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Course by: Sunil Kumar Singh. E-mail the author

# Inelastic collision (application)

Module by: Sunil Kumar Singh. E-mail the author

Summary: Solving problems is an essential part of the understanding process.

Questions and their answers are presented here in the module text format as if it were an extension of the treatment of the topic. The idea is to provide a verbose explanation, detailing the application of theory. Solution presented is, therefore, treated as the part of the understanding process – not merely a Q/A session. The emphasis is to enforce ideas and concepts, which can not be completely absorbed unless they are put to real time situation.

## Representative problems and their solutions

We discuss problems, which highlight certain aspects of the study leading to inelastic collision. The questions are categorized in terms of the characterizing features of the subject matter :

• Coefficient of restitution
• Successive collisions
• Kinetic energy and coefficient of restitution
• Motion subsequent to collision

## Coefficient of restitution

Problem 1 : A block of mass 1 kg, moving at a speed “v”, collides with another block of mass 10 kg at rest. The lighter block comes to rest after the collision. Find the coefficient of restitution.

Solution : Let us consider that lighter block is moving along x-axis. The initial velocities are :

v 1 i = v v 1 i = v

v 2 i = 0 v 2 i = 0

The final velocities are :

v 1 f = 0 v 1 f = 0

v 2 f = v v 2 f = v

From conservation of linear momentum,

1 X v + 10 X 0 = 1 X 0 + 10 X v 1 X v + 10 X 0 = 1 X 0 + 10 X v

v = v 10 v = v 10

The coefficient of restitution is :

e = velocity of separation velocity of appraoch e = velocity of separation velocity of appraoch

e = v 2 f v 1 f v 1 i v 2 i = v 0 v = v 10 v = 1 10 = 0.1 e = v 2 f v 1 f v 1 i v 2 i = v 0 v = v 10 v = 1 10 = 0.1

Problem 2 : A spherical ball hits “head – on” another identical ball, which is stationary. The velocity of second ball, immediately after the collision, is twice that of first ball. Find coefficient of restitution.

Solution : We solve this problem using equation of conservation of linear momentum and equation of restitution. Let velocity of first ball before collision be “v”, velocity of first ball after collision be “ v 1 v 1 ” and velocity of second ball after collision be “ v 2 v 2 ”.

Applying equation of conservation of linear momentum, we have :

m 1 v 1 i + m 2 v 2 i = m 1 v 1 f + m 2 v 2 f m 1 v 1 i + m 2 v 2 i = m 1 v 1 f + m 2 v 2 f

Putting values, we have :

m v + m X 0 = m v 1 + m v 2 m v + m X 0 = m v 1 + m v 2

v 1 + v 2 = v v 1 + v 2 = v

It is given that :

v 2 = 2 v 1 v 2 = 2 v 1

Substituting in the equation,

v 1 + 2 v 1 = 3 v 1 = v v 1 + 2 v 1 = 3 v 1 = v

v 1 = v 3 v 1 = v 3

Putting this value of “ v 1 v 1 ” in the equation of conservation of linear momentum, we get “ v 2 v 2 ” as :

v 2 = v v 1 = v v 3 = 2 v 3 v 2 = v v 1 = v v 3 = 2 v 3

Now, applying equation of restitution, we have :

e = v 2 f v 1 f v 1 i v 2 i = v 2 v 1 v 0 e = v 2 f v 1 f v 1 i v 2 i = v 2 v 1 v 0

e = 2 v 3 v 3 v e = 2 v 3 v 3 v

e = 1 3 e = 1 3

Problem 3 : In a “head – on” inelastic collision between two identical particles, the linear momentum of the projectile is “P”, whereas the target particle is stationary. If the target applies an impulse of magnitude “J” on the projectile, find coefficient of restitution.

Solution : In order to solve this problem, we need to evaluate the defining expression of coefficient of restitution. This means that we need to know final velocities.

We note here that we need to consider motion and collision in one dimension as the collision is “head – on”.

Let the mass of each particle is “m”. Let velocity of first ball before collision be “v”, velocity of first ball after collision be “ v 1 v 1 ” and velocity of second ball after collision be “ v 2 v 2 ”.

According to question,

v = P m v = P m

Now, we know that impulse is equal to change in linear momentum :

J = Δ P = P 1 f P 1 i = m v 1 f v 1 i = m v 1 v J = Δ P = P 1 f P 1 i = m v 1 f v 1 i = m v 1 v

However, v 1 > v v 1 > v . It means that impulse acts in opposite direction to that of reference direction of motion. Hence,

J = m v 1 v = m v 1 + m v J = m v 1 v = m v 1 + m v

v 1 = - J m + v v 1 = - J m + v

Putting value of “v”,

v 1 = - J m + P m v 1 = - J m + P m

Now, applying conservation of linear momentum,

m 1 v 1 i + m 2 v 2 i = m 1 v 1 f + m 2 v 2 f m 1 v 1 i + m 2 v 2 i = m 1 v 1 f + m 2 v 2 f

m v + m X 0 = m v 1 + m v 2 m v + m X 0 = m v 1 + m v 2

v = v 1 + v 2 v = v 1 + v 2

Substituting for “ v 1 v 1 ”, we have :

v = - J m + v + v 2 v = - J m + v + v 2

v 2 = J m v 2 = J m

The coefficient of restitution is :

e = v 2 v 1 v = J m + J m P m P / m = 2 J P 1 e = v 2 v 1 v = J m + J m P m P / m = 2 J P 1

## Successive collisions

Problem 4 : Three identical balls designated “1”, “2” and “3” are placed 16 meters apart on a smooth horizontal surface. The ball “1” is given an initial velocity 10 m/s towards “2”. The collision between “1” and “2” is inelastic with coefficient of restitution 0.6, whereas collision between “2” and “3” is perfectly inelastic. Find the time elapsed between two consecutive collisions between “1” and “2”.

Solution : To answer this question, we need to understand the collision sequence. The ball “1” moving with velocity 10 m/s strikes second ball in - elastically. As a consequence, depending on coefficient of restitution two balls move with different velocities. This collision constitutes the first collision.

Let the mass of each particle is “m”. Let velocity of first ball before collision be “v”, velocity of first ball after collision be “ v 1 v 1 ” and velocity of second ball after collision be “ v 2 v 2 ”. Applying conservation of linear momentum,

m v = m v 1 + m v 2 m v = m v 1 + m v 2

v = v 1 + v 2 v = v 1 + v 2

v 1 + v 2 = 10 v 1 + v 2 = 10

Applying equation of restitution,

v 2 v 1 = e v = 0.6 X 10 = 6 v 2 v 1 = e v = 0.6 X 10 = 6

2 v 2 = 16 2 v 2 = 16

v 2 = 8 m / s v 2 = 8 m / s

Substituting this value in the equation of conservation of linear momentum,

v 1 = 10 v 2 = 10 8 = 2 m / s v 1 = 10 v 2 = 10 8 = 2 m / s

Clearly, second ball covers the horizontal distance of 16 m, subsequent to collision at 8 m/s and strikes the third ball.

The second ball strikes third stationary identical ball elastically. As a result, the balls exchange their velocities. It means second ball becomes stationary and third ball moves off towards right with velocity at 8 m/s. First ball, however, keeps moving towards right at 2 m/s and covers the distance of 16 m to strike the second ball again, which has come to a standstill. The time taken by the first ball to hit the second ball again is :

t = 16 2 = 8 s t = 16 2 = 8 s

This is the time interval between required two consecutive collisions.

## Kinetic energy and coefficient of restitution

Problem 5 : A spherical ball of mass “m” moving at a speed “v” makes a head on collision with an identical ball at rest. The kinetic energy of the balls after collision is 3/4 th of the value before collision. Find coefficient of restitution.

Solution : The spherical balls have equal mass and initial velocity of “target” ball is zero.

Let the mass of each particle is “m”. Let velocity of first ball before collision be “v”, velocity of first ball after collision be “ v 1 v 1 ” and velocity of second ball after collision be “ v 2 v 2 ”.

For identical mass and stationary target, we have :

v 1 = 1 e 2 v v 1 = 1 e 2 v

and

v 2 = 1 + e 2 v v 2 = 1 + e 2 v

### Note:

Instead of remembering these relations, we can alternatively solve simultaneous equations of conservation of linear momentum and restitution to obtain these relations.

According to question,

K f = 3 4 K i K f = 3 4 K i

1 2 m v 1 2 + v 2 2 = 3 4 X 1 2 m v 2 1 2 m v 1 2 + v 2 2 = 3 4 X 1 2 m v 2

v 1 2 + v 2 2 = 3 4 m v 2 v 1 2 + v 2 2 = 3 4 m v 2

Substituting for final velocities, we have :

1 - e 2 4 v 2 + 1 + e 2 4 v 2 = 3 4 m v 2 1 - e 2 4 v 2 + 1 + e 2 4 v 2 = 3 4 m v 2

1 e 2 + 1 + e 2 = 3 1 e 2 + 1 + e 2 = 3

1 + e 2 2 e + 1 + e 2 + 2 e = 3 1 + e 2 2 e + 1 + e 2 + 2 e = 3

e 2 = 1 2 e 2 = 1 2

e = 1 2 e = 1 2

## Motion subsequent to collision

Problem 6 : The ram of a pile driver has a mass m and is released from rest at a height h above the top of the pile which has the mass 3m. To what height does the ram rebound if the coefficient of restitution between the ram and the pile is e and e>(1/3)? If the pile meets a constant resistance F from the material in which it is lodged, calculate how far into the material it penetrates before coming to rest.

### Note:

This question is as submitted by Baline Patrick through e-mail.

Solution : We need to know the velocities of ram and pile immediately after the inelastic collision. Let the velocity of the ram just before strike be "v", velocity of the ram after the strike be " v 1 v 1 " velocity of the pile after strike be " v 2 v 2 ".

Let the downward vertical y-direction be positive. We can determine velocity of the ram just before it hits the pile. Applying equation of motion for constant acceleration,

v 2 = 0 + 2 g h = 2 g h v 2 = 0 + 2 g h = 2 g h

In order to determine velocities immediately after collision, we shall apply conservation of linear momentum in vertical direction (downward direction as positive)

m 1 v 1 i + m 2 v 2 i = m 1 v 1 f + m 2 v 2 f m 1 v 1 i + m 2 v 2 i = m 1 v 1 f + m 2 v 2 f

Putting values, we have :

m v + 0 x 3 m = m v 1 + 3 m v 2 m v + 0 x 3 m = m v 1 + 3 m v 2

v = v 1 + 3 v 2 v = v 1 + 3 v 2

Now, applying equation of restitution, we have :

e = v 2 f v 1 f v 1 i v 2 i = v 2 v 1 v 0 e = v 2 f v 1 f v 1 i v 2 i = v 2 v 1 v 0

e v = v 2 v 1 e v = v 2 v 1

Thus, we have two unknowns “ v 1 v 1 ” and “ v 2 v 2 ”. Also we have two equations. Adding two equations,

1 + e v = 4 v 2 1 + e v = 4 v 2

The velocity of pile, therefore, is :

v 2 = 1 + e v 4 v 2 = 1 + e v 4

Putting this value in the first equation, we get the velocity of the ram after collision,

v 1 = v 3 v 2 = 1 3 e v 4 v 1 = v 3 v 2 = 1 3 e v 4

Putting expression of “v”, as obtained earlier, in above two expressions, we have :

v 2 = 1 + e v 4 v 2 = 1 + e v 4

v 2 = 1 + e 2 g h 4 v 2 = 1 + e 2 g h 4

and

v 1 = 1 3 e 2 g h 4 v 1 = 1 3 e 2 g h 4

Note that 1 - 3e < 0 as e > 1/3. It means that the velocity of ram is opposite to the assumed positive reference direction. This further means that it rebounds after the strike in upward direction.

v 1 = 3 e 1 2 g h 4 v 1 = 3 e 1 2 g h 4

Let h' be the height attained by the ram after the strike. For the upward motion of ram,

0 = v 1 2 2 g h 0 = v 1 2 2 g h

h ' = v 1 2 2 g = 3 e - 1 2 X g h 2 g X 16 = 3 e - 1 2 X h 16 h ' = v 1 2 2 g = 3 e - 1 2 X g h 2 g X 16 = 3 e - 1 2 X h 16

Now, we consider the motion of pile into the material. There are two forces acting on the pile (i) constant resistance “F” and (ii) force due to gravity “3mg". As given, the net force acts up to decelerate the pile. The deceleration is :

a = F 3 m g 3 m a = F 3 m g 3 m

Let "y" be the vertical distance moved by the pile. Applying equation of motion,

0 = v 2 2 2 X F 3 m g 3 m X y 0 = v 2 2 2 X F 3 m g 3 m X y

y = v 2 2 X 3 m 2 F 3 m g y = v 2 2 X 3 m 2 F 3 m g

y = 1 + e 2 X 6 m g h 2 F 3 m g X 16 = 1 + e 2 X 3 m g h 16 F 3 m g y = 1 + e 2 X 6 m g h 2 F 3 m g X 16 = 1 + e 2 X 3 m g h 16 F 3 m g

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