by 0 x
180 16
V 180 16x
The acceleration of the particle is m/s2 is
Sol.
(A)
– 16
(B)
–8
(C)
164
(D)
8 160 16x
v2 = 180 – 16x
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d of both sides dx
Taking
d 2 d v 180 16x dx dx
2v
vdv dx
in
or a =
dV 16 dx
19.
je
Hence acceleration is – 8 m/s2.
e.
= – 8 m/s2.
Three vectors of equal magnitude A are inclined at an angle of 60° with
zero
(B)
A
(C)
A 6
Sol.
cr
(D)
ac k
(A)
iit
each other. The magnitude of the resultant will be:
cannot be calculated
R A BC R= =
A B C . A B C A2 B2 C2 2A.B 2B.C 2A.C
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6A2 = 3A 2 2
=
6A2
=A 6 20.
A man in a lift ascending with an upward acceleration 'a' throws a ball vertically upwards with a velocity 'v' with respect to himself and
in
catches it after 't1' seconds. After wards when the lift is descending with the same acceleration 'a' acting downwards the man again throws
e.
the ball vertically upwards with the same velocity with respect to him
the acceleration of the ball w.r.t. ground is g when it is in air
(B)
the velocity v of the ball relative to the lift is
(C)
the acceleration 'a' of the lift is
cr
ac k
iit
(A)
(D) Sol.
je
and catches it after 't2' seconds ?
g t1 t2 t1t2
g t2 t1 t1 t2
the velocity 'v' of the ball relative to the man is
gt1t2 t1 t2
For first case (when lift is ascending) t1 =
2v ga
…(i)
for second case (when lift is descending)
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t2 =
2v ga
…(i)
on solving equation (i) and (ii) we get V=
gt1t2 t1 t2
t1 g a t2 g a
in
gt1 + at1 = gt2 – at2
je
A particle is moving along a straight line with constant acceleration. At
ac k
21.
g t2 t1 t1 t2
iit
a=
e.
a(t1 + t2) = g(t2 – t1)
the end of tenth second its velocity becomes 20 m/s and in tenth second it travels a distance of 10m. Then the acceleration of the
(A)
cr
particle will be–
(C) Sol.
10 m/s2 (B)
20 m/s2
1 m/s2 (D) 5
3.8 m/s2
V = u + at 20 = u + a × 10 Sn = u
…(1)
a 2n 1 2
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10 u
a 2 10 1 2
…(2)
on solving (1) and (2) we get a = 20 m/s2 22.
A particle moving along a straight line with a constant acceleration of – 4 m/s2 passes through a point A on the line with a velocity of + 8 m/s at some moment. Find the distance travelled by the particle in 5 seconds
e.
u = + 8 m/s
je
a = – 4 m/s2
ac k
0 = 8 – 4t
iit
V=0
or t = 2 sec.
displacement in first 2 sec.
cr
Sol.
in
after that moment.
S1 = 8 × 2 +
1 . (– 4). 22 2
= 8m S1 = 8 × 2 +
1 (– 4) (2)2 2
= 16 – 8 = 8 m displacement in next 3 sec.
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S2 = 0 × 3 +
1 (– 4) 32 2
= – 18 m. S2 = 0 × 3 +
1 (– 4) (3)2 2
= – 18 m. distance travelled = |S1| + |S2|
in
= 26 m.
e.
Ans. 26 m
= 8 + 18
23.
iit
cr
= 26 m
1 1 ×2×8+ × 3 × 12 2 2
ac k
total distance =
je
ALTER:
A person throws a ball vertically up in air. The ball rises to maximum height and then falls back down such that the person catches it. Neglect the friction due to air. While the ball was in air, three statements are given below. (g = 9.8 m/s2) Statement 1: Just after the bait leaves the persons hand, the direction of its acceleration is upwards.
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Statements 2: The acceleration of ball is zero when it reaches maximum height. Statement 3: The acceleration of ball is g = 9.8 m/s2 downwards while the ball is falling down. Then which of the above statement or statements are correct in the options below.
(B)
Statement 2 only
(C)
Statement 3 only
(D)
Both statement 2 and statement 3
(C)
The acceleration of ball during its flight is g = 9.8 m/s2
iit
je
e.
in
Statements 1 only
ac k
Sol.
(A)
downwards.
A stone is projected vertically upwards at t = 0 second. The net
cr
24.
displacement of stone is zero in times interval between t = 0 second to t = T seconds. Pick up the INCORRECT statement. (A)
From time t =
T 3T second to t = second, the average 4 4
velocity is zero.
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The change in velocity from time t = 0 to t =
(B)
as change in velocity from t =
T 3T second to t = second 8 8
The distance travelled from t = 0 to T =
(C)
T second is larger than 4
T 3T second t = second 4 4 T 3T second to t = second is 2 4
in
distance travelled from t =
T second is same 4
The distance travelled from t =
e.
(D)
je
half the distance travelled from t =
T 3T and t = , the stone is at same height, 4 4
ac k
iit
At t =
(D)
Hence average velocity in this time interval is zero. Change in velocity in same time interval is same for a particle moving
cr
Sol.
T second to t = T second. 2
with constant acceleration. Let H be maximum height attained by stone, then distance travelled from t = 0 to t =
From t =
T 3 T H is H and from t = to t = T distance travelled is . 4 4 4 2
T T 3T to t = T sec distance travelled is H and from t = to t = 2 2 4
distance travelled is
H . 4
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25.
The velocity displacement graph of a particle moving along a straight line is shown in figure. Then the acceleration displacement graph is.
Sol.
(C)
From the graph, we can write
V=–x+2 a=
dv dt
e.
in
vdv v 1 dx
je
=–v
iit
= – (– x + 2)
26.
ac k
=x–2
A point moves in a straight line under the retardation a v2, where 'a' is a positive constant and v is speed. If the initial velocity is u, the distance
(A)
Sol.
cr
covered in 't' seconds is: aut
(B)
1 log (a u t) a
(C)
1 log (1 + a u t) a
(D)
a log (a u t)
The retardation is given by
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dv av2 dt integrating between proper limits v
b
dv 2 a dt V u 0
dx
in
dt 1 at dx u
e.
1 1 at v u
u dt 1 aut
je
or
dx
u dt 0 1 aut
1 ln 1 aut a
cr
0
t
ac k
s
iit
integrating between proper limits
S 27.
A point moves rectilinearly with deceleration whose modulus depends on the velocity v of the particle as a = k v, where k is a positive constant. At the initial movement the velocity of the point is equal to v0. What distance will it traverse before it stops ? What time will it take to cover that distance ?
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Sol.
28.
t0 =
2 v0 , k
S=
2 3 /2 v0 3k
A particle starts moving rectilinearly at time t = 0 such that its velocity 'v' changes with 't' according to the equation v = t2 – 1 where t is in seconds and v is in m/s. Find the time interval for which the particle
e.
Acceleration of the particle a = 2t – 1
The particle retards when acceleration is opposite to velocity.
iit
ac k
(2t – 1) (t2 – 1) < 0
je
a.v < 0
t (2t – 1) (t – 1) < 0
now t is always positive (2t – 1) (t – 1) < 0
cr
Sol.
in
retards.
either 2t – 1 < 0 & t – 1 > 0 t<
1 &t>1 2
This is not possible. or 2t – 1 < 0 & t – 1 < 0
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1 t 1 Ans. 2
or 2t – 1 < 0 & t – 1 < 0 2t > 1 & t < 1
1 t 1 Ans. 2
in
1 2
A particle starts moving along a straight line at t = 0. Its acceleration -
e.
29.
t>
je
time graph is shown. Correct relation between magnitudes of displacements between time durations t = 2s to t = 3s and t = 7s to t =
S78 > S23
S23>S78 (B)
(C)
S23 = S78 (D)
(A)
At t = 6s, particle again is at rest having same acceleration but in
ac k
(A)
S78 > S23
cr
Sol.
iit
8s represented by S23 and S78 respectively is :
opposite direction. Thus S23 > S78. 30.
A particle is projected vertically upwards in vacuum with a speed u. (A)
When it rises to half its maximum height, its speed becomes
u . 2
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When it rises to half its maximum height, its speed becomes
(B)
u . 2 The time taken to rise to half its maximum height is half the
(C)
time taken to reach its maximum height. The time taken to rise to three - fourth of its maximum height
(D)
is half the time taken to reach its maximum height.
H 2
ac k
V2 = u2 – 2gh
iit
At height h =
je
e.
u2 H= & 2g
in
At maximum height, velocity = 0
u2 u2 V = u 2g 4g 2 2
2
cr
Sol.
u2 V 2 2
V
u 2
Time taken to rise to maximum height u T=
u g
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for height h =
H , 2
u u 2 t= g
2 1 u 2g
3 4
e.
Time taken to rise to
in
=
31.
iit
T T 2 2
ac k
=T
H 4
je
H = T – time taken to fall down by
ˆ and a constant force A particle has initial velocity, v 3iˆ 4j
Sol.
(A)
cr
ˆ acts on the particle. The path of the particle is : F 4iˆ 3j
(B)
parabolic
(C)
circular
(D)
elliptical
Straight line
(B)
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For constant acceleration if angle between initial velocity makes an oblique angle with acceleration then path will be parabolic. 32.
A particle moves rectilinearly with a constant acceleration 1 m/s 2. Its speed after 10 seconds is 5 m/s. The distance covered by the particle in this duration is (Initial & final velocities are in opposite direction)
Sol.
For the given problem, velocity time graph will be as below:
1 5 5 2
in
Form the graph distance = 2
A particle moves through the origin of an xy - co-ordinate system at t =
je
33.
e.
= 25 m
0 with initial velocity u = 4i – 5 j. The particle moves in the xy plane
iit
with an acceleration a = 2i m/s2. Speed of the particle at t = 4 second is
ac k
:
12 m/s (B)
(C) Sol.
cr
(A)
5 m/s
(D)
8 2 m/s 13 m/s
Using vx = ux + axt = 4i + (2i) 4 = 12 i
As ay = 0, velocity component in y - direction remains unchanged. Final velocity = 12 i - 5j speed at t = 4 sec. = 122 5 13 m / s. 2
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34.
Two particles at a distance 5m apart, are thrown towards each other on an inclined smooth plane with equal speeds 'V'. It is known that both particles moves along the same straight line. Find the value of v if they collide at the point from where the lower particle is thrown. Inclined plane is inclined at an angle of 30° with the horizontal. [take g = 10 m
5 m/sec
(C)
7.5 m/sec
(D)
10 m/sec
e.
(B)
je
2.5 m/sec
5 = V.t + t2
ac k
5 = Vt
…(1)
iit
Down the plane
1 g sin t2 2
up the plane
cr
Sol.
(A)
in
/s2]
0 = V – g sin q t1 =
V gsin
0=V– t' =
V gsin
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t 2t1 t = 2t1 =
2V gsin
5=
2.V2 1 gsin .4v2 gsin 2 g2 sin2
in
10 g sin = 8v2
=
je iit
100 16
ac k
=
e.
1 10 10 2 v= 8
10 2.5 sec. 4
cr
Alternate
2vt = 5 …(1)
2V t gsin
2v
2v 5 gsin
v
5gsin 5 4 4
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35.
A stone is dropped from the top of a tower. When it is gallen by 5 m from the top, another stone is dropped from a point 25 m below the top. If both stones reach the ground at the same moment, then height of the tower from ground is : (take g = 10 m/s2)
Sol.
(A)
45m
(B)
50m
(C)
60m
(D)
65m
Vel of 1st stone when passing at A
in
V2 = 0 + 2.10.5
je
S1 – S2 = 20 m.
e.
V = 10 m/s
t = 2s
1 .10.4 2
cr
S2 =
ac k
iit
1 1 2 2 10.t 2 10.t 2 .10.t 20
= 20 m
Ht = 25 + 20 = 45 m. 36.
Two bikes A and B start from a point. A moves with uniform speed 40 m/s and B starts from rest with uniform acceleration 2 m/s2. If B start at t = 0 and A starts from the same point at t = 10s, then the time during the journey in which A was ahead of B is:
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20s
(B)
8s
(C)
10s
(D)
A is never ahead of B
A will be ahead of B when XA > XB 40 (t – 10) > (0) t +
1 (2)t2 2
in
Sol.
(A)
as A is 10 sec. late than B.
je
(t – 20)2 < 0
e.
t2 – 40 t + 400 < 0
A tiger running 100 m race, accelerates for one third time of the total
ac k
38.
iit
Which is not possible? So A will never be ahead of B.
and then moves with uniform speed Then the total time taken by the
(A)
cr
tiger to run 100m if the acceleration of the tiger is 8m/s2 is:
(C) Sol.
3 5s
(B)
3 5s
12 s
(D)
9s
Let the total time of race be t seconds and the distance be S = 100 m. The velocity vs time graph is Area of OAD
s 5
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s 1 T a 5 2 3 1 T = 8 2 3
2
2
or T = 3 5 m/s. 40.
Velocity (v) versus displacement (S) graph of a particle moving in a
in
straight line, corresponding acceleration (a) versus velocity (v) graph will be
From the given relation between velocity (V) and displacement (S) is given
e.
Sol.
je
by
ac k
d 1 dS
iit
v=S
cr
Hence, acceleration a =
d dS
a=×1 a=
Therefore, graph between acceleration and velocity will be as shown. 42.
A particle is projected-at angle 60° with speed 10 3, from the point 'A'. At the same time the wedge is made with speed 10 3 towards
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right as shown in the figure. Then the time after which particle will strike with wedge is (g = 10 m/sec2): (A)
4 sec (D) 3
(C)
none of these
Suppose particle strikes wedge at height 'S' after time t. S = 15t –
1 10t2 2
in
= 15 t – 5 t2. During this time distance travelled by particle in horizontal
ac k
15t 5t2 = 1 3
je
S tan30
iit
X=
e.
direction = 5 3 t. Also wedge has travelled extra distance
Total distance travelled by wedge in time
cr
Sol.
2 3 sec
(B)
2 sec
t = 10 3 t =5
3 t + 3 (15 – 5t2)
t = 2 sec. Alternate (by Relative Motion)
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T=
2u1 g1
= 10 3 sin 30 5 3 =
2 10 3 1 10 3
= 2 sec. A particle is projected from a point (0, 1) on y - axis (assume + Y
in
43.
ground along x axis in 1 sec.
e.
direction vertically upwards) aiming towards a point (4, 9). It fell on
je
Taking g = 10 m/s2 and all coordinate in meters. Find the X - coordinate
(B)
(4, 0)
(C)
(2, 0)
(D)
( 2 5 , 0)
tan =
9 1 2 40
ac k
(3, 0)
cr
Sol.
(A)
iit
where it fell
now, – 1 = u sin (1) –
1 g(1)2 2
u sin = 4 and sin =
2 5
u2 5
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horizontal direction & Y axis is vertically upwards. Take g = 10 m/s2. Find Y component of initial velocity and
(ii)
X component of initial velocity
From graph (1) Vy = 0 at t =
1 sec. 2
in
Sol.
(i)
i.e., time taken to reach maximum height H is
uy 5 m / s
je
e.
uy 1 g 2
…Ans. (i)
iit
t=
ac k
from graph (2): Vy = 0 at X = 2m
i.e. when the particle is at maximum height, its displacement along
cr
horizontal X = 2 m
X ux t
2 ux
1 2
Ux 4 m / s 47.
…Ans. (ii)
A particle P is projected from a point on the surface of smooth inclined plane (see fig.) Simultaneously another particle Q is released on the
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smooth inclined plane from the same position. P and Q collide after t = 4 second. The speed of projection of P is
Sol.
(A)
5 m/s
(B)
10m/s
(C)
15 m/s (D)
20 m/s
(B) It can be observed from figure that P and Q shall collide if the initial component of P along incline. uII = 0 that is particle is projected
in
perpendicular to incline.
2u gcos
je
gT cos 2
cr
u
iit
=
2u gcos
ac k
T=
e.
Time of flight
= 10 m/s 49.
A stone is projected from a horizontal plane. It attains maximum height 'H' & strikes a stationary smooth wall & falls on the ground vertically below the maximum height. Assuming the collision to be elastic the height of the point on the wall where ball will strike is (A)
H 2
(B)
H 4
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3H 4
(C)
Sol.
H=
None of these
1 2 g 2t 2
= 2gt2 h=
(D)
…(1)
1 2 gt 2
…(2)
je
Some students are playing cricket on the roof of a building of height 20
ac k
50.
3H 4
iit
=
H 4
e.
h=H
in
By (1) & (2)
m. While playing, ball falls on the ground. A person on the ground
cr
returns their ball with the minimum possible speed at angle 45° with the horizontal find out the speed of projection. Sol.
Let us assume that person throws ball from distance x. Taking point of projection origin. By equation of trajectory
1 gx2 y = X tan – 2 U2 cos2
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1 10x2 20 = x tan 45° – 2 U2 cos2 45 10x2 U = x 20 2
…(1)
For 'U' to be minimum
du 0 dx
On differentiation w.r.t. 'X'
e.
in
2 du 10 2x x 20 10x 1 2U 0 2 dx x 20
je
= 10 (x – 40)
iit
x = 40
ac k
For x greater than 40, slope is positive & x less than 40, slope in negative. So at x = 40 There is minima
cr
Required minimum velocity from Eq. 1
U2min
10 402 40 20
Umin =
800
Umin = 20 2 m / s 51.
A train is standing on a platform, a man inside a compartment of a train drops a stone. At the same instant train starts to moves with constant
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acceleration. The path of the particle as seen by the person who drops the stone is: (A)
parabola
(B)
straight line for sometime & parabola for the remaining time
(C)
straight line
(D)
variable path that cannot be defined
(E)
Relative to the person in the train, acceleration of the stone is
1 2 at , 2
y=
1 2 gt 2
X a Y g
ac k
iit
je
x=
e.
According to him :
cr
Sol.
in
'g' downward, (acceleration of train backwards).
Y
g x a
straight line. 52.
Two men P and Q are standing at corners A & B of square ABCD of side 8 m. They start moving along the track with constant speed 2 m/s and 10 m/s respectively. Find the time when they will meet for the first time.
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Sol.
(A)
2 sec
(B)
3 sec
(C)
1 sec
(D)
6 sec
a=8m They meet when Q displace 8 × 3 m more that P [relative displacement = relative velocity × time.]
in
8 × 3 = (10 – 2) t
53.
e.
t = 3 sec. Ans.
A coin is released inside a lift at a height of 2 m from the floor of the
je
lift. The height of the lift is 10 m. The lift is moving with an acceleration
ac k
lift: (g = 10 m/s2)
Sol.
4s
4 s 21
cr
(A) (C)
iit
of 9 m/s2 downwards. The time after which the coin will strike with the
(B) (D)
2s
2 s 11
Relative to lift Initial velocity and acceleration of coin are 0 m/s and 1 m/s2 downward
2
1 1 t2 2
or t = 2 second
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54.
A particle is thrown up inside a stationary lift of sufficient height. The time of flight is T. Now it is thrown again with same initial speed v0 with respect to lift. At the time of second throw, lift is moving up with speed v0 and uniform acceleration g upwards (the acceleration due to gravity). The new time of flight is – (B)
T 2
(C)
T
(D)
2T
With respect to lift initial speed = v0
je
acc = – 2g
ac k 1 × 2g × T' 2 2
cr
0 = v0 T' +
1 2 at 2
iit
displacement = 0 S = ut +
in
T 4
e.
Sol.
(A)
T' =
= 55.
0 g
1 20 1 T 2 5 2
i ˆj and Two particles P and Q are moving with velocities of ˆ
ˆi 2jˆ respectively. At time t = 0, P is at origin and Q is at a point
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with position vector 2iˆ ˆj . Find the equation of the path of Q with respect to P. Sol.
ˆ t 2iˆ ˆj ˆi 2j
rP ˆi ˆj t rQ
rQP rQ rP
e.
je
rQP 2 2t ˆi 1 t ˆj
in
= 2iˆ ˆj 2iˆ t ˆj
iit
X = 2 – 2t
ac k
y=1+t x = 2 – 2 (y – 1)
55.
cr
x + 2y = 4 Ans.
When two bodies move uniformly towards each other, the distance between them diminishes by 16 m every 10s, If bodies move with velocities of the same magnitude and in the same magnitude and in the same direction as before the distance between then will decrease 3m every 5s. Calculate toe velocity of each body.
Sol.
Let velocity of bodies be V1 + V2. V1 + V2 =
16 10
…(1)
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V1 – V2 =
2V1 =
3 5
…(2)
16 3 22 10 5 10
= 2.2 m/s V1 = 1.1 m/s V2 = 1.6 – 1.1
in
= 0.5 m/s
e.
After solving we have
A weight W is supported by two strings inclined at 60° and 30° to the
ac k
56.
iit
and V2 = 0.5 m/s.
je
V1 = 1.1 m/s
vertical. The tensions in the strings are T1 & T2 as shown. If these tensions are to be determined in terms of W using a triangle of forces,
Sol.
cr
which of these triangles should you draw? (block is in equilibrium)
AB W,
BC T1, CA T2
AB BC CA 0 Ans. (E)
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58.
A 2 kg toy car can move along an x axis. Graph shows force F x, acting on the car which beings at rest a time t = 0. The velocity of the particle at T = 10 s is :
Sol.
(A)
– i m/s (B)
– 1.5 i m/s
(C)
6.5 i m/s
(D)
dp p
f
13 i m/s
Pi
in
= Fdt
e.
= Area under the curve.
iit
Net Area = 16 – 2 – 1
13 2
cr
=
ac k
= 13 N-s = Vf
je
Pi = 0
= 6.5 i m/s
[As momentum is positive, particle is moving along positive x axis.] 59.
A particle is moving in a straight line whose acceleration versus time graph is given. Assume that initial velocity is in the direction of acceleration. Then which of the statement is correct between time t = 0 to t = t0.
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(A)
Velocity first increases then decreases, displacement always increases Velocity and displacement both, first increases and then
(B)
decreases
Sol.
(C)
Displacement increases and velocity decreases
(D)
Displacement and velocity both always increases.
It is clear from the figure that acceleration does not change sign, i.e., does not change in direction. Only the magnitude of acceleration first
e.
Velocity keeps on increasing.
in
increases and then decreases.
Surface inside which on constant force F starts acting on the trolley as
iit
59.
je
Hence displacement also keeps on increasing.
ac k
a result of which the string stood at an angle of 37° from the vertical (bob at rest relative to trolley) then.
(B)
Sol.
acceleration of the trolley is
cr
(A)
40 m/sec2 3
force applied is 60 N
(C)
force applied is 75 N
(D)
tension in the string is 25 N.
F = (8 + 2) a a=
F 10
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T cos 37° = mg T sin 37° = ma tan 37° =
a g
a = g tan 37∟ = 10
3 4
in
= 7.5 m/s2
e.
F = 10 × 7.5
60.
iit
mg cos 37
ac k
T=
je
= 75 N
2 10 5 4
=
100 25 N 4
cr
=
A cylinder of mass M and radius R is resting R is resting on two corner edges A and B as shown. The normal reaction at the edges A and B are: (neglect friction).
(A)
NA 2 NB
(B)
NB 3 NA
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NB
(C) Sol.
2 3 mg 5
NA
(D)
Mg 2
(B) (D) For equilibrium NA cos 60° + NB cos 30° = Mg and NA sin 60° = NB sin 30°
in
NA NB 3 Mg 2 2
NA 3 NB
cr
61.
Mg . 2
3 NA ;
ac k
On Solving NB = NA =
iit
je
e.
NA 3 NB 2 2
An insect moving along a straight line, travels in every second distance equal to the magnitude of the elapsed. Assuming acceleration to be constant, and the insects starts at t = 0. Find the magnitude of initial velocity of insect.
Sol.
Distance travelled from time 't – 1' sec to 't' sec is S=u
a 2t 1 2
…(1)
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From given condition S=t
…(2)
2t = 2u + 2at – a
2u = 2t – 2at 2u = 2t (1 – a) (1) & (2)
in
a t 1 a . 2
e.
u
a 2t 1 2
je
t u
iit
Since u and a are arbitrary constants, and they must be constant for
ac k
every time.
coefficient of t must be equal to zero.
cr
1–a=0
a = 1 for a = 1, u=
1 unit 2
Initial speed is 62.
1 units. 2
In the figure if blocks A and B will move with same acceleration due to external agent, there is no friction between A and B then the magnitude of interaction force between the two blocks will be :
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63.
(A)
2 mg/ cos
(B)
2 mg cos
(C)
mg cos
(D)
none of these
Three rigid rods are joined to from an equilateral triangle ABC of side 1m. Three particles carrying charges 20 µC each are attached to the vertices of the triangle. The whole system is at rest always in an inertial frame.
3.6 3 N
Fnet ma
3.6 N
e.
(C)
(B)
(D)
je
zero
7.2 N
iit
Sol.
(A)
in
The resultant force on the charged particle at A has the magnitude.
ac k
a = acceleration of charge of particle at A = 0
64.
cr
Fnet 0.
Two stones are projected simultaneously from a tower at different angles of projection with same speeds 'u' the distance between two stones is increasing at constant rate 'u'. Then the angle between the initial velocity vectors of the two stone is:
Sol.
(A)
30°
(B)
60°
(C)
45°
(D)
90°
(B)
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To an observer who starts falling freely under gravity from rest at the instant stones are projected, the motion of stone A and B is seen as
dx u dt
….(1)
dl u dt
…(2)
x=l
A rope of negligible mass passes over a pulley of negligible mass
e.
65.
in
and BOA = 60°
je
attached to the ceilling, as shown in figure. Open end of the rope is held by student A of mass 70 Kg. Who is at rest on the floor. The
iit
opposite end of the rope is held by student B of mass 60 Kg. who is
ac k
suspended at rest above the floor. The minimum acceleration a0 with which the student B should climb up the rod to lift the student A
(A)
(C) Sol.
cr
upwards off the floor.
1 m/s2 (B) 3
2 m/s2 3
4 m/s2 (D) 3
5 m/s2 3
(D) For student A to just lift off the floor, tension 1 in string must The F.B.D. of student B is
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Applying Newton's second law T – mg = mg 700 – 600 = 60 a or a = 66.
Force exerted by the supporting table on the pulley (in Newton) Let T1 T2 and FS be the force exerted by the horizontal string, vertical
in
Sol.
5 m / s2 3
e.
string and the support on the mass-less respectively. Then
iit
or FS T1 T2
Tension in each string is T1 T2 2 mg )
cr
68.
ac k
= 2 2 mg (
je
T1 T2 FS 0
The system starts from rest and A attains a velocity of 5 m/s after it has moved 5 m towards right. Assuming the arrangement to be frictionless every where pulley & strings to be light, the value of the constant force F applied on A is:
Sol.
(A)
50 N
(C)
100 N
(B) (D)
75 N
96 N
(B)
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V2 a= 2S =
25 10
= 2.5 m/s2 For 6 kg: – F – 2T = 6A
in
For 2 kg:
e.
– T – 2g = 2 (2a)
ac k
A rod of length 2l is moving such that its ends A and b move in contact with the horizontal floor and wall respectively as shown in figure O is the intersection point of the vertical wall and horizontal floor velocity vector of the centre of rod C is always directed along tangent drawn at
cr
69.
iit
F = 75 N
je
Feon (1) & (2)
C to the (A)
circle of radius
(B)
circle of radius l whose centre lies at O
(C)
circle of radius 2l whose centre lies at O
(D)
None of these
2
whose centre lies at O
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Sol.
(B) At any instant of time the rod makes an angle
with horizontal, the x & y
coordinates of centre of rod are. x = l cos y = l sin x 2 + y 2 = l2
in
Hence the centre C moves along a circle of radius l with centre at O.
e.
velocity vector of C is always directed along the tangent drawn at C to the circle of radius l who centre lies at O.
A system is as shown in the figure. All speeds shown are with respect to
je
70.
(C)
15 m/s (D)
(B) 1
ac k
5 m/s
10 m/s
7.5 m/s
cr
Sol.
(B)
(A)
iit
ground. Then the speed of B with respect to ground is:
2
2
= constant
d 1 2d 2 0 dt dt
(5 + 5) + 2 (5 + V0) = 0 or VB = 10 m/s
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The acceleration of the cart is
(B)
The cart comes to momentary rest after 2.5 s.
(C)
The distance travelled by the cart in the first 5s is 17.5m.
(D)
The velocity of the cart after 5s will be same as initial velocity.
0.2 g = 0.7a
2g m / s2 7
in
a
e.
For the case, it comes to rest when V = 0
= 2.5 s
iit
49 2g
ac k
t
je
2g 0 7 t 7
cr
Sol.
2g towards right. 7
(A)
Distance travelled till it comes to rest
2g 0 72 2 s 7 S = 8.75 m So in next 2.5s, it covers 8.75 m towards right Total distance = 2 × 8.75 = 17.5 m
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After 5s, it speed will be same as that of initial (7 m/s) but direction will be reversed. 72.
A painter is applying force himself to raise him and the box with an acceleration of 5 m/s2 by a mass-less rope and pulley arrangement. Mass of painter is 100 kg and that of box is 50 kg. If g = 10 m/s2, then:
(B)
tension in the rope is 2250 N
(C)
force of contact between the painter and the floor is 375 N
(D)
none of these
e.
in
tension in the rope is 1125 N
…(1)
ac k
R + T = m(g + a)
iit
R + T – mg = ma
je
For painter:
For the system: 2T – (m + M) g
cr
Sol.
(A)
= (m + M) a
2T = (m + M) (g + a)
…(2)
where: m = 100 kg M = 50 kg a = 5 m/sec2
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T
150 15 2
= 1125 N and; R = 375 N. 73.
A person wants to raise a block laying on the ground to height h. In both the case time required is same then in which case he has to exert more force. Assume pulleys and strings light
(B)
(ii)
(C)
Same is both
(D)
Cannot be determined
1 2 at 2
a should be same in both cases, because h and t are
ac k
Since, h =
iit
je
e.
in
(i)
same in both cases as given In (i) F1 – mg = ma
cr
Sol.
(A)
F1 = mg + ma.
In (ii) 2 F – mg = ma F2 =
mg ma 2
F1 F2.
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74.
A particle is projected from the horizontal x – z plane, in vertical x – y plane where x - axis is horizontal and positive y - axis vertically upwards. The graph of 'y' coordinate of the particle v/s time. The range
(B)
403 m/s 4
(C)
2 5 m/s
(D)
28 m / s
in
3 m/s
je
(A)
ac k
From graph
iit
(D)
uy = tan 60° =
3 m/s
cr
Sol.
3 m. Then the speed of the projected particle is:
e.
of the particle is
range R =
or
3
2uxuy g
2 ux 3 g
or ux = 5 m/s
u u2x u2y
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=
(A)
3 m/s
(B)
2 m/s
(C)
2 3 m/s
(D)
3 m/s
Let AB = l, B = (x, y)
e.
in
vB vxˆi vyˆj vB 3 ˆi vyˆj x2 + y2 = l2
ac k
2x vx + 2y vy = 0
iit
je
…(i)
x vx + 2y vy = 0
3
y vy 0 x
cr
Sol.
28 m / s
3 tan 60 vy 0 vy = – 1 Hence from (i)
vB 3 ˆi ˆj Hence
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vB 2 m / s 76.
Mass m is in equilibrium. If it is displace further by x and released find its acceleration just after it is released. Take pulleys to be light &
4kx 5m
(B)
2kx 5m
(C)
4kx m
(D)
e.
none of these
(C)
je
Sol.
(A)
in
smooth and strings light.
iit
Initially the block is at rest under action of force 2T upward and mg down
ac k
wards. When the blocks is pulled downwards by x, the spring extends by 2x. Hence tension T increases by 2kx. Thus the net unbalanced force on
cr
block of mass m is 4kx.
acceleration of blocks is = 77.
4kx m
Assume that cylinder remains in contact with the wedges. The velocity of cylinder is – (A)
19 4 3
(B)
13 u m/s 2
u m/s 2
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3 u m/s
(D)
7 u m/s
(D) Method-I As cylinder will remains in contact with wedge A Vx = 2u
in
As it also remain in contact with wedge B
e.
u sin 30° = Vy
iit
sin30 usin30 cos30 cos30
ac k
vy = v x
je
= cos 30° – Vx sin 30°
vy = vx tan 30° + u tan 30° Vy = 3u tan 30° = V= =
cr
Sol.
(C)
3u
v2x v2y
7u
Method-II
In the frame of A
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3u sin 30° = Vy cos 30° Vy = 3 u tan 30° =
3u
and Vx = 2u
v2x v2y 7 u 78.
Two stones are thrown vertically upwards simultaneously from the
in
same point on the ground with initial speed u1 = 30 m/sec and u2 = 50
e.
m/sec. which of the curve represent correct variation (for the time
je
interval in which both reach the ground) of
iit
(x2 – x1) = the relative position of second stone with respect to first with
ac k
time (t).
(v2 – v1) = the relative velocity of second stone with respect to first with time (t).
cr
Assume that stones do not rebound after hitting.
Sol.
While both the stones are in flight, a1 = g and a2 = g So arel = 0 Vrel = constant xrel = (const) t
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Curve of xrel v/s t will be straight line. 78.
After the first particle drops on ground, the separation (xrel) will decrease parabolically (due to gravitational acceleration), and finally becomes zero. and Vrel = slope of xrel v/s t
Two springs are in a series combination and are attached to a block of
e.
79.
in
So
je
mass 'm' which is in equilibrium. The spring contents and the
iit
extensions in the springs are as shown in the figure. Then the force
Sol.
ac k
exerted by the spring on the block is:
k1k2 x1 x2 k1 k2
(B)
k1x1 k2x2
cr
(A)
(C)
k1
(D)
None of these
(A) Tension in both springs are same both the springs are in series,
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keq =
(A)
the acceleration of A will be more than g sin
(B)
the acceleration of A will be less than g sin
(C)
normal force on A due to B will be more than mg cos
(D)
normal force on A due to B will be less than mg cos
in
Sol.
A & B are free to move. All the surfaces are smooth.
ma0 sin + N = mg cos mg cos
je
N = mg cos – ma0 sin
e.
80.
k1k 2 k1 k 2
ac k
iit
N < mg cos Hence, (D) is true.
ma0 cos + mg sin = ma
cr
a = g sin + a0 cos
Hence acceleration of A = 81.
a a0 cos
2
a0 sin gsin 2
There are three particle A, B & C are lying in a horizontal plane the particle b is situated 5m due North from A and particle C is situated 30° East of North at a distance of 2 3 m from A. These three particles start moving simultaneously along straight lines and collide after 2 second of
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their start. Particle A moves with constant velocity 5 m/s due 30° South of East. Find out the velocity vectors (constant) of the particles B & C. Assume unit vector ˆi in the east and ˆj is the north. Taking A at the origin and east direction as the direction x - axis and positive y - axis along north.
vA
5 3ˆ i 2.5 ˆj 2
u, u uxˆi uyˆi
iit
je
i vyˆi and C as v, v vxˆ
e.
in
Assume velocity of B as
ac k
rB t 5ˆj uyt ˆj uxt ˆj
rB 2 2uxˆi 2uy 5 ˆj
…(1)
rA 2 5 3 ˆi 5ˆj 2
cr
Sol.
rC 2 3 ˆi 3ˆj 2vxˆi 2vyˆj
3 2vx ˆi 3 2vy ˆj
…(3)
(1) & (2)
ux
5 3 , 2
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uy = – 5 (2) & (3)
Vx 2 3,
Vy
…(4)
u
5 3ˆ i 5ˆj 2
Two inclined planes inter in a horizontal plane. Their inclinations to the
je
82.
e.
i 4ˆj respectively. and v 2 3 ˆ
in
the velocity vector of B and C are
iit
horizontal being and . If a particle is projected with velocity u at
ac k
right to the former from a point on it, find the time after which the velocity vector will become perpendicular to the other inclined plane. Taking x - axis parallel to second inclined, we can say that when x component of velocity will become zero, than velocity vector will become
cr
Sol.
perpendicular to the other inclined. ux = u sin ( + ) ax = – g sin vx – u x + a x t 0 = u sin ( + ) – g sin t
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t=
usin gsin
Ans. t = 83.
usin gsin
A man standing on the block is pulling the rope velocity of the point of string in contact with the hand of the man is 2 m/s downwards. The velocity of the block will be: [assume that the block does not rotate]
in
1 m/s (D) 2
e.
(C)
2 m/s 1 m/s
je
3 m/s
(B)
2
3
4
C
ac k
1
iit
Sol.
(B)
(A)
cr
d 1 d 2 d 3 d 4 0 dt dt dt dt –v–v+0+v+2=0 v = 2 m/s 84.
The velocity of lift is 2 m/s while string is winding on the motor shaft with velocity 2 m/ s and block A is moving downwards with a velocity of 2 m/s, then find out the velocity of block B.
(A)
2 m/s
(B)
2 m/s
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(C) Sol.
4 m/s
(D)
none of these
(D)
vB, 4 m / s
vB, vB , g v , g 4 m/s = vB , g 2 m / s
85.
in
vB , g 2 m / s Three stones A, B and C are simultaneously projected from same point
e.
with same speed. A is thrown upwards, B is thrown horizontally and C
je
is thrown downwards from a building. When the distance between
(C)
5 2 m
ac k
10 m
(B)
5m (D)
10 2 m
Let the stones be projected at t = 0 sec with a speed u form point O.
cr
Sol.
(A)
iit
stone A and C becomes 10m, Then distance between A and B will be –
Then an observer at rest at t = 0 and having constant acceleration equal to acceleration due to gravity, shall observe the three stones moves with constant velocity.
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Sol.
In the given time each ball shall travel a distance 5 metre as seen by this observer. Hence the required distance between A and B will be =
52 52 5 2 meter. 86.
A stone is projected horizontally with speed v from a height h above ground. A horizontal wind is blowing in direction opposite to velocity of projection and gives the stone a constant horizontal acceleration f (in direction opposite to initial velocity). As a result the stone falls on
in
ground at a point vertically below the point of projection. Then the
gv2 2f 2
(B)
je
(A)
iit
2v2 f2
(D)
ac k
(C)
gv2 f2
2gv2 f2
Time taken to reach the ground is given by h=
1 gt 2
cr
Sol.
e.
value of height h in terms of f, g, v is (g is acceleration due to gravity)
…(1)
Since horizontal displacement in time t is zero
t
2v f
…(2)
2gv2 h= f2
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87.
Two blocks A and B connected to an ideal pulley string system. In this system when bodies are released then: (neglect friction and take g = 10 m/s2)
(B)
Acceleration of block A is 2 m/s2
(C)
Tension in string connected to block B is 40 N
(D)
Tension in string connected to block B is 80 N
400 – 4T = 40 a
T = 80 N
cr
Comprehension
ac k
Solving a = 2 m/s2
iit
For 10 kg block T = 10.4 a
e.
Applying NLM on 40 kg block
in
Acceleration of block A is 1 m/s2
je
Sol.
(A)
A Weighing machine kept in a lift. Lift is moving upwards with acceleration of 5 m/s2. A block is kept on the weighing machine. Upper surface of block is attached with a spring balance. Reading shown be weighing machine and spring balance is 15 kg and 45 kg respectively. Answer the following question. Assume that the weighing machine can measure weight by having negligible deformation due to block, while the spring balance requires larger expansion: (take g = 10 m/s 2)
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88.
Mass of the object in kg is and the normal force acting on the block due to weighing machine are: 60 kg, 450 N
(B)
40 kg, 150 N
(C)
80 kg, 400 N
(D)
10, kg, zero
FBD of Block in ground frame:
in
Sol.
(A)
e.
applying N.L.
600 15
ac k
M
iit
15 M = 600
je
150 + 450 – 10 M = 5M
M = 40 Kg.
89.
cr
Normal on block is the reading of weighing machine i.e. 150 N. In lift is stopped and equilibrium is reached. Reading of weighing machine and spring balance will be:
Sol.
(A)
40 kg, zero
(B)
10 kg, 20 kg
(C)
20 kg, 10 kg
(D)
zero, 40 kg
If lift is stopped & equilibrium is reached then
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450 + N = 400 N = – 50 So block will loss the contact with weighing machine thus reading of weighing machine will be zero. T = 40 g So reading of spring balance will be 40 Kg. Find the acceleration of the lift such that the weighing machine shows
in
90.
22 m/s2 4
a= 91.
85 m/s2 4
(D)
60 m/s2 4
950 400 40
cr
Sol.
je
(C)
(B)
iit
45 m/s2 4
ac k
(A)
e.
its true weight.
Two messes m1 and m2 which are connected with a light string, are placed over a frictionless pulley. This set up is placed over a weighing machine, as shown. Three combination of masses m1 and m2 are used, in first case m1 = 6 kg and m2 = 2 kg, in second case m1 = 5 kg and m2 = 3 kg and in third case m1 = 4 kg and m2 = 4 kg. masses are held stationary initially and then released. If the readings of the weighing machine after the release in three cases are W1, W2 and W3 respectively then:
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Sol.
(A)
W1 > W2 > W3
(B)
W1 < W2 < W3
(C)
W1 = W2 = W3
(D)
W1 = W2 < W3
(B) Reading of the weighing machine = 2T + weight of the machine.
e.
2m1m2 m1 m2
je
T=
in
As weight of the machine is constant.
So reading is maximum for the case m1m2 is maximum as m1 + m2 is all
In the following arrangement the system is initially at rest. The 5 kg
ac k
92.
iit
cases is same.
block is now released. Assuming the pulleys and string to be massless
Sol.
(A)
cr
and smooth, the acceleration of block 'C' will be zero
(B)
2.5 m/s2
(C)
10 m/s2 7
(D)
5 m/s2 7
Block b will not move. 5 g – T = 5a 2T – 8g = 8
a 2
…(1) …(2)
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10g – 2T = 10a …From (1)
g 7
a=
A large cubical shaped block of mass M rests on a fixed horizontal
in
surface. Two blocks of mass m1 and m2 are connected by a light
e.
inextensible string passing over a light pulley as shown. Neglect friction every-where. Then the constant horizontal force of magnitude F that
iit
je
should be applied to M so that m1 and m2 do not move relative to M is:
F
m2 m1 m2 M g m1
(B)
F
m1 m1 m2 M g m2
(C)
F
m1 m1 M g m2
(D)
F
m2 m1 M g m1
ac k
(A)
cr
93.
a g 10 5 m / s2 2 14 14 7
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Sol.
Let a the acceleration of cubical block of mass M from ground frame. Then from the frame of block mass M, forces on m1 and m2 . Hence for m1 and m2 to remain at rest in frame of M.
m1 g m2
F=
m1 m1 m2 M g m2
All the contacts are smooth. Strings and spring are light litially 'A' is
e.
94.
or a =
in
m2a = m1g
je
held by someone and 'B' and 'C' are at rest and in equilibrium also. Find out the acceleration of each block just after the block 'A is released.
ac k
Before block A was released, the system was at rest, and all blocks were in equilibrium hence, tension is both the strings is equal to 2 Mg. When block A is released, it will have an unbalanced force on it and
cr
Sol.
iit
Masses of A, B and C are M, M and 2M respectively.
hence the tension is string (2) will change to say T2. Now the arrangement will be. Since, tension is spring does not change instantaneously, hence, tension is string 1 will remain same i.e. 2 Mg, Thus, Block C will remain at rest and ac = 0.
Newton's law along the string (2),
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2Mg – Mg = Ma + Ma
a
g 2
Hence acceleration of
A
g g , B , & C 0 2 2
Ans. Hence acceleration of
Match the column:
in
je
95.
g g , B , & C 0 2 2
e.
A
iit
In each of the situations assume that particle was initially at origin and there after it moved rectilinearly. Some of the graph in left column
ac k
represent the same motion as represented by graph in right column match these graphs.
cr
Column 1
Column 2
Ans. A – R, B – Q, C – S, D – P Sol.
(A)
a is +ve constant position will increase parabolicaly with time.
(B)
a is –
velocity will decrease linearly with time &
position will decrease parabolicaly with time.
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(C)
First acceleration is +ve & then –ve velocity will first increase & then decrease. Also, position of the particle will increase parabolicaly.
(D)
First acceleration is –ve & then +ve
velocity will decrease
linearly & then increase. 96.
A uniform rope of length L and mass M is placed on a smooth fixed wedge as shown. Both ends of rope are at same horizontal level. The
in
rope is initially released from rest, then the magnitude of initial
(B)
M (cos - cos )g
(C)
m (tan – tan )g
(D)
none of these
je
zero
ac k
iit
(A)
Let L1 and L2 be the portions (of length) of rope on left and right surface of wedge as shown
cr
Sol.
e.
acceleration of rope is
magnitude of acceleration of rope
M L1 sin L2 sin g L 0 a= m
L1 sin L2 sin
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97.
A bob is hanging over a pulley inside a car through a string. The second end of the string is in the hand of a person standing in the car. The car is moving with constant acceleration 'a' directed horizontally. Other end of the string is pulled with constant acceleration 'a' (relative to car)
m g2 a2
(B)
m g2 a2 ma
(C)
m g2 a2 ma
(D)
m(g + a)
ac k
(C)
(force diagram in the frame of the car) Applying Newton's law perpendicular to string
cr
Sol.
iit
je
e.
(A)
in
vertically. The tension in the string is equal to
mg sin = ma cos
tan
a g
Applying Newton's law along string
T m g2 a2 ma T m g2 a2 ma Ans.
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98.
All the surface are smooth. All the blocks A, B and C are movable, x axis is horizontal and y - axis vertical as shown just after the system is released from the position as shown. (A)
Acceleration of 'A' relative to ground is in negative y - direction
(B)
Acceleration of 'A' relative to positive x - direction
(C)
The horizontal acceleration of 'B' relative to ground is in negative x - direction. The acceleration of 'B' relative to ground directed along the
in
(D)
je
There is no horizontal force on block A, therefore it does not move in x-
iit
direction, whereas there is net downward force (mg – N) is acting on it, making its acceleration along negative y - direction. Blocks B moves
ac k
downward as well as in negative x - direction. Downward acceleration of A and B will be equal due to constraint thus w.r.t. B, A moves is positive x - direction.
cr
Sol.
e.
inclined surfaced of 'C' is greater than g sin
Due to the component of normal exerted by C on B, it moves in negative x - direction. The force acting vertically downward on block B are mg and N A (normal reaction due to block A). Hence the component of net force on block B along the inclined surface of B is greater than mg sin . Therefore the acceleration of 'B' relative to ground directed along the inclined surface of 'C' is greater than g sin .
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99.
In the figure shown blocks 'A' and 'B' are kept on a wedge 'C'. A, B, C each have mass m. All surfaces are smooth. Find the acceleration of C.
Sol.
Let c be acceleration of wedge C. a be acceleration of block A w.r.t. wedge c. b be acceleration of block B w.r.t. wedge c. Applying Newton's law in horizontal direction to system of A + B + C.
in
mc + m(c – a cos 37°) + m (c + b cos 53°) = 0 .................(1)
In case of A:
iit
In case of B:
…(2)
je
mg sin 37° = m (a – c cos 37°)
e.
Applying Newton's law to block A and B along the incline gives.
ac k
mg sin 53° = m (b + c cos 53°)
…(3)
solving (1), (2) & (3)
cr
we get c = 0 Ans. ac = 0 100.
A force F is applied on block A of mass M so that the tension in light string also becomes F when block B of mass m acquires an equilibrium state with respect to block A. Find the force F Give your answer in terms of m, M and g.
Sol.
Applying Newton's law on the system in horizontal direction F = (M + m)a.
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Now consider the equilibrium of block B w.r.t. blocks M F2 = (mg)2 + (ma)2
F = (mg) + m m M
2
2
2
F 1
m M
2
2
e.
F 1 m M
in
mg
A system is shown in figure. All contact surfaces are smooth and string
iit
101.
;
je
F=
m2g2 m2
ac k
is tight & inextensible wedge 'A' moves towards right with speed 10 m/s & 'B' relative to 'A' is in down ward direction along the incline having magnitude 5 m/s. Find the horizontal and vertical component of
Sol.
cr
velocity of Block 'C'.
Let vx and vy be the horizontal and vertical component of velocity of block C. The component of relative velocity of B and C normal to the surface of contact is zero. 10 + 5 cos 37° – Vx = 0
...(1)
Vx = 14 M/s l1 + l2 + l3 = constant
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d 1 d 2 d 3 0 dt dt dt
(– 10) + (– 5 – 10 cos 37°) + (– 5 sin 37° + Vy) = 0 Vy = 26 m/s Horizontal component of velocity is 14 m/sec and vertical component of velocity is 26 m/sec. 102.
Three equal balls 1, 2, 3 are suspended on springs one below the other.
in
OA is a weightless thread. The balls are in equilibrium If the thread is cut, the system starts falling. Find the
e.
(A)
je
acceleration of all the balls at the initial instant. Find the initial acceleration of all the balls if we cut the spring
iit
(B)
ac k
BC, which is supporting ball 3, instead of cutting the thread. [Ans. (a) 3g , 0, 0, (b) 0, g, g] (a)
cr
Sol.
TAB = 2mg, TBC = mg
For A, 2 mg + mg = maA aA = 3g For B, TAB – mg – TBC = maB maB = 0
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aB = 0 2mg – mg – mg = maB TBC – mg = maC aC = 0 TAB = 2mg
(b)
TAB – mg = maB
in
2mg – mg = maB
e.
Þ aB = g()
ac k
A block 'A' of mass 'm' is attached at one end of a light spring and the other end of the spring is connected to another block 'B' of mass 2m through a light string. 'A' is held and B is static equilibrium. Now A is released. The acceleration of A just after that instant is 'a'. In the next
cr
103.
iit
and ac = g
je
aA = 0
case, B is held and A is in static equilibrium. Now when B is released, its acceleration immediately after the release is 'b'. The value of (pulley, string and the spring are massless)
(A)
0
(B)
undefined
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84
a is: b
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(C) Sol.
2
1 2
(D)
For first case tension in spring will be Ts = 2mg just after 'A' is released. 2mg – mg = ma a=g
in
In second case Ts = mg 2mg – mg = 2mb
e.
g 2
je
b=
ac k
104.
iit
a =2 b
A particle is moving along a straight line with constant acceleration. At the end of tenth second its velocity becomes 20 m/s and in tenth
cr
second it travels a distance of 10 m. Then the acceleration of the particle will be –
Sol.
(A)
10 m/s2 (B)
20 m/s2
(C)
1 m / s2 5
(D)
3.8 m/s2
v = u + at 20 = u + a × 10
…(1)
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Sn = u
a 2n 1 2
10 = u +
a (2 × 10 – 1) 2
…(2)
on solving (1) and (2) we get a = 20 m/s2 105.
A block slides down an inclined plane of slope angle
with constant
in
velocity. If it is then projected up the same plane with an initial speed
mg sin = uk mg cos
ac k
02 = v02 – 2as,
iit
k cos = sin …(i)
je
Sol.
e.
v0 the distance in which it will come to rest is:
where a = g [k cos + sin ]
cr
Using (i), v02 = 2g [sin + sin ] s
s 106.
v20 . 4g sin
A block of mass 4 kg is kept on ground. The co-efficient of friction between the block and the ground is 0.80. An external force of magnitude 30 N is applied parallel to the ground. The resultant force exerted by the ground on the block is: (A)
40 N
(B)
30 N
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(C) Sol.
0N
50 N
(D)
N = mg = 40 N (fs)max = µN = (0.8) (40) = 32 N fs = ext.
in
force = 30
e.
R 2 = N2 + f s2
je
= (50)2
107.
iit
R = 50 Nt.
A body of mass m is kept on a rough fixed inclined plane of angle of
ac k
inclination . It remains stationary. Then magnitude of force acting on the body by the inclined plane is equal to:
(C) Sol.
mg
(B)
mg sin
mg cos
(D)
none
cr
(A)
N = mg cos , fs = mg sin R 2 = N2 + f s 2 R = mg
(A)
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108.
A body of mass 50 kg produces an acceleration of 2 m/s2 in a block of mass 20 kg by pushing it is horizontal direction. If the boy moves with the block without sliding, then find friction force (in Newton) exerted by the surface on the boy. Assume no friction between the block and the surface.
Sol.
For Block F = 20.2 = 40 N As man exerts 40 N force on block, block exerts 40 N force on the man, in
in
opposite direction. As man is also moving with same acceleration
e.
f – 40 = 50.2
je
f = 140 N.
iit
After:
ac k
Consider the man + block system.
The only external force is friction acting on man 'f'
cr
f = (Mman + Mblock) a = 140 N. 109.
A force F = 20 N is applied to a block (at rest) as shown in figure. After the block has moved a distance of 8 m to the right, the direction of horizontal component of the force F is reversed. Find the velocity with block arrives at its starting point.
16 7 m / s Ans. V 3
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Sol.
µN = 2, N = 8, a=
16 2 7 2
V2 = 2 (7) 8
…(i)
when the direction of horizontal component of the force F is reversed
in
a1 = 9 m/s2 () and distance covered by the block before it stops
Again s2 = 8 + s1
16 7 29
ac k
=8
je
e.
16 7 29
iit
S1 =
cr
and a2 = 7 ()
V2 = 0 + 2 (a2) (s2)
16 7 2 9
V
16 7 m/s 3
= 2 7 8
110.
A body of mass 10 kg lies on a rough inclined plane of inclination sin–1
=
3 with the horizontal, when a force of 30 N is applied on the 5
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block parallel to & upward the plane, the total reaction by the plane on the block is nearly along:
Sol.
(A)
OA
(B)
OB
(C)
OC
(D)
OD
Frictional force is along the upward direction –
in
10 g sin – 30 = 30 Nt
e.
N = 10 g cos = 80 Nt.
If the coefficient of friction between A and B is µ, the maximum
cr
111.
4 5
ac k
cos
iit
3 5
sin
je
Direction of R is along OA.
horizontal acceleration of the wedge A for which B will remain at rest w.r.t. the wedge is: (A)
µg
(B)
1 g 1
(C)
g
(D)
1 g 1
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Sol.
FBD of block B w.r.t. wedge A, for maximum 'a': Fy = (mg cos + ma sin – N) = 0. and Fy = mg sin + µN – ma cos = 0 (for maximum a) mg sin + µ (mg cos + ma sin ) – ma cos = 0
g sin g cos cos sin
in
a=
e.
for 0 = 45°
ac k
A bead of mass m is located on a parabolic wire with its axis vertical and vertex directed towards downwards as in figure and whose a
cr
112.
iit
1
a = g Ans. 1
je
tan 45
; a= cot 45
equation is x2 = ay. If coefficient of friction is µ, the highest distance above the x - axis at which the particle will be in equilibrium is.
(A)
µa
(B)
µ 2a
(C)
1 2 µa 4
(D)
1 a 2
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Sol.
(C) For the sliding not to occur when tan < µ
=
2x a
=
2 y y 2 a a
je
y a
iit
2
in
dy dx
e.
tan =
113.
ac k
a2 or y 4
Two blocks A and B are as shown in figure. The minimum horizontal
cr
force F applied on block 'B' for which slipping begins between 'B' and ground is:
Sol.
(A)
100 N
(C)
50 N
(B)
120N (D)
140 N
(A) The sliding shall start at lower surface first
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if F > 0.5 [10 + 10] g or F > 100 N
Comprehension In the situation a wedge of mass m is placed on a rough surface, on which a block of equal mass is placed on the inclined plane of wedge friction coefficient between plane and the block and the ground and
in
the wedge (µ). An external force F is applied horizontally on the wedge.
(C)
> 2 µ mg
je
µ mg
3
(B)
= µ mg 2
(D)
< µ mg
iit
(A)
Wedge will start slipping when
cr
Sol.
The value of F at which wedge will start slipping is:
ac k
114.
e.
Given that m does not slide on incline due to its weight.
F > µ (2 mg) 115.
The value of F at which not friction will act on block on inclined plane, is: (A)
2 µ mg
(B)
2 µ mg + 2 mg tan
(C)
2 µ mg + mg tan
(D)
2 µ mg + mg sin
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Sol.
If wedge start slipping, common acceleration ac =
F 2mg 2m
If mg sin = ma cos then no force along the plane will be felt by the block and hence friction will be zero.
F 2mg mgsin m cos 2m
The minimum value of acceleration of wedge for which the block starts
e.
116.
in
F = 2 mg tan + 2 µmg.
cos sin g sin cos
(B)
sin cos g cos sin
(D)
cr
ac k
iit
(A)
(C)
Sol.
je
sliding on the wedge, is:
sin cos sin cos
cos sin sin cos
Block will start sliding if ma cos > mg sin + µ (mg cos + ma sin ) ma cos – µ ma sin > mg sin + µ mg cos
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a (cos – µ sin ) > g (sin + µ cos )
sin cos
get a g cos sin
sin cos
amin = g cos sin 117.
A block kept on a rough inclined plane. The maximum external force down the incline for which the block remains at rest is 1 N while the
in
maximum external force up the incline for which the block is at rest is 7
3 2
(C)
ac k
(B)
1 6
(D)
4 3 3
iit
(A)
3
cr
Sol.
je
e.
N. The coefficient of static friction µ is:
When block is about to slide down 1 + mg sin = µ mg cos
....(1)
when block is about to slide up mg sin + µ mg cos = 7
....(2)
from (1) and (2)
4 3 3
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118.
A man is sitting inside a moving train and observes the objects outside of the train then choose the single correct choice from the following statements – (A)
all stationary objects outside the train will move with same velocity in opposite direction of the train with respect to the man.
(B)
stationary objects near the train will greater velocity & object
in
far from train will move with lesser velocity with respect to the
(C)
e.
man.
large objects like moon or mountains will move with same
ac k
VO, M VO VM
VO, M VO VTrain
cr
Sol.
all of these.
iit
(D)
je
velocity as that of the train.
VO, M = velocity of object with respect to man VO = velocity of object VM = velocity of man Here velocity of object is zero. So, VO, M VO VM
VO, M VM
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119.
An arrangement of the masses and pulleys is shown in the figure. String connecting masses A and B with pulleys are horizontal and ail pulleys and strings are light. Friction coefficient between the surface and the block B is 0.2 and between blocks A and B is 0.7. The system is released from rest. (use g = 10 m/s2) (A)
The magnitude of acceleration of the system is 2 m/s2 and there is no slipping between block A and block B. The magnitude of friction force between block A and block B is
in
(B)
Acceleration of block C is 3 m/s2 downwards.
(D)
Tension in the string connecting block B and block D is 12 N.
iit
je
(C)
ac k
Suppose block A and move together. Applying NLM on C, A + B, and D 60 – T = 6a
cr
Sol.
e.
42 N.
T – 18 – T' = 9a
Solving a = 2 m/s2
To check slipping between A and B, we have to find friction force in this case. If it is less than limiting static friction, then there will be no slipping between A and B. Applying NLM on A. T – f = 6(2)
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as T = 48 N f = 36 N and fs = 42 N hence A and B move together. and T' = 12N. 120.
Select the incorrect statement: A powerful man pushes a wall and the wall gets deformed. The
(A)
magnitude of force exerted by the wall on man will be equal to
in
the magnitude of forced exerted by man on the wall. Ten person are pulling horizontally an object on a smooth
e.
(B)
je
horizontal surface in different directions. The resultant
iit
acceleration of the object is zero. Then the pull of each man is
(C)
A massless object cannot exert force on any other object.
(D)
All contact forces are electromagnetic in nature.
cr
Sol.
ac k
equal to the pull of the other nine men.
(B) & (C)
The direction of pull of each man is in opposite direction to that of net pull by nine other people. A mass less body can exert force on other bodies if other bodies exert force on it. B & C are correct choices, marks shall be awarded for either correct choice.
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122.
The equivalent resistance of the circuit across points A and B is equal to:
Sol.
(A)
22.5 (B)
(C)
37.5 (D)
Equivalent circuit is
A bulb rated 200 W, 200 V is used at 100 V. Then the number of
e.
123.
in
= 37.5
zero
(C)
3.125 × 1018
(D)
6.25 × 1018
iit
(B)
ac k
3.125 × 1017
V2 W = VI = R
cr
Sol.
(A)
je
electrons passed through bulb in one second is:
200 200 =
2
R
R = 200 When used at 100 VI
V 100 0.5 A — - - - = 0.5A R 200
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Q = l × t = 0.5 C number of electron = 125.
0.5 1.6 1019
Two wedges, each of mass m, are placed next to each other on a fiat horizontal floor. A cube of mass M is balanced on the wedges. Assume no friction between the cube and the wedges, but a coefficient of static friction µ < 1 between the wedges and the floor. What is the largest M
je
m 1
iit
(C)
(B)
(D)
ac k
m 2
m 2
2m 1
The free body diagrams of all bodies are as shown. From FBD of block
cr
Sol.
(A)
e.
in
that can be balanced as shown with out motion of the wedges?
2N cos 45° = mg …(1)
For wedge to remain at rest N sin 45° < µN'
…(2)
and N' = mg + N cos 45°
…(3)
From 1, 2 and 3 we get
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M 126.
2m 1
A solid cube of mass 5 kg is placed on a rough horizontal surface, in xy plane as shown. The friction coefficient between the surface and the
ˆ N is applied on the ˆ 20k cube is 0.4. An external force F 6iˆ 8j cube. (use g = 10 m/s2) The block starts slipping over the surface
(B)
The friction force on the cube by the surface is 10 N.
(C)
The friction force acts in xy-plane at angle 127° with the
e.
in
(A)
The contact force exerted by the surface on the cube is 10 10
iit
(D)
je
positive x-axis in clockwise direction.
N = 50 – 20 = 30 N
Limiting friction force = µN = 12 N and applied force in horizontal
cr
Sol.
ac k
N.
direction is less than the limiting friction force, therefore the block will not slide.
For equilibrium in horizontal direction, friction force must be equal to 10 N. From the top view, it is clear that
= 37° i.e. 127° from the x-axis that is
the direction of the friction force. It is opposite to the applied force. Contact force =
N2 f 2
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cr
ac k
iit
je
e.
in
= 10 10 N
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Comprehension An observer having a gun observes a remotely controlled balloon. When he first noticed the balloon, it was at an altitude of 800 m and moving vertically upward at a constant velocity of 5 m/s. The horizontal displacement of balloon from the observer in 1600 m. shells fired the gun have an initial velocity of 400 m/s at a fixed angle
3 4 and cos = ). The observer having gun waits (for some time 5 5
in
=
(sin
after observing the balloon and fires so as to destroy the balloon.
2 sec
(C)
10 sec (D)
iit
(A)
ac k
Sol.
The flight time of the shell before it strikes the balloon is: (B)
5 sec
15 sec
The motion in the x-direction is a constant velocity motion. We find the
cr
130.
je
e.
Assume g = 10 m/s2 neglect air resistance.
flight time =
1600m ux
1600 5 sec 400 cos Flight time = 5 sec.
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131.
Sol.
The altitude of the collision above ground level is: (A)
1250 m (B)
1325 m
(C)
1075 m (D)
1200 m
From the flight time, the initial velocity in the y - direction and the acceleration in the y - direction, we can calculate the altitude of the shell:
iit
= 1200 – 125 = 1075 m
ac k
= 1075 m. 132.
in
1200 1 5 10 25 5 2
je
=
1 2 gt 2
e.
h uyt
After noticing the balloon, the time for which observer having gun waits before firing the shell is:
(B)
55 sec
(C) 60 sec
(D)
45 sec
cr
(A) 50 sec
Sol.
After the waiting time plus the flight time, the balloon should reach the same altitude as the shell. Let tw be the waiting time. t2 + 5 sec =
1075 800 5
or tw = 50 sec.
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134.
A bucket tide to a string is lowered at a constant acceleration of
g . If 4
the mass of the bucket is M and is lowered by a distance d, the work done by the string will be (assume the string to be massless)
3 Mg d 4
(C)
4 Mg d 3
(D)
4 Mg d 3
in
1 Mg d (B) 4
Let tension is string be T, then work done by tension T = – Td
e.
Sol.
(A)
3 Mg 4
ac k
or T =
iit
g 4
Mg – T = M
je
Applying Newton's second law on the bucket
cr
required work done = 135.
3 Mg d 4
A block of mass 10 kg is released on a fixed wedge inside a cart which is moved with constant velocity 10 m/s towards right. Take initial velocity of block with respect to cart zero. Then work done by normal reaction (with respect to ground) on block in two second will be: (g = 10 m/s2). (A)
zero
(B)
960 J
(C)
1200 J (D)
none of these
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Sol.
Because the acceleration of wedge is zero, the normal reaction exerted by wedge on block is N = mg cos 37°. The acceleration of the block is g sin 37° along the incline and velocity of the block is v = 10 m/s horizontally towards right as shown in figure. The component of velocity of the block normal to the incline is V sin 37°. Hence the displacement of the block normal to the incline in T = 2 second
e.
3 × 2 = 12 m. 5
je
= 10 ×
in
is S = V sin 37° × 2
= 100 ×
ac k
W = mg cos 37° S
4 × 12 5
cr
= 960 J 136.
iit
The work done by normal reaction
The components of a force acting on a particle are varying according to the graphs shown. When the particle moves from (0, 5, 12) to (4, 20, 0) then the work done by this force is:
Sol.
(A)
192 J
(B)
400 J 3
(C)
0
(D)
None of these
From given graphs:
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4 3 4 ˆ F x 10 ˆi 20 y ˆj z 16 k 4 3 3
W = F.ds 4,20,0
=
3 4 ˆ 4 ˆ x 10 20 y j z 16 k 4 3 3 0,5,12
in
ˆ 192 J dx ˆi dy ˆj dzk
e.
Alternating solution:
Work done can also be found by finding area under these curves. A block is attached with a spring and is moving towards a fixed wall
je
137.
iit
with speed v as shown in figure. As the spring reaches the wall, it starts
ac k
compressing. The work done by the spring on the wall during the process of compression is:
(C) Sol.
1 mv2 (B) 2
mv2
k mv
(D)
cr
(A)
zero
As point of application of force is not moving, therefore work done by the forced is zero.
138.
A particle of mass 'm' moves along the quarter section of the circular path whose centre is at the origin. The radius of the circular path is 'a'.
i x ˆj newton acts on the particle, where x, y denote A force F y ˆ
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the coordinates of position of the particle. Calculate the work done by this force is taking the particle from point A (a, 0) to point B (0, a) along the circular path. Work done by force F : F
W = F.dr =
yˆi xˆj . dxˆi dyˆj
=
ydx xdy
in
…(1)
e.
x2 + y2 = a2
je
xdx + y dy = 0
=
x
a
=
2
ac k
iit
ydy W y xdy x y2
x
dy
cr
Sol.
0
a
a2 y2
dy
a2 = 2 Alternate Method:
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It can be observed that the force is tangent to the curve at each point and the magnitude is constant. The direction of force is opposite to the direction of motion of the particle. work done = (force) × (distance)
x2 y2
= a
a 2
a 2
in
=
je
ac k
Two blocks of mass m and 2 m are slowly just placed in contact with each other on a rough fixed inclined plane as shown. Initially both the blocks are at rest on inclined plane. The coefficient of friction between
cr
139.
a2 J 2
iit
Ans. W =
e.
a2 J = 2
either block and inclined surface is µ. There is no friction between both the blocks. Neglect the tendency of rotation of blocks on the inclined surface. Column I gives four situation. Column II gives condition under which statements in column l are true. Match the statement in column I with corresponding conditions in column II and indicate your answer by darkening appropriate bubbles in the 4 × 4 matrix usually given in the OMR she e.
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(A)
The magnitude of acceleration
(p)
µ=0
(q)
µ>0
of both blocks are same if (B)
The normal reaction between both the blocks is zero if
(C)
The net reaction exerted by
(r)
inclined surface on each blocks
in
make same angle with inclined
The net reaction exerted by
(s)
je
(D)
e.
surface (AB) if
iit
inclined surface on block of
ac k
mass 2m is double that of net reaction exerted by inclined
(A) (C) Sol.
(A)
cr
surface on block of mass m if p, q, r, s (B)
p, q, r, s
p, q, r, s (D)
p, q, r, s
For µ > tan , the magnitude of acceleration of both is zero. Hence acceleration of both blocks is same. For µ > tan , the acceleration of both blocks is same and equal to (g sin + µ cos ). Hence whatever be the value of µ, the acceleration of both blocks shall be same.
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For µ > tan both blocks are at rest and their binding with inclined surface is not broken. Hence the blocks cannot exert force on each other. Therefore normal reaction between both blocks is zero. For µ > tan , both blocks will move down the incline with same acceleration when they are not in contact. Hence they have no tendency to approach.
e.
reaction no tendency to approach.
in
Hence when both blocks are in contact, they will not exert normal
je
Hence whatever be the value of µ, normal reaction between both blocks
iit
is zero.
ac k
(C & D) for µ > tan , both blocks are at rest. The normal reaction (N), friction (f) and net reaction on each blocks by inclined surface.
cr
(B)
It is obvious 1 = 2 and R2 = 2R1. For µ < tan , both blocks move down the incline. The normal reaction (N), friction (f) and net reaction on each blocks by inclined surface are as shown. Again it can be seen that 1 = 2 and r2 = 2r1. Hence whatever be the value of µ, R2 = 2R, and
2
=
1.
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140.
STATEMENT-1: For a particle moving along straight line with constant acceleration, magnitude of displacement is less than distance covered in same time interval. Then the magnitude of average velocity will be less than initial speed in that time interval. STATEMENT-2:
(A)
Total Distance covered Total time taken
je
and Average speed =
in
Total Displacement Total time taken
e.
Average velocity =
statement-1 is true, statement-2 is true; statement-2 is a
statement-1 is true, statement-2 is true; statement-2 is not a
ac k
(B)
iit
correct explanation for statement-1.
correct explanation for statement-1.
(D) Sol.
statement-1 is true, statement-2 is False.
cr
(C)
statement-1 is False, statement-2 is True.
Let the direction of initial velocity 'u' and constant acceleration 'a' of the particle be opposite. Then after the particle turns back and acquire a velocity of magnitude larger than 3u, the magnitude of average velocity
u V shall be greater than v. Hence statement-1 is false. 2
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142.
A block of mass m starts from rest at height h on a frictionless inclined plane. The block slides down the plane, travels a distance d on a rough horizontal surface (before coming to rest for the first time) with coefficient of kinetic friction µ & compresses a spring with force constant k through a distance x before momentarily coming to rest. Then the block travels back across the rough surface, sliding up the plane. The correct expression for the maximum height h' which the block reaches
(B)
h' = h + 2µ d
(C)
h' = h + 2 µ d +
(D)
h' = h – 2 µ d –
k 2 x mg
je
h' = h – 2µ d
iit
(A)
e.
in
on its return is:
ac k
WG + WSF + WFF + K.E.
cr
Sol.
k 2 x mg
mg (h – h') + 0 + µ mg (– 2d) = 0 h – h' – 2 µd = 0 h' = h – 2µd (A) 143.
A particle is projected vertically upwards with a speed of 16 m/s, after some time, when it again passes resistance is same during upward and
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downward motion. Then the maximum height attained by the particle is (Take g = 10 m/s2):
Sol.
(A)
8m
(B)
(C)
17.6 m (D)
12.8 m
4.8 m
(A) From work energy theorem for upward motion
in
1 m (16)2 = mgh + W (work by air resistance) 2
je
iit
1 2 m 8 mgh W 2
e.
for downward motion
ac k
1 2 2 16 8 2gh 2
144.
cr
or h = 8 m.
A horse drinks water from a cubical container of side 1 m. The level of the stomach of horse is at 2 m from the ground. Assume that all the water drunk by the horse is at a level of 2 m from the ground. Then minimum work done by the horse in drinking the entire water of the container is (take
Sol.
water
= 1 000 kg/m 3 = 10 m/s2):
(A)
10 kJ
(B)
15 kJ
(C)
20 kJ
(D)
zero
The mass of water is
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= 1 × 103 kg The increase is potential energy of water is = 1 × 103 × 10 × 1.5 = 15000 J = 15 kJ 145.
A man places a chain (of mass 'm' and length 'l') on a table slowly.
in
Initially the lower end of the chain just touch the table. The man drops
(C)
mg 4
mg 8
2
3mg 8
ac k
(C)
(B)
je
mg
(D)
The work done by man in negative of magnitude of decrease in potential
cr
Sol.
(A)
iit
by the man is this process is:
e.
the chain when half of the chain is in vertical position. Then work done
energy of chain.
U mg = 3mg
2
m g 2 4
8
W
3mg 8
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146.
A ball is suspended from the top of a cart by a string of length 1.0 m. The cart and the ball are initially moving to the right at constant speed V, The cart comes to rest after colliding and sticking to a fixed bumper, as in figure II. The suspended ball swings through a maximum angle 60°. The initial speed V is (take g = 10 m/s2)
(C)
5 2 m/s
(D)
2 5 m/s
in
(B)
4 m/s
e.
10 m / s
1 mV2 = mg (L – 2
2gL 1 cos
ac k
V=
je
As string does no work on the ball, energy conservation can be applied
iit
Sol.
(A)
147.
cr
on putting values V = 10 m / s S1: Path of a particle moving along a straight line with respect to the observer moving along another straight line must be straight line. S2: A man is standing inside a lift which is moving upwards. He shall fell more weight as compared to when lift was are rest. S3: A block moves down the inclined surface of a wedge. The wedge lies on smooth horizontal surface. The work done by normal reaction (exerted by wedge on the block) on the block must be zero.
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S4: When ever a block is placed on top of the rough horizontal surface of another block and lower block is moved horizontally, work done by friction on block is always positive.
Sol.
(A)
F F T T (B)
FFFT
(C)
F T F T (D)
TTFT
If the initial relative velocity and relative acceleration are neither parallel nor anti parallel, the particle shall move in any not straight line curve.
in
Hence S1 is false.
e.
If the lift is retarding upwards, he shall feel lighter. Hence S2 is false.
je
For moving wedge, the work done by normal reaction on block (exerted
iit
by wedge) is non zero. Hence S3 is false.
ac k
The friction on 'upper block is in the direction of its motion. w.d. is positive. Hence S4 is true. A light, inextensible string attached to a cart that can slide along a
cr
148.
frictionless horizontal rail aligned along an x axis. The left end of the string is pulled over a pulley, of negligible mass and friction and fixed at height h = 3m from the ground level. The cart slides from X 1 = 3 3 m to x2 = 4m and during the move, tension in the string is kept constant 50 N. Find change in kinetic energy of the cart in joules (Use
Sol.
3 = 1.7)
Displacement of the point of 'A of the string
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=
3 3
2
3 42 32 2
= 6 – 5 = 1m Change in kinetic energy = work done by tension = 50 × 1 = 50 Joule. Two identical blocks A and B are placed on two inclined planes neglect air resistance and other friction.
in
Read the following statements and choose the correct options.
e.
Statement-I:
je
Kinetic energy of 'A' on sliding to J will be greater than the kinetic
ac k
Statement-II:
iit
energy of B on falling to M.
Acceleration of 'A' will be greater than acceleration of 'B' when both are released to slide on inclined plane.
cr
149.
Statement-III:
Work done by external agent to move block slowly from position B to O is negative. (A)
only statement-I is true
(B)
only statement-II is true
(C)
only I and III are true
(D)
only II and III are true
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Sol.
Statement-I: Work done by gravity is same for motion from A to J and B to M for equal mass. So K.E. with be equal. Statement-II: Acceleration = g sin sin A > sin B
h 2
in
h
Wg + Wext = 0
ac k
Wext = – Wg
iit
(Because moved slowly)
je
e.
Statement-III:
From B to O :
150.
cr
Wg is positive so Wert < 0.
Block A is released from rest when the extension in the spring is x 0 (x0 < mg/k). The maximum downwards displacement of the block is (there is no friction): (A)
2Mg 2x0 K
(B)
Mg x0 2K
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2Mg x0 K
(D)
2Mg x0 K
1 kx20 mgh 2 1 2 x0 h 0 2
h
in
=
2Mg 2x0 K
e.
Sol.
(C)
iit
A small ball thrown at an initial velocity v0 at an angle
cr
Q.151.
2x0
ac k
2Mg
= K
je
Maximum downward displacement
horizontal strikes a vertical wall moving towards it at a horizontal velocity v and is bounced to the point from which it was thrown. Determine the time t from the beginning of motion to the moment of impact, neglecting friction losses. Sol.
Since the wall is smooth, the impact against the wall does alter the vertical component of the ball velocity. Therefore, the total time t 1 of motion of the ball is the total time of the ascent and descent of the body thrown upwards at a velocity v0
gravitational field.
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Consequently, t1 = 2v0 sin
. The motion of the ball along the g
horizontal is the sum of two motions. Before the collision with the wall, it moves at a velocity v0 cos . After the collision, it traverses the same distance backwards, but at a different velocity. In order to calculate the velocity of the backward motion of the ball, it should be noted that the velocity at which the ball approaches the wall (along the horizontal) is v0 cos + v. Since the impact is perfectly elastic, the ball moves away
in
from the wall after the collision at a velocity v0 cos + v. Therefore, the
e.
ball has the following horizontal velocity relative to the ground.
je
(v0 cos + v) + v = v0 cos + 2v.
iit
If the time of motion before the impact is t, by equating the distances
ac k
covered by the ball before and after the collision, we obtain the following equation:
cr
v0 cos . t = (t1 — t) (v0 cos + 2v) Since the total time of motion of the ball is t1 = 2v0 sin
t= Q.152.
, we find that g
v0 sin v0 cos 2v . g v0 cos v
A small ball move at a constant velocity v along a horizontal surface and at point A falls into a vertical well of depth H and radius r. The velocity v of the ball forms an angle with the diameter of the well
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drawn through point A (See Figure). Determine the relation between v, H, r, and for which the ball can "get out" of the well after elastic impacts with the walls. Friction losses should be neglected. The top view of the trajectory of the ball. Since the collisions of the ball with the wall and the bottom of the well are elastic, the magnitude of the horizontal component of the ball velocity remains unchanged and equal to v. The horizontal distances between points of two successive
in
collisions are AA1 = A1A2 = A1A2 = ....... = 2r cos . The time between two successive collisions of the ball with the wall of the well is t1 = 2r
e.
. v
je
cos
iit
The vertical component of the ball velocity does not change upon a
ac k
collision with the wall and reverses its sign upon a collision with the bottom. The magnitude of the vertical velocity component for the first impact against the bottom is
cr
Sol.
2gH, and the time of motion from the
top to the bottom of the well is t2 =
2H . g
The vertical plane development of the polyhedron A1A2A3...... On this development, the segments of the trajectory of the ball inside the well are parabolas (complete parabolas are the segments of the trajectory, between successive impacts against the bottom). The ball can "get out" of the well if the moment of the maximum ascent along the parabola
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coincides with the moment of an impact against the wall (i.e. at the moment of maximum ascent, the ball is at point An of the well edge). The time t1 is connected to the time t2 through the following relations: n1 = 2kt2, where n and k are integral and mutually prime numbers. Substituting the values of t1 and t2, we obtain the relation between v1 H, r, and for which the ball can "get out" of the well:
A cannon fires from under a shelter inclined at an angle to the
e.
Q.153.
in
nr cos 2H k . v g
je
horizontal. The cannon is at point A at a distance l from the base of the shelter (point B). The initial velocity of the shell is v0, and its
iit
trajectory lies in the plane of the figure. Determine the maximum
From all possible trajectories of the shell, we choose the one that touches
cr
Sol.
ac k
range Lmax of the shell.
the shelter. Let us analyze the motion of the shell in the coordinate system with the axes directed. The "horizontal" component (along the axis Ax) of the initial velocity of the shell in this system is v0x = v0 cos ( – ), where is the angle formed by the direction of the initial velocity of the shell and the horizontal. Point C at which the trajectory of the shell touches the shelter determines the maximum altitude h' of the shell above the horizontal. Figure shows that h'
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= l sin . The projection of the total velocity v of shell on the axis Ay is zero at this point, and h' =
v20 y g'
,
Where g' = g cos is the "free-fall" acceleration in the coordinate system xAy. Thus,
e.
Hence it follows, in particular, that if
in
v20 sin2 2gl cos sin .
je
v20 2gl cos sin gl sin2
By hypothesis, none of the shell trajectories will touch the shelter, and
iit
the maximum range Lmax
to the horizontal. 4
ac k
=
cr
v20 Here Lmax = . g
If v20 gl sin2 by hypothesis, the angle at which the shell should be fired to touch the shelter will be
t arc sin
gl sin2 . v0
If, moreover, the inequality
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v20 sin2 v20 2gl is satisfied by hypothesis, which means that the condition 1 >
should be fired will be equal to
holds 4
v2 , and Lmak = 0 . If, however, the 4 g
in
inverse inequality
je
e.
v20 sin2 v20 2gl
iit
is known to be valid, which in turn means that 1 <
v20 sin2 g
cr
L max
gl sin2 , v0
ac k
t arc sin
, we have 4
gl sin2 v20 = sin2 arc sin . g v0 Q.154.
The slopes of the windscreen of two motorcars are 1 = 30° and 2 15° respectively. At what ratio
=
v1 of the velocities of the cars will v2
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their drivers see the hailstones bounced by the windscreen of their cars in the vertical direction? Assume that hailstones fall vertically. Sol.
Let us suppose that hall falls along the vertical at a velocity v. In the reference frame fixed to the motorcar, the angle of incidence of hailstones on the windscreen is equal to the angle of reflection. The velocity of a hailstone before it strikes the windscreen is v – v1 Since hailstones are bounced vertically upwards (from the viewpoint of
in
the driver) after the reflection the angle of reflection, and hence the
e.
angle of incidence, is equal to 1(1 is the slope of the windscreen of
v = 21 = cot 21 and 2 v1
je
the motorcar). Consequently, + tan
ac k
two motor cars.
iit
cot 21. Therefore, we obtain the following ratio of the velocities of the
Q.155.
cr
v1 cot 22 3. v2 cot 21
A car must be parked in a small gap between the cars parked in a row along the pavement. Should the car be driven out forwards or backwards for the manoeuvre if only its front wheels can be turned?
Sol.
Let a car get in a small gap between two other cars. It is parked relative to the pavement. Is it easier for the car to be driven out of the gap by forward or backward motion? Since only the front wheels can be turned, the centre O of the circle along which the car is driven out for any
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manoeuvre (forward or backward) always lies on the straight line passing through the centres of the rear wheels of the ear. Consequently, the car being driven out is more likely to touch the hind car during backward motion then the front car during forward motion (the centre of the corresponding circle is shifted backwards relative to the middle of the car). Obviously, a driving out is a driving in inversed in time. Therefore, the car should be driven in a small gap by backward
Q.156.
in
motion. An aeroplane flying along the horizontal at a velocity v0 starts to
e.
ascend, describing a circle in the vertical plane. The velocity of the
je
plane changes with height h above the initial level of motion
iit
according to the law v2 = v20 2a0h. The velocity of the plane at the
ac k
upper point of the trajectory is v1 =
v0 . Determine the acceleration a 2
of the plane at the moment when its velocity is directed vertically
Sol.
cr
upwards.
Let us consider the motion of the plane starting from the moment it goes over to the circular path. By hypothesis at the upper point B of the path, the velocity of the plane is v1 =
v0 , and hence the radius r of 2
the circle described by the plane can be found from the relation
v20 v20 2a0.2r, 4
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Which is obtained from the law of motion of the plane for h = 2r. At point C of the path where the velocity of the plane is directed upwards, the total acceleration will be the sum of the centripetal acceleration aC
v2C (v2C v20 2a0r, where vC is the velocity of the plane at point = r C) and the tangential acceleration at (this acceleration is responsible for the change in the magnitude of the velocity).
in
In order to find the tangential acceleration, let us consider a small displacement of the plane from point C to point C'. Then
e.
v2C' v2C 2a0 r h . Therefore, v2C v2C 2a0h, where
je
h is the change in the altitude of the plane as it goes over to point C'.
iit
Let us divide both sides of the obtained relation by the time interval t
ac k
during which this change takes place.
v2C' v2C 2a h 0 . t t
cr
Then making point C' tend to C and t to zero, we obtain
2a1vC 2a0vC.
Hence a1 = – a0. The total acceleration of the plane at the moment when the velocity has the upward direction is then
v2C 2 a = a0 r
2
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= Q.157.
a0 100 . 3
A bobbin rolls without slipping over a horizontal surface so that the velocity v of the end of the thread (point A) is directed along the horizontal. A board hinged at point B leans against the bobbin. The inner and outer radii of the bobbin are r and R respectively. Determine the angular velocity of the board as a function of an
e.
Let the board touch the bobbin at point C at a certain instant of time. The velocity of point C is the sum of the velocity v0 of the axis O of
je
bobbin and the velocity of point C (relative to point O), which is
iit
tangent to the circle at point C and equal in magnitude to v 0 (since
ac k
Were is no slipping). If the angular velocity of the board at this instant is , the linear velocity of the point of the board touching the bobbin
(See figure). Since the board remains in contact 2
will be R tan–1
cr
Sol.
in
angle .
with the bobbin all the time, the velocity of point C relative to the
= v0 sin . 2
board is directed along the board, whence R tan–1
Since there is no slipping of the bobbin over the horizontal surface, we can write
v0 v . R R r
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Therefore, we obtain the following expression for the angular velocity .
v sin .tan R r 2
Q.158.
in
2v sin2 2 . = R r cos 2 A magnetic tape is wound on an empty spool rotating at a constant
e.
angular velocity. The final radius rf of the winding was found to be
je
three times as large as the initial radius ri The winding time of the
iit
tape is t1. What is the time t2 required for winding a tape whose
The area of the spool occupied by the wound thick tape is S 1 =
rf2 ri2 8ri2. Then the length of the wound tape is l =
cr
Sol.
ac k
thickness is half that of the initial tape?
ri2 S1 8 , where d is the thickness of the thick tape. d d The area of the spool occupied by the wound thin tape is S 2 =
r '2f ri2 , where r 'f the final radius of the wound part in the latter case. Since the lengths of the tapes are equal and the tape thickness in the latter case is half that in the former case, we can write
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l=
2 r '2f ri2 d
, r' r 2 f
2 i
4ri2.
Consequently, the final radius r'f of the wound part in the latter case is r'f =
5 r1.
The numbers of turns N1 and N2 of of the spool for the former and latter winding can be written as
Q.159.
d 2
5 1 t1.
,
in
5 1 r1
iit
where t2 =
e.
2r1 , N2 d
je
N1 =
It was found that the winding radius of a tape on a cassette was
ac k
reduced by half in a time t1 = 20 min of operation. In what time t2 will the winding radius be reduced by half again? Let the initial winding radius be 4r. Then the decrease in the winding
cr
Sol.
area as a result of the reduction of the radius by half (to 2r) will be S = (10 r2 – 4 r2) = 12 r2, Which is equal to the product of the length l1 of the wound tape and its thickness d. The velocity v of the tape is constant during the operation of the tape-recorder, hence l1 = vt2, and we can write 12r2 = vt1d
…(i)
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When the winding radius of the tape on the cassette is reduced by half again (from 2r to r), the winding area is reduced by (4 r2 – r2) = 3 r2, i.e. 3r2 = vt2d
…(ii)
Where t2 is the time during which the winding radius will be reduced in the latter case. Dividing equation (i) and (ii) term wise, we obtain
Two rings O and O' are put on two vertical stationary rods AB and
e.
Q.160.
t1 5 min. 4
in
t2 =
je
A'B' respectively. An inextensible thread is fixed at point A' and on ring O and is passed through ring O' (See figure). Assuming that ring
iit
O' moves downwards at a constant velocity v1, determine the velocity
Let us go over to the reference frame fixed to ring O' c . In this system
cr
Sol.
ac k
v2 of ring O if AOO' = .
the velocity of ring O is
v1 and is directed upwards since the cos
thread is inextensible and is pulled at a constant velocity v 1 relative to ring O c. Therefore, the velocity of ring O relative to the straight line AA' c (which is stationary with respect to the ground) is
2 sin2 2 v1 v1 v1 v2 = cos cos
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and is directed upwards. Q.161.
A weightless inextensible rope rests on a stationary wedge forming an angle with the horizontal (See figure). One end of the rope is fixed to the wall at point A. A small load is attached to the rope at point B. The wedge starts moving to the right with a constant acceleration a. Determine the acceleration a, of the load when it is still on the wedge.
in
By the moment of time t form the beginning of motion, the wedge
e.
at2 covers a distance s = and acquires a velocity vwed = at. During this 2
je
time, the load will move along the wedge over the same distance s, and
iit
its velocity relative to the wedge is vrel = at and directed upwards along
vwed, i.e.
ac k
the wedge. The velocity of the load relative to the ground is v 1 = vrel +
v1 = 2vwed sin
2 sin t, 2 2
cr
Sol.
and the angle formed by the velocity v1 with the horizontal is
= constant. 2
Thus, the load moves in the straight line forming the angle =
2
with the horizontal. The acceleration of the load on the
wedge relative to the ground is
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a1 2a sin Q.162.
. 2
An ant runs from an ant-hill in a straight line so that its velocity is inversely proportional to the distance from the centre of the ant-hill. When the ant is at point A at a distance l1 – 1 m from the centre of the ant-hill its velocity v1 = 2 cm/s. What time will it take the ant to run from point A to point B which is at a distance l1 = 2 m from the
The velocity of the ant varies with time according to a nonlinear law.
e.
Therefore, the mean velocity on different segments of the path will not
je
be the same, and the well-known formulas for mean velocity cannot be
iit
used here.
ac k
Let us divide the path of the ant from point A to point B into small segments traversed in equal time intervals t. Then t =
l l , vm
where vm (l) is the mean velocity over a given segment l. This
cr
Sol.
in
centre of the ant-hill?
formula suggests the idea of the solution of the problem, we plot the dependence of
1 (l) on 1 for the path between points A and B. The vm
graph is segment of a straight line. The hatched area S under this segment is numerically equal to the sought time. Let us calculate this area.
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1 1 v v2 S= 1 l2 l1 2 1
1 l
2 = l2 l1 2v1 2v1 l1
l22 l12 = 2v1l1
in
1l 1 2. v2 v1 l1
e.
Since
je
Thus, the ant reaches point B in the time
= 75 s.
Three school boys, Sam, John, and Nick are on merry-go-rounds-Sam and John occupy diametrically opposite points on a merry-go-round
cr
Q.163.
ac k
iit
4m2 1m2 t= 2 2 m / s 102 1m
of radius r. Nick is on another merry-go-round of radius R. The positions of the boys at the initial instant are shown in figure. Considering that the merry-go-round touch each other and rotate in the same direction at the same angular velocity , determine the nature of motion of Nick from John's point of view and of Sam from Nick's point of view.
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Sol.
Let the merry-go-round turn through a certain angle . We construct a point O (OA = R) such that points 0, S, A and J lie on the same straight line. Then it is clear that ON = R + r at any instant of time. Besides, point O is at rest relative to John (Sam is always opposite to John). Therefore, from John's point of view, Nick translates in a circle of radius R + r with the centre at point O which moves relative to the ground in a circle of radius R with the centre at point A. From Nick's point of view, Sam translates in a circle of radius R + r with the centre
in
at point O which is at rest relative to Nick. However, point O moves
A hoop of radius r rests on a horizontal surface. A similar hoop moves
je
Q.164.
e.
relative to the ground in a circle of radius r with the centre at point B.
iit
past it at a velocity v. Determine the velocity VA of the upper point of
ac k
"intersection" of the hoops as a function of the distance d between their centres, assuming that the hoops are thin, and the second hoop is in contact with the first hoop as it moves past the latter. Since the hoop with the centre at point O is at rest, the velocity VA of
cr
Sol.
the upper point A of "intersection" of the hoops must be directed along the tangent to the circle with centre O1 at any instant of time. At any instant of time, the segment AB divides the distance d = OO1 between the centres, of the hoops into two equal parts, and hence the horizontal projection of the velocity VA is always equal to
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v . 2
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Consequently, the velocity VA forms an angle =
with the 2
horizontal and is given by vA =
v v . 2 cos 2 sin
Since sin = 1 cos2 2
in
d 1 , 2r
.
je
d 1 2r
2
iit
v
cr
ac k
vA =
e.
the velocity of the upper point of "intersection" of the hoops is
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For the free-fall acceleration g (wherever required) should be taken equal to 10 m/s2.
Q.1.
A body with zero initial velocity moves down an inclined plane from a height h and then ascends along the same plane with an initial velocity such that it stops at the same height h. In which case is the time of motion longer?
in
Let us first assume that there is no friction. Then according to the
e.
energy conservation law, the velocity v of the body sliding down the
je
inclined plane from the height h at the foot is equal to the velocity
iit
which must be imparted to the body for its ascent to the same height h. Since for a body moving up and down an inclined plane the
ac k
magnitude of acceleration is the same, the time of ascent will be equal to the time of descent.
If, however, friction is taken into consideration the velocity v 1 of the
cr
Sol.
body at the end of the descent is smaller than the velocity v (due to the work done against friction), while the velocity v2 that has to be imparted to the body for raising it along the inclined plane is larger than v for the same reason. Since the descent and ascent occur with constant (although different) accelerations, and the traversed path is the same, the time t1 of descent and the time t2 of ascent can be found from the formulas
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s=
v1t1 , 2
s=
v2 t 2 , 2
where s is the distance covered along the inclined plane. Since the inequality v1 > v2 is satisfied, it follows that t1 > t2. Thus, in the presence of sliding friction, the time of ascent to the
in
same height.
e.
While solving the problem, we neglected an air drag. Nevertheless, it can easily be shown that if an air drag is present in addition to the
je
force of gravity and the normal reaction of the force of gravity and the
iit
normal reaction of the inclined plane, the time of descent is always
ac k
longer than the time of ascent irrespective of the type of this force. Indeed, if in the process of ascent the body attains an intermediate height h', its velocity v' at this point, required to reach the height h in
cr
the presence of drag, must be higher than the velocity in the absence of drag since a fraction of the kinetic energy will be transformed into heat during the subsequent ascent. The body sliding down from the height h and reaching the height h' will have (due to the work done by the drag force) a velocity v" which is lower than the velocity of the body moving down without a drag. Thus, while passing by the same point on the inclined plane, the ascending body has higher velocity than the descending body. For this reason, the ascending body will
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cover a small distance in the vicinity of point A' in a shorter time than the descending body. Dividing the entire path into small regions, we see that each region will be traversed by the ascending body in a shorter time than by the descending body. Consequently, the total time of ascent will be shorter than the time of descent. Q.2.
At a distance L = 400 m from the traffic light, brakes are applied to a locomotive moving at a velocity v = 54 km/h. Determine the position
in
of the locomotive relative to the traffic light 1 min after the
Since the locomotive moves with a constant deceleration after the
je
Sol.
e.
application of brakes if its acceleration a = – 0.3 m/s2.
v 50 s, during which a
iit
application of brakes, it will come to rest in t =
ac k
v2 375 m. Thus, in 1 min after the it will cover a distance s = 2a application of brakes, the locomotive will be at a distance l = L – s = 25
Q.3.
cr
m from the traffic light.
A helicopter takes off along the vertical with an acceleration a = 3 m/s2 and zero initial velocity. In a certain time r1, the pilot switches off the engine. At the point of take-off, the sound dies away in a time t1 = 30 s. Determine the velocity v of the helicopter at the moment when its engine is switched off, assuming that the velocity c of sound is 320 m/s.
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Sol.
At the moment the pilot switches off the engine, the helicopter is at an
at12 . Since the sound can no longer be heard on the altitude h = 2 ground after a time t2, we obtain the equation t2 = t1
at12 , 2c
where on the right-hand side we have the time of ascent of the
in
helicopter to the altitude h and the time taken for the sound to reach
equation, we find that 2
je
iit
t1 =
c c c a 2 a t2 a .
e.
the ground from the altitude h. Solving the obtained quadratic
ac k
We discard the second root of the equation since it has no physical meaning. The velocity v of the helicopter at the instant when the
cr
engine is switched off can be found from the relation
c 2 c c v = at1 = a 2 t2 a 2 a = Q.4.
c2 2act2 c 80 m / s.
A point mass starts moving in a straight line with a constant acceleration a. At a time t1 after the beginning of motion, the acceleration changes sign, remaining the same in magnitude.
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Determine the time t from the beginning of motion in which the point mass returns to the initial position. Sol.
During a time t1, the point mass moving with an acceleration a will
at12 and will have a velocity v = at1. Let us choose cover a distance s = 2 the x-axis. Here point O marks the beginning of motion, and A is the point at
in
which the body is at the moment t1. Taking into account the sign reversal of the acceleration and applying the formula for the path
e.
length in uniformly varying motion, we determine the time t2 in which
je
the body will return from point A to point O;
ac k
iit
at12 at22 at1t2 , O= 2 2
Where t2 = t1 1
2 .
cr
The time elapsed from the beginning of motion to the moment of return to the initial position can be determined from the formula
t = t1 + t2 = t1 2 2 . Q.5.
Two bodies move in a straight line towards each other at initial velocities v1 and v2 and with constant accelerations a1 and a2 directed against the corresponding velocities at the initial instant. What must be the maximum initial separation lmax between the bodies for which they meet during the motion?
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Sol.
We shall consider the relative motion of the bodies from the view point of the first body. Then at the initial moment, the first body is at rest (it can be at rest the subsequent instants as well), while the second body moves towards it at a velocity v1 + v2. Its acceleration is constant, equal in magnitude to a1 + a2, and is directed against the initial velocity. The condition that the bodies meet indicates that the distance over which the velocity of the second body vanished must be longer than the separation between the bodies at the
in
v1 v2 . lmax = 2 a1 a2
je
Small balls with zero initial velocity fall from a height H =
iit
Q.7.
e.
2
R near the 8
ac k
vertical axis of symmetry on a concave spherical surface of radius R. Assuming that the impacts of the balls against the surface are perfectly elastic prove that after the first impact each ball gets into
Sol.
cr
the lowest point of the spherical surface (the balls do not collide). Let us consider the motion of a bail falling freely from a height H near the symmetry axis starting from the moment it strikes the surface. At the moment of impact, the ball has the initial velocity v0 =
2gh
(since the impact is perfectly elastic), and the direction of the velocity v0 forms an angle 2 with the vertical. Let the displacement of the ball along the horizontal in time t after the impact be s. Then v0 sin 2. t = s. Hence we obtain t =
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s
2gh sin2
, where v0 sin 2 is the horizontal component of
the initial velocity of the bail (the ball does not strike the surface any more during the time t). The height at which the ball will be in time t is
gt2 h h0 v0 cos2.t , 2 Where v0 cos 2 is the vertical component of the initial velocity of the
in
ball. Since the ball starts falling from the height H near the symmetry 2, cos
1, and s R. Taking into account these and other relations
je
2
0, sin 2
e.
axis (the angle is small), we can assume that h0
obtained above, we find the condition for the ball to get at the lowest
iit
point on the spherical surface.
s R , 2gH sin2 2 2gH
ac k
t=
cr
gt2 R R2 h v0t 0. 2 2 16H Hence H = Q.8.
R . 8
A hinged construction consists of three rhombs with the ratio of sides 3 : 2 : 1. Vertex A3 moves in the horizontal direction at a velocity v. Determine the velocities of vertices A1, A2, B2 at the instant when the angles of the construction are 90°.
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By hypothesis, the following proportion is preserved between the lengths l1, l2, and l3 of segments A0A1, A0A2, and A0A3 during the motion l1 : l2 : l3 = 3 : 5 : 6. Therefore, the velocities of points A1, A2, and A3 are to one another as
vA1 : vA2 : vA3 = 3 : 5 : 6, and hence
v 5v , vA2 . 2 6
in
vA1
Let us now consider the velocity of the middle link (A1B2A2C2) at the
e.
instant when the angles of the construction are equal to 90°. In the
je
reference frame moving at a velocity v A1 , the velocity v 'B2 of point B2
iit
is directed at this instant along B2A2. The velocity of point A2 is directed
ac k
along the horizontal and is given by
v 'A2 vA2 vA1
v . 3
cr
Sol.
From the condition of inextensibility of the rod B2A2, it follows that
v 'B2 v 'A2 sin
v 2 . 4 6
We can find the velocity of point B2 relative to a stationary reference frame by using the cosine law.
2 2 17 2 vB22 v2A1 v 'B22 v A1 v 'B2 v. 36 2
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vB2 Q.9.
17 v. 6
The free end of a thread wound on a bobbin of inner radius r and outer radius R is passed round a nail A hammered into the wall. The thread is pulled at a constant velocity v. Find the velocity v 0 of the centre with the vertical, assuming that the bobbin rolls over the horizontal
e.
If the thread is pulled as shown in figure, the bobbin rolls to the right, rotating clockwise about its axis.
je
For point B, the sum of the projections of velocity v0 of translatory
iit
motion and the linear velocity of rotary motion (with an angular
ac k
velocity ) on the direction of the thread is equal to v. v = v0 sin – r.
Since the bobbin is known to move over the horizontal surface without
cr
Sol.
in
surface without slipping.
slipping, the sum of the projections of the corresponding velocities for point C is equal to zero. v0 – R = 0. Solving the obtained equations, we find that the velocity v0 is v0 =
vR . R sin r
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It can be seen that for R sin = r (which corresponds to the case when points A, B and C lie on the same straight line), the expression for v0 becomes meaningless. It should also be noted that the obtained expression describes the motion of the bobbin to the right (when point B is above the straight line AC and R sin > r) as well as to the left (when point B is below the straight line AC and R sin < r). Q.10.
A rigid ingot is pressed between two parallel guides moving in
in
horizontal directions at opposite velocities v1 and v2. At a certain instant of time, the points of contact between the ingot and the
e.
guides lie on a straight line perpendicular to the directions of
je
velocities v1 and v2.
The velocities of the points of the ingot lying on a segment AB at a given instant uniformly vary from v1 at point A to v2 at point B.
cr
Sol.
ac k
v2 at this instant?
iit
What points of the ingot have velocities equal in magnitude to v 1 and
Consequently, the velocity of point O (See figure) at a given instant is zero. Hence point O is an instantaneous centre of rotation. (Since the ingot is three dimensional, point O lies on the instantaneous rotational axis which is perpendicular to the plane of the figure). Clearly, at a given instant, the velocity v1 corresponds to the points of the ingot lying on the circle of radius OA, while the velocity v2 to points lying on the circle of radius OB.
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(In a three dimensional ingot, the points having such velocities lie on cylindrical surfaces with radii OA and OB respectively). Q.11.
A block lying on a long horizontal conveyer belt moving at a constant velocity receives a velocity v0 = 5 m/s relative to the ground in the direction opposite to the direction of motion of the conveyer. After t = 4 s, the velocity of the block becomes equal to the velocity of the belt. The coefficient of friction between the block and the belt is µ =
In order to describe the motion of the block, we choose a reference
e.
Sol.
in
0.2. Determine the velocity v of the conveyer belt.
frame fixed to the conveyer belt. Then the velocity of the block at the
je
initial moment is vb = v0 + v, and the block moves with a constant
iit
acceleration a = – µg. For the instant of time t when the velocity of the
ac k
block vanishes, we obtain the equation 0 = v0 + v – µgt. Hence the velocity of the conveyer belt is v = µgt – v0 = 3 m/s. A body with zero initial velocity slips from the top of an inclined plane forming an angle with the horizontal. The coefficient of friction µ
cr
Q.12.
between the body and the plane increases with the distance l form the top according to the law µ = bl. The body stops before it reaches the end of the plane. Determine the time t from the beginning of motion of the body to the moment when it comes to rest.
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Let us write the equation of motion for the body over the inclined plane. Let the instantaneous coordinate (the displacement of the top of the inclined plane) be x, then ma – mg sin – mg cos . bx. Where m is the mass of the body, and a is its acceleration. The form of the obtained equation of motion resembles that of the equation of vibratory motion for a body suspended on a spring of rigidity k = mg
in
b in the field of the "force of gravity" mg sin . This analogy with
e.
the vibratory motion helps solve the problem.
je
Let us determine the position x0 of the body for which the sum of the forces acting on it is zero. It will be the "equilibrium position" for the
iit
vibratory motion of the body. Obviously, mg sin – mg cos . bx0 = 0.
1
ac k
Hence we obtain x0 = tan . At this moment, the body will have b the velocity v0 which can be obtained from the law of conservation of
cr
Sol.
the mechanical energy of the body.
mv20 kx20 = mg sin . x0 – 2 2 = mg sin . x0 –
mgb cos 2 x0 , 2
v20 = 2gx0 sin - gb x20 cos
g sin2 . = b cos
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In its further motion, the body will be displaced again by x0 (the "amplitude" value of vibrations, which can easily be obtained from the law of conservation of mechanical energy). The frequency of the corresponding vibratory motion can be found from the relation
k = m
gb cos - 20.
1
in
Therefore, having covered the distance x0 = tan after passing b the "equilibrium position", the body comes to rest. At this moment, the
e.
restoring force "vanishes" since it is just the force of sliding friction. As
je
soon as the body stops, sliding friction changes the direction and
iit
becomes static friction equal to mg sin . The coefficient of friction
ac k
between the body and the inclined plane at the point where the body stops is µst = b.2x0 = 2 tan , i.e. is more than enough for the body to remain at rest.
cr
Using the vibrational approach to the description of this motion, we find that the total time of motion of the body is equal to half the "period of vibrations". Therefore,
t
=
T 2 2 20
. gb cos
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Q.13.
A loaded sledge moving over ice gets into a region covered with sand and comes to rest before it passes half its length without turning. Then it acquires an initial velocity by a jerk. Determine the ratio of the braking lengths and braking times before the first stop and after the jerk. The friction Ffr (x) of the loaded sledge is directly proportional to the length x of the part of the sledge stuck in the sand. We write the
in
equation of motion for the sledge decelerated in the sand in the first
x , t
je
ma mg
e.
case.
iit
where m is the mass, a the acceleration, l the length, and µ the
ac k
coefficient of friction of the sledge against the sand. The deceleration
mg µ) having l
to the motion of a load on a spring (of rigidity k =
acquired the velocity v0 in the equilibrium position. Then the time
cr
Sol.
dependence x (t) of the part of the sledge stuck in the sand and its velocity v (t) can be written as x(t) = x0 sin 0t, v(t) = v0 cos 0t, where x0 =
v0 , 0
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0
k m
g . t
It can easily be seen that the time before the sledge comes to rest is
2 0
=
2 . l g
e.
t1 =
in
equal to quarter the "period of vibrations". Therefore,
je
In the second case (after the jerk), the motion can be regarded as if the
iit
sledge stuck in the sand had a velocity v1 > v0 and having traversed the
ac k
distance x0 were decelerated to the velocity v0 (starting from this moment, the second case is observed). The motion of the sledge after the jerk can be represented as a part of the total vibratory motion
cr
according to the law x (t) = x1 sin 0t, v (t) = v1 cos 0t
starting from the instant t2 when the velocity of the sledge becomes equal to v0. As before, x1 =
v1 . Besides, 0
mg 2 mv12 mv20 x0 , 2l 2 2
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whence v1 = x00 2. The distance covered by the sledge after the jerk is x1 – x0 =
1 v1 v0 0
= x0
2 1 .
in
=
v1 v0 0 0
je
iit
x1 x0 2 1. x0
e.
Consequently, the ratio of the braking lengths is
ac k
In order to determine the time of motion of the sledge after the jerk, we must find the time of motion of the sledge from point x0 to point x1 by using the formula x (t) = x1 sin 0t. For this purpose, we determine t2
cr
from the formula x0 = x1 sin 0t2. Since x1 =
2 x0 , 0 t 2
t . Consequently, t2 = 1 . Since 4 40 2
t3 = t1 – t2, where t3 is the time of motion of the sledge after the jerk, we obtain the required ratio of the braking times
t3 1 . t1 2
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Q.14.
A rope is passed round a stationary horizontal log fixed at a certain height above the ground. In order to keep a load of mass m = 6 kg suspended on one end of the rope, the maximum force T1 = 40 N should be applied to the other end of the rope. Determine the minimum force T2 which must be applied to the rope to lift the load. The force of gravity mg = 60 N acting on the load is considerably stronger than the force with which the rope should be pulled to keep
in
the load. This is due to considerable friction of the rope against the log.
e.
At first the friction prevents the load from slipping under the action of the force of gravity. The complete analysis of the distribution of friction
je
acting on the rope is rather complex since the tension of the rope at
iit
points of its contact with the log varies from F1 to mg. In turn, the force
ac k
of pressure exerted by the rope on the log also varies, being proportional at each point to the corresponding local tension of the rope. Accordingly, the friction acting on the rope is determined just by the force of pressure mentioned above. In order to solve the problem,
cr
Sol.
it should be noted, however, that the total friction Ffr. (whose components are proportional to the reaction of the log at each point) will be proportional (with the corresponding proportionality factors) to the tensions of the rope at the ends. In particular, for the maximum tension to the minimum tension is constant for a given arrangement of the rope and the log :
mg 1 since T1 = mg – kmg. T1 1 k
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When we want to lift the load, the ends of the rope as if change places. The friction is now directed against the force T 2 and plays a harmful role. The ratio of the maximum tension (which is now equal to T 2) to the minimum tension (mg) is obviously the same as in the former case :
T2 1 mg . Hence we obtain mg 1 k T1 90 N.
Why is it more difficult to turn the steering wheel of a stationary
e.
Q.15.
T1
2
in
T2
mg =
Let us see what happens when the driver turns the front wheels of a
iit
stationary car (we shall consider only one tyre). At the initial moment,
ac k
the wheel is undeformed (from the point of view of torsion), and the area of the tyre region in contact with the ground is S. By turning the steering wheel, the driver deforms the stationary tyre until the
cr
Sol.
je
motorcar than of a moving car?
moment of force MS applied to the wheel and tensing to turn it becomes larger than the maximum possible moment of static friction acting on the tyre of contact area S. In this case, the forces of friction are perpendicular to the contact plane between the tyre and the ground. Let now the motorcar move. Static frictional forces are applied to the same region of the tyre of area S. They almost attain the maximum values and lie in the plane of the tyre. A small moment of force ~MS
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applied to the wheel is sufficient to turn the wheel since it is now counteracted by the total moment of "oblique" forces of static friction which is considerably smaller than for the stationary car. In fact, in the case of the moving car, the component of the static friction responsible for the torque preventing the wheel from being turned is similar to liquid friction since stagnation is not observed for turning wheels of a moving wheel, and the higher the velocity (the closer the static friction
A certain constant force starts acting on a body moving at a constant
e.
Q.16.
in
to the limiting value), the more easily can the wheel be turned.
je
by half and after the same time interval, the velocity is again reduced
iit
by half. Determine the velocity vf
Let us choose the reference frame as shown in figure. Suppose that vector OA is the vector of the initial velocity v. Then vector AB is the
cr
Sol.
ac k
from the moment when the constant force starts acting.
t. Since the force acting on
the body is constant, vector equal to vector AB is the change in velocity during the next time interval t. Therefore, in the time interval 3t after the beginning of action of the force, the direction of velocity will be represented by OD and AB BC CD. Let the projections of vector AB on the x and y-axis be vx and vy. Then we obtain two equation.
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v2 v , 4
v vx
2
2 y
v 2vx
2
2vy
2
v2 . 16
Since the final velocity satisfies the relation
v2f v 3vx 3vy , 2
2
A heavy ball of mass m is tied to a weightless thread of length l. The
je
Q.17.
17 v. 4
e.
vf
in
Using the above equations, we get
iit
friction of the ball against air is proportional to its velocity relative to
ac k
the air. Ffr = µv. A strong horizontal wind is blowing at a constant velocity v.
cr
Determine the period T of small oscillations, assuming that the oscillations of the ball attenuate in a time much longer than the period of oscillations. Sol.
It can easily be seen that the ball attains the equilibrium position at an angle of deflection of the thread from the vertical, which is determined from the condition tan =
v . mg
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During the oscillatory motion of the ball, it will experience the action of a constant large force F =
mg
2
v and a small drag force. 2
Consequently, the motion of the ball will be equivalent to a weakly attenuating motion of a simple pendulum with a free-fall acceleration g' given by
mg
2
v
2
in
=g
g cos
mg
e.
g' =
ac k
iit
je
2
v . = g 1 mg
The period of small (but still damped) oscillations of the ball can be
cr
determined from the relation T=
2
g' 2 l 4m2
=
2 2
v 2 g 1 l mg 4m2
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Q.18.
For the system at rest, determine the accelerations of all the loads immediately after the lower thread keeping immediately after the lower thread keeping the system in equilibrium has been cut. Assume that the threads are weightless and inextensible, the springs are weightless, the mass of the pulley is negligibly small, and there is no friction at the point of suspension. We assume that the condition m1 + m2 > m3 + m4 is satisfied, otherwise
in
the equilibrium is impossible. The left spring was stretched with the force T1 balancing the force of gravity m2g of the load, T1 = m2g. The
e.
equilibrium condition for the load m3 was
je
m3g + T2 – Ften = 0,
iit
where T2 is the tension of the right spring and Ften is the tension of the
ac k
rope passed through the pulley. This rope holds the loads of mass m1 and m2, whence
Ften = (m1 + m2) g.
cr
Sol.
We can express the tension T2 in the following way. T2 = (m1 + m2 – m3) g. After cutting the lower thread, the equations of motion for all the loads can be written as follows, m1a1 = m1g + T1 – Ften, m2a2 = m2g – T1, m3a3 – T2 + m3g – Ften,
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– m4a4 = m4g – T2, Using the expressions for the forces T1, T2 and Ften obtained above, we find that a1 = a2 = a3 = 0, a4 = Q.19.
m3 m4 m1 m2 g . m4
A person hoists one of two loads of equal mass at a constant velocity
in
v. At the moment when the two loads are at the same height h, the
e.
upper pulley is released (is able to rotate without friction like the
je
lower pulley).
iit
Indicate the load which touches the floor first after a certain time t,
ac k
assuming that the person continues to slack the rope at the same constant velocity v. The masses of the pulleys and the ropes and the elongation of the roes should be neglected. Immediately after releasing the upper pulley, the left load has a
cr
Sol.
velocity v directed upwards, while the right pulley remains at rest. The accelerations of the loads will be as if the free end of the rope were fixed instead of moving at a constant velocity. They be found from the following equations. ma1 = T1 – mg, ma2 = T2 – mg, 2T1 = T2,
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a1 = – 2a2 Where m is the mass of each load, and T1 and T2 are the tensions of the ropes acting on the left and right loads. Solving the system of
2
1
equations, we obtain a1 = – g and a2 = g. Thus, the 5 5 acceleration of the left load is directed downwards, while that of the right load upwards. The time of fall of the left load can be found from
je
6.25 v2 5h . g2 g
iit
2.5 v t= g
e.
0.4 gt2 h vt 0, whence 2
in
the equation
ac k
During this time, the right load will move upwards. Consequently, the left load will be the first to touch the floor. A block can slide along an inclined plane in various directions. If it
cr
Q.20.
receives a certain initial velocity v directed downwards along the inclined plane, its motion will be uniformly decelerated, and it comes to rest after traversing a distance l1. If the velocity of the same magnitude is imparted to it in the upward direction, it comes to rest after traversing a distance l2. At the bottom of the inclined plane, a perfectly smooth horizontal guide is fixed.
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Find the distance l traversed by the block over the inclined plane along the guide if the initial velocity of the same magnitude is imparted to it in the horizontal direction? Each time the block will move along the inclined plane with a constant acceleration, the magnitudes of the accelerations for the downward and upward motion and the motion along the horizontal guide will be respectively
in
a1 = µg cos – g sin ,
e.
a2 = µg cos + g sin ,
je
a = µg cos
iit
Here is the slope of the inclined plane and the horizontal, and µ is
a=
ac k
the coefficient of friction. Hence we obtain
a1 a2 . 2
The distances traversed by the block in uniformly varying motion at the
cr
Sol.
initial velocity v before it stops can be written in the form l1 =
v2 , 2a1
v2 l2 = , 2a2
v2 . l= 2a
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Taking into account the relations for the accelerations a1, a2 and a, we can find the distance l traversed by the block along the horizontal guide. l=
2l1l2 . l1 l2
Q.21. horizontal. The time of the ascent of the block to the upper point was
in
found to be half tile time of its descent to the initial point.
e.
Determine the coefficient of friction µ between the block and the
We shall write the equations of motion for the block in terms of
iit
projections on the axis directed downwards along the inclined plane.
ac k
For the upwards motion of the block, we take into account all the forces acting on it, the force of gravity mg, the normal reaction N, and friction Ffr, and obtain the following equation.
cr
Sol.
je
roof.
mg sin + µmg cos = ma1. The corresponding equation for the downward motion is mg sin – µmg cos = ma2. Let the distance traversed by the block in the upward and downward motion be s. Then the time of ascent t1 and descent t2 can be determined from the equations
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a1t12 , s= 2 a2t22 . s= 2 By hypothesis, 2t1 = t2, whence 4a2 = a1. Consequently, g sin + µg cos = 4 (g sin – µg cos ), and finally µ = 0.6 tan .
in
Two balls are placed on a "weightless" support formed by tow
e.
Q.22.
je
horizontal. The support can slide without friction along a horizontal
iit
plane. The upper ball of mass m1 is released. Determine the condition under which the lower ball of mass m2 starts
If the lower ball is very light, it starts climbing the support. We shall find its minimum mass m2 for which it has not yet started climbing but
cr
Sol.
ac k
"climbing" up the support.
has stopped pressing against the right inclined plane. Since the support is weightless, the horizontal components of the forces of pressure (equal in magnitude to the normal reactions) exerted by the balls on the support must be equal, otherwise, the "support" would acquire an infinitely large acceleration. N1 sin = N2 sin , N1 = N2.
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Moreover, since the lower ball does not ascend, the normal components of the accelerations of the balls relative to the right inclined plane must be equal (there is no relative displacement in this directions). Figure shows that the angle between the direction of the normal reaction N2 of the support and the right inclined plane is
2, and hence the latter condition can be written in the form 2
e.
in
m1gcos N1 m2gcos N1 cos2 , m1 m2
Whence m2 = m1 cos 2. Thus, the lower ball will "climb" up if the
A cylinder of mass m and radius r rests on two supports of the same
ac k
Q.23.
iit
m2 < m1 cos 2.
je
following condition is satisfied.
height. One support is stationary, while the other slides from under
cr
the cylinder at a velocity v.
Determine the force of normal pressure N exerted by the cylinder on the stationary support at the moment when the distance between points A and B of the supports is AB = r 2, assuming that the supports were very close to each other at the initial instant. The friction between the cylinder and the supports should be neglected. Sol.
As long as the cylinder is in contact with the supports, the axis of the cylinder will be exactly at the midpoint between the supports.
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Consequently, the horizontal component of the cylinder velocity is
v . 2
Since all points of the cylinder axis move in a circle with the centre at point A, the total velocity u of each point on the axis is perpendicular to the radius OA = r at any instant of time. Consequently, all points of
u2 . the axis move with a centripetal acceleration ac = r
projections on the "centripetal" axis.
e.
mg cos – N = mac
in
We shall write the equation of motion for point O in terms of
iit
je
mu2 ....(i) = r
ac k
Where N is the normal reaction of the stationary support. The condition that the separation between the supports is r 2 implies
cr
that the normal reaction of the movable support gives no contribution to the projections on the "centripetal" axis. According to Newton's third law, the cylinder exerts the force of the same magnitude on the stationary support. From equation (i), we obtain
mu2 N = mg cos – . r At the moment when the distance between points A and B of the supports is AB = r 2 , we have
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cos
r 2 1 . 2r 2
The horizontal component of the velocity of point O is u cos = whence u = v 2. Thus, for AB = r
v , 2
2 , the force of normal pressure
exerted by the cylinder is
in
mg mv2 N= . 2r 2
e.
For the cylinder to remain in contact with the supports until AB
Q.24.
gr 2.
ac k
i.e. v
iit
je
g v2 must be satisfied, becomes equal to r 2 , the condition 2 2r
A cylinder and a wedge with a vertical face, touching each other, move along two smooth inclined planes forming the same angle
cr
with the horizontal. The masses of the cylinder and the wedge are m1 and m2 respectively. Determine the force of normal pressure N exerted by the wedge on the cylinder, neglecting the friction between them. Sol.
The cylinder is acted upon by the force of gravity m1g, the normal reaction N1 of the left inclined plane, and the normal reaction N3 of the wedge (force N3 has the horizontal direction). We shall write the
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equation of motion of the cylinder in terms of projections on the x 1axis directed along the left inclined plane. = m1a1 = m1g sin – N3 cos
…(i)
where a1 is the projection of the acceleration of the cylinder on the x1axis. The wedge is acted upon by the force of gravity m2g, the normal reaction N2 of the right inclined plane, and the normal reaction of the cylinder, which according to Newton's third law, is equal to – N3. We
in
shall write the equation of motion of the wedge in terms of projections
....(ii)
je
m2g2 = – m2g sin + N cos
e.
on the x2-axis directed along the right inclined plane.
During its motion, the wedge is in contact with the cylinder. Therefore,
iit
if the displacement of the wedge along the x2-axis is x, the centre of
ac k
the cylinder (together with the vertical face of the wedge) will be displaced along the horizontal by x cos . The centre of the cylinder
cr
will be thereby displaced along the left inclined plane (x 1-axis) by x. This means that in the process of motion of the wedge and the cylinder, the relation a1 = a2 ....(iii) is satisfied. Solving Equations (i)-(iii) simultaneously, we determine the force of normal pressure N = N3 exerted by the wedge on the cylinder.
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N3 = Q.25.
2m1m2 tan . m1 m2
A weightless rod of length l with a small load of mass m at the end is hinged at point A and occupies a strictly vertical position, touching a body of mass M. A light jerk sets the system in motion. For what mass ratio
with the 6
M m
in
horizontal at the moment of the separation from the body? What will be the velocity u of the body at this moment? Friction should be
je
As long as the load touches the body, the velocity of the latter is equal
iit
to the horizontal component of the velocity of the load, and the
ac k
acceleration of the body is equal to the horizontal component of the acceleration of the load.
Let a be the total acceleration of the load. Then we can write a = at + ac,
cr
Sol.
e.
neglected.
where ac is the centripetal acceleration of the load moving in the circle
v2 , where v is the velocity of the load. The of radius l, i.e. ac = l horizontal component of the acceleration is
v2 cos . ah = at sin – l The body also has the same acceleration. We can write the equation of motion for the body.
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v2 cos , N = Mah = Mat sin – M l Where N is the force of normal pressure exerted by the load on the body. At the moment of separation of the load, N = 0 and at sin =
v2 l
cos . The acceleration component at at the moment of separation of the
in
load is only due to the force of gravity.
e.
at = g cos .
gl sin , and the velocity of the body at the same moment is u = gl sin .
ac k
v sin = sin
iit
v=
je
Thus, the velocity of the load at the moment of separation is
According to the energy conservation law, we have
cr
mv2 Mv2 sin2 . mgl = mgl sin + 2 2 2 Substituting the obtained expression for v at the moment of separation and the value of sin = sin
1 into this equation, we obtain the 6 2
ratio.
M 2 3 sin 4. m sin3 The velocity of the body at the moment of separation is
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u = v sin = Q.26.
1 2
gl . 2
A ball moving at a velocity v = 10 m/s hits the foot of a football player. Determine the velocity u with which the foot should move for the ball impinging on it to come to a halt, assuming that the mass of the ball is much smaller than the mass of the foot and that the impact is perfectly elastic. If the foot of the football player moves at a velocity u at the moment of
in
Sol.
e.
kick, the velocity of the ball is v + u (the axis of motion is directed along the motion of the ball) in the reference frame fixed to the point of the
je
player. After the perfectly elastic impact, the velocity of the ball in the
iit
same reference frame will be – (v + u), and its velocity relative to the
ac k
ground will be – (v + u) – u. If the ball comes to a halt after the impact, v + 2u = 0, where u = –
v = – 5 m/s. The minus sign indicate that the 2
cr
foot of the sportsman must move in the same direction as that of the ball before the impact. Q.27.
A body of mass m freely falls to the ground. A heavy bullet of mass M shot along the horizontal hits the falling body and sticks in it. How will the time of fall of the body to the ground change? Determine the time t of fall if the bullet is known to hit the body at the moment it traverses half the distance, and the time of free fall
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from this height is t0. Assume that the mass of the body (M > m). The air drag should be neglected. Since in accordance with the momentum conservation law, the vertical component of the velocity of the body-bullet system decreases after the bullet has hit the body, the time of fall of the body to the ground will increase. In order to determine this time, we shall find the time t 1 of fall of the
in
body before the bullet hits it and the time t2 of the motion of the body
e.
with the bullet. Let t0 be the time of free fall of the body from the height h. Then the time in which the body falls without a bullet is t 1 =
iit
je
h t 0 . At the moment the bullet of mass M hits the body of mass g 2
mv =
ac k
m, the momentum of the body is directed vertically downwards and is
mgt0 . 2
cr
Sol.
The horizontally flying bullet hitting the body will not change the vertical component of the momentum of the formed system, and hence the vertical component of the velocity of the body-bullet system will be
u
v=
m mM
m mM
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g
t0 . 2
The time t2 required for the body-bullet system .to traverse the remaining half the distance can be determined form the equation
h gt22 ut2 . 2 2
m2 m M m 2
mM
.
e.
t t= 0 2
in
This gives
Q.28.
m2 m M M
iit
2
mM
t0 2
ac k
t t= 0 2
je
Thus, the total time of fall of the body to the ground (M > m) will be
Two bodies of mass m1 = 1 kg and m2 = 2 kg move towards each other
cr
in mutually perpendicular directions at velocities v1 = 3 m/s and v2 = 2 m/s. As a result of collision, the bodies stick to together. Determine the amount of heat Q liberated as a result of collision. Sol.
In order to solve the problem, we shall use the momentum conservation law for the system. We choose the coordinate system, the x-axis is directed along the velocity v1 of the body of mass m1 and the y-axis is directed along the velocity v2 of the body of mass m2. After the collision, the bodies will stick together and fly at a velocity u. Therefore,
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m1v1 = (m1 + m2) ux, m2v2 = (m1 + m2) uy, The kinetic energy of the system before the collision was
m1v12 m2v22 W'k . 2 2 The kinetic energy of the system after the collision (sticking together) of the bodies will become
je
m1v12 m2v22 . 2 m1 m2
in
iit
=
m1 m2 2 ux u2y 2
e.
W"k
ac k
Thus, the amount of heat liberated as a result of collision will be Q = W'k – W"k
Q.29.
m1m2 u12 u22 2 m1 m2
cr
=
4.3 J.
The inclined surface M are smoothly conjugated with the horizontal plane. A washer of mass m slides down the left wedge from a height h.
To what maximum height will the washer rise along the right wedge? Friction should be neglected.
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Since there is no friction, external forces do not act on the system under consideration in the horizontal direction. In order to determine the velocity v of the left wedge and the velocity u of the washer immediately after the descent, we can use the energy and momentum conservation laws.
Mv2 mu2 = mgh, 2 2
in
Mv = mu.
e.
Since at the moment of maximum ascent hmax of the washer along the right wedge, the velocities of the washer and the wedge will be equal,
je
the momentum conservation law can be written in the form
iit
mu = (M + m) V,
ac k
where V is the total velocity of the washer and the right wedge. Let us also use the energy conservation law.
mu2 M m 2 V mghmax. 2 2
cr
Sol.
The joint solution of the last two equations leads to the expression for the maximum height hmax of the ascent of the washer along the right wedge. hmax = h
M2
M m
2
.
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Q.30.
A symmetric block of mass m1 with a notch of hemispherical shape of radius r rests on a smooth horizontal surface near the wall. A small washer of mass m2 slides without friction from the initial position. Find the maximum velocity of the block. The block will touch the wall until the washer comes to the lowest position. By this instant of time, the washer has acquired the velocity v which can be determined from the energy conservation law : v 2 = 2gr.
in
During the subsequent motion of the system, the washer will "climb"
e.
the right-hand side of the block, accelerating it all the time in the rightward direction until the velocities of the washer and the block
je
become equal.
iit
Then the washer will slide down the block; the block being accelerated
ac k
until the washer passes through the lowest position. Thus, the block will have the maximum velocity. Thus, the block will have the maximum velocity at the instants at which the washer passes through the lowest position during its backward motion relative to the block.
cr
Sol.
In order to calculate the maximum velocity of the block, we shall write the momentum conservation law for the instant at which the block is separated from the wall.
m2 2gr m1v1 m2v2, and the energy conservation law for the instant at which the washer passes through the lowest position.
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m1v12 m2v22 m2gr . 2 2 This system of equations has two solutions. v1 = 0,
(2)
v1 =
2m2 2gr, m1 m2
v2 =
m2 m1 2gr. m1 m2
in
2gr,
v2 =
e.
(1)
je
Solution (1) corresponds to the instants at which the washer moves
iit
and the block is at rest. We are interested in solution (2) corresponding
Q.31.
2m2 2gr . m1 m2
cr
v1max =
ac k
to the instants when the block has the maximum velocity:
A ball is made to fall freely under gravity from a height 80 m on a ground. It makes inelastic collision with the ground having coefficient of restitution e = 0.4. Find the height attained by the ball after first impact.
Ans.
h = e2 H = (0.4)2 × 80 = 12.8 m Ans.
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Q.32.
A horizontal weightless rod of length 3l is suspended on two vertical strings. Two loads of mass m1 and m2 are in equilibrium at equal distances from each other and from the ends of the strings. Determine the tension T of the left string at the instant when the right strings snaps. At the moment of snapping of the right string, the rod is acted upon by the tension T of the left string and the forces N1 and N2 of normal
in
pressure of the loads of mass m1 and m2. Since the rod is weightless (its
e.
mass is zero), the equations of its translatory and rotary motions will
N1l = 2N2l.
iit
– T + N1 – N2 = 0,
je
have the form
ac k
The second equation (the condition of equality to zero of the sum of all moments of force about point O) implies that N1 = 2N2
cr
Sol.
…(1)
Combining these conditions, we get T = N 1 – N 2 = N2
…(2)
At the moment of snapping of the right string, the accelerations of the loads of mass m1 and m2 will be vertical (point O is stationary, and the rod is inextensible) and connected through the relation a2 = 2a1 …(3) Let us write the equations of motion for the loads at this instant
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m1g – N'1 = m1a1, m2g + N'2 = m2a2 where N'1 and N'2 are the normal reactions of the rod on the loads of mass m1 and m2. Since N'1 = N1 and N'2 = N2, we have m1g – 2N2 = m1a1,
in
m2g + N2 = 2m2a1.
e.
Hence we can find N2, and consequently the tension of the string
A ring of mass m connecting freely two identical thin hoops of mass
ac k
Q.33.
m1m2 g. m1 4m2
iit
=
je
T = N2
M each starts sliding down. The hoops move apart over a rough
cr
horizontal surface.
Determine the acceleration a of the ring at the initial instant if AO1O2 = , neglecting the friction between the ring and the hoops. Sol.
Let the ring move down from point A by a distance x during a small time interval t elapsed after the beginning of motion of the system and acquire a velocity v. The velocity of translatory motion of the hoops at this moment must be equal to u = v tan (t is so small that the angle
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practically
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remains unchanged). Consequently, the linear velocity of all points of the hoops must have the same magnitude. According to the energy conservation law, we have
mv2 mg x = 2Mu + 2 2
= 2Mv2 tan2 +
mv2 , 2
in
where Mu2 is the kinetic energy of each hoop at a given instant. From
iit
1 g. M 2 1 4 tan m
ac k
=
je
v2 m g 2x 4Mtan2 m
e.
this equality, we obtain
cr
As x 0, we can assume that v2 = 2a x, where a is the acceleration of the ring at the initial instant of time. Consequently, a=
Q.34.
1 g. M 2 1 4 tan m
A smooth washer impinges at a velocity v on a group of three smooth identical blocks resting on a smooth horizontal surface. The mass of each block is equal to the mass of the washer.
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The diameter of the washer and its height are' equal to the edge of the block. Determine the velocities of all the bodies after the impact. Sol.
It is clear that at the moment of impact, only the extreme blocks come in contact with the washer. The force acting on each such block is perpendicular to the contact surface between the washer and a block and passes through its centre (the diameter of the washer is equal to the edge of the block!). Therefore, the middle block remains at rest as
in
a result of the impact. For the extreme blocks and the washer, we can write the conservation law for the momentum in the direction of the
je
2mu 2 mv '. 2
iit
mv =
e.
velocity v of the washer.
ac k
Here m is the mass of each block and the washer, v' is the velocity of the washer after the impact, and u is the velocity of each extremes
cr
block. The energy conservation law implies that v2 = 2u2 + v'2.
As a result, we find that u = v 2 and v' = 0. Consequently, the velocities of the extreme blocks after the impact form the angles of 45° with the velocity v, the washer stops, and the middle block remains at rest. Q.35.
Several identical balls are at rest in a smooth stationary horizontal circular pipe. One of the balls explodes, disintegrating into two fragments of different masses.
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Determine the final velocity of the body formed as a result of all collisions, assuming that the collisions are perfectly inelastic. Sol.
In this case, the momentum conservation law can be applied in a peculiar form. As a result of explosion, the momentum component of the ball along the pipe remains equal to zero since there is no friction, and the reaction forces are directed at right angles to the velocities of the fragments. Inelastic collision do not change longitudinal
in
momentum component either. Consequently, the final velocity of the
Q.36.
e.
body formed after all collisions is zero.
Three small bodies with the mass ratio 3:4:5 (the mass of the lightest
je
body is m) are kept at three different points on the inner surface of a
iit
smooth hemispherical cup of radius r. The cup is fixed at its lowest
ac k
point on a horizontal surface. At a certain instant, the bodies are released.
Determine the maximum amount of heat Q that can be liberated in
cr
such a system. At what initial arrangement of the bodies will the amount of liberated heat be maximum? Assume that collisions are perfectly inelastic. Sol.
For the liberated amount of heat to be maximum, the following conditions must be satisfied: (1)
the potential energy of the bodies must be maximum at the
initial moment.
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(2)
the bodies must collide simultaneously at the lowest point of
the cup. (3)
the velocity of the bodies must be zero immediately after the
collision. If these conditions satisfied, the whole of the initial potential energy of the bodies wilt be transformed into heat. Consequently, at the initial instant the bodies must be arranged on the brim of the cup at a height
in
r above the lowest point The arrangement of the bodies must be such
e.
that their total momentum before the collision is zero (in this case, the body formed as a result of collision from the bodies stuck together will
je
remain at rest at the bottom of the cup). Since the values of the
iit
moment a of the bodies at any instant are to one another as 3 : 4 : 5,
ac k
the arrangement of the bodies at the initial instant. After the bodies are left to themselves, the amount of heat Q liberated in the system is maximum and equal to 4 mgr.
Why is it recommended that the air pressure in motorcar tyres be
cr
Q.37.
reduced for a motion of the motorcar over sand? Sol.
The tyres of a motorcar leave a trace in the sand. The higher the pressure on the sand, the deeper the trace, and the higher the probability that the car gets stuck. If the tyres are deflated considerably, the area of contact between the tyres and the sand increases. In this case, the pressure on the sand decreases, and the track becomes more shallow.
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Q.38.
An inextensible rope tied to the axle of a wheel of mass m and radius r is pulled in the horizontal direction in the plane of the wheel. The wheel rolls without jumping over a grid consisting of parallel horizontal rods arranged at a distance l from one another (l < r). Find the average tension T of the rope at which the wheel moves at a constant velocity v, assuming the mass of the wheel to be concentrated at its axle.
in
Let us suppose that at a certain instant, the wheel is in one of the
e.
positions such that its centre of mass is above a rod, and its velocity is v. At the moment of the impact against the next rod, the centre of
je
mass of the wheel has a certain velocity v' perpendicular to the line
iit
connecting it to the previous rod. This velocity can be obtained form
ac k
the energy conservation law.
mv2 mv '2 . mgh + 2 2
cr
Sol.
l2 l2 . Therefore. Here h = r – r 4 8r 2
gl2 . v' = v 1 4rv2 By hypothesis (the motion is without jumps), the impact of the wheel against the rod is perfectly inelastic. This means that during the impact, the projection of the momentum of the wheel on the straight line
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connecting the centre of the wheel to the rod vanishes. Thus, during each collision, the energy
m v ' sin W , 2 2
Where sin
l , is lost (converted into heat). For the velocity v to r
remain constant, the work done by the tension T of the rope over the
e.
gl2 l2 1 , whence 4rv2 r2
je
mv2 Tl = 2
in
path l must compensate for this energy loss. Therefore,
Q.39.
ac k
iit
mv2l gl2 mv2l T= 1 , 2r2 4rv2 2r2
Two coupled wheels (i.e. light wheels of radius r fixed to a thin heavy axle) roll without slipping at a velocity v perpendicular to the
cr
boundary over a rough horizontal plane changing into an inclined plane of slope .
Determine the value of v at which the coupled wheels roll from the horizontal to the inclined plane without being separated from the surface. Sol.
Since the wheels move without slipping, the axle of the coupled wheels rotates about point O while passing through the boundary between the planes. At the moment of separation, the force of pressure of the
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coupled wheels on the plane and the force of friction are equal to zero,
found from the condition
mv12 . mg cos = r From the energy conservation law, we obtain
in
mv2 mv12 mgr 1 cos . 2 2
je
not smaller than , and hence
e.
No separation occurs if the angle determined from these equations is
iit
cos < cos .
v<
ac k
Therefore, we find that the condition
gr 3 cos 2
cr
is a condition of crossing the boundary between the planes by the
2
wheels without separation. If 3 cos –2 < 0, i.e. > arc cos , the 3 separation will take place at any velocity v. Q.40.
Two small balls of the same size and of mass m1 and m2 (m1 > m2) are tied by a thin weightless thread and dropped from a balloon. Determine the tension T of the thread during the flight after the motion of the balls has become steady-state.
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Sol.
The steady-state motion of the system in air will be the falling of the balls along the vertical at a constant velocity. The air drag F acting on the lower (heavier) and the upper ball is the same since the balls have the same velocity and size. Therefore, the equations of motion for the balls can be written in the form m1g – T – F = 0, m2g + T – F = 0.
e.
2
A ball is tied by a weightless inextensible thread to a fixed cylinder of
iit
Q.41.
m1 m2 g .
je
T=
in
Solving this system of equations, we obtain the tension of the thread.
radius r. At the initial moment, the thread is wound so that the ball
ac k
touches the cylinder. Then the ball acquires a velocity v in the radial direction, and the thread starts unwinding.
cr
Find the length l of the unwound segment of the thread by the instant of time t, neglecting the force of gravity. Sol.
At each instant of time, the instantaneous axis of rotation of the ball passes through the point of contact between the thread and the cylinder. This means that the tension of the thread is perpendicular to the velocity of the ball, and hence it does no work. Therefore, the kinetic energy of the ball does not change, and the magnitude of its velocity remains equal to v.
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In order to determine the dependence l (t), we mentally divide the segment of the thread unwound by the instant t into a very large number N of small equal pieces of length l =
l each. Let the time N
during this time, the end of the thread has been displaced by a n,
and the thread has turned through an angle n =
v tn . The radius drawn to the point of contact between the thread n l
n l tn . Then vr
ac k
iit
2
e.
l , whence r
je
n = =
in
and the cylinder has turned through the same angle, i.e.
t = t1 + t2 + ……… + tn
l l 2 l N l ........ = vr vr vr 2
2
cr
2
l =
N N 1 . vr 2 2
Since N is large, we have
l t=
2
N2 l2 , 2vr 2vr
l=
2vrt.
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Q.42.
Three small balls of the same mass, white (w), green (g), and blue (b), are fixed by weightless rods at the vertices of the equilateral triangle with side l. The system of balls is placed on a smooth horizontal surface and set in rotation about the centre of mass with period T. At a certain instant, the blue ball tears away from the frame. Determine the distance L between the blue and the green ball after the time T.
in
During the time T, the distance covered by the blue ball is
e.
2 2l l is the rotational frequency. T , where 3 T 3
je
During the same time, the centre of mass of the green and the white
l l . T 2 3 3
ac k
iit
ball will be displaced by a distance
The rod connecting the green and the white ball will simultaneously
cr
Sol.
turn through and angle 2 since the period of revolution of the balls around their centre of mass coincides with the period T. Therefore, the required distance is 2
3 1 L= l 3 , 4 2 or for another arrangement of the balls (the white and the green ball change places),
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2
3 1 L= l 3 , 4 2 Q.43.
Find the minimum coefficient of friction µmin between a thin homogeneous rod and a floor at which a person can slowly lift the rod from the floor without slippage to the vertical position, applying to its end a force perpendicular to it. Let us assume the equilibrium conditions for the rod at the instant
in
when it forms an angle with the horizontal. The forces acting on the
e.
rod. While solving this problem, it is convenient to make use of the
je
equality to zero of the sum of the torques about the point of intersection of the force of gravity mg and the force F applied by the
iit
person perpendicular to the rod (point O) since the moments of these
ac k
force about this point are zero.
If the length of the rod is 2l, the arm of the normal reaction N while the arm of the friction is
cr
Sol.
l l sin , and the equilibrium sin
condition will be written in the form
1 sin sin
Nl cos = Ffrl
1 sin2 , whence = Ffrl sin
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Ffr = N
cos sin 1 sin2
=N
cos sin 2 sin2 cos2
=N
1 . 2 tan cot
On the other hand, the friction cannot exceed the sliding friction µN,
1 . 2 tan cot
e.
in
and hence
je
This inequality must be fulfilled at all values of the angle .
iit
Consequently, in order to find the minimum coefficient of friction µmin,
ac k
1 we must find the maximum of the function 2x2 2 x
1
, where x2
2
1 1 = tan . The identity 2x 2 2x 2 2 implies that x x
cr
2
the maximum value of
1 2 1 is and is 4 2 tan cot 2 2
attained at x2 = tan =
2 . Thus, the required minimum coefficient 2
of friction is µmin =
2 . 4
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Q.44.
Three weightless rods of length l each are hinged at points A and B lying on the same horizontal and joint through hinges at points C and D. The length AB = 2l. A load of mass m is suspended at the hinge C. Find the minimum force Fmin applied to the hinge D for/which the middle rod remains horizontal. Since the hinge C is in equilibrium, the sum of the forces applied to it is zero. Writing the projections of the forces acting on the hinge C on the
…(1)
e.
(m + mhin) g sin = T cos
in
axis perpendicular to AC, we get
je
where mhin is the mass of the hinge. Similarly, from the equilibrium condition for the hinge D and from the condition that the middle rod is
iit
horizontal, we obtain
ac k
T cos = F cos + mhin g sin
…(2)
Solving Equation (1) and (2) together, we find that
T cos mhin gsin cos
cr
Sol.
F=
=
mgsin mgsin . cos
Thus, the minimum force Fmin for which the middle rod retains its horizontal position is Fmin = mg sin
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=
mg 2
and directed at right angles to the rod BD. Q.45.
Two balls of mass m1 = 56 g and m2 = 28 g are suspended on two threads of length l1 = 7 cm and l2 = 11 cm at the end of a freely hanging rod. at which the rod should be rotated
e.
Let the right and left threads be deflected respectively through angles
je
and from the vertical. For the rod to remain in the vertical position,
…(1)
ac k
T1 sin = T2 sin
iit
the following condition must be satisfied.
where T1 and T2 are the tension of the relevant threads. Let us write the equations of motion for the two bodies in the vertical and horizontal directions.
cr
Sol.
in
about the vertical axle so that it remains in the vertical position.
T1 sin = m12l1 sin , T1 cos = m1g, T2 sin = m22l2 sin , T2 cos = m1g. Solving this system of equations and taking into account Equation (1), we obtain
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1 2
m2 m22 g 2 12 m1 l1 m22 l22 14 rad / s. Q.46.
The dependence of the kinetic energy WK of a body on the displacement s during the motion of the body in a straight line. The force FA = 2N is known to act on the body at point A. Determine the forces acting on the body at points B and C. The change in the kinetic energy Wk of the body as a result of a small
in
Sol.
e.
displacement s can be written in the form
je
Wk = F s,
iit
where F is the force acting on the body. Therefore, the force at a
ac k
certain point of the trajectory is defined as the slope of the tangent at the relevant point of the curve describing the kinetic energy as a function of displacement in a rectilinear motion. Using the curve given
N. Q.47.
cr
in the condition of the problem, we find that F C
– 1 N and FB
–3
A conveyer belt having a length l and carrying a block of mass m moves at a velocity v. Find the velocity v0 with which the block should be pushed against the direction of motion of the conveyer so that the amount of heat liberated as a result of deceleration of the block by the conveyer belt
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is maximum. What is the maximum amount of heat Q if the coefficient of friction is µ and the condition v <
The amount of liberated heat will be maximum if the block traverses the maximum distance relative to the conveyer belt. For that purpose, it is required that the velocity of the block relative to the ground in the vicinity of the roller A must be zero. The initial velocity of the block relative to the ground is determined from the conditions
in
– v0 + at = 0,
je
e.
at2 , l = v 0t – 2
iit
where a = µg is the acceleration imparted to the block by friction.
ac k
Hence
v0 2gl.
The time of motion of the block along the conveyer belt to the roller A is
cr
Sol.
2 lg is satisfied?
t=
2l . g
The distance covered by the block before it stops is s1 = l + vt =l v
2l . g
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Then the block starts moving with a constant acceleration to the right. The time interval in which the slippage ceases is
v v . The a g
distance by which the block is displaced relative to the ground during this time is
a2 s= 2
2gl by hypothesis the block does not slip from the
je
Since v
e.
in
v2 = . 2g
conveyer belt during this time, i.e. s < l. The distance covered by the
cr
v2 . = 2g
ac k
v2 s2 = v 2a
iit
block relative to the conveyer belt during this time is
The total distance traversed by the block relative to the conveyer belt is s = s1 + s2
2l v2 =lv g 2g
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=
v 2gl 2g
2
.
So the amount of heat liberated at the expense of the work done by friction is given by Q = µmgs
m v 2gl 2
.
A heavy pipe rolls from the same height down two hills with different
e.
Q.48.
in
=
2
profiles. In the former case, the pipe rolls down without slipping,
je
while in the latter case, it slips on a certain region.
In the former case (the motion of the pipe without slipping), the initial amount of potential energy stored in the gravitational field will be
cr
Sol.
ac k
lower?
iit
In what case will the velocity of the pipe at the end of the path be
transformed into the kinetic energy of the pipe, which will be equally distributed between the energies of rotatory, and translatory motion. In the latter case (the motion with slipping), not all the potential energy will be converted into the kinetic energy at the end of the path because of the work done against friction. Since in this case the energy will also be equally distributed between the energies of translatory and rotatory motions, the velocity of the pipe at the end of the path will be smaller in the latter case.
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Q.49.
The masses of two stars are m1 and m2, and their separation is l. Determine the period T of their revolution in circular orbits about a common centre.
Sol.
After the spring has been released, it is uniformly stretched. In the process, very fast vibrations of the spring emerge, which also attenuate very soon. During this time, the load cannot be noticeably displaced, i.e. if the middle of the spring has been displaced by a distance x in
in
doing the work A, the entire spring is now stretched by x. Therefore, the potential energy of the spring, which is equal to the maximum
je
e.
kx2 , kinetic energy in the subsequent vibratory motion, is Wk = 2
iit
where k is the rigidity of the entire spring. When the spring is pulled downwards at the midpoint, only its upper half (whose rigidity is 2k) is
ac k
stretched, and the work equal to the potential energy of extension of
cr
x2 the upper part of the spring is A = 2k = kx2. Hence we may 2 conclude that the maximum kinetic energy of the load in the subsequent motion is Wk = Q.50.
A . 2
The masses of two stars are m1 and m2, and their separation is l. Determine the period T of their revolution in circular orbits about a common centre.
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Since the system is closed, the stars will rotate about their common centre of mass in concentric circles. The equations of motion for the stars have the form
m112 l1 F,
m222 l2 F
…(1)
Here 1 and 2 are the angular velocities of rotation of the stars, l1 and l2 are the radii of their orbits, F is the force of interaction between the
in
Gm1m2 , where l is the separation between the stars, l2
e.
stairs, equal to
je
and G is the gravitational constant. By the definition of the centre of
ac k
m1l1 = m2l1,
iit
mass,
l1 + l2 = l …(2)
Solving Equation (1) and (2) together, we get
1 2
cr
Sol.
=
G m1 m2 l3
= l1
G m1 m2 , l
and the required period of revolution of these stars is
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T = 2l Q.51.
l . G m1 m2
A meteorite approaching a planet of mass M (in the straight line passing through the centre of the planet) collides with an automatic space station orbiting the planet in the circular trajectory of radius R. The mass of the station is ten times as large as the mass of the meteorite. As a result of collision, the meteorite sticks in the station
in
which goes over to a new orbit with the minimum distance
R from 2
je
Let v1 be the velocity of the station before the collision, v2 the velocity
iit
of the station and the meteorite immediately after the collision, m the
ac k
mass of the meteorite, and 10 m the mass of the station. Before the collision, the station moved around a planet in a circular orbit of radius R. Therefore, the velocity v1 of the station can be found
cr
Sol.
e.
the planet. Find the velocity u of the meteorite before the collision.
from the equation
10m12 10mM G . R R2 Hence v1 =
GM . R
In accordance with the momentum conservation law, the velocities u, v1, and v2 are connected through the following relation. mu + 10 mv1 = 11 mv2.
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We shall write the momentum conservation law in projections on the x and y-axis. 10 mv1 = 11 mv2x
…(1)
mu = 11 mv2y
…(2)
After the collision, the station goes over to an elliptical orbit. The energy of the station with the meteorite stuck in it remains constant during the motion in the elliptical orbit. Consequently,
11mM 11m 2 V R 2 2
in
…(3)
iit
= G
e.
11mM 11m 2 2 v2x v2y R 2
je
G
ac k
where V is the velocity of the station at the moment of the closest proximity to the planet. Here we have used the formula for the potential energy of gravitational interaction of two bodies (of mass m 1
Gm1 . According to Kepler's second law, the velocity m2r
cr
and m2): Wp = –
V is connected to the velocity v2 of the station immediately after the collision through the relation
VR v2xR 2
…(4)
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Solving Equation (1)-(4) together and considering that v1 =
GM , we R
determine the velocity of the meteoritge before the collision.
58 GM . R
u= Q.52.
The cosmonauts who landed at the pole of a planet found that the force of gravity there is 0.01 of that on the Earth, while the duration
in
of the day on the planet is the same as that on the Earth. It turned
e.
out besides that the force of gravity on the equator is zero. Determine the radius R of the planet.
je
For a body of mass m resting on the equator of a planet of radius R,
ac k
form
iit
which rotates at an angular velocity , the equation of motion has the
m2R = mg' – N,
Where N is the normal reaction of the planet surface, and g' = 0.01 g is
cr
Sol.
the free-fall acceleration on the planet. By hypothesis, the bodies on the equator are weightless, i.e. N = 0. Considering that w =
2 , 2jt/r, T
where T is the period of rotation of the planet about its axis (equal to the solar day), we obtain R=
T2 g'. 42
Substituting the values
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T= 8.6 × 104 s and g'
0.1 m/s2, we get
1.8 × 107 m
R
= 18000 km. Q.53.
The radius of Neptune's orbit is 30 times the radius of the Earth's orbit. Determine the period TN of revolution of Neptune around the Sun.
in
We shall write the equation of motion for Neptune and the Earth
e.
around the Sun (for the sake of simplicity, we assume that the orbits
iit
GMmN , RN2
ac k
mN N2 RN
je
are circular).
mEE2 RE
GMmE . RE2
Here mN, mE, N, E, RN and RE are the masses, angular velocities, and
cr
Sol.
orbital radii of Neptune and the Earth respectively, and M is the mass of the Sun. We now take into account the relation between the angular velocity and the period of revolution around the Sun.
N
2 , TN
E
2 . TE
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Here TN and TE are the periods of revolution of Neptune and the Earth. As a result, we find that the period of revolution of Neptune around the Sun is TN =
165 years. A similar result is obtained for elliptical orbits from Kepler's third law. Q.54.
A homogeneous rod of length 2l leans against a vertical wall at one
in
end and against a smooth stationary surface at an other end. What
e.
function y (x) must be used to describe the cross section of this
je
surface for the rod to remain in equilibrium in any position even in the absence of friction? Assume that the rod remains all the time in
ac k
Let the cross section of the surface be described by the function y (x) shown in figure. Since the rod must be in equilibrium any position, the equilibrium can be only neutral, i.e. the centre of mass of the rod must
cr
Sol.
iit
the same vertical plane perpendicular to the plane of the wall.
be on the same level for any position of the rod, If the end of the rod leaning against the surface has an abscissa (x) the wall can be found from the condition. l2 = [y(x) – y0]2 + x2, y0 = y(x)
l2 x2 .
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Since the rod is homogeneous, its centre of mass is at the midpoint. Assuming for defmiteness that the ordinate of the centre of mass is zero, we obtain
y0 y x 0, whence 2
l2 x2 . 2
y(x) =
in
Only the solution with the minus sign has the physical meaning.
e.
Therefore, in general, the cross section of the surface is described by
l2 x2 , 2
iit
y(x) = a
je
the function
Q.55.
ac k
where a is an arbitrary constant.
A ball of mass m falls from a certain height on the pan of mass (M >>
cr
m) of a spring balance. The rigidity of the spring is k. Find the displacement x of the point about which the pointer of the balance will oscillate, assuming that the collisions of the ball with the pan are perfectly elastic. Sol.
Let the ball of mass m falling from a height h elastically collide with a stationary horizontal surface. Assuming that the time of collision of the bail with the s between two consecutive collisions, we obtain
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2h , g
t 2
As a result of each collision, the momentum of the ball changes by p = 2m = 2m
2gh . Therefore, the same momentum p = 2m 2gh is
transferred to the horizontal surface in one collision. In order to determine the mean force exerted by the ball on the horizontal surface we consider the time interval
in
>> t. The momentum transferred to the horizontal surface during
je mg.
ac k
2gh 2h 2 g
cr
= 2m
t
iit
P = p
e.
the time is
Consequently, the force exerted by the jumping ball on the horizontal can be obtained from
the relation Fm =
P mg.
By hypothesis, the mass M of the pan of the balance is much larger than the mass m of the ball. Therefore, slow vibratory motion of the balance pan will be superimposed by nearly periodic impacts of the
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ball. The mean force exerted by the ball on the pan is F m = mg. Consequently, the required displacement x of the equilibrium position of the balance is
x Q.56.
mg . k
Two blocks having mass m and 2m are connected by a spring of rigidity k lie on a horizontal plane. Find the period T of small
e.
It follows from the equations of motion for the blocks
je
ma1 = Fe1,
iit
2ma2 = – Fe1,
ac k
where Fe1 is the elastic force of the spring, that their accelerations at each instant of time are connected through the relation a 2 = –
a1 . 2
Hence the blocks vibrate in antiphase in the inertial reference frame
cr
Sol.
in
longitudinal oscillations of the system, neglecting friction.
fixed to the centre of mass of the blocks, and the relative displacements of the blocks with respect to their equilibrium positions are connected through the same relation as their accelerations.
x2
x1 . Then 2
3
Fe1 = k x1 3k x2. 2
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Consequently, the period of small longitudinal oscillations of the system is T = 2 Q.57.
2m . 3k
A load of mass M is on horizontal rails. A pendulum made of a ball of mass m tied to a weightless inextensible thread is suspended to the load. The load can move only along the rails. Determine the ratio of
in
T2 of small oscillations of the pendulum in vertical T1
e.
the periods
planes parallel and perpendicular to the rails.
je
The period of oscillation of the pendulum in the direction
l . g
ac k
T1 = 2
iit
perpendicular to the rails is
(l is the length of the weightless inextensible thread) since the load M
cr
Sol.
is at rest in this case. The period of oscillations in the plane parallel to the rails ("parallel" oscillations) can be found from the condition that the centre of mass of the system remains stationary. The position of the centre of mass of the system is determined from the equation ml1 = M (l – l1). Thus, the ball performs oscillations with point O remaining at rest and is at a distance l1 =
Ml from point O. M m
Hence the period of "parallel" oscillations of the pendulum is
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T2 2
Ml . m M g
Consequently,
T2 T1 Q.58.
M . mM
Four weightless rods of length l each are connected by hinged joints and form a rhomb. A hinge A is fixed, and a load is suspended to a
in
hinge C. Hinges D and B are connected by a weightless spring of
e.
length 1.5 l in the undeformed state. In equilibrium, the rods form
je
angles 0 = 30° with the vertical.
ac k
The force exerted by the rods on the load is F1 = 2Ften cos , while the force exerted on the spring is F2 = 2Ften sin . According to Hooke's law, F2 = (1.5l – 2l sin ) k, where k is the rigidity of the spring. As a result F1 = 1.5lk cot – 2lk cos .
cr
Sol.
iit
Find the period T of small oscillations of the load.
In order to determine the period of small oscillations, we must determine the force F acting on the load for a small change h in the height of the load relative to the equilibrium position h0 = 2l cos 0. We obtain F = 1.5lk (cot ) – 2lk (cos ). Where
dcot cot d 0
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, sin20
(cos ) = – sin 0 . Consequently, since h = – 2l sin 0 , we find that F = – 1.5 k + 2kl sin 0 = – 5 kl
1 . 2
e.
Because sin 0 =
in
= – 5k h
iit
m , where m is the mass of the load determined 5k
ac k
formula T = 2
je
The period of small oscillations of the load can be found form the
from the equilibrium condition:
cr
1.5 kl cot 0 – 2kl cos 0 = mg,
3 m = 2 . Thus, kl g T = 2 Q.59.
3l . 10 g
A weightless rigid rod with a load at the end is hinged at point A to the wall so that it can rotate in all directions. The rod is kept in the
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horizontal position by a vertical inextensible thread of length l, fixed at its midpoint. The load receives a momentum in the direction perpendicular to the plane. Determine the period T of small oscillations of the system. It should be noted that small oscillations of the load occur relative to the stationary axis AB. Let DC
AB. Then small oscillations of the load
are equivalent to the oscillations of a simple pendulum of the same
in
mass, but with the length of the thread
L l2 2
2
, and the free-fall acceleration
je
l
iit
=L
e.
l' = L sin
=g
ac k
g' = g cos
l 2
,
cr
Sol.
L l2 2
2
where L = AD. Thus, the required period of small oscillations of the system is T = 2
l' g'
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2l . g
= 2 Q.60.
One rope of a swing is fixed above the other rope by b. The distance between the poles of the swing is a. The lengths l1 and l2 of the ropes are such that l12 l22 = a2 + b2. Determine the period T of small oscillations of the swing, neglecting the height of the swinging person in comparison with the above
e.
In order to solve the problem, it is sufficient to note that the motion of
je
the swing is a rotation about an axis passing through the points where
iit
the ropes fixed, i.e. the system is "tilted simple pendulum". The component of the force of gravity mg along the rotational axis does not
ac k
influence the oscillations, while the normal component mg fact the restoring force.
Therefore, using the formula for the period of a simple pendulum, we
cr
Sol.
in
lengths.
can write T = 2
h=
h , where gsin
l1 l2 2 1
2 2
l l
sin =
l1 l2
2
a b
a 2
2
2
a b
,
.
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Consequently, the period of small oscillations of the swing is T = 2 Q.61.
l1 l2 . ag
A cone of mass 'm' falls from a height 'h' and penetrates into sand. The resistance force R of the sand is given by R = kx2. If the cone penetrates upto a distance x = d where d < l, then find the value of 'k'.
Sol.
Force on cone while it is penetrating the sand is shown in F.B.D. below
in
Applying work energy theorem to the cone as x changes from o to d
d
d
iit
0
je
Kfinal – Kinitial = mgd – kx2 dx
e.
KE = Work done by mg + Work done by resistive force R
ac k
0 – mgh = mgd – kx2 dx 0
cr
kd3 mgd mgh 3 K= Q.62.
3mg h d d3
The potential energy U of a particle is plotted against its position 'x' from origin. Then which of the following statement is correct A particle at:
(A)
x1 is in stable equilibrium
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Sol.
(B)
x2 is in stable equilibrium
(C)
x3 is in stable equilibrium
(D)
none of these
(D) x = x1 and x = x3 are not equilibrium positions because
du 0 at dx
these points.
Initially spring is in unscratched state & blocks are at rest. Now 100 N
e.
Q.63.
in
x = x2 is unstable, as U is maximum at this point.
je
force is applied on block A & B as shown in figure. After some time
iit
velocity of 'A' becomes 2 m/s & that of 'B' 4 m/s block A displaced by amount 10 cm, spring is stretched by amount 30 cm. Then work done
9 J 3
(B)
–6J
6J
(D)
None of these
cr
(A)
ac k
by spring force on A will be:
(C) Sol.
Wspring + W100N 2 10 1 2 2 100 2
Wspring + 100 Wspring = 4 – 10 = – 6 J.
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Q.64.
M is a man of mass 60 kg standing on a block of mass 40 kg kept on ground. The co-efficient of friction between the feet of the man and the block is 0.3 and that between B and the ground is 0.1. If the person pulls the string with 100 N force, then:
(B)
A and B may move together with acceleration 1 m/s2
(C)
the friction force between A & B may be 40 N
(D)
The friction force acting between A & B may will be 180 N
in
B may slide on ground
FBD of M
e.
Sol.
(A)
ac k
T + fs – fk = 40 a
iit
FBD of B…(1)
je
100 – fs = 60 a
…(2)
fk = (0.1) (60 + 40)g
cr
From (1) and (2) 100 – fs = 60 a fs = 40 Nt Q.65.
A bar of mass m resting on a smooth horizontal plane starts moving due to the force F =
mg of constant magnitude. In the process of its 3
rectilinear motion the angle between the direction of this force and the horizontal varies as = as, where a is constant, and s is the
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distance traversed by the bar from its initial position. Find the velocity of the bar as function of the angle . Sol.
dv mv F cos as ds
dv mg mv cos as ds 3
g cos as ds 3
in
vdv
ac k
c=0
2g sin as 3a 2g sin 3a
cr
V2 =
je iit
v = 0 for s = 0
e.
v2 g sin as C 2 3a
V=
Q.66.
A block surface for which the coefficient of friction is µ =
3 . The 4
block of mass m is initially at rest. The block of mass M is released from rest with spring is unstretched state. The minimum value of M required to move the block up the plane is (neglect mass of string and pulley and friction in pulley.)
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Sol.
(A)
3 m 5
(B)
4 m 5
(C)
6 m 5
(D)
3 m 2
As long as the block of mass m remains stationary, the block of mass M released from rest comes down by
2mg (before coming it rest K
in
momentaly again).
2mg K
…(1)
je
X=
e.
Thus the maximum extension is spring is
iit
For block of mass m to just move up the incline
ac k
kx = mg sin + µ mg cos
3 3 4 mg 5 4 5
cr
2Mg = mg ×
….(2)
or M =
3 m Ans. 5
Comprehension A van accelerates uniformly down an inclined hill going from rest to 30 m/s in 6s. During the acceleration, a toy of mass m = 0.1 kg hangs
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by a light string from the van's ceiling. The acceleration is such that string remains perpendicular to the ceiling. (Take g = 10 m/s2)
30°
(B)
60°
(C)
90°
(D)
45°
in
(A)
Acceleration of the van =
e.
Sol.
The angle of the incline is:
30 6
je
Q.67.
iit
= 5 m/s2
ac k
g sin = a
1 2
cr
sin =
= 30° Q.68.
The tension in the string is (A) (C)
Sol.
1.0 N
3 N 2
(B) (D)
0.5 N
3N
Tension T = mg cos
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3 N 2
=
The friction force on the van is (A)
Zero
(B)
mg cos
(C)
mg sin
(D)
mg tan
in
Q.69.
Since acceleration the van is g sin , there is no friction.
Q.70.
A football is kicked with a speed of 22 m/s at an angle 60° to the
e.
Sol.
je
positive x direction taken along horizontal. At that instant, an
iit
observer moves past the football in a car that moves with a constant
ac k
speed of 11 m/s in the positive x direction, Take +ve y direction vertically upwards. (g = 10 m/s2) The initial velocity of the ball relative to the observer in the
cr
(A)
car is 11 3 m/s in the + y direction
(B)
The initial velocity of the ball relative to the observer in the car is 17 m/s at 60° to the +x direction.
(C)
According to the observer in the car, the ball will follow a path that is straight up and down in the y direction.
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According to the observer in the car, the ball will follow a
(D)
straight line that is angled (less than 90°) with respect to the observer. Sol.
w.r.t. to observer (Vx) football/ obs. = 11 – 11 = 0 So Motion of the ball as seen from observer will be purely vertical with initial velocity 11 3 upwards. A juggler throws ball into air. He throws one whenever the pervious
in
Q.71.
e.
one is at its highest point. He throws two balls per second. Find the
ac k
at its highest point,
iit
As the juggler is throwing n balls each second and 2nd when the first is
So the time taken by one ball to reach the highest point,
1
t= n
cr
Sol.
je
height to which each ball goes (in centimeter). (g = 10 m/s2)
V = u – gt putting V = 0, at highest point, t=
1 n 1 n
v = u g
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u
g n
V2 = u2 – 2gh
h
u2 2g
Putting value of u,
e.
iit
Consider the following statements: (1)
Static friction may do positive work.
(2)
It may be possible that on a body kinetic friction act in the
ac k
Q.72.
1000 125 cm. 24
je
=
g 2n2
in
h=
direction of motion of body. A lift going down with retardation increases the weight of an
cr
(3)
object measure by a spring balance.
The correct order of True / False of above statement is:
Sol.
(A)
TTT
(B)
TFT
(C)
TTF
(D)
FFT
Statement (1): It is possible in a two block system moving together.
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Statement (2): It is possible in a case of two block system sliding on each other. Statement (3): The acceleration of the lift is upward, hence N = m (g + a). Q.73.
A man is standing on a cart of mass double the mass of the man. Initially cart is at rest on the smooth ground. Now man jumps with relative velocity 'v' horizontally towards right with respect to cart.
e.
Let the velocity of man after jumping be 'u' towards right, then speed
je
of cart is v – u towards left. From conservation of momentum mu = 2m
2v 3
ac k
u
iit
(v – u)
work done =
1 1 2 mu2 2m v u 2 2
cr
Sol.
in
Find the work done by man during the process of jumping.
2
1 2V 1 V 2m = m 3 2 3 2
2
4V2 2V2 1 = m 2 9 9 =
1 2 6V m 2 9
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ac k
iit
je
e.
in
1 mV2 Ans. 3
cr
=
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