Kinemat Th

Part-I-KM-PH-1

1.

IIT- JEE Syllabus

Displacement, velocity, acceleration, kinematics in one and two dimensions, projectiles, circular motion, concept of relative motion.

2.

Linear Kinematics

2.1

Rest and motion

Motion means change of position with time. When a body is at rest, its position does not change with time. But how do we describe the position of a body? We describe it relative to another body for reference. Thus, rest and motion are relative. A man sitting in a car moving at 55 mph on a highway is at rest relative to a co-passenger, while he is in motion relative to a person standing on the highway. In order to describe rest and motion, we select a frame of reference and then describe rest or motion relative to this frame of reference. Motion can be of two types - translational & rotational. When a body moves such that it always remains parallel to itself throughout the motion it undergoes translation. When a body moves so that each point in the body maintains a constant distance relative to a fixed axis in space, the motion is rotation. 2.2

Position & frames of reference

Position: If a particle is moves along a given straight line (assumed along x-axis), its position is represented by the x-coordinate relative to a fixed origin. If the particle moves in a plane (let x-y plane) its position is completely known when the x-and y coordinates of its position are known with respect to the given coordinate axis ox and oy.

y

P (x, y)

B

x O

A

Similarly for a particle moving in space, three coordinates (x, y, z) are required.  In vector notation, the position vector OP  r in the three cases mentioned above are represented as    r  x ˆi , r  x ˆi  yˆj , r  x ˆi  yˆj  zkˆ Several types of coordinate frames may be used to describe position.

Frames of reference: Two of the commonest kinds of coordinate system in use are (a) Rectangular Cartesian Coordinates (b) Polar coordinates

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(I) 2 dimensions Cartesian: A point P with coordinates (x, y) in a rectangular cartesian coordinate system (fig.1) is  described by its position vector OP , also  represented by r .

fig.

1

y P(x,y)=(r,)

 L

O  r  ˆi x  ˆj y ,

x

Where ˆi , ˆj are unit vectors along x, y respectively. . . . (1)

Polar: If the initial ray is chosen to be OX, the point P can be described by the  coordinates (r, ) in the polar coordinate system. The position vector OP , is   . . . (2) r  rrˆ , rˆ being the unit vector along OP Equation (1) and (2) relate the two coordinate systems. From figure 1, x = r cos  r = x2  y2 y = r sin  tan  = y/x Putting (3) in (1) and (2) and comparing,

. . . (3)

rˆ  ˆi cos   ˆj sin 

(II) 3 - dimensions The simplest system of coordinate used in 3 dimensions is the rectangular cartesian coordinate system which consists of an origin O and three mutually perpendicular axis x, y & z.

z P(x,y,z)  O

A point P having coordinates (x, y, z) in this system has the position vector

y



 r  OP  ˆi x  ˆj y  kˆz

x

Illustration 1:

L

A particle is kept at the point of intersection of the diagonals. Find its position w.r.t. point O.

y

C

B P

4m x O

Solution:

Here x = (2.5m) y = (2 m)    OP = x i  y   = (2.5 i +2 j ) m

Exercise 1:

A

5m

y P

 j



= (2.5m i +2 m

 j )

2m

O

2.5m

x

In the above illustration, a plane mirror is placed along x – z plane. Find the position of the image of particle w.r.t. point O.

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2.3

Displacement

If a particle moves from its initial position A to a final position B, the vector joining A & B  and directed along the line AB is known as the displacement vector  r . The actual distance covered may have any value which is greater than or equal to the length of the straight line joining the two points A and B. Suppose that the position vector of A = x ˆi  y ˆj  z kˆ 1

1



= OA =

2

r2

A



= OB =

 r1

2

 r2

Using vectors, 

A



OA + AB =



B

 r

Y

1

Suppose that the position vector of B = x ˆi  y ˆj  z kˆ 2

r1





OB



 r

=

AB

O



= r2  r1

x

=

 x 2  x1  ˆi   y 2  y1  ˆj   z 2  z1 kˆ = x ˆi  yˆj  zkˆ Here x, y and z represent the components of  displacement vector  r along x, y and z directions. Illustration 2 :

In the above problem, if the ball shifts from points P to point C, find the displacement of the ball.

y

C

B P

4m x O

Solution:

5m

A

`

= OP = (2.5 ˆi +2 ˆj ) m



r1

r2

= OC = (4 ˆj ) m

`

 Displacement

  Δr  r2  r1

= { 4 ˆj – (2.5 ˆi +2 ˆj )}m = (– 2.5 ˆi +2 ˆj )m. Exercise 2 :

In the above illustration, if the ball moves through 2m from P to point Q such that line PQ is perpendicular to the plane OABC, then find its displacement.

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2.4

Distance

The distance covered by the particle increases with time whereas the magnitude of the displacement (the shortest distance between the initial and final position) increase, decrease or may be zero (e.g. when it passes through its initial position). Therefore the distance travelled is never smaller than the magnitude of the displacement.  d  S. Illustration 3:

A car moves from A to B on a straight road and returns from B to the mid-point of AB. AB = 10 m. Find the displacement and distance covered.

Solution:

Let A be the origin and C be the mid-point of A B. Y The displacement       s   r  rC  rA  rC ( rA  0 )

 s  (rC )ˆi  ( AC)ˆi  ( AB  BC)ˆi

 s  (10  5)ˆi  5ˆi m

P

A rA

X C

B

The distance covered = d= AB + BC = 10+5 = 15 m. Exercise 3 :

2.5

Referring to previous illustration, what are the magnitudes of the displacement and distance covered when the car returns to A?

Speed

(a) Average Speed A particle covers a distance by s = 10 m during a time interval t = 5 seconds. Therefore s/t is the ratio of distance covered s and the time interval t. This ratio is termed as time rate of change of distance. It is common known as Speed of the particle over that particular time interval t. That is why this speed is known as “Average Speed” uav denoted mathematically as uav =

s 10 m   2 m / sec t 5 sec

(b) Instantaneous Speed If the particle is not moving with uniform speed, the average speed completed in the above section depends on the time interval t and the instant when it was computed. It becomes necessary to define a quantity, called instantaneous speed which does not depend on the time interval t. We take the time interval t as it becomes very small (infinitesimal) and the corresponding distance travelled s (which is also infinitesimal).t  0 (infinitesimal time)

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s  0 (infinitesimal distance) As t  0, s  0 t=0

t=2 sec t=(2+0.1)sec. t  0.1 sec.

s = 2m

s = .25m

t = (2 + 0.01)sec t  0.01sec

s = .015m

t = (2 + 0.001)sec t  0.001sec

s = .001m

t = (2 + 0.0001)sec t  0.0001sec t = (2 +t 0)sec t  0

s  0

 

 

s ds   v at time t (or instantaneous velocity) t  0 t dt ds  v insant  v . dt Lim

 ds 

Here v at the time t = 2 sec is represented by  .   dt  t  2 Instantaneous speed is equal to the magnitude of instantaneous velocity.

The ratio of the infinitesimal distance covered and the corresponding infinitesimal time yields a finite value. This finite value is known as the instantaneous speed of the particle. s  u  Lim t  0 t (c) Uniform Speed If the particle covers equal distances in equal intervals of time, it is said to be moving with uniform speed. Student Notes     

The slope of s-t graph at any time t gives the speed at that time t. The slope of x-t graph at any time t gives the instantaneous velocity (velocity at that time) For positive slope v is positive, for negative slope v is negative. The area of u-t graph during time t (= 2t 0, say) gives the distance traversed during the time 2t0. The area of v-t graph during time t (= 2t 0, say) gives the displacement of the particle during time t. Total displacement snet = (positive displacement-negative

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2.6

displacement)  =  (s+ - s-) We see that snet = vt0 + (-vt0) = 0 where as d = 2ut 0 during time t = 2t0. Velocity

(a) Average Velocity  As the displacement of a particle changes with time, the ratio of change s of  displacement vector s in a time interval t to that interval of time is known as time rate of change of displacement (vector), this is known as the Velocity of the particle averaged over the time interval t or the average velocity, v av   s v av = t

(b) Instantaneous Velocity Following the previous argument of instantaneous speed, instantaneous velocity    dx ˆ dy ˆ dz ˆ ds  s i j k v   = v  Lim dt dt dt dt t  0 t  * Instantaneous speed is the magnitude of instantaneous velocity v .  dv  Since a = , dt The slope of the v-t graph gives the acceleration. If the slope is positive the acceleration is positive; if the slope is negative then the acceleration is negative (retardation or deceleration). Since v =

t

 0 a dt (discuss later),

the area of a-t graph gives the change in velocity. If the area remains (above x- axes) positive, the change in velocity is positive and vice versa. (c) Uniform Velocity A particle is said to be moving with uniform velocity if its velocity is independent of time. Illustration 4 :

A particle moves in a straight line obeying the relation x = t (t-1) where x = displacement in m and t = time in sec. Find the velocity of the particle when its displacement is zero.

Solution:

x = t (t-1). If x = 0, t(t-1) = 0  t = 0 and t = 1 The instantaneous velocity dx v  2 t 1 dt Putting t = 0 and t = 1 we obtain v t  0   1 m / sec & v t 1   1 m / sec .

Exercise 4 :

Referring to the previous illustration, (a) What is the significance of positive and negative values of instantaneous velocities?

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(b) What are the distance covered and displacement during t = 2 sec. ?

2.7

Acceleration

(a) Average Acceleration If the velocity (magnitude, direction or both) of a particle changes with time, its motion is said to be non-uniform.  Suppose that the velocity of particle changes by v over a time interval t, the time rate  of change of velocity is given by v /t, this is known as the average acceleration of the   v a  particle over the time interval t. av . t (b) Instantaneous Acceleration For an instant (infinitely small or infinitesimal time interval), the change in velocity of the particle is infinitely small, but the ratio of infinitesimal change in velocity and the infinitesimal time  is finite.  This finite ratio is known as instantaneous acceleration:  v dv  a  Lim  t  0 t dt (c) Uniform Acceleration When a particle undergoes constant acceleration (vector) for some time interval we say that it is moving with constant or uniform acceleration over that time interval. 2.8

Average speed and Velocity

Average Speed When a particle moves with different uniform speeds u 1, u2, u3 etc. in different (finite) time intervals t1, t2, t3 etc. respectively, its average speed over the total time of journey is given as s1  s2  s3  ... total dis tan ce cov ered uav  = total time elapsed t1  t 2  t 3  ... where s1 = u1 t1, s2 = u2 t2 etc. u1 t1  u2 t 2  ....  uav = t1  t 2  ... 

i n

uav =

in

 u t  t i

i 1

i

i

i 1

Average Velocity    When a particle moves with different velocities v 1, v 2 , v 3 etc. in different time intervals t1, t2, t3 etc. respectively, its average velocity over the total time of motion can be given as   Net displaceme nt vector s v av   Total time t

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   s1  s2  s3  ... = t1  t 2  t 3  ... Where the displacement of the particle during time interval t1, t2 etc. are given by 

  s2  v 2 t 2 etc .



 s1  v1t1,

   v1 t1  v 2 t 2  ...  v av  t1  t 2  ...   v av 

i n



in

 v i t i

 t

i 1

Illustration 5:

Solution :

Exercise 5 :

i

i 1

A particle moves with a velocity v(t) = (1/2)kt2 along a straight line. Find the average speed of the particle in a time t.

1 uav = T

t



v( t )dt =

0

1 T

t

1

 2 kt 0

2

dt =

1 kT 2 . 6

Find the average speed of a particle whose velocity is given by v = v0 sin  t.

Time Average  v av ( t ) 



T

0

 v dt T

 1 aav ( t )  T

T

0



1 T



T

0

 v dt

 a dt

  Here, v and a are the functions of t. Illustration 6:

A passenger is standing ‘d’ m away from a bus. The bus begins to move with constant acceleration a. To catch the bus, the passenger runs at a constant speed v towards the bus. What must be the minimum speed of the passenger so that he may catch the bus ?

Solution :

Let the passenger catch the bus after time t, the distance travelled by the bus 1 2 s1 = 0 + at . . . . (1) 2 and the distance travelled by the passenger s2 = ut . . . . (2) Now passenger will catch the bus if d + s1 = s2 . . .. (3)  Substituting the values of s1 and s2 from equation (1) and (2) in (3),

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Part-I-KM-PH-9

we get d+ i.e.

1 at2 = ut 2

1 2 at – ut + d = 0 2

2 or t = u  u  2ad

a

So the passenger will catch the bus if t is real, i.e. u2  2ad or u  2ad So the minimum speed of passenger for catching the bus is

2ad

.

Exercise 6 :

A man standing 9m behind a bus starts running uniformly to catch the bus. At the same instant the bus starts from rest with an acceleration of 2m/s2. Find the speed of the man if he is just able to get in.

Illustration 7:

A float is overcome by a motor boat going downstream at a point A. T hours later it turns back and after some time it passes the float at a distance d km from point A. What is the velocity of stream if the speed of motor remains constant w.r.t. water ?

Solution :

Let u = speed of stream, V (V+u) v = speed of motor boat ( v > u) u Let t1 = time of motion of motor boat t during up stream. T + t1 The motion of float and motor boat is – (V– u) shown in the following velocity time graph. As float has no speed of its own hence it will move for the complete time (T + t1) hrs. with the flow speed. For motor boat, velocity down stream = (v + u) and velocity up stream = – v – u) Negative sign represents the opposite direction. As area of velocity time graph gives the displacement and from the question, displacement of float = d km displacement of boat = d km Hence, we get the following equations. For float, u (T + t1) = d …(i) For motor boat (v + u) (T) + { – (v – u) (t1)} = d …(ii) Solving (i) & (ii), we get d km / hr u= 2T

.

Exercise 7 :

A car accelerates from rest with  = 2 m/sec2 in a straight track, then it comes to rest applying its brakes after a time T = 20 seconds covering a

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distance of 100 m. Using a graphical method, find the maximum speed of the car.

3.

Accelerated Motion

3.1

Uniformly accelerated motion

Suppose that a particle passes the origin of the coordinate system with a velocity v 0 at the time t = 0. Suppose that its displacement is s and its velocity v after time t. Then, applying the definitions of average acceleration for constant value, average value, we get the following expressions. 

= v 0 + at

v

s

 v0  v  1 2  t at =  = v 0 t+  2 2  

v2 = v 02  2 a . s v t 1  v 0  a t  1 ,

s(tth sec) = v 0 + (2t – 1) a

2

Illustration 8 :

A particle moves in a straight line with constant acceleration. If it covers 10 m in first second and extra 10 m in next second, find its initial velocity.

Solution:

Let the initial speed be V0 and the acceleration be a 1 s  v 0 t  at 2 2 Putting t = 1 sec. and s = 10 m, we obtain 10 = v0 + 5 a …(1) Again in next second, that means t = (1+1) sec [t is the time measured from the initial position (when t = 0, s = 0)] the displacement of the particles = 10 + 10 = 20 m. Putting t = 2 sec and s = 20 m, we obtain 10 = 2 v0 + 2 a  v+a=5 …(2) Solving (1) and (2), we obtain 10 = v0 + 5 (5-v0) 15  3.75 m / sec  10 =  4 v 0  25  v 0  4

Exercise 8 :

Referring to the previous illustration, (a) What is the acceleration of the particle? (b) What is the velocity of the particle at the end of 3rd second? (c) What is the displacement of the particle in 3 seconds? (d) What is the displacement of the particle in 3rd second?

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3.2

Non-uniformly accelerated motion in a straight line

If the acceleration a is a function of time (t) or function of distance (s) the equation of instantaneous velocity and acceleration can be written as a = f(t) t t ds dv v  s   vdt and a   v   a dt  v 0 dt dt 0 0 

a

d  ds  d2 s    2 dt  dt  dt

a = f(s) dv dv ds  . Again a  , considering the magnitudes we obtain a scalar equation. dt ds dt ds   dv  v .  av  dt ds   v

Therefore,



S

vdv 

v0

2



v  2

v 02

 ads 0

S

 ads



0

s





v 2  v 02  ads . 0

Illustration 9:

The acceleration of a particle moving rectilinearly varies with the magnitude of its velocity as a = -  v. Find its initial speed, if it stops after t0 = 1 sec from starting.

Solution:

a=- v dv  v dt v





v0

dv

dv v

  dt

  dt

v

 2( v   t2





v0 )   t

v0 

v



At t = t0, it stops; Putting v = 0, we obtain t 0  2 v 0  v0  Exercise 9 :

4.

t 02 ; putting t0= 1, we obtain v0 = 0.5 m/sec. 2

Referring to the previous illustration, (a) Find the velocity-time equation. (b) Find the displacement-time equation.

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Consider a body P projected vertically upwards from the surface of the earth with an initial velocity, u (see figure). The body rises to a maximum height at B and then returns (motion BC).

B P +ve g

Even through we have shown the motion of the body as ABC, the actual path is always along the same line AB. The picture has been slightly modified for clarity.

u A C

Let the displacement of the body (at P) at time t, measured from its initial position A, be denoted by h. We can now apply the equations before. The acceleration, a= -g (note the positive direction in the figure, any vector in the opposite direction is negative); x = h and the initial velocity is u. After time t, v = u - gt 1 1 2 h = ut + (-g)t2 = ut gt 2 2 v2 = u2 + 2 (-g)h = u2 - 2gh

. . . . (1) . . . . (2) . . . . (3)

Other interesting questions that may be posed are: (a) What's the maximum height to which it rises? (b) What’s the time of flight? Let us note that at the point of maximum elevation, B, vB = 0 (it's got to be zero, if it were not the body would have risen further). vB = u - gtAB = 0 ; tAB represents the time taken for motion AB. or, tAB = u/g . . . . (4) If the maximum elevation is H (at the pt. B of course) 2 2 v B = 0 = u - 2gH u2 2g

or, H =

. . . . (5)

When the body reaches the ground again (at the pt C), we can write, 1 h = 0 = utAC g t 2AC ; Where tAC represents the time taken for motion AC 2 1 1 = (u gtAC) tAC  u gtAC = 0 2 2 or tAC =

2u = 2tAB g

. . . . (6)

The velocity at the point C, VC = u - gtAC = u - g 

2u g

= -u, which is equal in magnitude to u (the velocity of projection) but opposite in direction. In both the examples considered above, the acceleration is constant (or uniform). This may not always be true. Particle projected from the top of a cliff

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For the motion of a particle projected vertically up or down from some height (h) we have assign the directions of displacement vector, velocity vector and acceleration vector with reference to a system of coordinates. Let us consider the following cases : (i) When the particle is projected with an initial velocity v0 in the upward direction and its vertical distance from the point of projection O at a time t be h, then 1 h = v0t - gt 2 (if it is above the point O) 2 1 -h = v0t - gt 2 (if it is below the point O) 2 We obtain the value of time by putting the given values of h and v0. If the origin is shifted to the ground level, which is at a depth H below the point of projection, then h and –h mentioned above are replaced by (H + h) and (H – h) respectively. Similar procedure is applicable for the following equations v = u  gt v2 = u2  2gh Now using the following equations 1 2 gt  v 0 t  h 2 

2v 0 4v 02 8h v0 v 02 2h   2 t= g =   g g g g g2 2

If the particle is released from rest then v0 = 0. Illustration 10:

Solution:

A stone is dropped from a balloon ascending with v 0 = 2 m/sec, from a height h = 4.8 m. Find the time of flight of the stone (g = 10 m/sec).

h= 

1 2 gt  v 0 t 2 t2 

2v 0 h g 0 g g

v0 t=0



t

v0 g



 1 

 2gh  1 2 v 

S

Putting the values of v0 and h etc. we obtain, 

t 

2 10

 

 1

 2(10 ) ( 4.8)  1  1.2 sec . 2x 2 

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v0

h t=t

Part-I-KM-PH-14

Exercise 10 :

Referring to the previous illustration, if the stone reaches the ground after t = 2 second from same initial height of release, find the (a) speed of the balloon at the time of releasing the stone, (b) total distance covered by the stone till it reaches the ground level, (c) the average speed and (d) average velocity of the stone for the total time of its flight, one from the bottom and the other from the top.

(i) Particle from the Top Projected Down and Particle at the Bottom Projected Up 

Refer Figure . We see that s1 is downward  s1 = v1t  s2 = v 2 t 

1 2 gt 2

…(1) s1

1 2 gt 2

…(2) v

(1) + (2) gives

s2

 s1 + s2 = (v1 + v2)t  t =

h v1  v 2

…(3)

v1 + v2 = Relative velocity and h = initial relative distance of separation. Now putting t=h/ (v2 – v1) in (a) or in (b) find the position of meeting in terms of distance s 1 from the top or from the bottom (ground) in terms of distance s2. Illustration 11 :

Solution:

A stone is released form the top of a cliff. Another particle is simultaneously projected with v = 10 m/sec. from the bottom of the cliff. If they meet after 2 seconds, find the height of the cliff.

 Let after a time t, they meet si is downward  sI 

1 2 gt 2

…(1)

 1 Since, s2 is upward,  s2  v 2 t  gt 2 2

Putting v 2  v, we obtain s2  vt 

1 2 gt 2

…(2)

(1) + (2)  s1+s2 = v t  h = v t. Putting v = 20 m/sec and t = 2 sec. We obtain h = (20) (2) = 40 m. Exercise 11 :

A body is released from a height and falls freely towards the earth. Exactly 1 sec. later another body is released. What is the distance between the bodies 2 sec. after the released of the second body if g = 9.8 m/s2 ?

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Part-I-KM-PH-15

(ii) Particle from the Top Projected Up and Particle at the Bottom Projected Up Refer Figure. 



Case 1: Let the particles meet above the top of the cliff. Now s1 & s2 are upwards. Therefore,

s1  v1t 

and s2 = v 2 t 

1 2 gt 2

1 2 gt 2

…(1) …(2)

 s2 - s1 = (v2 - v1)t  h = (v2 - v1)t t=

h

.

v 2  v1

Now put the value of t in any one equation (a) or (b) to find the position of meeting. t =t v1

v1

s1

t=0 s1

h v2

h

S22

t=t v2

Ss22

t=0

t=0

Case II: Let the particles meet below the point of projection as shown in Figure Hence   the s1 is downwards and s 2 is upwards.  s1 =

1 2 gt  v1t 2

…(1)

1 2 gt 2

…(2)

and s2 = v 2 t 

 (a) + (b): s1 + s2 = (v2 - v1)t, Setting s1 + s2 = h. We obtain t = h/(v2 – v1)  The expression for time remains equal for any point of meeting. From the previous analysis we know that t = h/(v 2 – v1) for any point of meeting. Therefore we should not bother about the position of point of meeting at the beginning of analysis of a problem. We may take it above or below the top of the cliff. Then find t and substitute in 1 the equation s2 = v2t – gt2 . If it is greater than h, meeting point is above the top of the 2 cliff and vice versa. Illustration 12:

Two particles are simultaneously projected upwards from the top and bottom of a cliff of height h = 20 m. If the speed of one is double that of the other and they meet after a time t = 2 second, find their speed of projection.

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Part-I-KM-PH-16

h v 2  v1 h h  Putting v2 = 2v1, we obtain t  2v 1  v 1 v1 h 20  10 m / sec .  v1   t 2 Therefore, v1 = 2 v1 = 20 m/sec.

Solution:

From the obtained general formula, t 

Exercise 12:

Referring to the previous illustration, what is the position of meeting of the particles?

Non-Simultaneous Projection of Particles Take any case, as shown given in the Figure. Let a particle be projected up from the top of the cliff and another particle be projected up from the bottom of the cliff t 0 second later. Let they meet after a time t from the instant of projection of the first particle. Therefore times of flights of the particles are t and (t-t0) respectively. 1 2 gt  v1t  s1 = …(a) 2 1 and s2 = v 2 ( t  t 0 )  g( t  t 0 )2 …(b) 2 (a) + (b)  s1 + s2 = h

v1 t=0

h

s1 t=t t =(t - t0) S s22 v2 t = t0

 1 2  gt 0  v 2 t 0   2 

 h = ( v 2  v1 )  gt 0  t  

(h  v 2 t 0  ½ gt 2 )  t= ( v 2  v1 )  gt 0

…(3)

Now obtain the position of meeting point by substituting t in (a) or (b). Remember that, since equation (a) is simpler, substitute t in equation (a) for quicker result. Similarly you can use this concept for other cases as discussed earlier. Illustration 13:

A body is dropped from a height h = 50 m. Another body is projected with vertically up a speed V = 10 m/sec after a time t 0 = 2 sec from the instant of release of the first body. Find the time of their meeting.

Solution:

Let the time of fall of the first body be t, before meeting at time = t. The displacement of the first body v =0 1 2 s s1`  gt …(a) 2 t=t h t = (t-t ) The 2nd body is projected from the ground s after t0 = 2 sec from the instant of v =v projection of the first body. Therefore the t =2 time of its motion before its meeting =(t - t 0) = (t-2) sec.  The displacement of the 2nd body is given as 1

1

0

2

2

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Part-I-KM-PH-17

s 2  v ( t  2) 

1 g( t  2)2 2

…(b)

(a) + (b)  sl  s2  v( t  2) 



1 g t 2  ( t  2)2 2



g ( 2 ) ( 2t  2) 2 Putting h = 50 m, g = 10 m/sec2 and v = 10 m/sec., we obtain

 h = v ( t  2) 

50 = 10 ( t  2) 

20 ( t  1) 2

 5 = t2 + 2t2 = 3t4  3t = 9  t = 3 sec. Exercise 13 :

5.

Referring to the previous illustration, (a) Where do the particles meet? (b) What are the speeds of the particle sat the time of meeting?

Relative Velocity

Suppose that two particles A & B are at the two points position vectors of the particles with respect to an   rA & rB respectively. Z The relative separation between the    r  r r particles is given by BA B A Differentiating both sides w.r.t time, we obtain    d rBA d rB d rA  dt  dt   dt  v BA  v B  v A 

where rBA is the position of B with  respect to A & v BA is the velocity of B relative to A.

as shown in the figure. The inertial reference frame be

A  rA  rA

B  rB

0

Y

Z

Physical Significance of Relative Velocity Two persons A & B are in the two vehicles moving in the same direction as shown in the figure. Assume, vA = 10m/sec & vB = 4m/sec VB The person A notices the person B moving towards VA him with a speed of (10 - 4) m/s = 6 m/sec. This is the  velocity of B with respect to ( or relative to) A. v BA is directed from B to A. Similarly A seems to move towards B with a speed of 6 m/sec. Therefore the velocity of A  relative to B  v AB  has a magnitude of 6 m/sec & is directed from A to B as shown in the figure. Bhatnagar IIT-JEE/PMT Academy. C-5, Rama Park,Uttamnagar, New Delhi

Part-I-KM-PH-18

   Therefore v AB = - v BA .    In general, v BA = v B - v A

 v BA  v AB

   v BA  v AB

v AB  v  v  2v A v B cos   ( v B sin )  &   tan 1    ( v A  v B cos )  2 A

  vB  v A

2 B

 vB



 vA

   vA

Illustration 14 :

A train moves due east with a velocity v1 = 20m/sec. and a car moves due north with a velocity v2 = 15 m/sec. Find the velocity of the car as observed by a passenger sitting in the train.

Solution :

The passenger observes the velocity of the car w.r.t. himself. That is, the velocity of the car relative  to thetrain is v ct  v c - v t     v 21  v 2 - v 1  Magnitude of v 21

N   . vc  v2 W

  v 2 - v1

v21 =

S N

 v 21  v12  v 22 

 . v 21 

( 20 )2  (15)2

v21 = 25 m/sec  Direction of v 21 : 20 4   tan1  tan -1 west of north 15 3 Relative Motion between Rain and Man We know that vr  vrg = velocity of rain w.r.t. ground, vm  vmg velocity of ground  man w.r.t.   v rm  v r  v m     v r  v rm  v m That means the vector addition of the velocity of rain with  v respect to man ( rm ) and the velocity of man (vehicle)   v m yields the actual velocity of rain v r . The magnitude  and direction of v r can be given as

E

  . v t  v1

W

 . v2 E

 .  v1 S

Vm  v rm

 Direction of rain fall as seen by the moving man

Vm

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Part-I-KM-PH-19

2 v r  v rm  v m2  2v rm v m cos   with horizontal ( v m ) v rm sin   1     tan   v rm cos   v m 

 vm

Vm Direction of rain fall as seen by the moving man (Actual direction of rainfall)

Vm= 0

Solution :

 vr

 v rm



Illustration 15:





   v rm  v m  v r

 vr

A man holds his umbrella vertically up while walking due west with a constant velocity of magnitude vm = 1.5 m/sec in rain. To protect himself from rain, he has to rotate his umbrella through an angle  = 30° when he stops walking. Find the velocity of the rain.

     v r  v rm  v m .Since v rm  v m , we obtain, sin  

vm vr

 vr = vm cosec   vr = (1.5 m/sec) cosec 30° = 3m/sec . Exercise 14 :

A boy is running on a horizontal road with a velocity of 5 m/s. At what angle should he hold his umbrella in order to protect himself from the rain, if it is raining with a velocity of 10 m/s vertically downward ?

Relative Motion of a Swimmer in Flowing Water  v m can be found by the velocity addition of      v mw & v w . Since v mw  v m  v w .     v m  v mw  v w

 v mw  vm   v mw  v w  vw

Illustration 16:

Solution :

A man swims at an angle  = 120° to the direction of water flow with a speed vmw = 5km/hr relative to water. If the speed of water v w = 3km/hr, find the speed of the man.

   v mw  v m  v w    v m  v mw  v w

 v mw



vm  vw

 vm

  2  v mw  v w  v mw  v 2w  2v mw . v w cos 

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Part-I-KM-PH-20

 v m  5 2  3 2  2(5) (3) cos 120  v m  25  9  15  19m / sec . Exercise 15 :

To a man walking at 7 km/ hr due west the wind appears to blow from north west, but when he walks at 3 km/hr due west the wind appears to blow from the north. What is the actual direction of the wind & what is its velocity ?

Crossing of the river with Minimum drift CASE -1 : v mw  v w A man intends to reach the opposite bank at the point directly opposite to the stationary point. He has to swim at angle  with a given speed vmw w.r.t.  water, such that his actual velocity v m will direct along AB, that is  perpendicular to the bank (or velocity of water v w ).    For minimum drift, v m  v w

B

 v mw

d   (v mw  v w )

 A

 vw

You can realise the situation by a simple example. If you want to reach the directly opposite point or cross the river perpendicularly, a man, that is to say, Hari, must report you that, you are moving perpendicular to  the shore. What does this report signify? Since  v Hari observes your actual velocity ( m ) to be perpendicular to the bank v m is  perpendicular to v w . Observing the vector-triangle vw = vmw sin  & vm = vmw cos 

 v



1 w     sin  v  mw 

2 & v m  v mw  v 2w

 The time of crossing = t =  t

d vm

d v

2 mw

 v 2w

Illustration 17:

The speed of a man in a pond is double that of the water in a river. The man starts swimming from a point P on the bank. What is the angle at which the man should swim so as to get directly to the opposite bank.

Solution :

Let the speed of the man w.r.t. water be vmw When the man swims in a river (moving water), the velocity of the man is directly  across  the river, i.e. v  v  m w ; if the swims at an angle  w.r.t. water,

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Part-I-KM-PH-21

sin  

vw v mw

1    sin

vw v mw

River  v mw

  vm

…(1)

 vw

Since the speed of water is half of that of the man relative to water (speed of the man in still water) vw 1  vmw = 2vw  …(2) v mw 2 1  30 2  The angle of swimming =  = 90 +  = 90° + 30° = 120° to the direction of flow of water.

Using (a) & (b) we find  = sin-1

Exercise 16:

Referring to the previous illustration, find the velocity of the swimming man, if speed of water is 3km/hr.

CASE 2 : vmw < vw

Let the man head at an angle  with normal to the bank for minimum drift. Suppose the drift is equal to zero. For zero drift, the velocity of the man along the bank is zero.  vm = vw – vmw sin  = 0 This gives, vw sin   , sin ce v w  v mw v mw

t=t

B y d

 v mw



t=0 A

C

 vm  vw

x

sin > 1 which is impossible. Therefore the drift can not be zero. Now let the man head at an angle  with normal to the bank to experience minimum drift. Suppose that the drifting of the man during time t when he reaches the opposite bank is BC = x. x = (vm)x ( t) …(1) Where t =

AB d  …(2) ( v m ) y cos  v m cos 

& (vm)x = component of velocity of man along the water flow  (vm)x = vw – vmw sin …(3) Using (1), (2) & (3) we obtain

x  ( v w  v mw sin )

 v  d   w sec - tan   d v mw cos   v mw 

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Part-I-KM-PH-22

 vw  x   sec  - tan   d  v mw 

…(4)

For x to be minimum,

 vw  dx   sec  tan   sec 2   d  0 d  v mw  vw v tan   (sec )  sin   mw v mw vw  v mw  vw

y d



  sin 1 

 . Substituting the value of  



2  v 2w  v mw

in (4) we obtain, x    

t=t

B

 v mw



C

 vm

t=0 A

 vw

x

 d 

v mw 

Illustration 18:

The speed of water is double that of (swimmer) relative to water. What should be the direction of the swimmer so as to experience minimum displacement assuming vm = 5m/sec ?

Solution:

Putting the value of

  sin 1

v mw vw

  sin 1 Exercise 17 :

v mw 1  in the derived expression vw 2 we obtain

1  30 2

Referring to previous illustration, What is the time of crossing of the man assuming width of the river is 500 m?

Crossing of the River in the Minimum Possible Time Case 1: To reach the opposite bank for a given vmw Let the man swim at an angle  with AB. We know that the component of the velocity of man along shore is not responsible for its crossing the river. Only the component of velocity of man (vm) along AB is responsible for its crossing along AB. The time of crossing = t 

AB v mw cos 

Time is minimum when cos  is maximum The maximum value of cos  is 1 for  = 0. That means the man should swim perpendicular to the shore    v mw  v w  Then

t min 

d v mw cos 

 0

d v mw



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Part-I-KM-PH-23

t min 

d v mw

Illustration 19:

A man crosses the river in shortest time at an angle  = 60° to the direction of flow of water. If the speed of water is v w = 4km/hr, find the speed of the man.

Solution :

Referring to the theory, we know that for minimum time of crossing the man should to the shore  head perpendicular  v  v  mw w vw Therefore, cos   vm   v mw vm 4 0 cos 60   vm   vm = 8km/hr  vw

Exercise 18 :

Referring to previous illustration, (a) If the width of the river is ½ km, find the displacement of the man in crossing the river. (b) What is the drift of water on the man ?

CASE 2: To reach directly opposite point on the other bank for a given v mw & velocity v of walking along the shore. To attain the direct opposite point B in the minimum time. Let the man swim at an angle  with the direction AB. The total time of journey t = the time taken from A to C+ the time taken from C to B AB  t = tAC + tCB Where tAC = v mw cos  BC & t CB  where v = walking speed of the man from C to B. v B AB BC C   t v mw cos v Again BC = (vm)xt    AB  BC  (v w - v mw sin)   v mw cos 

Using (a) & (b) we obtain,

t 

 v mw

d  A

 vm  vw

(v w - v mw sin) AB AB  v mw cos v(v mw cos)

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Part-I-KM-PH-24



v w  sec  tan     v  v mw v   v w  sec  d  tan     t    1  v mw   v  v mw v   t  AB   1 

Putting

dt  0 for minimum t we obtain d

v  sec dt d  d  tan    1 w   d d  v mw  v  v mw v   sec  tan  v  sec 2      1 w   0 v mw v  v    v  tan  sec  1  w   v mw  v  v 

v





v



-1 mw mw   .  sin      sin   v  vw   v  vw  This expression is obviously true when vmw < v + vw.

Illustration 20:

A man can walk on the shore at a speed v 1 = 6km/hr & swim in still water is a speed v2 = 5km/hr. If the speed of water is v 3 = 4 km/hr, at what angle should he head in the river in order to reach the right opposite shore in shortest time including his swimming & walking?

Solution :

Directly using the previous result we obtain the angle of swimming v mw   sin -1 v  vw Putting v = v1 = 6km/hr, vmw = v2 = 5km/hr vw = v3 = 4km/hr we obtain  = sin-1 ½ = 30° The man should head to an angle  = 90 +  = 120° with the direction of flow of water.

Exercise 19 :

Referring to previous illustration, What is the time elapsed for swimming, if the width of the river is ½ km?

6.

Projectile Motion

6.1

Oblique projection on a horizontal surface

 Let a particle (body) be projected with certain velocity v 0 at an angle 0 to the horizontal.  The horizontal component of its velocity v 0 = (vo)x = v0 cos0 and the vertical component  of v 0 = (v0)y = v0sin0. The particle moves simultaneously in both horizontal and vertical directions under earths gravitational field (no other external forces like wind drag are small and therefore their effect neglected.

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Part-I-KM-PH-25

Change in Position Vector (Displacement) y

Let the particle acquire a position P having the coordinates (x, y) just after time t from the instant of projection. The corresponding position vector of the particle at time  t is r as shown in the Figure

 v0 P(x,y)  r 0

t=0 O



 r  x ˆi  y ˆj

  v



x

…(1)

Since no external force acts upon the particle horizontally, its horizontal acceleration is zero, that means, the particle moves horizontally with constant velocity of magnitude (v 0)x = v0cos0.  vx = v0cos0. …(2)  The horizontal distance covered during time t is given as x = (v0)xt  x = (v0cos0)t …(3) If we consider the vertical motion of the particle, the external force acting on the particle is gravitational force mg. Consequently the particle accelerates downwards (towards the centre of earth) with an acceleration of magnitude g = 9.8 m/sec 2. In the other words we can say that the particle decelerates upward with g = 9.8 m/sec2. Consequently the vertical velocity of the particle at time t is given as vY = (v0)y – gt, putting (v0)y = v0 sin 0 we obtain vy =v0 sin0-gt …(4) The vertical velocity of the particle at any height y is given as vy2 = (v02) – 2gy, putting (v0)y = v0 sin we obtain 2

v y  v 0 sin 2  0  2gy

…(5)

Now the vertical displacement y is given as y = (v0)y t –1/ 2 gt2 1  y = ( v 0 sin  0 ) t  gt 2 …(6) 2 Putting the values of x and y from equation (3) and equation (6) in equation (1) we obtain the position vector at any time t as  1   r  ( v 0 cos  0 )t ˆi   ( v o sin  0 ) t  gt 2  ˆj 2  



r=

1   ( v 0 t cos  0 ) 2   ( v 0 t sin  0 )  gt 2 )  2  



r=

v 0 t 2 cos 2  0  v 0 t 2 sin 2  0 

2

2

2

1 2 4 g t  v 0 gt 3 sin  0 4

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Part-I-KM-PH-26



 gt r  v 0 t 1    2v 0

2

  



gt sin  0 v0

and

 = tan (y/x) 2 1 ( v 0 t sin  0  1 / 2gt ) = tan ( v 0 t cos  0 )



  tan 1 

-1

 2v 0 sin  0  gt   .  2v 0 cos  0 

Illustration 21:

A boy throws a stone with an speed V0 = 10 m/sec at an angle  0 = 30o to the horizontal. Find the position of the stone w.r.t. the point of projection just after a time t = 1/2 sec.

Solution:

The position of the stone is given by

y u0

t=t P (x, y)

 r  xi  yj

where x=(v0 cos 0)t

r t=0 O

 1 = (10 cos 30 )    2

0

y



x

x

= 4.33 m. and y = ( v 0 sin  0 )t 

1 2 gt 2

1  1  1    10    2 2    2

2

= (10 sin 30 ) 

= 1.25 m   r  ( 4.33i  1.25 j) m . Exercise 20 :

Referring to the previous illustration find the position of the particle as t = v0/g.

Average Velocity   Therefore the average velocity of the particle during time t can be found as v av =  r /t. We have assumed the point of projection as the origin of the coordinate system. That   means, the initial position vector of the particle has a magnitude equal to zero  r = r . 

Putting t = t we obtain, v av v av  v 0

And

 gt  2v 0

1  





2





 r t

gt sin  v0

 2 v sin  0  gt   tan 1 0 2v 0 cos  0



. Substituting the obtained value of | r | we obtain.

.

Instantaneous Velocity  The velocity v of the particle at time t is equal to the vector sum of the velocity components along x and y axis-+

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

v  vx i  vy j  …(7) Putting the  values of vx and vy using equations (2) and (3) in Equation (7), we obtain v  v 0 cos  0 i  ( v 0 sin  0  gt ) j



v=

( v 0 cos  0 ) 2  ( v 0 sin  0  gt ) 2



v=

v 0 cos 2  0  v 0 sin 2  0  g 2 t 2  2 v 0 gt sin 



v=

v 0  g 2 t 2  2 v 0 gt sin  0



v  v0

2

2

2



 gt   1   v    0 

2



 2gt sin  0  v0  

 vy   v sin  0  gt  1   tan 1  0   = tan    v 0 cos     vx    = tan-1 {tan0 – (gt/v0)sec0} Substituting the values of vx and vy by equations (2) and (5) in equation (7) we obtain and

  2 v  ( v 0 cos  0 ) i  ( v 0 sin 2  0  2gy ) j 

v=



v 

and

  tan 1

( v 0 cos  0 ) 2  v 02 sin 2  0  2gy v 02  2gy .



vy vx

 tan 1   

v 20 sin 2  0  2gy   . v 0 cos  0 

Equation of trajectory Substituting ‘t’ by equation (3) in equation (6), we obtain 

 x 1   g  v cos  2 0   0

Y = ( v 0 sin 0 )  

y  (tan 0 )x 



 x   v cos   0   0

gx 2 sec 2 0 2v 02

This is the equation of a parabola. Therefore the path of the particle is parabolic when the particle passes the level of projection. The horizontal distance covered is known as its range R and the time of motion is known as time of flight T. For this range R = Vxt, t = T; X = R, y = 0 (as shown in Figure). Illustration 22:

2



y v0

t = T/2

V0cos

(R/2, H) H 0



t=0 O

(R, 0) t=T

y

0



R

v

A ball is projected so that it follows a curve y = ax – bx 2. Find the initial velocity of projection.

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Part-I-KM-PH-28

Comparing the given equation y = ax – bx 2 with the general trajectory equation 1 gx 2 y = (tan 0) x – (1 + tan2 0), we obtain, 2 v 02

Solution:

g (1 + tan2 0) = b 2v 02 g 0 = tan-1 a and (1 + a2) = b (since tan 0 = a) 2v 02

tan 0 = a and  

v0 =

g(1  a 2 ) and 0 = tan-1 (a). 2b

Time of Flight Substituting these values in equation (6), we obtain. 2 V0 Sin0 1 0 = ( V0 Sin0 ) T  gT 2  T . g 2 Maximum Height When the particle is at the highest position, Vy = 0 and y = H, putting Vy = 0 and y = H in equation (5), we obtain 0= V02 Sin2 0  2gH 

H=

V02 Sin 2  0  H is maximum when 0 = 90o 2g



Hmax = V02/2g

Range When x = R, y =0, putting this value in locus equation or equation of trajectory we obtain, 0 = (tan0) R -

g R2 2

2V0 Cos 0 2



tan 0 =

gR 2

2V0 Sin0 Cos0

2

 

V 2 Sin20 2V0 Sin0 Cos0  R  0 g g R is maximum when Sin2o is maximum  0 = 45o.

R=

Illustration 23:

A body is projected up such that its position vector varies with time as  r = 6t iˆ + (8t-5t2) ˆj . Find the (a) initial velocity (b) time of flight (c) range of the body.

Solution:

(a) The position of the body at any time t is given as  r = 6t ˆi + (8t-5t2) ˆj . When t = 0, r = 0. That means the body is projected from the origin of the coordinate system. Differentiating both sides w.r.t. time ‘t’, we obtain  dr = 6 ˆi + (8-10t) j dt   v = 6 ˆi + (8-10t) j. Putting t = 0, we obtain the initial velocity

(velocity of projection) given as   v |t=0 = v 0 = 6 ˆi + 8 ˆj

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Part-I-KM-PH-29

 v0 = 10 m/sec; (b) The time of flight T = T=

2( v y )0

2v 0 Sin 0 g

where (vy)0 = 8

g 28 T= = 1.6 sec. 10 Exercise 21:

Referring to the previous illustration, (a) Find the maximum height attained by the body. (b) Find the equation of trajectory of the body.

Angle of Projection for Given Ratio of Range and Maximum Height When the range R is times greater than the maximum height H. Then R = H 2v 02 sin 0 cos 0 v 2 sin 2   0  g 2g   tan 0  4 /   cot0 = 4 4H  76o (when H = R or  = 1).  0 = tan-1 R Therefore, the angle of projection will be 76° for H = R. Illustration 24:

A cricketer of height 2.5 m throws a ball at an angle 30° with the horizontal such that it is received by another cricketer of same height standing at a distance of 50 m from the first one. Find the maximum height attained by the ball.

Solution:

Let the ball rise to an additional height H. The distance between the cricketer is equal to R = 50 m. The angle of projection, 0 = 30°, we know that 4H 4H tan 0 =  tan 30° = R 50  H=

50 4 3

m

u0 H

0 2.5 m

h

 7.2 m.

Therefore the maximum height attained by the ball =H =2.5m + 7.2m.  H´ = 9.7 m. Exercise 22:

Referring to the previous illustration, find the time of flight of the ball.

Projectile Passing through Two Different Points at Time t1 and t2

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Part-I-KM-PH-30

Y

If the particle passes two points situated at equal height y at t = t1 and t = t2, then referring to the Figure we obtain 1 y = (v0 sin0) t1  gt12 2 1 = (v0sin0) t2 gt22 …(1) 2

V

h t1

0

t2 y

y

X

O

1 g (t22-t12) 2 g( t1  t 2 ) 2v 0 sin 0 ; Since  T v0sin0 = 2 g



H

v0 sin0 (t2 – t1) =



 t1+ t2 = T

g( t1  t 2 ) in 2

Substituting v0sin0 =

gt12 we obtain 2

1 y  gt1t 2 2 Speed and Angle of Projection so that projectile Passes through Two Given Points

y = (v0 sin) t1 

Y

Referring the Figure, we see that the particle passes through two points P and Q having coordinates (x1, y1) and (x2, y2) respectively. Setting these values of the coordinates in the equation of locus of the particle (trajectory equation) we obtain, y1 = (tan0) x1 -

1 2

gx12 v 02 cos2

v0

P(x1, y1)

Q(x2, y2)

0 = ?

X

O

0 …(1)

and y2 = (tan0 ) x2 -

gx 22

…(2)

2v 02 cos2 0

 y1 x22 - y2 x12 = tan0 (x1 x22 – x2 x12)

 y1 x 22  y 2 x12    x x2  x x2  . 2 1   1 2

1  0  tan 

The speed of projection Similarly y1x2 – y2x1 =

2v 02

g cos2 0

x

2 2

x1  x12 x 2



g x1 x 2 ( x 2  x1 ) (1  tan2 0 )  v0 = 2 ( x 2 y1  x1 y 2 ) 2

 y1 x 22  y 2 x12   , we can obtain V0. Putting the value of tan0 =  2 2   x1 x 2  x 2 x1 

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Part-I-KM-PH-31

Minimum Velocity of Projection required to Pass through a given point When the particle is projected with velocity v at an angle 0 so as to pass through a point P(x,y) as shown in figure, the trajectory equation is written as gx 2 y = (tan0) x (1+tan2 0) 2 2v gx2 (1+tan20) – 2v2x tan0 + 2v2y = 0  tan20 -

P(x,y) 

y

 x

x

2v 2 ( 2v 2 y  gx 2 ) tan 0  0 gx gx 2

For real value of 0, the discriminant  of this quadratic equation is greater than or equal to zero. 2

 2v 2  ( 2v 2 y  gx 2 )  4 0   =   gx  gx 2  v4 ( gx 2  2v 2 y ) 0  2 2  g x gx 2  v4 = g2 x2 – 2v2y g  0 Putting v2 = K we obtain

( 2gy ) 

k2 –(2gy) k - g2x2  0



k

k  g (y +



v2  g (y +

 v

x2  y2 )

g( y 

 v min 

4g2 y 2  4g2 x 2 2 x2  y2 )

x2  y2 )

g( y 

x2  y2 )

.

To find the corresponding angle of projection, known as critical angle of projection (0). We have to differentiate the trajectory equation w.r.t.  to obtain. dx d gx 2  d (1  tan2 ) d  2 x (tan 0 )  ( 1  tan  )  (x2 ) dy /d = tan0   0 d d d 2v 2  d 2v 2  dx  0 for x  Constant we obtain Setting dy/d = 0 for constant y and d v2 gx 2 2 0 = x sec20 ( 2 tan  sec  )  tan  = 0 0 gx 2v 2 Putting 0 = (0) critical for v = vmin = g( y  x 2  y 2 ) we obtain, tan(0)c = 

g( y 

x2  y2 ) gx

 y 

( 0 )c  tan 1  

x 2  y 2   x 

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Part-I-KM-PH-32

Position, Time and Speed at Any Angular Elevation  Y

Let a particle be projected with v 0 at an angle 0. After a time t it moves with V at an angle  with horizontal as shown in the Figure. Since the horizontal components of velocity of a projectile remains constant (vx)0 = V0 cos0 = (Vx) = Vcos  v

v 0 cos 0 cos 

vy v0

v vx

P(x, y)

Q(x2, y2)

0 = ?

X

O

As we know that v2 = v02 – 2gy, substituting the obtained value of V in this equation we obtain  2  v cos   2  V2  cos2 0  0   2g y =  v0   0  y  0  1  2g   cos    cos 2    We know that Vy = (Vy)0 - gt  V sin = V0sin0 – gt, substituting V =

 V0 cos  0   cos   t=

v0 g

V0 cos 0 we obtain, cos 



 sin   v 0 sin  0  gt 



 sin  0 

cos  0 sin   cos  

 t

v 0 sin(  0  ) g cos 

 The horizontal distance x covered during the time t is given as x = (V0 cos0) t  x = ( v 0 cos  0 )  x

v 0 sin(  0  ) g cos 

v 02 sin(  0 ) cos 0 g cos 

Referring to the adjoining figure  When the velocity vector V becomes  perpendicular to the initial velocity vector V0   V. V0  0 where V0 = V0 cos0 ˆi +V0 sin0

Y 900

v0

ˆj

and V = (V0cos0) ˆi + (gt – V0 sin0) (- ˆj )

0 = ?

O

P

vx vy

v

 = /2+0

 {(V0cos0) ˆi + (V0sin0-gt) ˆj }. {(V0cos0 ˆi + V0sin0 ˆj } = 0 (V0cos0)2 + (V0 sin0 – gt) sin0 = 0  V0 (sin20 + cos20) = gt sin0

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X

Part-I-KM-PH-33

 t

V0 . gSin0

Illustration 25 :

A particle is projected at an angle  0 = 60° with horizontal. If after

3

seconds the angle between its velocity vector is reduced to half of its initial value, find the speed of projection (g = 10 m/sec 2).

Solution:

Directly from the derived expression, t =

v 0 sin ( 0  ) ; since  is g cos 

reduced to half of 0, putting  = 0/2, we obtain t =

v 0 sin ( 0 / 2) g cos (  0 / 2)

 v0 = gt tan (0/2). Putting 0 = 60° and t = 3, we obtain v0 = 10 m/sec. Horizontal Projection from a given height

6.2

Referring to the Figure, let a particle be projected with a horizontal velocity V 0, which remains constant along horizontal line due to the absence of any horizontal force. Due to earth’s gravitation the particle acquires vertical velocity V y at any time t and at any position P(x,y).  Let the position vector of this point be r . Vy = (Vy)0 + gt. Since there is no vertical component of V0 initially, (Vy)0 = 0.  Vy = gt and the vertical displacement is y = (Vy)0t +

1 2 1 2 gt = gt 2 2

Again, Vy2 = (Vy)02 + 2gy Putting

(Vy)0 = 0 Vy =

2gy

.

Displacement Now the horizontal displacement x = V0t and the vertical displacement y = 1/2 gt 2. Since  the position vector r  x ˆi  yˆj , putting the values of x and y, we obtain,  1 r  V0 t ˆi  ( gt 2 )ˆj . 2

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Part-I-KM-PH-34

Therefore,  gt 2     2 

 Again r  V0 t ˆi  yˆj 



2

 gt    Vo t 1     2V0 

putting t  2y ˆ i  yˆj g

we obtain, r  V0

 (gt 2 / 2)  1   tan ( V t ) 0  

&  tan1 





gt    2V  0   

2y g

 r

 2y    y 2  g 

2 V0 

    2V02 y   tan 1   1 &   tan 1    V 2y / g  gy    0 

ry



2

( V0 t )2  

r

 gy V0   2 

Velocity ˆ V = Vx ˆi - Vy j   V  V0 ˆi  gt ˆj Therefore

V 

V02

2

 ( gt )  V0

 V  V0 ˆi 

Again Therefore

V 

ˆ  V = V0 ˆi – Vy j (V  gt ) . y

 gt   1     V0 

2gy ˆj

2

 gt &   tan 1   V0 ( Vy 

   

2gy ) .

V02  2gy  2gy     V0 

&   tan 1 

Range If y = H (height of the cliff or height of fall of the projectile), the corresponding horizontal distance (Range R) can be found by putting the values of time of fall t =

2H in the g

equation x = V0t.  x  R  V0

2H g

Putting x = R and t =

2H etc. g

We can find the average velocity and displacement of the particle during the motion when projected horizontally from the top of a cliff of height H with a speed V0. Equation of Trajectory The locus of the path of the particle is given as 1 2 x g x2 y y = gt where t   It is a parabola. 2 V0 2V02 Illustration 26:

A ball is thrown from ground level so as to just clear a wall 4 metres high at a distance of 4 metres and falls at a distance of 14 metres from the wall. Find the magnitude and direction of the velocity of the ball.

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Part-I-KM-PH-35

Solution :

Referring to figure, let P be a point on the trajectory whose u sin  coordinates are (4, 4). As the ball strikes the ground at a distance 14 metre from the wall, the range is 4 + 14 = 18 metre.  The equation of trajectory is 1 x2 y = x tan  g 2 2 u cos 2  gx   or y = x tan   1  2 2 2u cos . tan    

or y = x tan   1  



= x tan   1  

v sin 

u v cos  h u cos 

X



x 

2

2u sin  cos  / g 

x R 

. . . (1)

Here x = 4, y = 4 and R = 18.  4 = 4 tan  [1 – 4/18] = 4 tan  (7/9) or tan = 9/7, sin  =

9 130

and cos  =

7 130

Again R = (2u2 sin  cos /g) =

2 9 7  u2   9. 8 130 130

u2 = 182.  u = Exercise 23 :

(182)

m/s.

A particle is thrown over a triangle from one end of a horizontal base and grazing the vertex falls on the other end of the base. If  and  be the base angle and  the angle of projection, prove that tan  = tan  + tan  .

6.3 Calculation of Radius of curvature at any point on the path of a projectile  Y Let at any time t, the velocity vector V be inlined at an angle  with horizontal at a point, say, P as shown in Figure . Since v  gravitational acceleration g is always v0 P  y=x tan  - gx (1+tan  )/2v acting vertically downwards, the vx  components of g perpendicular to the a = g cos   velocity vector V can be treated as a Effective circle 0 X radial acceleration towards, the centre of a O 2 circular path of radius (let ) as shown in   v , when ar = g cos  Figure;  is known as radius of curvature v  ar : ar  r of the parabola at P. 0

2

r

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2

0

2 0

Part-I-KM-PH-36

V2 V2  is given as where ar =  ar 

V2  = where ar = g cos  ar

V2 g cos 

where v and  can be found in terms of t as v=

v 20  gt 2  2V0 gt sin  0

 v 0 sin   gt    v 0 cos  0 

1 and  = tan 

or v and  can be found in terms of y as v=

V02  2gy





and  = tan 1 v 0 2 sin 2  0  2gy v 0 cos  . Illustration 27:

Solution:

A boy is standing inside a train moving with a constant velocity of magnitude 10 m/sec. He throws a ball vertically up with a speed 5 m/sec relative to the train. Find the radius of curvature of the path of the ball just at the time of projection.

The ball continues to move horizontally with (vx)0 = 10 m/sec. It begins to move up with (vy)0 = 5 m/sec. Therefore 0 is given as, 0 = tan-1 (v y0 )

(v x0 )  0 = tan-1 (5/10) = tan-1 (1/2). Now the required radius of curvature is given as

y

v2 = g cos 

putting v = v0 =

0

(v x )  (v y ) 2 0

2 0

2

= 10  5 = 5 5 m/sec, g = 10 m/sec and  = tan-1 (1/2), we obtain   14 m. 2

Exercise 24:

6.4

2

v0

g

x 0 g cos  0

Referring to the previous illustration, if the boy releases the ball from rest, what will be the radius of curvature of the path at the instant of its release?

Projectile on an inclined plane

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Part-I-KM-PH-37

y Let a particle be projected up with a speed x v0 at an angle  to horizontal onto an v0 P inclined plane of inclination . Hence the component of initial velocity (velocity of projection) parallel and perpendicular to the g g cos  g sin  plane are equal to v0 cos ( – ) and v0 sin  ( – ) respectively. The component of g  O along the plane is gsin and perpendicular to the plane is gcos as shown in the Figure. Therefore, the particle decelerates at a rate of g sin as it moves from O to P. Suppose that the particle hits P after a time T from the instant of projection. During this time the particle moves up from O to P along the incline with a deceleration g sin and moves to and fro perpendicular to the incline. Considering the motion along Y-axis, the displacement y of the particle during time t (= T) perpendicular to the plane is zero. Time Using y = v0 sin (-) t – 1/2 (g cos) t2 and putting y = 0 when t = T we obtain,

T

2v 0 sin (   ) . g cos 

Range Now considering the motion along X-axis: x = v0 cos ( – ) t – 1/2 (gsin) t2 putting X = R for t = T and substituting the obtained value of T we write  2v 0 sin(   )   2v 0 sin(   )  1 R = v 0 cos(  )    ( g sin )   g cos  2 g cos     

2

2v 02 sin(  )  cos(   ) sin  sin(   )     g cos  cos   

 R=

2v 02 sin(   )

 R=

g cos 2 

 cos(   ) cos   sin  sin(  )

2v 02 sin(   ) cos(   )  

 R=

g cos 2  2v 02 sin(   ) cos 

 R=



g cos 2 



v 02 sin(     )  sin(     )

 R=  R

v 02

2

g cos 

 sin(2  )  sin  g cos 2 

.

Maximum Range Range R is maximum when sin(2 – ) is maximum, that is equal to one: Bhatnagar IIT-JEE/PMT Academy. C-5, Rama Park,Uttamnagar, New Delhi

Part-I-KM-PH-38

 Rmax =

v 02 (1  sin )

 R max 

g cos 2 

=

v 02 (1  sin ) . g(1  sin ) (1  sin )

v 02 . g(1  sin )

Similarly when the particle is projected down the plane the corresponding range is given as v 02  R max  . g(1  sin ) Finding the angle  for maximum range when projected up and down the plane, for 1 1 1   = (/4 + /2), (/4 – /2) it can be found that = . Rmax Rmax R Where R = maximum range of the projectile on the horizontal plane for same speed of projection. Condition for retracing the path of a projectile on an inclined plane Referring Figures and we see that when a particle strikes the inclined plane perpendicularly or when it strikes the inclined plane at an angle  (equal to the angle of inclination of the plane) to the normal to the inclined plane. In first case, the particle retraces its path just after the collision with the inclined plane and in the second case, the particle moves vertically up and retraces its path from highest position Q. Consider the Case 1

The motion perpendicular the plane (along Y-axis) y = (v0 Sin)T- ½ g T2 = 0. Putting y = 0

 T

2v 0 sin  g cos 

…(1)

The motion parallel to the plane (x axis) (vp ) x = -(v0)x + {-(gsin)}T Since at P velocity v is P perpendicular to the plane, the component (v p)x of velocity Vp(=V) parallel to the plane is zero.  (vp)x = 0 and (v0)x = v0cos  0 = (v0 cos) – g sin T  T=

v 0 cos  g sin 

…(2)

Using equations (1) and (2), we obtain,

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Part-I-KM-PH-39

2v 0 sin  v cos   0 g cos  g sin 

 2tan = cot cot   cot        tan 1   2  2   cos    .   =  + tan-1  2  

  = tan-1

Following the same procedure we can find the necessary condition for the second case referring Figure. 1. 2. 3. 4.

as  = tan-1 (3 cot), we can find the times of flight of the particle for one complete cycle in both cases, we can find the range and maximum height attained by the particle in both cases, and we can find the velocity of striking of the particle in both cases.

Conditions for same range for a given speed of projection (a) Horizontal Plane When a particle is projected on horizontal plane with relating V 0(say) at any angle  with v 02 sin 2 . horizontal. Then the range of the particle is given as R = g When range is maximum, sin 2 is maximum. Because sin 2 max = 1  2 = /2   = /4 or 45o Referring to Figure We see that for a particular value of range we have two angles of projection. Let it be 1 and 2 with that it will satisfy the relation v 20 v 02 sin 2   sin 2 2 R= 1 g g 

v 20 sin 2 g

 sin21 = sin (-22)

 1+2 = /2 Hence the angles of projection are complementary.  If one is , the other becomes /2- Relation between the parameters of two possible paths for constant range for given speeds of projection Times of flights for path (1), (2) and (3) are given as 2v 0 sin 1 2v sin  2 2v sin 45 o , T2  0 , T 0 T1 = g g g Maximum Heights for Path (1), (2) and (3) are given as 2

v 0 sin 2 1 , H1 = 2g

2

v 0 sin 2  2 H2  , 2g

2

v 0 sin 2 45 o H 2g

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Part-I-KM-PH-40

Ranges for (1), (2) and (3) 2

v sin 21 , R1 = 0 g

2

v sin 2 2 R2  0 , g

2

v sin(2  45 o ) R 0 . g

Maximum possible time and height for any speed of projection are given as 2 2 V0 V0 Tmax = , and Hmax = g 2g Remember that if 1 =  than 2 = (90-). Illustration 28 :

If R be the range of a projectile on horizontal plane and H 1, H2 be the maximum heights for its two possible trajectories, find the relation between the given parameters.

Solution:

Referring to the theory, v 02 sin 2  0 v 02 sin 2 (90   0 ) H1 = ; H2 = 2g 2g  

v 02 cos 2  0 H2 = g

H1H2 =

Range R =

v 02 sin 2  0 . v 02 cos 2  0 4g 2

v 02 sin 2 0 g

…(1) …(2)

From (1) and (2) we obtain 

H1H2 = 1/16 R2

R=4 Exercise 25 :

H1H2 .

Referring to the previous illustration, if the maximum heights for two possible paths are given as H 1 = 3 m and H2 = 9 m, find the corresponding angles of projection.

(b) Inclined Plane The expression for range is given as R = v 02  sin (2  )  sin  / g cos 2  For maximum range sin (2-) = 1  2 -  = /2   = /4 + /2 When projected down the plane R = v02 [sin (2+) – sin]/gcos2 and  = /4 - /2. For R to be same for two values of  (let 1 and 2). Bhatnagar IIT-JEE/PMT Academy. C-5, Rama Park,Uttamnagar, New Delhi

Part-I-KM-PH-41

sin (21 - ) = sin(22- ) or sin [ - (22 - )]  21- =  - (22-  )  1 + 2 =/2-. That means if one angle is  the other will be /2-(+)  1 – (/4 +/2) = (/4 + /2) -2 Since (/4 + /2) is the angle of elevation 0 for greatest range, the direction of projection for a given range is equally inclined to the direction of projection for the maximum range for the same velocity of projection as shown in the Figure. When  = 0, the results will be reduced to that obtained for a horizontal plane. This projection on an inclined plane also can be analysed by using the formula  x 2  y 2 by putting (y/x) = v0 = [g( y  x 2  y 2 )]; H int; put ( y / x )  tan  & R tan and R =

[g( y 

x 2  y 2 )];

H int; put ( y / x )  tan  & R



x2 y2

.

Illustration 29 :

A particle is projected up an inclined plane of inclination  at an elevation  to the horizontal show that (a) tan  = cot  + 2 tan  , if the particle strikes the plane at right angles (b) tan  = 2 tan  , if the particle strikes the plane horizontally.

Solution :

Let t be the time of flight from O to A. Then t=

u sin ( - )

u

A u cos ( - )

2(u sin ) 2u sin(      g g cos 

O

. . . (1)





Now we shall consider the motion of the particle along OA. Here initial velocity = u cos ( - ) final velocity = 0 (The particle strikes the plane A at right angles acceleration due to gravity = - g sin   0 = u cos ( - ) – g sin  t or, t =

u cos(     g sin 

From equation (i) & (ii)

. . . . (2) 2u sin(     u cos(     = g sin  g cos 

or, 2 tan ( - ) = cot   tan   tan  

 = cot  or, 2   1  tan  tan   or, 2 tan  - 2tan  = cot  + tan  or, tan  = cot  + 2 tan 

(b) When the particle strikes the plane horizontally In this case, t =

2u sin(     g cos 

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Part-I-KM-PH-42

and 0 = u sin  - gt  2u sin(      g cos   

or, u sin  = g 

or, sin  cos  = 2 [ sin cos  - cos  sin ] or, 2 cos  sin  = sin  cos   Exercise 26 :

2 tan  = tan .

A ball starts falling with zero initial velocity on a smooth inclined plane forming an angle  with the horizontal. Having fallen the distance h, the ball rebounds elastically off the inclined plane. At what distance from the impact point will the ball rebound for the second time ?

7.

Kinematics of Circular Motion

7.1

Uniform circular motion 

Displacement (r ) When particle moves in a circular path describing an angle  during time t as shown in the Figure from the position 1 to the position 2, we see that the magnitude of the position    vector r (that is equal to the radius of the circle) remains constant  r1  r2  r and the direction of the position vector changes from time to time. The change of position  vector or the displacement S of the particle from position 1 to the position 2 is given by referring the Figure 56. As     r  r2  r1

 r   r 

 r



2

2

  r2  r1

r1  r2  2r1 r2 Cos 

Putting r1 = r2 = r we obtain r 

 r 

r 2  r 2  2r. r Cos 

2r 2 (1  Cos ) 

  2r 2  2 Sin2  2  

 2  *Find the direction of the displacement vector S .



r  S  2r Sin

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Part-I-KM-PH-43







OA  AB  OB







  r  r2  r1

Average Velocity

 V  2 V sin (  / 2) Vav   t t

Ratio of Distance and Displacement The distanced covered by the particle during the time t is given as d = length of the arc AB  d = r . Therefore, the ratio of distance to the magnitude of displacement for any angular distance is given as d r d    Co sec( / 2)  S 2r Sin  / 2 S 2 Angular and Linear Speed If the particle moves with constant linear speed V the angular speed of the particle also remains constant Angular Speed = 

w 

Angular dis tan ce cov ered Time elapsed

 t

Since r = length of the Arc AB and linear speed = V 

length of the arc AB time t

 V = r/t V r   Find the ratio of magnitude of average velocity (  r / t   r / t ) to the  instantaneous velocity V .   Find the ratio of magnitude change in velocity V to instantaneous velocity V .

Eliminating t from the previous equation we obtain w   

Change in Velocity We want to know the magnitude and direction of the change in the velocity of the particle as it moves from A (position 1) to B (position 2) during time t as shown in Figure. The   change in velocity vector is given as V  V2  V1 

 V 

  V2  V1

Therefore V1 = V2 = V

 V 

V1  V2  2 V1 V2 Cos  .

 V 

2 V 2 (1  Cos )

2

2

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Part-I-KM-PH-44

 2   gives the magnitude of V and the direction of V is shown in Figure that can be 180     90   / 2 . given as   2

  V  2V Sin













OA  AB  OB   r  r2  r1

Centripetal Acceleration If  tends to zero,   90o, that means, the direction of change of velocity vector 

becomes perpendicular to the instantaneous tangential velocity vector V . In the other 

words, we can say that, at any instant the change of velocity vector V of the particle executing uniform circular motion is always directed radially. That means the acceleration vector

  V   a   t  

of the particle is radially inwards. Therefore the particle is accelerating

towards the centre. Hence this type of acceleration of the particle in a circular path is known as Centripetal Acceleration. The magnitude of centripetal acceleration ar is given  V 2v sin (  / 2) as ar =  t

t

when  is very small during very small time interval  sin   / 2 ` 2 2v (  / 2) v    ar = t t v v2 Putting  / t  w  we obtain a r   w 2r . r r Illustration 30 :

Find the magnitude of average acceleration of the tip of the second hand during 10 seconds.

Solution:

Average acceleration has the magnitude a = v/t, where v = 2v sin /2

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Part-I-KM-PH-45



a =

2V sin  / 2 t

Putting v = /300 m/sec (obtained earlier), t = 15 seconds and  = 60°, we obtain 2(  / 300 ) sin 30 a = 15  a =  m/sec2. 4500

Exercise 27 :

7.2

An astronaut is rotating in a rotor of radius 4 m. If he can withstand upto acceleration of 10 g, then what is the maximum number of permissible revolutions ? (g = 10 m/s2)

Non-uniform Circular Motion

(a) Constant Angular Acceleration (Uniform Angular Acceleration) If the magnitude of velocity of the revolving particle in a circular path changes uniformly, we say that the particle is accelerating tangentially.  Let the tangential acceleration at any time be a t , the corresponding instantaneous velocities (as   shown in the Figure) at 1 and 2 are v 1 & v 2 respectively and the respective instantaneous angular speeds at 1 and 2 be 1 and 2 respectively. Since  = v/r and v = 1/2 (v1 + v2) the average angular velocity during time t can be given as

1  2 2



and the angular acceleration  = =

2  1  2  1  t t

 . t

Putting 1 = 0, 2 =  and t = t we obtain   0   t Since the angular distance  covered during time t is given as  1  2   t  = t   2   2 2  0     t, putting t    0 we obtain     0 2    2

 

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Part-I-KM-PH-46

 2  0 2  2 Putting  = 0 + t we obtain    (0  t ) t = 0 2 1    0 t  t 2 2 Since  = v/r V   r  ( v / t )  when t  0  t r d 1  dv  d     putting dt r  dt  dt d d and   Since   dt dt d(d / dt ) d2    = dt dt 2 d d d d   Again =  . dt d dt d dv  a t we obtain r  a t . and dt



Exercise 28:

Referring to the previous illustration, find the (a) average angular speed of the disc (b) total number of revolutions ‘N’ of the disc before coming to rest.

(b) Circular Motion in a Vertical Plane (Non-uniform Angular Acceleration) If w and  are not constant in magnitude, that means  = f (t) or  = f() we write the kinematical equations as   w

w

t

 0  dt ,

 d 

0

Exercise 29 :

7.3



0



 d 

t

 0  dt  0 2  0 2 



0

 d .

Referring to the previous illustration, if, initially the stone is hanging from the string at its lowest position and it is pushed horizontally with v = 2 m/sec., find the linear velocity of the stone when the string will be horizontal.

Circular Motion-an alternative treatment

A particle P moves in a circle of radius r (constant). We calculate the direction and magnitude of the velocity as well as the acceleration of the particle.

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Part-I-KM-PH-47

Let the position of particle relative to a cartesian coordinates system with its origin at the centre of the circle (the path of the particle) be given by P  P (x, y) Also, let POX be represented by . Since the particle moves in a circle of fixed radius r,  | OP | = r = constant or, x 2  y 2 = r or, x2 + y2 = r2 . . . (1) which is the equation of the path. Further, x = r cos  y = r sin , where  is the angle (POX) . . . (2)

Y P(x, y)

O

 X

Thus, the only other quantity that can vary is . Thus,  = (t), a function of time. The position vector,  r  x ˆi  yˆj = (rcos ) ˆi  (r sin )ˆj . . . (3)   dr d v   (r cos  ˆi + r sin  ˆj ) The velocity, dt dt d d  cos .ˆj = r (-sin . ˆi. ) dt dt  d   (-sin  ˆi  cos  ˆj ) = r  dt  = r (-sin  ˆi + cos  ˆj ) . . . (4) d Where  = , represents the angular velocity. dt

The velocity is, in magnitude,  v = | v | = r (sin2   cos2 ) = r, d Where = . dt   dv The acceleration, a  dt =

d {r( sin  ˆi  cos  ˆj ) dt

=r

. . . (5)

d d (  sin  ˆi  cos  ˆj )  r(  cos  ˆi  sin  ˆj ) dt dt

d (  sin  ˆi  cos  ˆj )  r2 (cos  ˆi  sin  ˆj ) dt d ˆ - r2 rˆ =r . . . . (6)  dt where we have defined two new vectors

=r

ˆ   sin  ˆi  cos  ˆj  rˆ  cos  ˆi  sin  ˆj

( since  =

d ) dt

. . . . (7)

Both having magnitude unity

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Part-I-KM-PH-48



ˆ is perpendicular to The vector rˆ is in the same direction as OP ( rˆ ) while the vector   ˆ is in the tangential direction and directed towards the rˆ or ( OP ) while the vector  ˆ sense of increasing  . (see figure).  dv   ˆ =r  ˆ   The velocity, v = t   dt 

= r

2

d  ˆ  d   -    dt  dt 2



and the acceleration, a = r

d ˆ  - 2 r rˆ dt

2

rrˆ

. . . (8)

 The velocity v is tangential (as it should be ) while the acceleration has both a normal (centripetal) component  2r and a tangential component r. Exercise 30 :

Find the total acceleration of a particle moving in a circular track of radius 2 m, with a constant angular acceleration of 1 rad/sec 2 at time t = 2 seconds from the starting.

Bhatnagar IIT-JEE/PMT Academy. C-5, Rama Park,Uttamnagar, New Delhi

Part-I-KM-PH-49

8.

Solution to the Exercises

1.

Image of P will be formed at P such that distance of P from plane mirror = distance of P from the plane mirror.      r   x i  y  j = 2. 5 i  2 j

2.

P

x

P

   r1 = ( 2.5 i +2 j ) m    r2 = (2.5 i +2 j +2 k ) m

    Displacement  r  r2  r1 = = –2 k m

 3.

y

    Displacement = | r2  r1 | 0 , (| r2 |  | r1 | 0 ) 



Distance = | AB |  | BA | 20m. 4.

5.

(a) Velocities are mutually opposite in direction; + ve means along +ve X-axis and –ve means along – x axis. (b) Distance covered = d = AB + BC + CD + DE + EF 1 1 1 1 =     2  3m 4 4 4 4 Displacement = 2m 1 vav = T

T

 v( t )dt 0

1 = 2 / 

1  = 2 /   6.

2 / 

v

0

sin tdt

0

/

 0

v 0 sin tdt 

 v 0 sin tdt  ; (sin  = negative for  =  to 2)  / 

2 / 



Let the man catch the bus at a further distance x from the initial position of the bus after a time t.  9 + x = ut . . . . (1) Where u is the uniform speed of the man. For the bus : distance travelled = x 1  x= (2) t2 = t2 . . . (2) 2 From (1) and (2) 9 + t2 = ut  t2 – ut + 9 = 0 As the man was just able to get in, that implies there is only a single meeting between the bus and the man. Bhatnagar IIT-JEE/PMT Academy. C-5, Rama Park,Uttamnagar, New Delhi

Part-I-KM-PH-50

 For real t u2 – 4  9 = 0  u = 6 m/s 7.



The area of V/t graph gives the displacement. Since the car moves in one direction, its total distance covered will be numerically equal to its displacement. v S = 100 m (given) 1 A (OB ) ( AB ) s 2 v0 1 2 1  (BC) ( AB ) 2 S2 1 S1 ( OB  BC ) ( AB ) = 2 B C

1 (OC) ( AB) 2 1 = ( t1  t 2 ) V0 , where V0 = maximum 2 speed of the car

O

t1

t2

(t1 +t2) = T = 20 sec and s = 100 m 1  100  (20 ) V0  V0  10 m / sec . 2 8.

(a) a = 5- v0 = 5 – 3.75 = 1.25 m/sec2 (b) v = v0 + at = 3.75 + (1.25)(3) = 7.5 m/sec. (c) s = v0t + ½ at2 = (3.75)(3) + ½ (1.25) (3)2 m (d) s = v0 +

9.

a 1.25 (2t  1)  (3.75 )  [( 2)(3 )  1] m 2 2

(a) v0 = v + t/2  v = v0 – t/2 (b) v =



v = v0 +

t2  4

v0 t

dx dt t

x=

 vdt

put the obtained expression of v.

0

10.

(a) 4.8 = -v0t + ½ gt2 putting t = 2 sec. we obtain 2v0 = 20 – 4.8 = 15.2 m/sec.



v0 = 7.6 m/sec.

Bhatnagar IIT-JEE/PMT Academy. C-5, Rama Park,Uttamnagar, New Delhi

t

Part-I-KM-PH-51

(b) D = 4.8 + 2  (c) & (d) U  _

v

2 v0 v2 = 4.8 + 0 2g g

where v0 = 7.6 m/sec.

D , put the values of d and t = 2 sec. t

s put s = 4.8 and t = 2 sec. to obtained v  2.4m / sec . t

11.

According to given problem, second body falls for 2s, so that, 1 h2 = g (2)2 2 while 1st has fallen for 2 + 1 = 3 sec. 1 h1 = g (3)2 2  Separation between two bodies 2s. after the release of 2nd body 1 d = h1 – h2 = g(32 – 22) = 4.9  5 = 24.5 m. 2

12.

y = v1t + (1/2)gt2 =  10  2 + (1/2) (10) (22) = 0 at the top of the cliff

13.

(a) s1 = (1/2)gt2 = (1/2) (10) (32) = 45 m from the top (b) v1 = gt1 = 10  3 = 30 m/sec. v2 = 10 – 10  (3 – 2) = 0

14.

vb = 5 m/s

vB



–5 m/s O

C



10 m/s

vr

= 10 m/s A

B

To obtain the relative velocity of rain w.r.t. boy, a velocity triangle is formed between – v B and v r as shown. Let  be the angle made by the resultant with the horizontal. 10 tan  = =2   = tan-1(2)  5 15.

  Case – I v m  7ˆi , v   v x ˆi  v y ˆj    v m  v   v m  ( v x  7)ˆi  v y ˆj

Bhatnagar IIT-JEE/PMT Academy. C-5, Rama Park,Uttamnagar, New Delhi

Part-I-KM-PH-52

Case II :

  v m  3ˆi , v   v x ˆi  v y ˆj

 v m  ( v x  3)ˆi  v y ˆj

Since  the wind appears to blow from north v  m must have no x – component  vx + 3 = 0  vx = = -3 km/hr & vy = 4 km/hr  wind speed = 5km / hr, and it is blowing 530 north of east. 16.

vw 1  v mw 2  vmw = 2vw = 2(3) = 6 km/hr d v mw cos  500 200  2 400  sec . = = 3 3 2.5 1  1 / 4

17.

t=

18.

(a) AC =

AB 2  BC2

where AB = ½ km and BC = put vm = 4 km/hr, vmw =  AB 

(b) BC =  v  v w   mw 

AB .v w v mw

2 vm  v 2w  82  42 

48 km/hr

( AB )( v w ) 2 vm  v 2w

put AB = (1/ 2) km, vw = 4 km/hr, vm = 8 km/hr d v mw cos  1/ 2 1 2 1    hr. t= 0 10 3 5 3 5 cos 30

19.

Time of swimming = t =

20.

x = (v0cos 0)t = (v0 cos 300 ) =

v0 g

3 v 02 3  10  10  5 3 m 2g 2  10

y = (v0 sin  )t = (1/2) gt2 = (v0sin 300) (v0 /g) – (1/2)g (v0/g)2

Bhatnagar IIT-JEE/PMT Academy. C-5, Rama Park,Uttamnagar, New Delhi

Part-I-KM-PH-53

 v 02 v 02   0  r  x ˆi  yˆj 2g 2g

=

 r  5 3 ˆi .

 21.

(a) h =

( v 0 sin 0 )2 ( 8 )2  = 3.2 m 2g 2  10

1 gx 2 (1  tan2 ) 2 v2 Where v0 cos  = 6 & sin 0 = 8 tan 0 = 8/6 = 4/3

(b) y = x tan  -

 y = (4/3)x -

22.

 2v 0 sin 0  v 02 sin2  8   g 2g g  

(T)2 =  T2 =

8H g 8H  2 g

T= 23.

9x2 . 500

2H g

put H = 7.2 m.

The situation is shown in the figure. Y

P(x, y)

u  O

 x

 B (R-x)A )

X

From figure, tan  + tan  =

y y  x (R  x )

where R is the range. y(R  x )  xy  tan  + tan  = (R  x ) or tan  + tan  =

y R  x (R  x )

. . . (1)

we know 

x  R  y R or tan  =  x (R  x )

y = x tan   1 

. .. . (2)

Bhatnagar IIT-JEE/PMT Academy. C-5, Rama Park,Uttamnagar, New Delhi

Part-I-KM-PH-54

from equation (1) and (2), we have tan  = tan  + tan . 24.

The radius of curvature  =

v2 g cos 0

0 = angle between v & horizontal = 0 when the ball is released, it starts moving horizontally with a speed v = speed of the train = 10 m/sec. v2 (10)2  = g cos 0 g cos 0 0 = angle between V & horizontal = 0 when the ball is released, it starts moving horizontally with a speed of the train = 10 m/sec. = 25.

v2  10m . g cos 0

H1/H2 =

v 02 sin2  v 02 sin2 (90  )   tan2  2g 2g

  = tan1 26.

3 1 1  300 .   tan 9  3

On impact at point P, the velocity of the ball = v = 2gh . Since the ball rebounds elastically, it gets deflected with the same velocity at an angle . From the geometry,  = . The distance  of the point P where it makes second impact is then its range with acceleration g sin . The motion along the direction normal to the plane is with velocity v cos  and acceleration –g cos . Hence the time of flight T is given by 2v 1 0 = v cos T g cos T2 or T = g 2 1 Now  = v sin  T + g sin T2 2 = v sin  =

27.

 H1  tan1  H2 

 2v  2v 1   g sin  g 2  g 

v

 

P P 

2

4 v 2 sin  4 sin   . 2 gh = 8 h sin . g g

In case of uniform circular motion  v2 

 = r2 ar =   r 

ar = (2f)2 r

v

[ as v = r] [ as  = 2f]

Bhatnagar IIT-JEE/PMT Academy. C-5, Rama Park,Uttamnagar, New Delhi

Part-I-KM-PH-55

1 ar 2 r Here, ar < 10 g, so f=

f<

5 10  10 i.e. fmax = rev/sec. 2 4

1 2

w 0  w 10  0   5rad / sec . 2 2 2 2 (b)  = 0  2N = 0 2 2 2 2 (10) 25 0  N= N= . ( 4  )( 1 )  4

28.

(a) w 

29.

v = l = 

 

 2g sin   l  l 

= 2gl sin  putting  = /2 rad for lowest portion . we obtain v = 2gl  2  10  1  4.5 m/sec. 30.

The total acceleration is given as a = a2t  ar2 = (r ) 2  (r2  2 ) 2 =

r 2 2  r 24

=

r 2 2  r 2 (  t )4

= r 1   2 t 4 = (2) (1) 2 = 2 17 m/sec .

1  (1)(2)2

Bhatnagar IIT-JEE/PMT Academy. C-5, Rama Park,Uttamnagar, New Delhi