Skip to content Skip to navigation Skip to collection information

OpenStax-CNX

You are here: Home » Content » Physics for K-12 » Inelastic collision (application)

Navigation

Table of Contents

Recently Viewed

This feature requires Javascript to be enabled.
 

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 :

Figure 1: Velocities before and after collision
Inelastic collision
 Inelastic collision  (icq1.gif)

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 ”.

Figure 2: Velocities before and after collision
Inelastic collision
 Inelastic collision  (icq2.gif)

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 ”.

Figure 3: Velocities after collision
Inelastic collision
 Inelastic collision  (icq3.gif)

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,

Figure 4: Direction of impulse
Inelastic collision
 Inelastic collision  (icq4.gif)

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”.

Figure 5: Three identical balls are placed on a smooth horizontal surface.
Successive collisions
 Successive collisions  (icq5.gif)

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.

Figure 6: First collision is inelastic.
First collision
 First collision  (icq6.gif)

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

Adding two resulting equations,

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 :

Figure 7: Second collision is perfectly elastic.
Second collision
 Second collision  (icq7a.gif)

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 ”.

Figure 8: Velocities before and after collision
Inelastic collision
 Inelastic collision  (icq2.gif)

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.

Figure 9: The ram of a pile driver is released from rest at a height "h".
Inelastic collision
 Inelastic collision  (icq8.gif)

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.

Figure 10: Velocities of colliding bodies.
Inelastic collision
 Inelastic collision  (icq9.gif)

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,

Figure 11: The pile is decelerated due to resistance offered by the material.
Motion of pile
 Motion of pile  (icq10.gif)

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

Collection Navigation

Content actions

Download:

Collection as:

PDF | EPUB (?)

What is an EPUB file?

EPUB is an electronic book format that can be read on a variety of mobile devices.

Downloading to a reading device

For detailed instructions on how to download this content's EPUB to your specific device, click the "(?)" link.

| More downloads ...

Module as:

PDF | More downloads ...

Add:

Collection to:

My Favorites (?)

'My Favorites' is a special kind of lens which you can use to bookmark modules and collections. 'My Favorites' can only be seen by you, and collections saved in 'My Favorites' can remember the last module you were on. You need an account to use 'My Favorites'.

| A lens I own (?)

Definition of a lens

Lenses

A lens is a custom view of the content in the repository. You can think of it as a fancy kind of list that will let you see content through the eyes of organizations and people you trust.

What is in a lens?

Lens makers point to materials (modules and collections), creating a guide that includes their own comments and descriptive tags about the content.

Who can create a lens?

Any individual member, a community, or a respected organization.

What are tags? tag icon

Tags are descriptors added by lens makers to help label content, attaching a vocabulary that is meaningful in the context of the lens.

| External bookmarks

Module to:

My Favorites (?)

'My Favorites' is a special kind of lens which you can use to bookmark modules and collections. 'My Favorites' can only be seen by you, and collections saved in 'My Favorites' can remember the last module you were on. You need an account to use 'My Favorites'.

| A lens I own (?)

Definition of a lens

Lenses

A lens is a custom view of the content in the repository. You can think of it as a fancy kind of list that will let you see content through the eyes of organizations and people you trust.

What is in a lens?

Lens makers point to materials (modules and collections), creating a guide that includes their own comments and descriptive tags about the content.

Who can create a lens?

Any individual member, a community, or a respected organization.

What are tags? tag icon

Tags are descriptors added by lens makers to help label content, attaching a vocabulary that is meaningful in the context of the lens.

| External bookmarks