30432279 Jr M1 Kinematics

INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 04064606657

KINEMATICS SYLLABUS : Motion in one and two dimension, Projectile motion, Relative Velocity & Circular Motion.

POSITION VECTOR If the coordinates of a particle are given by (x2, y2, z2) its position vector with → j +(z2 - z1) k . respect to (x1,y1,z1) is given by r = (x2 - x1) i + (y2 - y1) Usually, position vector with respect to the origin (0,0,0) is specified and is j given by → r =x i +y +z k DISPLACEMENT Displacement is a vector quantity. It is the shortest distance between the final and initial positions of a particle. If

→

r 1 is the initial position vector and r 2

is the final position vector, the displacement vector →

→

∆r

→

→

→

→

r1

is given by ∆ r = r 2− r 1 .

→

r2

The magnitude of the displacement is given by

( x 2 − x 1) 2 + ( y2 − y1) 2 + ( z 2 − z1) 2 This is nothing but the straight line distance between two points (x1,y1,z1) and (x2,y2,z2). The displacement is independent of the path taken by the particle in moving from (x1 , y1, z1) to (x2, y2, z2) DISTANCE : If a particle moves along a curve, the actual length of the path is the distance. Distance is always more than or equal to displacement.

Q( x 2 , y 2 , z 2 )

P( x 1 , y 1 , z 1 )

Illustration – 1 : A car travels along a circular path of radius (50 /π )m with a speed of 10 m/s. Find its displacement and distance after 17.5 sec. Solution : Distance = (speed) time = 10 (17.5) = 175 m Perimeter of the circular path = 2 π (50/π ) = 100 m ⇒ The car covers 1

3 rounds of the path 4

A

If the car starts from A, it reaches B and the displacement is the shortest distance between A and B ⇒ Displacement =

R2 + R2

=

2 R

=

50 2 π

m.

B

INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 04064606657 INSTANTANEOUS AND AVERAGE VELOCITY : →

If ∆ r is the displacement of the particle in time ∆ t, the average velocity is given by →

→

→

r 2− r 1 ∆ r = V average = t2 − t1 ∆t

→

∆ = Find value - Initial value. The above definition is valid for any magnitude of ∆ large or small. But when ∆ is infinitesimally small, the instantaneous velocity is obtained. →

V instantaneous =

Lt ∆t →0

→

→

∆r = d r ∆t dt

In normal notation, velocity refers to the instantaneous velocity.

SPEED : Speed =

Dis tan ce time

When the time under consideration is very small, distance becomes equal to the displacement and speed becomes the magnitude of instantaneous velocity. speed is represented only by its magnitude where as velocity is represented by magnitude as well as direction.

INSTANTANEOUS AND AVERAGE ACCELERATION : →

If ∆V is the change in velocity in time ∆ t, average acceleration is given by →

a average

→

= ∆V . ∆t

When ∆ becomes infinitesimally small,

→

Lt ∆t → 0

→

∆V = d V ∆t dt

which gives the instantaneous acceleration. In normal notation, acceleration refers to the instantaneous acceleration. →  → dV d d r  a = =   dt dt  dt   

→

→

2 = d r dt 2

→

2 d 2r It may be noted here that magnitude of d r is not equal to always (as in 2

dt 2

the case of circular motion)

dt

INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 04064606657 Illustration – 2 : A bus shuttles between two places connected by a straight road with uniform speed of 36 kmph. If it stops at each place for 15 minutes and the distance between the two places is 60 km, find the average values of (a) Speed (b) Velocity (c) acceleration between t = 0 and t = 2 hours and the instantaneous values of (d) Velocity (e) acceleration at t = 2 hrs. Solution : Time taken for forward trip =

60 5 = hrs. 36 3

Time of stoppage = 15 min = 0.25 hrs. Time available for return trip = 2 - 5/3 - 0.25 = 1/12 hrs. Distance travelled in the return trip = (36) 1/12 = 3 km. d = 60km

d 1 = 3km Total distance Total time

60+ 3 = 31.5 kmph. 2 60− 3 Displacement b) Average velocity = = = 28.5 kmph 2 Time changein velocity c) Average acceleration = Time

a) Average speed =

→

=

→

( − 36) − ( + 36) = V2 − V1 = = - 36 km/H2 2 t

 1   m/s2  360

=- 

d) Velocity at t = 2 hours = - 36 kmph e) Acceleration at t = 2hours = 0 as there is no change in velocity Illustration – 3 : A car travels towards North for 10 minutes with a velocity of 60 Kmph, turns towards East and travels for 15 minutes with a velocity of 80 kmph and then turns towards North East and travels for 5 minutes with a velocity of 60 kmph. For the total trip, find a) distance travelled b) displacement c) average speed d) average velocity and e) average acceleration. Solution : Total time taken = (10 + 15 + 5)min = 1/2 hour a) Distance travelled = d1 + d2 + d3  10  15  5  + 80   + 60    60  60  60

= 60 

= 10 + 20 + 5 = 35 km

→

S3

→

S2 → → → → → b) displacement S = S1+ S 2 + S 3 S1 0 = 10 j + 20 i +5 cos 45 i + 5 sin 45 j = 23.5 i + 13.5 j Magnitude of displacement = ( 23.5) 2 + ( 13.5) 2 ~ 27 km

→

S

INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 04064606657 Total di stancetravelled 35km = Time  1 c) Average speed = = 70 kmph.  hr 2   Displaceme nt d) Average velocity = = Time

2 3.5i + 1 3.5j = ( 4 7i + 2 7j) 1 2

at an angle θ with the East given by Tan θ = Magnitude of average velocity =

472 + 272

kmph.

13.5 23.5

= 54 kmph

changein velocity Time Final velocity − Initial velocity = Time

e) Average acceleration =

=

=

(6 c0o4s 05i + 6 s0i 4n j5) − ( 6 j0)

( 2 i1− 9j)

2

km/H2 −9 21

at an angle θ with the East given by Tanθ = Magnitude of average acceleration =

212 + 92

~ 23 km/H2 ~ 1.8 x 10-3

m/s2 Illustration – 4 : A car moving along a circular path of radius R with uniform speed covers an angle θ during a given time. Find its average velocity and average acceleration during this time. Solution : Let V be the speed of the car V=

Distance Rθ = where θ is in radians. time t

Displacement=

R

2

+R

2

− 2R

2

V

A

cos θ

θ

B V

o

from the triangle

OAB = 2R sin θ /2 Average

2V sin θ

velocity

=

Diplacemnt Time

θ 2

Average acceleration =

2R sin

=

 Rθ     V 

θ 2

=

Change in velocity ∆V = time t

V θ V

∆V

INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 04064606657 ∆V=

V 2 + V 2 − 2V 2 cos θ

= 2 V sin

θ 2

θ θ 2 2V 2 sin ⇒ Average acceleration =  Rθ  = 2   Rθ  V  When θ is small sinθ ~ θ and θ  θ 2V  2V sin Average velocity = 2 = 2 =V θ θ ⇒ Average velocity = Instantaneous velocity for small angular displacements θ 2V 2 θ  2   V2 Average acceleration = 2V sin 2 2 =  = R Rθ Rθ ⇒ Average acceleration = Instantaneous acceleration for small angular displacements. 2V sin

KINEMATICAL EQUATIONS : ( CONSTANT ACCELERATION ) : V = u + at V2 - u2 = 2aS S= ut + 1/2 at2 The above equations are valid only for constant acceleration and in a particular direction. u,v and s must be taken with proper sign. Usually the direction of u is taken as positive and the sign of other variables are decided with respect to this direction. Displacement during the nth second = Sn - Sn-1 = u+

a (2n - 1) 2

It may be noted here that this is not the distance travelled in the nth second. Illustration – 5 : A particle is vertically projected upwards with an initial velocity of 22.5 m/s. Taking g=10 m/s2 find a) velocity b) displacement c) distance travelled in t = 4 sec and d) displacement and distance travelled in 3rd second Solution : Taking the upward direction positive a) V = u + (-g) t = 22.5 - 10 (4) = -17.5 m/s ⇒ 17.5 m/s down wards b) S = ut + 1/2 (-g) t2 = 22.5 (4) - 1/2 (10) 42 = 10 m c) Time to reach the top most point = t0 and at the top d2 most point velocity becomes zero. V = u - gt0 ⇒ 0 = 22.5 - 10 (t0) ⇒ t0 = 2.25 sec d1 Distance travelled in 4 sec = d1 + d2 2 d1 = u t0 - 1/2 g t 0 = 22.5 (2.25) - 1/2 (10) (2.25) 2 = 25.3 m d1 can be found from V2 - u2 = 2a S also.

INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 04064606657 0 - (22.5)2 = 2(-10) d1 ⇒ d1 =

d)

( 22.5) 2 = 25.3 m 20

d2 can be found from S = ut + 1/2 at 2 applied along the down ward direction starting from the top most point d2 = 0 (t - t0) + 1/2 g (t - t0)2 = 1/2 (10) (4 - 2.25)2 = 15.3 m ⇒ Distance travelled in 4 sec = 25.3 + 15.3 = 40.6 m Displacement in 4 sec = d1 - d2 = 25.3 - 15.3 = 10 m Displacement can also be found directly by applying S = ut + 1/2 at2 along the vertical Displacement in 4 sec = 22.5 (4) - 1/2 (10) (4)2 = 10M 3rd second is from t = 2 sec to t = 3 sec. a (2n - 1) 2 10 = 22.5 (6 - 1) = -2.5 m 2

Displacement in the 3rd second = u +

When there is no change in the direction of the motion along a straight line, distance will be equal to displacement. When the particle reverses its direction during the time under consideration, distance will be more than the displacement and the time at which the reversal is taking place must be found. When the particle reverses its direction, its velocity becomes zero. using V = u + at, 0 = 22.5 - 10 (t0) ⇒ t0 = 2.25 sec d = d1 + d2 t = 2.25sec using the formula S = ut + 1/2 at2 d1 d1 = [22.5 (2.25) - 1/2 (10) (2.25) 2 ] - [22.5 (2) - 1/2 (10) t = 2 sec (2)2] = 0.31 m Along the downwards vertical starting from the top d2 = 0 (3 - 2.25) + 1/2 (10) (3 - 2.25)2 = 2.81 m ⇒ d = 0.31 + 2.81 = 3 .12 m

d2 t = 3 sec

KINEMATICAL EQUATIONS ( VARIABLE ACCELERATION ) : When the acceleration is variable, the kinematical equation take the form V=

dx dt

a=

VdV  dV   dx     = dx  dx   dt 

V=u+

∫ ∫

d 2x dt 2

∫a

dt

0

t

x = ut +

= t

a= 

t   a dt  dt   0 0 

dV dt

x

and V - u = 2 2

2

∫a dx 0

Illustration – 6 : The position coordinate of a particle moving along a straight line is given by x = 4 t3-3t2+4t+5. Find a) Velocity and acceleration as a function of time b) Displacement as a function of time c) the time at which velocity becomes zero and the acceleration at this time d) the time at which acceleration becomes zero and the velocity at this time.

INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 04064606657

INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 04064606657 Solution : d( x − x 0 ) dS dx a) V = = = Where x0 is the initial position coordinate dt dt dt which is a constant d (4t3 - 3t2 + 4t + 5) = 12 t2 - 6t + 4 dt dv d a= = (12t 2 - 6t + 4) = 24t - 6 dt dt

=

b)

Displacement = (position coordinate at time t) - (position coordinate at t

= 0) = (4t3 - 3t2 + 4t + 5) - (5) = 4t3 - 3t2 + 4t c)

⇒t =

When V = 0, 12 t2 - 6t + 4 = 0

3 ± 9 − 48 12

since this value is imaginary, the velocity never becomes zero. d)

When a = 0, 24t – 6 =0

and

t=

6 1 = units and the velocity of 24 4

2

1 1 13 the particle at this time, V= 12   − 6  + 4 = units  4

 4

4

Illustration – 7 : The velocity of a particle moving in the positive direction of the x axis varies as V = x where α is a positive constant. Assuming that at the moment t = 0 the α particle was located at the point x = 0, find a) the time dependence of the velocity and the acceleration of the particle b) the mean velocity of the particle averaged over the time that the particle takes to cover the first S meters of the path. Solution : a)

V= 2

dx =α dt x =α t

b)

∫ 0

and

t

dx x

x=

=

∫α

dt

0

α2t 2 4

dV α2 = dt 2 Displaceme nt Mean velocity = time

⇒V =

α2t 2

x

x ⇒

and

a=

Displacement = S, and the time taken for this displacement t =

( S)

α S Mean velocity =  2 S  = 2

 α   

Mean velocity can also be found from the following formulae V mean =

∫V dx ∫dx

when V is a function of x

2 S α

INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 04064606657 and V mean =

∫V dt ∫dt

when V is a function of time

KINEMATICAL EQUATIONS IN VECTOR FORM ( CONSTANT ACCELERATION ) : →

→ →

V = u+ a t

→ → → →

→ →

V . V− u . u = 2 a . S → → 1→ S = u t + a t2 2

The above equations are useful in 2 and 3 dimensional motion.

Illustration – 8 : A particle moving on a horizontal plane has velocity and acceleration as shown in the diagram at time t = 0. Find the velocity and displacement at time 't'. Solution : METHOD - I →

u =

y

u cos300 i + u cos 60

u j 2 →

a = - a cos 45 i

→

→ →



 V = u+a t =  

j

=

3 u 2

−a a - a cos45 j = i 2 2 a  u 3 a  u− t i +  − t j  2 2  2  2

u

i +

300 a 450

j

2

The magnitude of the velocity = →

→

S=ut +

1→ 2 at = 2

2  3 at  u a    +  − u − t   2 2  2  2 

 3 u 1 a  2  1 a  2       2 ut − 2  2  t  i +  2 t − 2  2  t  j = Sx i + Sy j    

The magnitude of the displacement =

Sx 2 +S 2 y

x

INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 04064606657 METHOD - II This can be solved by vector addition method also. It may be → → → noted here that u t will be along the direction of u , a t and

→

u

1→ 2 → a t will be along the direction of a 2 →

450

300

→

a t

→

→ →

V

V = u+ a t

→

→

Since the angle between u and a t is 1650, the magnitude of the velocity is → u 2 + ( at ) 2 + 2u ( at ) cos 1650 ut →

→

S =ut+

1→ 2 at 2

1 →

→

2 is 1650, the u t and 2 a t magnitude of the displacement is

Since the angle between

( ut ) 2 +  1 at 2  2

450

300

→

S

1→ 2 a t 2

2

1  + 2( ut )  at 2  cos 1650  2 

KINEMATICAL EQUATIONS IN RELATIVE FORM ( CONSTANT ACCELERATION ) : When two particles A and B move simultaneously with initial velocities → u A →

and u B , at any time 't' →

→

→

VAB = VA - VB →

→

→

→

→

→

→

;

→

→

S AB = S A − S B

;

→

→

→

a AB = a A − a B

U AB = U A − U B VAB = u AB + a AB t →

→

S AB = u AB t +

1 → 2 a AB t 2

→

where X means parameter X of A with respect to B. AB →

Similarly if → r is the position coordinate at time 't' and r0 is the initial position coordinate at time 't' = 0, →

→

→

1 2 → → → 1 rB = r0B + u B t + 2

rA = r0A + u A t +

→

→

→

→

aA t2 →

aB t2 1 →

2 rAB = r0AB + u AB t + 2 a AB t gives the position coordinate of A with respect to B

at any time.

→

rA B

gives the distance between A and B at any time 't'.

Illustration – 9 : A loose bolt falls from the roof of a lift of height 'h' moving vertically upward with acceleration 'a'. Find the time taken by the bolt to reach the floor of the lift and the velocity of impact.

INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 04064606657 Solution : → S b

as the bolt travels a distance 'h' down wards before hitting the

= −hj

floor →

→

→

→

→

→

→

→

a b = a b − a  = (-g j ) - (a j ) = - (g + a) j

u b = u b − u  = uj - uj = 0 as they have the same initial velocity upwards

1→ a b t 2 2 1 -h j =0(a + g) t2 j 2 S b = u b t +

⇒t =

2h a +g

Velocity of impact is nothing but the relative velocity of the bolt with respect to the lift →

→

→

Vimpact = V b = u b + a b t = - (a+g)

2h a +g

j

=

− 2h ( a + g )

j

Illustration – 10 : Two particles A and B move on a horizontal surface with constant velocities as shown in the figure. If the initial distance of separation A between them is 10 m at t=0, find the distance between them at t = 2 sec

u B = 10m / s

600

10m 450

B

u A = 10 2 m / s

Solution : Distance between them =

→ r AB

Taking the origin at the initial position of A →

→

→

r AB = r0 AB + u AB t +

1 → a AB t 2 2

→

r0 AB = −10 i →

→

→

(

) (

u AB = U A − U B = 10 2 cos 450 i − 10 2 cos 450 j − 10cos 600 i + 10cos 300 j = 5 i - 18.7 j

→

→

→ r AB

= ( − 10 i ) + ( 5i − 18.7 j) t

)

→

a AB = a A − a B = 0

At t = 2 sec ,

→ r AB

= −37.4 j

and

→ r AB

= 37.4 units

DISPLACEMENT - TIME GRAPHS : The displacement is plotted along 'y' axis and the time along 'x' axis. The dS gives the instantaneous velocity at that point. The average dt ∆S slope between two points gives the average velocity between these points. ∆t

slope of the curve

INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 04064606657 Rate of change of slope gives the acceleration. If the slope is positive and decreases with time, the particle is under retardation. If the slope is positive and increases with time, the particle is under acceleration, constant slope implies zero acceleration. Illustration – 11 : The displacement - time graph of a particle moving along a straight line is given below. Find a) the time at which the velocity is zero b) the velocity at time t = 1 sec c) the average velocity between t = 2 sec and t = 4 sec

2m x

Semi circle 0

2

4 t

Solution : a) Velocity is zero when the slopoe is zero which happens at t = 2 sec b) Since any point (x,t) lies on the circle of radius 2 m and centre (2,0), (x-0)2 + (t - 2)2 = 22 ⇒ x = ± 4 − ( t − 2) 2 velocity is given by the slope V=

d dt

dx =V dt

 1  4 − ( t − 2) 2  = ±        2 4 − ( t − 2) 2 



( − 2( t − 2) )   

 1 

 = +   3

Since the slope is +ve between t = 0 and t = 2, v = c)

Average velocity =

Displaceme nt time

=

O −2 = -1 m/s 4− 2

1 m/s 3

VELOCITY - TIME GRAPH : If velocity is plotted on 'y' axis and time is plotted on x axis, the slope of the dv given instantaneous acceleration. dt ∆v between two points gives average acceleration. ∆t

curve at any point

The average slope

A3 The total area between the curve and the time axis V A1 gives distance where as algebraic sum of the areas A2 gives displacement. Distance = A1 + A2 + A3 Displacement = A1 - A2 + A3 The nature of acceleration can be found from the rate of change of slope

t

Illustration – 12 : The velocity time graph of a particle moving along a straight line has the form of a parabola v = (t2 - 6t + 8) m/s . Find a) the distance travelled between t = 0 second t = 3 sec b) the velocity of the particle when the acceleration is zero

INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 04064606657 c) the acceleration of the particle when the velocity is zero d) the velocity of the particle when the acceleration is zero Solution : a) Distance = area OAB + area BCF which can be obtained by the method of integration. Since at the points B and D, velocity becomes V zero t2 – 6t + 8 = 0 ⇒ t = 2 sec and 4sec Since F is in between B and D, the time O 2 +4 corresponding to F is = 3 sec. Similarly A

A

E

A1 B

F A3

A2

D A4

C

2

corresponds to t = 0 and E corresponds to t = 6 sec 2

∫

Area OAB = A1 =

2

V dt =

0 3

Area BCF = A2 = -

∫ V dt 2

b)

2 ∫ (t − 6t + 8)dt 0

2

 t 3 6t 2  = − + 8t   3  2  0

20 2 22 + = m Distance = 3 3 3

displacement between t = 3 sec and t = 6 sec = A4 - A3 = A1 - A2 =

a=

dv = 2t - 6 dt

When V = 0 ; ⇒ a = 2(2) - 6 = - 2 m/sec2 d) When a = 0, V = 32 - 6(3) + 8 PROJECTILE MOTION :

t = 2sec and 4 sec and 2(4) - 6 and 2 m/sec2 2t - 6 = 0 and t = 3 sec = -1 m/sec

At the top most point Vy = 0 and Vx = u cosθ From Vy = uy + ay t, 0 = usinθ gt

u sin θ ⇒ t= g

Time of flight = 2t = From

20 m 3

 t 3 6t 2  2 = − − + 8t  = m  3  3 2  

20 2 − = 6m 3 3

c)

=

Vy2 − u 2 y

T u H θ

R 2u sin θ g

= 2ay Sy, 0 - (u sinθ )2 = 2 (-g) H and H =

u 2 sin 2 θ 2g

Range = (Time of flight) (horizontal velocity) =  2u sin θ  u 2 sin 2θ   ( u cos θ ) =   g g  

Range is maximum when θ = 450 and Rmax =

u2 g

t

INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 04064606657 R = H

 2u 2 sin θcos θ      g    u 2 sin 2 θ      2g  

= 4 cot θ →

The velocity of the particle at any time 't' is given by V = Vx i + Vy j →

j V = (ucosθ ) i + (usinθ - gt)

The magnitude of the velocity = ( u cos θ) 2 + ( u sin θ − gt ) 2 If α is the angle made by the velocity at any time 't' with the horizontal, Tan α =

u sin θ − gt u cos θ

Taking the origin at the point of projection, the 'x' and 'y' coordinates at any time 't' are given by and y = usinθ t −

x = u cosθ t

Eliminating 't' from x and y  x    u cos θ 

y = u sinθ 

2 −1  x  g   2  u cos θ 

gx 2

= x tanθ -

1 gt2 2

2u 2 cos 2 θ

which is the equation of a parabola.

It may be noted here that the velocity of the projectile will be always tangential to its path. The equations of projectile motion derived above are valid only for constant acceleration due to gravity 'g'. Illustration – 13 : A particle is projected from the horizontal at an inclination of 600 with an initial velocity 20 m/sec. Assuming g = 10 m/sec2 find a) the time at which the energy becomes three fourths kinetic and one fourth potential b) the angle made by the velocity at that time with the horizontal c) the x and y coordinates of the particle taking the origin at the point of projection. Solution : a)

Let V be the velocity when the given condition is fulfilled

1 3 mV2 = ( 2 4

1 mu2) 2 3u 2

V=

= 10

3

m/sec

→

V = u cos θ i + ( u sin θ − gt ) j

= 20 cos 600 i + →

V

= 10

Solving t =

3

(

(20 sin 600 −10t )

⇒ 102 + 3± 2

) sec

(10

j

3 − 10 t

= 10 i

)2

=

+ (10 3 − 10 t ) j

(10 3)2

INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 04064606657 t = 3 − 2 while rising up and t = 3 + 2 while coming down b)

Tanα

= =

c)

u sin θ − gt u cos θ

(

10 3 − 10 3 ± 2 10

)

=+

x = ucos θ t = 10 ( 3 ± 2) = 10

(

and y = u sinθ t -

1 2

= 10

-5

3

(

3± 2

)

2

)

3 − 2 m or 10

gt

(

(

)

3+ 2 m

2

3± 2

)2 = 5m

PROJECTILE MOTION ON AN INCLINED PLANE : y'

y

Let α be the inclination of the plane and x' the particle is projected at an angle θ with the u inclined plane. It is convenient to take the reference frame with x' along the plane and y' g cosα θ g sin α perpendicular to the plane. gcosα will be α the component of the acceleration along the x downward perpendicular to the plane and g sin α will be the component of the acceleration along the downward direction of the inclined plane. Along the plane, the kinematical equations take the form ⇒

⇒

Vx ' = u x ' + a x ' t Vx ' = u cos θ − g sin α t 1 Sx' = u x ' t + a x' t 2 2 1 S x ' = ucosθ t gsinα t2 2

Vx ' 2 - u x ' 2 = 2a x 'S x ' ⇒ Vx ' 2 - (ucosθ )2 = 2 (-g sinα ) S x ' Similarly perpendicular to the plane, the kinematical equations take the form Vy ' = u y ' + a y ' t ⇒ Vy ' = u sin θ − g cos αt S y' = u y' t +

1 1 a y ' t 2 ⇒ S y ' = u sin θ t − g cos α t 2 2 2

Vy ' 2 − u 2y ' = 2 a y ' S y ' ⇒ Vy2' − ( u sin θ) 2 = 2 ( − g cos α) S y '

Here it may be noted that, When the particle strikes the inclined plane S y ' = 0 When the particle strikes the inclined plane perpendicular to it,

S y' = 0

and

Vx ' = 0

When particle strikes the inclined plane horizontally S y ' = 0 and Vy = 0 Illustration – 14 : y' From the foot of an inclined plane of inclination α , a projectile is shot at an angle β with the inclined plane. Find the relation between α and β if the projectile strikes the inclined plane a) perpendicular to the plane

u

β

α

x'

INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 04064606657 b) horizontally Solution : a) Since the particle strkes the plane perpendicularly S y ' = 0 and Vx ' = 0 u sinβ t -

1 g cosα 2

2u sin β

⇒ t = g cos α 2u sin β

u cos β

and

u cosβ

- g sinα t = 0

u cos β

and

⇒ g cos α = g sin α b)

t2 = 0

t = g sin α

⇒ 2 Tan β = cot α

Since the particle strikes the plane horizontally S y ' = 0 and Vy = 0 1 g cos α t2 = 0 2 u sin ( α + β) 2u sin β t = g cos α = g

u sinβ t -

and ⇒

u sin (α + β ) - gt = 0

2u sin β u sin ( α + β) = ⇒ g cos α g

2 sin β = sin cos α

(α + β ) CIRCULAR MOTION : When a particle moves in a circle of radius R with constant speed V, its called uniform circular motion. V ∆V V

θ

V

θ V

When the particle covers θ , the direction of velocity also changes by θ without change in magnitude. Change in velocity ∆ V will be towards the centre of curvature of the circular path which causes centripetal acceleration. θ is called the angular position (or) angular displacement. Centripetal acceleration, a r =

∆V ∆t

The rate of change of angular position is known as angular velocity (ω ) Time period of circular motion T =

2πR V

In the same time the particle covers an angle 2 π from which angular velocity can be found as 2π 2πV V = = T 2πR R θ Rθ ⇒ ∆t= = and ∆ V = ω V

ω =

When θ is small sinθ

~ θ

V 2 + V 2 − 2V 2 cos θ

⇒ ∆ V = Vθ ( Vθ) ∆V V2 centripetal acceleration = =  Rθ  =   ∆t R  V 

= 2V sin

θ 2

INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 04064606657 When speed of the particle continuously changes with time, the tangential acceleration is given by a t =

dV dt

The rate of change of angular velocity is called the angular acceleration (α ) since a r and a t are perpendicular to each other, the resultant ar2 + a t2

acceleration is given by a =

Angle made by the resultant with radius vector Tanθ =

at ar

Illustration – 15 : The speed of a particle in circular motion of radius R is given by V = Rt2. Find the time at which the radial and the tangential accelerations are equal and the distance traveled by the particle upto that moment. Solution : ar = at V 2 dV = 2Rt = R dt

⇒t =

1 23

1

Distance travelled =

23

∫

V dt

0

=

1 3 2 3

 Rt    3   

=

2R 3

INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 04064606657 RADIUS OF CURVATURE When a particle is moving in a plane

ar =

V2 R

where V is the instantaneous

velocity and R is the radius of curvature at that point. V2 ar

Radius of curvature =

If the path of the particle is given by y = f(x), radius of curvature can also be 3

found from the formula R =

2 2   1 +  dy     dx    d 2y dx 2

Illustration – 16 : A particle is projected with initial velocity 'u' at angle θ with the horizontal. Find the radius of curvature at a) point of projection b) the top most point Solution :

a)

at the point of projection P,

at the topmost point T, ⇒ R=

2

V ar

θ p V = u and a r = g cosθ

V2 u2 = g cos θ ar

⇒R = b)

T

u

=

2

V = ucosθ and a r = g

2

u cos θ g

SHORTEST DISTANCE OF APPROACH : When two particles A and B are moving simultaneously, their position coordinates at any time 't' are given by (when the accelerations are uniform) →

→

→

r A = r0 A + u A t +

→ → → 1→ 1→ a A t 2 and r B = r0 B + u B t + a B t 2 2 2

The distance between them at any time 't', S = →

→

→

Where r AB = r0 AB + u AB t +

→

r

AB

1→ a AB t 2 2

The distance between them becomes minimum when

dS =0 from which the dt

time at which it becomes minimum can be found. Substituting the value of →

time so obtained in r AB , S min can be found. Illustration – 17 : Two ships A and B move with constant velocities as shown in the figure. Find the closest distance of approach between them

North

A

30o

VA = 20kmph

10km O

450

20km

B

VB = 10 2 km ph East

INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 04064606657

Solution : →

→

r0 A = 10 j → VA

r0 B = 20 i

=10 i −10 → aA = 0 →

→

= 20cos 600 i − 20cos 30j

→

V B = 10 2 cos 45i + 10 2 cos 45 j = 10 i + 10 j

3 j →

aB = 0

→

→

r A = r0 A + V A t

→

= 10 t i + (10 −10 3 t ) j →

→

= (20 + 10 t) i

= - 20 i + (10 −10 3 t − 10 t ) j

→

r AB = r A − r B S =

→

r B = r0 B + V B t

→

r

AB

=

( − 20) 2 + (10 − 10

3 t − 10 t

)

+ 10 t

2

When the distance between A and B is minimum    2 

1

( − 20) 2 + (10 −10

10 - 10

3

3 t −10 t

t - 10 t = 0

)

2

    

(2(10− 10 

j

dS =0 dt

)(

))

3 t − 10 t − 10 3 − 10 = 0

  hr  1+ 3 

t = 

1

Substituting this value of time in the expression for S, S min = 20 km CYCLIC MOVEMENT OF PARTICLES : When three or more particles located at the vertices of a polygon of side l move with constant speed V such that particle 1 moves always towards particle 2 and particle 2 moves always towards 3 particle etc., they meet at the centre of the polygon following identical curved paths. Initial seperation

Time of meeting = Velocityof approach Velocity of approach is the component of the relative velocity along the line joining the particles.

INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 04064606657 Illustration – 18 : Six particles located at the six vertices of a hexagon of side l move with constant speeds V such that each particle always targets the particle in front if it. Find the time of meeting and the distance travelled by each particle before they meet Solution : Initial seperation

t = Velocity of approach

600 V

V

 2 = 0 = V V − V cos 60

V

Since they move with constant speed V, the distance travelled by each particle in time t =

2 is V

V

d = Vt =

V

2 V =2l V

RIVER PROBLEMS : If

→

→

B

V r is the velocity of the river and V b

C •

is the velocity of the boat with respect to still water, the resultant velocity of the →

→

Vr

→

boat V = V b + V r R Only the perpendicular component of the resultant velocity helps in crossing the river. Time of crossing, t =

Vb

θ

A

w where Vb cos θ

'w' is the width of the river. The boat crosses the river in the least time when θ = 0 The parallel component of the resultant velocity determines the drift. Drift is the displacement of the boat parallel to the river by the time the boat crosses the river 

 w   Vb cos θ 

Drift , x = ( Vr − Vb sin θ) 

Zero drift is possible only when Vr = Vb sinθ . When Vr > Vb zero drift is not possible. Illustration – 19 : A river of width 100 m is flowing towards East with a velocity of 5 m/s. A boat which can move with a speed of 20 m/s with respect to still water starts from a point on the South bank to reach a directly opposite point on the North bank. If a wind is blowing towards North East with a velocity of 5 2 m/s, find the time of crossing and the angle at which the boat must be rowed. Solution : → Vb

= −20sin θ i + 20cos θ j

→

Vr = 5 i

Vw

θ Vb

45

Vr

INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 04064606657 → V w = 5 2 cos 45 i + 5 2 cos 45 j = 5 i + 5 j →

→

→

→

V R = Resultant velocity of the boat = V b + V r + V w = ( - 20 sin θ + 5 + 5) i + (20 cos θ + 5)

j

For reaching directly opposite point, the component of the resultant velocity parallel to the river must be zero - 20 sinθ + 10 = 0

⇒ sinθ =

1 and θ = 300 2

Since time of crossing depends only on the perpendicular component of the resultant velocity. t=

100 w = = 4.48 sec 20 cos 300 + 5 20cos θ + 5

WORKED OUT OBJECTIVE PROBLEMS EXAMPLE : 01 A point moves along 'x' axis. Its position at time 't' is given by x 2 = t Its acceleration at time 't' is A)

1 x3 −t2 x3

B)

1 1 − x x2

C)

−t x2

2

+ 1.

D)

Solution : x= a=

t2 +1 2

d x xt 2

1 dx = (2t) = dt 2 t2 +1

;

=

t  t 2 + 1 − (2t )     2 t 2 +1 2

 t 2 + 1    

=

(

t t2 +1

1

)

t2 +1

3

=

1 x3

EXAMPLE : 02 A body thrown vertically up from the ground passes the height 10.2m twice at an interval of 10 sec. Its initial velocity was (g = 10 m/s2) A) 52 m/s B) 26 m/s C) 35 m/s D) 60 m/s Solution : Displacement is same in both cases s = ut + 1/2 at2 10.2 = ut -

1 (10) t2 2

2 t1 = u − u − 204 and

10

u 2 − 204

EXAMPLE : 03

= 50

⇒t =

u ± u 2 − 204 10

2 t2 = u + u − 204 ⇒ ∆ t = t2 - t1 = 10 sec

10

⇒ u2 = 2500 + 204

⇒ u = 52 m/s

INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 04064606657 A car starts from rest moving along a line, first with acceleration a= 2 m/s2, then uniformly and finally decelerating at the same rate and comes to rest. The total time of motion is 10 sec. The average speed during this time is 3.2 m/s. How long does the car move uniformly A) 4 sec B) 6 sec C) 5 sec D) 3 sec Solution : Let the car accelerate for time 't' and move uniformly with v = at for time t 1 Since the magnitudes of acceleration and deceleration are same, the time of deceleration is also 't'. t + t 1 + t = 10 sec Average speed =

 1 2  1 2 Distance  at  + ( at ) t 1 +  at  = 2  2  = 3.2 time 10 2

 10− t 1   10− t 1   +2   t 1 = 32 ⇒2   2   2 

2t + 2tt1 = 32 2

Solving t1 = 6 sec

This problem can be solved using velocity time graph also.

EXAMPLE : 04 A particle has an initial velocity of (3 i +4j) m/s and a constant acceleration (4i −3j) m/s2. Its speed after 1 sec will be equal to A) zero B) 10 m/s C) 5 2 m/s D) 25 m/s Solution :

V = u + a t = (3i +4j) +(4i −3j`) (1) →

→

→

→

Speed = magnitude of V =

72 + 12 = 5

=7 i -

j

2 m/s

EXAMPLE : 05 An aeroplane flies along a straight line from A to B with air speed V and back again with the same air speed. If the distance between A and B is l and a steady wind blows perpendicular to AB with speed u, the total time taken for the round trip is A)

2 V

D)

B)

2 2

V +u

2

2V

C)

V 2u 2

2 V2 − u2

Solution : The resultant velocity of the plane must be along AB during forward journey. 



t1 = V = R V2 − u2

u

V

B VR A

INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 04064606657 During return journey, the resultant velocity of the plane must be along BA 



t2 = V = R V2 − u2 Total time t = t1 + t2 =

B

u

VR V

2 V2 − u2

A

EXAMPLE : 06 A particle is thrown with a speed 'u' at an angle θ with the horizontal. When the particle makes an angle φ with the horizontal its speed changes to V. Then A) V = u cosθ B) V = ucosθ cosφ C) V = u cosθ secφ D) V = usecθ cosφ Solution : Since the horizontal component of the velocity of a projectile always remains constant u cosθ =V cosφ ⇒ V=ucosθ secφ EXAMPLE : 07 Two shells are fired from a cannon with same speed at angle α and β respectively with the horizontal. The time interval between the shots is T. They collide in mid air after time 't' from the first shot. Which of the following conditions must be satisfied. A) α > β B) t cosα = (t -T) cos β C) (t-T) cosα =cosβ (t-T)-

D) (usinα )t -

1 gt2 =(usinβ ) 2

1 g(t-T)2 2

Solution : When they collide, their 'x' and 'y' components must be same ucosα t = u cosβ (t-T) ⇒ cosα t = cos β (t-T) 1 1 gt2 = (usinβ ) (t-T) g (t-T)2 2 2 T  Since cosα = cos β  1 −  and T < t t  

(usinα ) t -

cos α < cosβ

and α > β

EXAMPLE : 08 A particle is projected from a point 'p' with velocity 5 2 m/s perpendicular to the surface hollow right angle cone whose axis is vertical. It collides at point Q normally on the inner surface. The time of flight of the particle is A) 1 sec B) 2 sec C) 2 2 sec D) 2 sec Solution :

INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 04064606657 → → It can be seen from the diagram that V becomes perpendicular to u . → 0 u = ucos45 i + u sin45 j →

→

→

V = u + a t = (ucos 45 i + usin45 →

j

) - (gt)

→

When V becomes perpendicular to u , u2 cos2 45 + u2 sin2 45 - (usin45) gt = 0

j

→ →

V. u =0 u ⇒ t = g sin 45 = 1 sec

EXAMPLE : 09 A man walking Eastward at 5 m/s observes that the wind is blowing from the North. On doubling his speed Eastward he observes that the wind is blowing from North East. The velocity of the wind is A) (5i+5j) m/s B) (5i - 5j) m/s C) (-5i +5j) m/s D) (-5i - 5j) m/s Solution : →

let V w = V i + V j 1 2 In the first case

(

)

V wm = V w − V m = V1i + V2 j - (5i )

→

→

→

Since no component along East is observed V1 - 5 = 0 ⇒ V1 = 5 m/s In the second case →

→

→

V wm = V w − V m = (V1 i + V2

j

) - (10 i )

= ( V1 − 10)i + V2 j

Since the wind is observed from North East the components along North and East must be same V1 - 10 = V2 ⇒ V2 = - 5 m/s → V w = (5i - 5j) m/s EXAMPLE :10 From a lift moving upward with uniform acceleration 'a', a man throws a ball vertically upwards with a velocity V relative to the lift. The time after which it comes back to the man is 2V

2V

V

A) g −a Solution :

B) g +a

C) g +a

D) →

2Vg 2

g −a2

Since the velocity of the ball is given relative to the lift V bl = V j When the ball comes back to the man, its displacement relative to the lift is →

zero S bl = 0 →

→

→

a bl = a b − a l = (-g) j - a j = - (g + a) j Applying S = ut + 1/2 at2 in relative form →

1

→

0 = ( Vt ) j +

1 2

(− ( g + a ) j)

→

2 S bl = V bl t + 2 a bl t

t2

2V

⇒ t = g +a

INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 04064606657

INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 04064606657 ASSIGNMENT SINGLE ANSWER TYPE QUESTIONS LEVEL – I 1. The greatest acceleration or deceleration that a train may have is a. The minimum time in which the train can go from one station to the next at a distance S is A) 2.

3.

4.

6.

1 g t1t 2 2

2a

8.

9.

s a

C) 2

D)

s 2a

B) g t1 t2

C) 2 g t1 t2

D) 4 g t1 t2

From a 20m high tower one ball is thrown upward with speed of 10m/s and another is thrown vertically downward at the same speed simultaneously. The time difference of their reaching the ground will be (take g = 10m/s2) A) 12s B) 6s C) 2s D) 1s A particle X moving with a constant velocity u crosses a point O. At the same instant another particle F starts from rest from O with a constant acceleration a. The maximum separation between them before they meet is 2 A) u

7.

2S a

B)

A car accelerates from rest at a constant rate α for sometime and attains a velocity of 20 m/s. Afterwards it decelerates with a constant rate α /2 and comes to a halt. If the total time taken is 10s, the distance travelled by the car is A) 200m B) 100m C) 10m D) 20m A particle starts from the position of rest under a constant acceleration. It travels a distance x in the first 10 seconds and distance y in the next 20 seconds. Then A) y = x B) y = 2x C) y = 8x D) y = 4x A body is projected vertically upwards. If t1 and t2 be the times at which it is at height h above the point of projection while ascending and descending respectively, then h is A)

5.

S a

2 B) u

a

2 C) 2u

a

2 D) u

4a

A bird flies in straight line for 4s with a velocity v = (2t-4) m/s. What is the distance covered by the bird in returning to the place from where it started its journey ? A) 0 B) 8m C) 4m D) 2m A ball is thrown vertically upwards. Which of the following plots represents the speed-time graph of the ball during its flight if the air resistance is not ignored

A) B) C) D) If a ball is thrown vertically upwards with speed 'u', the distance covered during the last t second of its ascent is A) (u+gt)s

B) ut

C)

1 gt2 2

D) ut -

1 gt2 2

INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 04064606657 10. A particle has an initial velocity of 9m/s due east and a constant acceleration of 2m/s2 due west. The distance covered by the particle in the fifth second of its motion is A) 0 B) 0.5m C) 2m D) none of these 11. Two particles are projected simultaneously in the same vertical plane from the same point, with different speeds u1 and u2, making angles θ 1 and θ 2 respectively with the horizontal , such that u1 cosθ 1 = u2cosθ 2. The path followed by one, as seen by the other (as long as both are in flight), is : A) a horizontal straight line B) a vertical straight line C) a parabola D) a straight line making an angle |θ 1 - θ 2| with the horizontal 12. A particle starts from the origin of coordinates at time t = 0 and moves in the xy plane with a constant acceleration α in the y-direction. Its equation of motion is y =β x2. Its velocity component in the x-direction is A) variable

13.

15.

16.

17.

2α β

1

1

α

α

C) 2β D) 2β A train starts from station A with uniform acceleration α for some distance and then goes with uniform retardation β for some more distance to come to rest at station B. The distance between station A and B is 4 km and the train takes 4 minute to complete this journey. If α and β are in km (min) -2 then 1

14.

B)

1

1

1

1

1

1

1

A) α+β=2 B) α+β=4 C) α+β= 2 D) α+β= 4 The driver of a train moving with a speed v1 sights another train at a distance d, ahead of him moving in the same direction with a slower speed v2. He applies the breaks and gives a constant de-acceleration 'a' to his train. For no collision, d is 2 v − v2 ( v1 − v 2 ) ( ( v1 − v 2 ) 2 v1 − v 2 ) 2 A) = B) > C) < D) < 1 2a 2a 2a 2a In the case of a moving body, pick the correct statement A) if speed changes with change in direction, velocity does not change B) if velocity changes, speed may or may not change but acceleration does change C) if velocity changes, speed also changes with same acceleration D) if speed changes without change in direction, the velocity may remain constant. Particle 1 is in one dimensional motion with uniform velocity whereas particle 2 is accelerating in a straight line. The graph representing path of 2 with respect to 1 is

A) B) C) D) Person A walking along a road at 3ms-1 sees another person B walking on another road at right angles to his road. Velocity of B is 4ms-1when he is 10m off. They are nearest to each other when person A has covered a distance of A) 3.6m B) 8m C) 6.3m D) 0.8m

INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 04064606657 18. The given graph shows the variation of velocity with displacement. Which one of the graphs given below correctly represents the variation of acceleration with displacement.

19.

20.

21.

A) B) C) D) 2 v versus s-graph of a particle moving in a straight line is as shown in figure. From the graph some conclusions are drawn. State which statement is wrong : A) the given graph shows a uniformly accelerated motion. B) initial velocity of particle is zero C) corresponding s-t graph will be a parabola D) none of the above A body dropped from the top of the tower covers a distance 7x in the last second of its journey, where x is the distance covered in first second. How much time does it take to reach the ground ? A) 3s B) 4s C) 5s D) 6s A body is projected with a velocity u. It passes through a certain point above the ground after t1 sec. The time interval after which the body passes through the same point during the return journey is 



u 2 A)  g − t 1   

22. 23. 24. 25.

26.

27.

u



u2

 − t1    g 

 u2



D) 3  2 − t1  g  The area of the acceleration-displacement curve of a body gives : A) impulse B) change in momentum per unit mass C) change in KE per unit mass D) total change in energy A body thrown vertically up from the ground passes the height 10.2m twice at an interval of 10s. What was its initial velocity (g = 10m/s2) A) 52m/s B) 26 m/s C) 35 m/s D) 60 m/s An insect crawls a distance of 4m along north in 10 seconds and then a distance of 3m along east in 5 seconds. The average velocity of the insect is : A) 7/15 m/sec B) 1/5 m/sec C) 1/3 m/sec D) 4/5 m/sec A particle returns to the starting point after 10s. If the rate of change of velocity during the motion is constant in magnitude, then its location after 7 seconds will be same as that after : A) 1 second B) 2 seconds C) 3 seconds 4) 4 second Two particles P and Q simultaneously start moving from point A with velocities 15m/s and 20m/s respectively. The two particles move with accelerations equal in magnitude but opposite in direction. When P overtakes Q at B then its velocity is 30m/s. The velocity of Q at point B will be A) 30m/s B) 5m/s C) 20m/s D) 15m/s Let v and a denote the velocity and acceleration respectively of a particle in one dimensional motion: A) the speed of the particle decreases when v • a < 0 B) the speed of the particle increases when v • a > 0 C) the speed of the particle increases when v • a = 0 D) the speed of the particle decreases when v t2 B) R' < R, H' < H, t 1' > t1 and t '2 < t2 C) R' < R, H' > H, t 1' > t1 and t '2 < t2 D) R' < R, H' < H, t 1' < t1 and t '2 > t2 Speed of a particle moving in a circle varies with time as, v = 2t. Then :

INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 04064606657 A) angle between velocity vector and acceleration vector is increasing with time. B) a is constant while ar is increasing with time. C) both A and B are correct D) both A and B are wrong. 36. Initial velocity and acceleration of two particles are as shown in fig. Assuming the shown direction as the positive, vBA versus time graph is as :

37.

38.

39.

40.

41.

42.

A) B) C) D) A graph is plotted between velocity (v) and displacement (s) of particle moving in a straight line. Here v is plotted along y-axis and 's' along x-axis. Choose the correct option. A) slope of this graph at any point always gives us the ratio of velocity and displacement at that point.7 B) slope represents a/v under all the conditions. (a = acceleration) C) both A and B are correct D) both A and B are wrong. In a projectile motion if a person wants to increase the maximum height to 2 times but simultaneously want to decrease the range same number of time. He can achieve it by increasing tan of angle of projection by ....... times. A) 2 B) 4 C) 3 D) 2 The velocity of a particle moving in a straight line varies with time in such a manner that v versus t graph is represented by one half of an ellipse. The maximum velocity is ν m and total time of motion is t0 i) Average velocity of particle is π /4 ν m ii) Such motion can not be realized in practical terms A) Only (i) is correct B) Only (ii) is correct C) Both (i) and (ii) are correct D) Both (i) and (ii) are wrong Starting from rest, a particle rotates in a circle of radius R = 2 m with an angular acceleration α = π /4 rad/ s2. The magnitude of average velocity of the particle over the time it rotates quarter circle is A) 1.5 m/s B) 2 m/s C) 1 m/s D) 1.25m/s In a car race car A takes t0 time less to finish than car B and passes the finishing point with a velocity v0 more than car B. The cars start from rest and travel with constant accelerations a1 and a2. Then the ratio ν 0/t0 is equal to a 1 +a 2 a 12 a 22 a a A) B) C) D) 1 2 a2 a1 2 A rod of length 1 leans by its upper end against a smooth vertical wall, while its other end leans against the floor. The end that leans against the wall moves uniformly downward. Then A) The other end also moves uniformly B) The speed of other end goes on decreasing C) The speed of other end goes on increasing D) The speed of other end first decreases and then increases

INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 04064606657 43. A particle is moving along a circular path of radius 5 m and with uniform speed 5 m/s. What will be the average acceleration when the particle completes half revolution? A) zero B) π m/s2 C) 10 π m/s2 D) 10/π m/s2 44. The velocity displacement graph of a particle moving along a straight line is shown The most suitable acceleration-displacement graph will be

A)

B)

C)

D)

1

2

3

4

5

6

7

8

9

C 2 1 B 4 1 C

B 2 2 C 4 2

C 2 3 A 4 3

A 2 4 C 4 4 A

C 2 5 C

A 2 6 B

B 2 7 A

C 2 8 C

C 2 9 A

1 0 B 3 0 A

1 1 B 3 1 C

1 2 D 3 2 D

1 3 A 3 3 B

1 4 B 3 4 D

1 5 B 3 5 C

1 6 A 3 6 A

1 7 A 3 7 B

1 8 A 3 8 B

1 9 B 3 9 C

2 0 B 4 0 C

LEVEL – II 1. A particle starts from rest at time t = 0 and moves on a straight line with an acceleration which varies with time as shown in fig. The speed of the particle will be maximum after how many seconds A) 4s C) 8s 2.

3.

B) 6s D) 10s

Due to air a falling body faces a resistive force proportional to square of velocity v, consequently its effective downward acceleration is reduced and is given by a = g - kv2 where k = 0.002m-1. The terminal velocity of the falling body is A) 49m/s B) 70m/s C) 9.8m/s D) 98m/s A balloon is rising with a constant acceleration of 2m/s 2. At a certain instant when the balloon was moving with a velocity of 4m/s, a stone was dropped from it in a region where g = 10m/s2. The velocity and acceleration of stone as it comes out from the balloon are respectively. A) 0, 10m/s2 B) 4m/s, 8m/s2 C) 4m/s, 12m/s2 D) 4m/s, 10m/s2

INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 04064606657 4. A stone is thrown vertically up from the top of a tower with some initial velocity and it arrives on the ground after t1 seconds. Now if the same stone is thrown vertically down from the top of the same tower with the same initial velocity, it arrives on ground after t2 seconds. How much time will the stone take to reach the ground if it is dropped from the same tower ? t +t t −t A) 1 2 B) 1 2 C) t 1 + t 2 D) t 1t 2 2 2 5. Two particles A and B start from the same point and slide down through straight smooth planes inclined at 300 and 600 to the vertical and in the same vertical plane and on the same side of vertical drawn from the starting point. The acceleration of B with respect to A is A) g/2 in vertical direction B) g 3 / 2 at 450 to vertical 0 C) g/ 3 at 60 to vertical D) g in vertical direction 6.

7.

8.

9.

10.

A particle starts from rest at the origin and moves along X-axis with acceleration a = 12-2t. The time after which the particle arrives at the origin is A) 6s B) 18s C) 12s D) 4s Figure represents position (x) versus time (t) graph for the motion of a particle. If b and c are both positive constants, which of the following expressions best describes the acceleration (A) of the particle ? A) a = +b B) a = -c C) a = b + ct D) a = b-ct Two particles instantaneously at A and B are 5m, apart and they are moving with uniform velocities, the former towards B at 4m/s and the latter perpendicular to AB at 3m/s. They are nearest at the instant A) 2/5s B) 3/5s C) 1s D) 4/5s Three particles start from the origin at the same time, one with a velocity 'u1', alone X-axis, the second along the Y-axis with a velocity u2 and the third alogn x = y line. The velocity of third particle so that the three may always lie on the same plane is u 1u 2 u + u2 2 u 1u 2 A) 1 B) u 1u 2 C) D) u1 + u 2 u1 + u 2 2 A ball is shot vertically upwards from the surface of a planet in a distant solar system. A plot of the y versus t for the ball is shown in fig. The magnitude of the free fall in m/s2 on the planet is A) 4 B) 8 C) 12 D) 16

INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 04064606657 11. The horizontal range of a projectile is R and the maximum height attained by it is H. A strong wind now begins to blow in the direction of the motion of the projectile, giving it a constant horizontal acceleration = g/2. Under the same conditions of projection, the horizontal range of the projectile will now be A) R + 12.

H 2

B) R + H

C) R +

3H 2

A particle moves int he xy plane with a constant acceleration g in the negative y-direction. Its equation of motion is y = ax - bx2, where a and b are constants. Which of the following are correct ? A) The x-component of its velocity is constant B) at the origin, the y-component of its velocity is a

13.

14.

15.

16.

D) R + 2H

g 2b

C) At the origin, its velocity makes an angle tan-1(a) with the x-axis. D) the particle moves exactly like a projectile. Two bodies are projected simultaneously from the same point, in the same vertical plane, one towards east and other towards west with velocities 8 ms-1 and 2 ms-1 respectively. The time at which their velocities are perpendicular to each other is A) 2/5 s B) 5/2s C) 1/5 s D) 5 s Two stones are projected so as to reach the same distance from the point of projection on a horizontal surface. The maximum height reached by one exceeds the other by an amount equal to half the sum of the heights attained by them. Then the angles of projection for the stones are A) 450, 1350 B) 00, 900 C) 300, 600 D) 200, 700 Velocity and acceleration of a particle at some instant of time are v =(3ˆi +4ˆj) m/s and a =−(6ˆi +8ˆj) m/s2 respectively. At the same instant particle is at origin. Maximum x-co-ordinate of particle will be A) 1.5m B) 0.75m C) 2.25m D) 4.0m a-t graph for a particle moving in a straight line is as shown in figure. Change in velocity of the particle from t=0 to t=6s is : A) 10m/s B) 4m/s C) 12m/s D) 8m/s

17.

Speed time graph of two cars A and B approaching towards each other is shown in figure. Initial distance between them is 60m. The two cars will cross each other after time. A) 2sec B) 3sec C) 1.5sec D) 2 sec

18.

The position of a particle along x-axis at time t is given by x = 2 + t - 3t2. The displacement and the distance travelled in the interval t = 0 to t = 1 are respectively A) 2, 2 B) -2, 2.5 C) 0, 2 D) -2, 2.16

19.

The acceleration time graph of a particle moving along a straight line is as shown in figure. At what time the particle acquires its initial velocity ?

INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 04064606657 A) 12s B) 5s C) 8s D) 16s 20.

A graph between the square of the velocity of a particle and the distance s moved by the particle is shown in the figure. The acceleration of the particle in kilometre per hour square is : A) 2250 B) 225 C) -2250 D) -225

21.

A particle starts from rest and traverses a distance l with uniform acceleration, then moves uniformly over a further distance 2l and finally comes to rest after moving a further distance 3l under uniform retardation. Assuming entire motion to be rectilinear motion the ratio of average speed over the journey to the maximum speed on its way is : A) 1/5 B) 2/5 C) 3/5 D) 4/5 Two stones are thrown up simultaneously with initial speeds of u 1 andu2 (u2 > u1). They hit the ground after 6s and 10s respectively. Which graph in figure correctly represents the time variation of ∆ x = (x2 - x1), the relative position of the second stone with respect to the first upto t = 10s ? Assume that the stones do not rebound after hitting the ground.

22.

A)

B)

C)

D)

23.

Figure shows the position-time (x-t) graph of the motion of two boys A and B returning from their school O to their homes P and Q respectively. Which of the following statements is true ? A) A walks faster than B B) Both A and B reach home at the same time C) B starts for home earlier than A D) A overtakes B on his way to home

24.

A ball is projected with a velocity 20 3 m/s at angle 600 to the horizontal. The time interval after which the velocity vector will make an angle 300 to the horizontal is (take g = 10m/s2) A) 4sec B) 2 sec C) 1 sec D) 3 sec

25.

The equation of motion of a projectile is : y = 12x -

3 2 x 4

Given that g = 10ms-2, what is the range of the projectile ? A) 12m B) 16m C) 20m D) 24m

INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 04064606657 26. A projectile is thrown with an initial velocity of (aˆi +bˆj)ms −1 . If the range of the projectile is twice the maximum height reached by it, then : A) a = 2b B) b=a C) b = 2a D) b = 4a 27.

A particle moves along a parabolic path y = 9x2 in such a way that the x component of velocity remains constant and has a value acceleration of the particle is A)

28.

29.

1ˆ jms −2 3

B)

2 3 ˆjm s −

C)

2ˆ jms −2 3

D)

1 ms −1 . The 3

2ˆjms −2

Two projectiles are projected with the same velocity. If one is projected at an angle of 300 and the other at 600 to the horizontal. The ratio of maximum heights reached, is : A) 1 : 3 B) 2 : 1 C) 3 : 1 D) 1 : 4 A particle is projected from the ground with velocity u at angle θ with horizontal. The horizontal range, maximum height and time of flight are R, H and T respectively. They are given by, R=

u 2 sin 2θ u 2 sin 2 θ ,H = g 2g

and T =

2u sin θ g

Now keeping u as fixed, θ is varied from 300 to 600. Then, A) R will first increase then decrease, H will increase and T will decrease B) R will first increase then decrease, while H and T both will increase C) R will decrease while H and T will increase D) R will increase while H and T will decrease 30.

Velocity and acceleration of a particle at some instant of time are ˆ ) m/s2 . Then, the speed of the particle ˆ ) m/s and a = (ˆi + 6ˆj − k v = (2ˆi − ˆj + 2k is ....... at a rate of ....... m/s2 A) increasing, 2 B) decreasing ,2 C) increasing, 4 D) decreasing, 4

31.

x and y coordinates of a particle moving in xy plane at some instant are : x = 2t2 and y =

3 2 t 2

The average velocity of particle in a time interval from t = 1 second to t = 2 second is : A) 32.

33.

(8ˆi +5ˆj) m / s

B)

(1 2ˆi + 9ˆj) m / s

C)

(6ˆi +4.5ˆj) m / s

D)

(1 0ˆi + 6ˆj) m / s

A particle is projected upwards with some velocity. At what height from ground should another particle be just dropped at the same time so that both reach the ground simultaneously. Assume that first particle reaches to a maximum height H. A) 6H B) 8H C) 4H D) 10H Particle A moves with 4m/s along positive y-axis and particle B in a circle x2 + y2 = 4(anticlockwise) with constant angular velocity ω = 2 rad/s. At time t = 0 particle is at (2m, 0). Then : A) magnitude of relative velocity between them at time t is 8 sin 2t B) magnitude of relative velocity between them is maximum at t = π /4 second.

INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 04064606657 C) both A and B are correct D) both A and B are wrong 34.

35.

36.

37.

38. 39.

1 A 2 1 C

An armored car 2 m long and 3 m wide is moving at 10 ms -1 when a bullet hits it in a direction making an angle tan-1(3/4) which the length of the car as seen by a stationary observer. The bullet enters one edge of the car at the corner and passes out at the diagonally opposite corner. Neglecting any interaction between the car and the bullet, the time for the bullet to cross the car is A) 0.20 s B) 0.15 s C) 0.10 s D) 0.50 s The V - t graph for the rectilinear motion of a particle is represented by a parabola as shown in fig. Find the distance traveled by the particle in time T/2. 2 2 Vmax T Vmax T 3 2 A) B) Vmax T Vmax T 2 3 2 C) D) A glass wind screen whose inclination with the vertical can be changed is mounted on a car. The car moves horizontally with a speed of 2 m/s. At what angle α with the vertical should the wind screen be placed so that the rain drops falling vertically downwards with velocity 6 m/s strike the wind screen perpendicularly. A) tan-1 (1/3) B) tan-1 (3) C) cos-1(3) D) sin-1(1/3) Two stones are thrown up simultaneously from the edge of a cliff with initial speeds v and 2 v. The relative position of the second stone with respect to first varies with time till both the stones strike the ground as A) Linearly B) First linearly then parabolically C) Parabolically D) First parabolically then linearly There are two values of time for which a projectile is at the same height. The sum of these two times is equal to A) 3T/2 B) 4T/3 C) 3T/4 D) T A particle is projected from a horizontal plane with 8 2 m/s at an angle. At highest point its velocity is found to be 8 m/s. Its range will be A) 6.4 m B) 3.2 m C) 5 m D) 12.8 m 1 1 1 1 1 1 1 1 1 1 2 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 B D D A B D D D B D A A C B B B D C C 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 A B B B C D A B B C C D A C D

LEVEL - III 1. Two particles A and B are connected by a rigid rod AB. The rod slides on perpendicular rails as shown in fig. The velocity of A to the left is 10m/s. What is velocity of B when α = 600? A) 10m/s B) 5.8m/s C) 17.3m/s D) 9.8m/s

INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 04064606657 2. In the figure , the pulley P moves to the right with a constant speed u. The downward speed of A is vA, and the speed of B to the right is vB. A) vB = vA B) vB = u + vA C) vB + u = vA D) The two blocks have accelerations of the same magnitude 3. A marble starts falling from rest on a smooth inclined plane forming an angle α with horizontal. After covering distance 'h' the ball rebound off the plane. The distance from the impact point where the ball rebounds for second time is A) 8h cosα B) 8h sinα C) 2h tanα D) 4h sinα 4. From the top of a tower of height 40m, a ball is projected upwards with a speed of 20m/s at an angle of elevation of 300. The ratio of the total time taken by the ball to hit the ground to its time of flight (time taken to come back to the same elevation) is (take g = 10m/s2) A) 2 : 1 B)3 : 1 C) 3 : 2 D) 1.5 : 1 5.

If time taken by the projectile to reach Q is T, then PQ is equal to : A) Tv sin θ B) Tv cos θ C) Tv sec θ D) Tv tanθ

6.

A particle is thrown with a speed u at an angle θ with the horizontal. When the particle makes an angle φ with the horizontal, its speed changes to v : A)v = u cos θ B) v = u cosθ cos φ C) v = u cosθ sec φ D) v = u sec θ cos φ

7.

A stone is projected from a point on the ground so as to hit a bird on the top of a vertical pole of height h and then attain a maximum height 2h above the ground. If at the instant of projection the bird flies away horizontally with a uniform speed and if the stone hits the bird while descending then the ratio of the speed of the bird to the horizontal speed of the stone is : A)

2 2 +1

B)

2 2 −1

C)

1 2

+

1 2

D)

2 2 +1

8.

Shots are fired simultaneously from the top and bottom of a vertical cliff with the elevation α = 300, β = 600 respectively and strike the object simultaneously at the same point. If a = 30 3 m is the horizontal distance of the object from the cliff, then the height of the cliff is : A) 30m B) 45m C) 60m D) 90m

9.

A particle if projected up an inclined plane of length 20m and inclination 30 0 (with horizontal). What should be the value of angle θ (with horizontal) with which the projectile be projected so that it strikes the plane exactly at midpoint, horizontally :

INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 04064606657  2       3

10.

 3    2   

3

A) θ = tan-1 B) θ = tan-1 (2) C) θ = tan-1 ( )D) θ = tan-1 A stone is projected at an angle θ with the horizontal with velocity u. It executes a nearly circular motion near its maximum height for a short time. The radius of circular path is : A)

u 2 cos 2 θ 2g

B)

u 2 sin 2 θ g

C)

u 2 cos 2 θ g

D)

u2 g

11.

Two projectiles A and B are fired simultaneously as shown in figure. They collide in air at point P at time t. then : A) t(u1 cosθ 1 - u2 cosθ 2) = 20 B) t(u1 sinθ 1 - u2 sinθ 2) = 10 C) both (A) and (B) are correct D) both (A) and (B) are wrong

12.

Two stones are projected simultaneously with equal speeds from point on an inclined plane along the line of its greatest slope upwards and downwards repectively. The maximum distance between their points of striking the plane is double that of when they are projected on a horizontal ground with same speed. If one strikes the plane after two seconds of the other, the angle of inclination of plane is A) 300 B) 450 C) 350 D) 150

13.

2 a g from a point at a A particle is projected under gravity with velocity height h above the level plane. The maximum range R on the ground is A) (a 2 +1) h V= θ B) a 2 h C) a h D) 2 a (a +h )

14.

The co-ordinates of a particle moving in a plane are given by x = a cos pt and y = b sin pt where a, b (< a) and p are positive constants of appropriate dimensions. Then A) The path of the particle is an ellipse B) The velocity and acceleration of the particle are normal to each other at t = π /2p C) The acceleration of the particle is always directed towards a fixed point D) The distance traveled by the particle in time interval t = 0 to t = π /2p is a.

1 B

2 BD

3 B

4 A

5 D

6 C

7 D

8 C

9 A

10 C

11 B

12 B

13

14 D

MULTIPLE ANSWER TYPE QUESTIONS 1.

A particle of mass m is thrown up vertically with velocity u. As air exerts a constant force F, the particle returns back at the point of projection with velocity v after attaining maximum height h, then

INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 04064606657

2.

3.

4.

5.

6.

u2 2( g + F / m )

h=

2h

A) t1t2 = g 7.

v2 2( g − F / m )

(g − F / m )

(g + F / m )

C) v = u (g + F / m) D) v = u (g − F / m) A particle of mass m moves on X-axis as follows; it starts from rest at t = 0 from the point x = 0, and comes to rest at t = 1 at the point x = 1. No other information is available about its motion at intermediate times (0 < t < 1). If α denotes the instantaneous acceleration of the particle, then A) α cannot remain positive for all t in the interval 0 ≤ t ≤ 1 B) |α | cannot exceed 2 at any point in its path C) |α | must be ≥ 4 at some point or points in its path D) α must change sign during the motion, but no other assertion can be made with the information given Two trains are travelling along a straight track one behind the other. The first train is travelling at 12 m/s. The second train, approaching from the rear is travelling at speed v > 12 m/s when the second train is 200 m behind the first, the driver of second train applies brakes producing a uniform deceleration of 0.20 m/s2. Then A) If v = 20 m/s, the trains will not collide B) If v = 20 /s, the trains will collide after about 20 s C) If v = 27 m/s, the trains will not collide D) If v = 27 m/s, the trains will collide after about 15s A particle of mass m and charge q starts from rest from origin along X-axis in a region where an electric field E = E0 - a x exists. Here E0 and a are constant and x is the distance from the starting point. Then in the region between x = 0 to x = 2 E0/a. A) the speed of particle first increases, then decreases B) the particle comes to rest at x = 2 E0/a C) the particle has maximum speed at x = E0/a D) the particle is subjected to an acceleration which changes sign at x = E0/a Two particles are projected from the same point with the same speed v0 at the different angles α 1 and α 2 with the horizontal. Their respective times of flights are T1 and T2. If they have the same horizontal range, and their maximum heights are H1 and H2 respectively, then T1 = tan α1 D) A) α 1 + α 2 = 900 B) H1 + H2 = v02/2g C) T2 H1 = tan 2 α1 H2 A projectile thrown on a level surface attains a height h after t 1 seconds and again after t2 seconds. If the maximum height attained by the projectile is H after t seconds, then A)

B)

B) t1 + t2 =

h=

8H g

C) t2 - t1 =

8( H − h ) g

D) t2 - t = t - t1

Which of the following statements are true for a moving body? A) If its speed changes, its velocity must change and it must have some acceleration B) If its velocity changes, its speed must change and it must have some acceleration

INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 04064606657 C) If its velocity changes, its speed may or may not change, and it must have some acceleration D) If its speed changes but direction of motion does not change, its velocity may remain constant 8. The figure shows the velocity (υ ) of a particle plotted against time (t) A) The particle changes its direction of motion at some point B) The acceleration of the particle remains constant C) The displacement of the particle is zero D) The initial and final speeds of the particle are the same 9.

A particle starts from the origin of coordinates at time t = 0 and moves in the xy plane with a constant acceleration α in the y-direction. Its equation of motion is y = β x2. Its velocity component in the x-direction is A) variable

10.

11.

13.

1

+θ

2

= 900

C)

α 2β

D)

α 2β

B)

t1 = tan θ1 t2

C)

t1 = tan θ 2 t2

D)

t1 t2 = sin θ1 sin θ 2

A large rectangular box falls vertically with an acceleration a. A toy gun fixed at A and aimed towards C fires a particle P. A) P will hit C if a = gB) P will hit the roof BC if a > g C) P will hit the wall CD or the floor AD if a < g D) May be either A, B, or C depending on the speed of projection of P Two shells are filed from a cannon with speed u each, at angles of α and β respectively with the horizontal. The time interval between the shots is T. They collide in mid air after time t from the first shot. Which of the following conditions must be satisfied? A) α > β B) t cos α = (t - T) cos β C) (t - T) cos α = t cos β D) (usin α ) t -

14.

2α β

Two particles A and B start simultaneously from the same point and move in a horizontal plane. A has an initial velocity u1 due east and acceleration a1 due north. B has an initial velocity u2 due north and acceleration a2 due east A) Their paths must intersect at some point B) They must collide at some point C) They will collide only if a1u1 = a2u2 D) If u1 > u2 and a1 < a2, the particles will have the same speed at some point of time Two particles are projected from the same point with the same speed, at different angels θ 1 and θ 2 to the horizontal. They have the same horizontal range. Their times of flight are t1 and t2 respectively A) θ

12.

B)

1 1 gt2 = (u sin β ) (t - T) g (t - T)2 2 2

A particle moving with a speed v change in speed A) The change in the magnitude of B) The change in the magnitude of C) The magnitude of the change in

changes direction by an angle θ , without its velocity is zero its velocity is 2v sin θ /2 its velocity is 2 v sin θ /2

INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 04064606657 D) The magnitude of the change is its velocity is v (1 - cos θ ) 15. A particle of mass 'm' moves on the x-axis as follows: It starts from rest at t = 0 from the point x = 0 and comes to rest at t = 1 at the point x = 1. No other information is available about its motion at intermediate times ( 0 < t < 1) If α denotes the instantaneous acceleration of the particle, then A) α cannot remain positive for all t in the interval 0 ≤ t ≤ 1 B) α cannot exceed 2 at any point on its path C) α must be ≥ 4 at some point or points in its path D) α must change sign during the motion, but no other assertion can be made with the information given 16. A point moves such that its displacement as a function of time is given by x2 = t2 + 1. Its acceleration at time t is

17.

A) 18.

1

t

B) −

C)

1 t2 − x x3

C)

1 gt

1

t

D) x − 2 x3 x2 x A body falls from a large height 'h' in 't' second. The time taken to cover the last metre is A)

1

B)

gh

1 2 gh

D)

1 2 gt

A bead is free to slide down a smooth wire tightly stretched between the points P1 and P2 on a vertical circle of radius R. If the bead starts from rest from P1, the highest point on the circle and P2 lies anywhere on the circumference of the circle. Then, A) time taken by bead to go from P1 to P2 is dependent on position of P2 and equals 2

R g

cos θ

B) time taken by bead to go from P1 to P2 is independent of position of P2 and equals 2

19.

20.

R g

C) acceleration of bead along the wire is g cos θ D) velocity of bead when it arrives at P2 is 2 gR cos θ The position of a particle travelling along x axis is given by x t = t3 - 9t2 + 6t where xt is in cm and t is in second. Then A) the body comes to rest firstly at (3 - 7 ) s and then at (3 + 7 )s B) the total displacement of the particle in travelling from the first zero of velocity to the second zero of velocity is zero C) the total displacement of the particle in travelling from the first zero of velocity to the second zero of velocity is -74 cm D) the particle reverses its velocity at (3 - 7 )s and then at (3 + 7 )s and has a negative velocity for (3 - 7 ) < t < (3 + 7 ) The velocity of a particle moving along a straight line increases according to the linear law v = v0 + kx, where k is a constant. Then A) the acceleration of the particle is k(v0 + kx) B) the particle takes a time

v 1 log e  1 k  v0

C) velocity varies linearly with displacement curve equal to k

  to attain a velocity v1  

displacement

with

slope

of

velocity

INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 04064606657 D) data is insufficient to arrive at a conclusion 21. A particle moves with an initial velocity v0 and retardation α v, where v is velocity at any istant t. Then v A) the particle will cover a total distance 0 α B) the particle continues to move for a long time span 1 1 v 0 at t = α 2 1 D) the particle comes to rest at t = α

C) the particle attains a velocity

22.

An aeroplane flies along a straight line from A to B with a speed v 0 and back again with the same speed v0. A steady wind v is blowing. If AB = l then 2 v0l A) total time for the trip is 2 , if wind blows along the line AB v0 − v2 B) total time for the trip is

23.

24.

2l v 02

−v2

, if wind blows perpendicular to the line AB

C) total time for the trip decreases because of the presence of wind D) total time for the trip increases because of the presence of wind At the instant a motor bike starts from rest in a given direction, a car overtakes the motor bike, both moving in the same direction. The speed time graphs for motor bike and car are represented by OAB and CD respectively. Then A) at t = 18 s the motor bike and car are 180 m apart B) at t = 18 s the motor bike and car are 720 m apart C) the relative distance between motor bike and car reduces to zero at t = 27 s and both are 1080 m far from origin D) the relative distance between motor bike and car always remains same A particle having a velocity v = v0 at t = 0 is decelerated at the rate a = α v , where α is a positive constant A) The particle comes to rest at t =

2 v0 α

B) The particle will come to rest at infinity C) The distance traveled by the particle is D) The distance travelled by the particle is 25.

3 2

2 v0

α 3 2

2 v0 3 α

Two particles P and Q move in a straight line AB towards each other. P starts from A with velocity u1 and an acceleration a1. Q starts from B with velocity u2 and acceleration a2. They pass each other at the midpoint of AB and arrive at the other ends of AB with equal velocities A) They meet at midpoint at time t =

2 (u 2 − u 1 ) (a 1 − a 2 )

B) The length of path specified i.e. AB is l =

4 ( u 2 − u 1 ) (a 1 u 2 − a 2 u 1 ) (a 1 − a 2 ) 2

INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 04064606657 C) They reach the other ends of AB with equal velocities if (u2 + u1) (a1- a2) = 8 (a1u2 - a2u1) D) They reach the other ends of AB with equal velocities if (u2 - u1) (a1 + a2) = 8 (a2u1 - a1u2) 26. A body is moving along a straight line. Its distance xt from a point on its path at a time t after passing that point is given by xt = 8t2 - 3t3, where xt is in metre and t is in second A) Average speed during the interval t = 0 s to t = 4 s is 20.2 l ms-1 B) Average velocity during the interval t = 0 s to t= 4 s is -16 ms-1 C) The body starts from rest and at t =

27.

28.

at xt = 8.43 m from the start. D) It has an acceleration of - 56 ms-2 at t = 4 s Two second after projection, a projectile is traveling in a direction inclined at 300 to the horizon. After one more second it is traveling horizontally. Then A) the velocity of projection is 20 ms-1 B) the velocity of projection is 20 3 ms-1 C) the angle of projection is 300 with vertical D) the angle of projection is 300 with horizon A shot is fired with a velocity u at an angle (α + θ ) with the horizon from the foot of an incline plane of angle α through the point of projection. If it hits the plane horizontally then A) tan θ

=

2 tan α 1 +2 tan 2 α

29.

16 s it reverses its direction of motion 9

tan α

B) tan θ = 2 tan α

1 +2 tan 2 α

D) tan θ =

tan

θ

=

sin αcos α 1 +sin 2 α

A particle is projected with a velocity 2 hg so that it just clears two walls of equal height h at horizontal separation 2h from each other. Then the A) angle of projection is 300 with vertical B) angle of projection is 300 with horizon 2h g

C) time of passing between the walls is D) time of passing between the walls is 2 30.

C)

h g

A ball starts falling freely from a height h from a point on the inclined plane forming an angle α with the horizontal as shown. After collision with the incline it rebounds elastically off the inclined plane. Then A) it again strikes the incline at t =

8h g

after it strikes the

incline at A B) it again strikes the incline at t =

31.

2h g

after it strikes the incline at A

C) it again strikes the incline at a distance 4h sin α from A along the incline D) it again strikes the incline at a distance 8h sin α from A along the incline Two particles projected from the same point with same speed u at angles of projection α and β strike the horizontal ground at the same point. If h 1 and h2 are the maximum heights attained by projectiles, R be the range for both and t1 and t2 be their time of flights respectively then

INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 04064606657 t1 h1 π A) α + β = B) R = 4 h 1h 2 C) = tan α D) tan α = t h 2 2 2 32. Two shells are fired from a cannon successively with speed u each at angles of projection α and β respectively. If the time interval between the firing of shells is t and they collide in mid air after a time T from the firing of the first shell. Then A) T cos α = (T - t) cos β B) α > β C) (T - t) cos α = t cos β 1 1 gT2 = ( u sin β ) (T - t) g(T - t)2 2 2

D) ( u sin α ) T 33.

34.

Two guns situated at the top of a hill of height 10 m, fire one shot each with -1 3 ms the same speed of 5 at some interval of time. One gun fires horizontally and other fires upwards at an angle of 600 with the horizontal. The shots collide in mid air at the point P. Taking the origin of the coordinate system at the foot of the hill right below the muzzle, trajectories in x - y plane and g = 10 ms-2 then A) the first shell reaches the point P at t1 = 1 s from the start B) the second shell reaches the point P at t2 = 2 s from the start C) the first shell is fired 1 s after the firing of the second shell D) they collide at P whose coordinates are given by (5 3 , 5) m A radar observer on the ground is watching an approaching projectile. At a certain instant he has the following information: i) The projectile has reached the maximum altitude and is moving with a horizontal velocity v; ii) The straight line distance of the observer to the projectile is l; iii) The line of sight to the projectile is at an angle θ above the horizontal Assuming earth to be flat and the observer lying in the plane of the projectile's trajectory then, A) the distance between the observer and the point of impact of the projectile is D=

v 2 sin θ g

- l cos θ

B) the distance between the observer and the point of impact of the projectile is D=v

35.

2 l sin θ g

- l cos θ

C) the projectile will pass over the observer's head for l <

2 v 2 tan θsec θ g

D) the projectile will pass over the observer's head for l >

2 v 2 tan θsec θ g

A projectile is thrown with an initial velocity u, at an angle of projection θ first from the equator and then from the pole. The fractional decrement in the range of projectile is A)

dR 1 =− R 291

 1 dR 1 =−g  − g p R g e 

B)    

dR 1 = R 291

C)

 1 dR 1 =g  −  R ge g p

   

D)

INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 04064606657 36. A boat is moving directly away from a cannon on the shore with a speed v1. The cannon fires a shell with a speed v2 at an angle α and the shell hits the boat. Then, A) the shell hits the boat when the time equal to B) the boat travels a distance

2 v 2 sin α is lapsed g

2 v1 v 2 sin α from its original position g

C) the distance of the boat from the cannon at the instant the shell is fired is 2 g

(v2 sin α ) (v2 cos α - v1)

D) the distance of the boat from the cannon when the shell hits the boat is 2 g

37.

(v2 sin α ) (v2 cos α )

A projectile is fired upward with velocity v0 at an angle θ and strikes a point P (x, y) on the roof of the building (as shown). Then, A) the projectile hits the roof in minimum time if θ + α =

π 2

B) the projectile hits the roof in minimum time if θ + α =

π 4

C) the minimum time taken by the projectile to hit the roof is tmin = v 0 − 38.

2 −2gh cos 2 α v0

g cos α

D) the projectile never reaches the roof for v0 < 2gh cos α . A shell fired along a parabolic path explodes into two fragments of equal mass at the top of the trajectory. One of the fragments returns to the point of firing having retraced its original path. If v is the velocity of projectile at highest point, then A) after explosion the other fragment has a velocity v along + x axis B) after explosion the other fragment has velocity 3v along + x axis C) after explosion both the fragments reach the ground with separation 2R between them, where R is the range of the projectile D) after explosion both the fragments hit the ground simultaneously at t = R 2v

INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 04064606657 39. Identify the correct statements for a moving particle along any arbitrary path is/are:     A) Vav =Vav B) Vav ≤Vav C) Vav ≥Vav D) Vav tα = T C) t β < tα = T D) cannot predicted If the time taken by the object to travel from A to B along γ is t γ , then :

be

19.

A) t γ = tα = T B) t γ > tα = T C) t γ < tα = T D) cannot predicted If the time taken by the object to travel from A to B along δ is t δ , then :

20.

A) tδ = tα = T B) tδ > tα = T C) tδ < tα = T D) cannot be predicted If we approximate the time taken to travel from A to B along the deep path ' δ' by the time taken for a vertical return path of depth 'H', i.e., from A and back to A, by assuming the initial velocity to be negligible as compared with that for the major portion of the path, then the total time taken is: A)

2H g

B)

4H g

C)

8H g

D) T

Passage VI (Q.No: 21 to 23): A point moves rectilinearly in one direction. Figure shows the distance s traversed by the point as a function of time t. Using this graph, answer the following questions. 21. The average velocity of the point during the time of motion is A) 10 cm/s B) 15 cm/s C) 20 cm/s D) 25 cm/s 22. The maximum velocity is A) 15 cm/s B) 20 cm/s C) 25 cm/s D) 30 cm/s 23. The time moment t 0 at which the instantaneous velocity is equal to the mean velocity averaged over the first t 0 second is A) 10s B) 16s C) 18s D) 20s Passage VII (Q.NO 24 to 26): Two particles are initially located at points A and B a distance d apart as show in figure. They start moving at time t = 0 such that the velocity u of B is always along the horizontal direction and velocity v of A is continually aimed at B as shown in figure.At time t = 0, u is perpendicular to v. 24. About the velocities u and v, we can say that A) both the velocities are constant

INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 04064606657 B) both the velocities are changing C) the velocity u is constant while the velocity v is changing D) the velocity v is constant while the velocity u is changing 25. The relative velocity of approach of A towards B is A) (v - u cos θ ) B) (v + u cos θ ) C) (u - v cos θ ) D) (u + v cos θ ) 26. The particles A and B will meet after a time A)

vd v +u 2

2

B)

(v

vd 2

−u

2

)

C)

(v

+u2 vd

2

)

D)

(v

−u2 vd

2

)

Passage - VIII (Q.No: 27 to 29) : Figure shows the velocity time graph of a particle moving along a straight line. Answer the following questions. 27. The region in which the rate of change of velocity of the particle is maximum. A) 0 to 2 s B) 2 to 4 s C) 4 to 6 s D) 6 to 8 s 28. The particle comes to rest at time A) 0 sec B) 4.67 sec C) 5 sec D) 8 sec 29. The maximum displacement of particle is A) 33.3 m B) 20.2 m C) 26.6 m D) zero PASSAGE - IX (Q.No: 30 to 33): On the bank of a river two swimmers made a challenge as "who will reach the point B on the other bank early?" So both of them start from point 'A' on one bank of the river to reach the winning point B on the other bank, lying directly opposite to point A. The stream velocity was known to be 2 km/hr and the velocity of both the swimmers in still water was 2.5 km/hr. Both of them follow different paths to reach point B the swimmer 'S 1' crosses the river along straight line AB, while the other swimmer 'S2' swims at rights to the stream and then walks the distance which he has been carried away by the stream to go to the point 'B'. Assume the velocity (uniform) of his walking as (2/3) km/hr and the width of the river as 'W'. 30. The value of angle 'θ' shown in the figure is: A) cos-1 (4/5) B) cos-1 (3/4) C) sin-1(4/5) D) sin-1(3/4) ' shown in the figure is : 31. The value of angle ' φ -1 A) sin (4/5) B) cos-1 (4/5) C) tan-1 (4/5) D) data insufficient

INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 04064606657 32. Match column I with column II in reference to the passage. Column I Column II I Time for S1 to reach (A  4W    B )  5  II

Time for S2 to reach (B C )

 6W     5 

III

Drift 'x' for S2

(C )

 2W     3 

IV

Time for S2 to reach (D B from C )

 8W     15 

V

33.

1.

Difference in time, (E)  2W    ∆t , for S1 and S2 to  5  reach B A) I-C, II-E, III-A, IV-B,V-D B) I-A, II-B, III-D,IV-C, V-E C) I-E, II-C, III-B, IV-D, V-A D) I-B, II-A, III-D, IV-C, V-E What should be the velocity (assume uniform) of walking of swimmer 'S 2' such that both the swimmers reach at the point B simultaneously? A) 1.2 km/hr B) 2.7 km/hr C) 3 km/hr D) needs more information *** MULTIPLE MATCHING TYPE QUESTIONS: Match the following: List - I List - II a) Range = height (Max) e) acceleration perpendicular velocity 0 b) For θ = 45 , at the f) R = 8 17 g highest point 2 1 c) y = px + qx g) Kmin = (Kmax) d) At t = T/2 (for projectile)

to

2

h) θ = Tan-1 (4) T

2P

i) u = g P 2 +1 2.

Match the following: List - I a) one dim. motion b) for no air resistance c) for projectile H = R/4 d) Range =

3.

u2 g

Match the following: List - I

List - II e) motion of one projectile w.r.t another projectile f) time of ascent = Time of decent for a vertically projected up body g) T = 2u/g, for a body vertically projected up i) θ = 450

List - II

INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 04064606657 a) v - t graph e) Area gives displacement b) x - t - graph f) slope gives velocity c) a - t graph g) slope gives acceleration h) Area gives change in velocity d) a α x h) slope of v - x graph is constant t) non uniform acceleration 4. Match the following: List - I List - II a) Magnitude of acceleration is constant e) uniform circular motion b) tangential acceleration is zero f) non-uniform circular motion c) speed is constant g) projectile motion d) angle between radial acceleration and h) accelerated straight line velocity is 900 motion ***

INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 04064606657 KEY Multiple Answer Type Questions: 1. 2. 3. 4. 5. 6. 7. 8. 9. 1 0. 1 1. 1 2. 1 3. 1 4.

ABC AC ABCD ABCD ABCD ABCD AC ABCD D ACD ABD

2 AC 3 .

1 BC 7 . 1 BCD 8 . 1 ACD 9 . 2 ABC 0 .

ABC ABD

1 AC 5 .

2 ABD 2 .

1 AC 6

2 ABCD 6 .

2 AD 8 . 2 AD 9

Comprehension Type Questions: 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26.

2 ABC 5 .

2 BC 7 .

2 AB 1 .

AC

2 AD 4 .

B D B D A C A A D E C

A C B C A B

27. 28. 29. 30. 31. 32. 33.

C A,B, D A

3 AD 0 .

3 ACD 7 .

3 ABCD 1 .

3 BCD 8 .

3 ABD 2 . 3 ABCD 3 . 3 BC 4 . 3 BD 5 . 3 ABCD

3 9. 4 0. 4 1. 4 2. 4 3. 4 4. 4 5. 4 6.

AC AC BCD CD ACD BCD ABD AD

INSIGHT IIT JEE CLASSES NALLAKUNTA HYDERABAD 04064606657 Multiple Matching Type Questions: a → f, h 1. b → e, g c →i d →e 2. a → e, h, g, f b → f, g

3.

c →i d →i a → e, g b →f c →h d → i, j

4.

a → e, g, h b →e c →e d → e, f