# Collisions and Impulse in Physics

These are my notes on collision and impulse in physics.

Momentum provides a new analysis technique. With force analysis and work-energy

analysis, application of the appropriate analysis technique was reasonably

straightforward. Knowing when to use work-energy analysis and when to use

momentum analysis adds a whole new dimension to problem solving.

There are certain types of collisions where conservation of momentum and

conservation of energy can be applied and certain problems where where only

conservation of momentum can be applied. This is nowhere better illustrated than

in the ballistic pendulum problem where conservation of momentum is applied in

one part of the problem and conservation of energy in another part of the

problem. Learning where to apply these two conservation laws is difficult. In

the upcoming problems, be aware of which law is being applied and why.

Before doing specific problems, take a general look at collision in one

dimension. A and B will designate two particles, and 1 and 2 will designate

before and after the collision.

There are two types of collisions. In elastic collisions, both momentum and

energy are conserved. Examples of elastic collisions are billiard balls or any

collision where the participants bounce. In inelastic collisions, only momentum

is conserved. Example of inelastic collisions are railroad cars coupling or a

steel ball thrown into a piece of clay or any collision where the participants

stick together and do not bounce.

It is this sticking or non-sticking that determine if energy is conserved or not.

When we say that energy is not conserved in a collision where the participants

stick together, we mean only tat mechanical energy \(1/2mv^2\) is not conserved.**Inelastic Collisions**

Consider first inelastic collisions, where the particles stick together. Using

the notation:

\[v_{A2}=v_{B2}=v_{2}\]

This equation states that after the collision, the velocity of the A particle is

the same as the velocity of the b particle. Applying the law of conservation of

momentum and this condition:

\[m_{a}v_{A1}+m_{B}v_{B1}=(m_{A}+m_{B})v_{2}\]

The left side of the equation is the momentum before collision, and the right

side is the momentum of the two masses stuck together at the same velocity after

the collision.

Take the special case where the B particle is initially stationary. While

mechanical energy is not conserved in the collision, the kinetic energies can be

written as:

\[K_{1}=(1/2)m_{A}v_{A1}^2 \text{ and } K_{2}=(1/2)(m_{A}+m_{B})v_{2}^2\]

\(K_{2}\) can be rewritten using the conservation of momentum statement.

\[K_{2}=1/2(m_{A}+m_{B})\frac{m_{A}v_{A1}}{m_{A}+m_{B}}^2=1/2

\frac{m_{A}^2}{m_{A}+m_{B}} v_{A1}^2\]

Comparing these two equations gives a relationship for the energy before and

after collision.

\[K_{2}=\frac{m_{A}}{m_{A}+m_{B}} K_{1}\]**Example 1**

A ballistic pendulum, a device for measuring the speed of a bullet, consists of

block of wood suspended by cords. When the bullet is fired into the block, the

block is free to rise. How high does a 5.0 kg block rise when a 12 g bullet

traveling at 350 m/s is fired into it?

Solution:

This is a most interesting and instructive problem. The collision between the

bullet and the block is clearly inelastic(the bullet comes to rest in the

block). Part of the kinetic energy of the bullet goes into friction as the

bullet burrows its way into the block. Therefore, mechanical energy is not

conserved.

Because the collision is inelastic, apply conservation of momentum to the

collision. Before the collision, all the momentum is in the mv of the bullet.

After the collision, the momentum is in the (m+M)V of the block and bullet. We

assume that the bullet comes to rest(transfers all its momentum) before there

is appreciable motion of the bullet-block combination.

\[mv=(m+M)V\]

After the collision, the rise of the block is determined by energy analysis. The

kinetic energy of the block goes into potential energy.

\[(1/2)(m+M)V^2=(m+M)gh\]

Substituting for V from mv =(m+M)V

\[(1/2)(\frac{m}{m+M})^2 v^2=gh\]

\[v=\frac{m+M}{m} \sqrt{2gh}\]

This gives the relation between the velocity of the bullet and the height the

block and bullet rise.

\[h=\frac{(350m/s)^2}{2*9.8m/s^2}*(\frac{0.012}{5.012})^2=3.6cm\]

In this problem, the 0.012 can be neglected in comparison with 5.0. This is not

always the case, so we write m+M as 5.012 as a reminder to include both m+M in

the calculation.**Example 2**

A 6.0 g bullet is fired horizontally into a 2.8 kg block resting on a horizontal

surface with a coefficient of friction of 0.30. The bullet comes to rest in the

block, and the block slides 0.65 m before coming to a stop. What is the velocity

of the bullet?

Solution:

Assume that the bullet comes to rest in the block before the block moves

appreciably and that all the momentum in the bullet is transferred to the

bullet-block combination.

\[mv=(m+M)V\]

Once the bullet-block combination is moving at V, the kinetic energy

\((1/2)(m+M)V^2\) goes into work to overcome friction, \(\mu (m+M)gx\).

\[(1/2)(m+M)V^2-\mu (m+M)gx\]

Substituting:

\[1/2(\frac{mv}{m+M})^2=\mu

gx=\frac{2.806}{0.006}*\sqrt{2(0.30)(9.8m/s^2)0.65m}=914m/s\]**Elastic Collisions**

Now look at elastic collisions, where the particles bounce. In elastic

collisions no energy is lost to permanent deformation of the particles. Write a

conservation of momentum statement for the collision.

\[m_{A}v_{A1}+m_{B}v_{B1}=m_{A}v_{A2}+m_{B}v_{B2}\]

Now write a conservation of energy statement.

\[1/2m_{A}v^2_{A1}+1/2m_{B}v^2_{B1}=1/2m_{A}v^2_{A2}+1/2m_{B}v^@_{B2}\]

These two statements can be rewritten as:

\(m_{A}(v_{A1}-v_{A2})=m_{B}(v_{B2}-v_{B1})\) and:

\(m_{A}(v^2_{A1}-v^2_{A2})=m_{B}(v^2_{B2}-v^2_{B1})\)

Now divide the rewritten conservation of energy statement by the conservation of

momentum statement to find:

\[v_{A1}+v_{A2}=v_{B2}+v_{B1}\]

The term \(v_{A1}-v_{B1}\) is the speed of the approach, the speed of A relative

to B measured by an observer on B. The term \(v_{A2}-v_{B2}\) is the speed of

departure. So, in an elastic collision, the speed of approach is equal to the

speed of departure.

Now find expressions for the velocities after the collision in terms of the

masses and the velocities before the collision. From the velocity statement:

\[v_{B2}=v_{A1}+v_{A2}-v_{B1}\]

From the conservation of momentum statement, substitute the equations for

\(v_{B2}\):

\[m_{A}v_{A1}+m_{B}v_{B1}=m_{A}v_{A2}+m_{B}(v_{A1}+v_{A2}-v_{B1})\]

With a little algebra:

\[v_{A2}=\frac{m_{A}-m_{B}}{m_{A}+m_{B}}*v_{A1}+\frac{2m_{B}}{m_{A}+m_{B}}*v_{B1}\]

With these two statements, \(v_{A2}\) and then \(v_{B2}\) can be predicted from

the initial masses and velocities.

Look at some special cases: If \(m_{A}=m_{B}\) the masses are equal and then:

\(v_{A2}=v_{B1} \text{ and } v_{B2}=v_{A1}\)

The particles exchange velocities. This is what happens in billiards.

If \(v_{B1}=0\) the struck mass is at rest and then:

\[v_{A2}=\frac{m_{A}-m_{B}}{m_{A}+m_{B}}*v_{A1}\]

and:

\[v_{B2}=v_{A1}+\frac{m_{A}-m_{B}}{m_{A}+m_{B}}*v_{A1}=\frac{2m}{m_{A}+m_{B}}*v_{A1}\]

For some realistic cases where the masses are equal or one particle is at rest,

the resulting expressions for the velocities are easily calculated.**Example 3**

A 1.5 kg block traveling at 6.0 m/s strikes a 2.5 kg block at rest. After an

elastic collision, what are the velocities of the blocks?

Solution:

This is the special case where the struck block is at rest, so:

\[v_{A2}=\frac{m_{A}-m_{B}}{m_{A}+m_{B}}=\frac{1.5-2.5}{4.0}*6.0m/s=-1.5m/s\]

\[v_{B2}=\frac{2m_{A}}{m_{A}+m_{B}}*v_{A1}=\frac{2*1.5}{4.0}*6.0m/s=4.5m/s\]

The striking block rebounds at -1.5 m/s and the stuck block moves off at 4.5

m/s.**Example 4**

A 1.0 kg steel ball is attached to a lightweight 1.0 m long rod pivoted at the

other end. The ball is released at the horizontal and strikes a 3.0 kg steel

block resting on a surface with a coefficient of friction of 0.25. How far does

the block travel?

Solution:

This problem is similar to the ballistic pendulum problem in that conservation

of energy and conservation of momentum and then work energy have to be applied

correctly. From the description of the collision, assume that it is elastic.

First, calculate the velocity of the ball as it hits the block. Potential energy

goes into kinetic energy.

\[mgh=(1/2)mv^2\]

\[v=\sqrt{2gh}=\sqrt{2(9.8m/s^2)1.0m}=4.4m/s\]

The collision is elastic and \(v_{B1}=0\), so:

\[v_{B2}=\frac{2m_{A}}{m_{A}+m_{B}}*v_{A1}=\frac{2m_{A}}{m_{A}+m_{B}}*\sqrt{2gh}=1/2\sqrt{2gh}=2.2m/s\]

This gives the initial velocity of the struck block. The kinetic energy goes

into work against friction.

\[1/2m_{B}v^2_{B2}=\mu m_{B}gL\]

\[L=\frac{v^2_{B2}}{2\mu g}=\frac{1}{2\mu

g}*\frac{2gh}{4}=\frac{h}{4\mu}=\frac{1.0m}{4*0.25}=1.0m\]

Go back over this problem and see where energy analysis is applied, conservation

of momentum is applied, and work-energy analysis is applied. Knowing what laws

to apply where in the problems is the hard part of collision problems.**Impulse**

Impulse is the name given to a force that acts for a very short period of time.

A struck baseball or golf ball are examples of impulses or impulse forces. In

most impulses, it is impossible to graph forces versus time, though we can often

estimate how the force varies with time.

Force is defined in terms of change in momentum as

\[F=\frac{\Delta (mv)}{\Delta t}\]

If the force is time dependent, F(t),then:

\[\Delta (mv)=F(t)\Delta t\]

With calculus notation this statement would be written as:

\[\int dp=\int F(t) dt\]

The left side is the change in momentum, and the right side is the area under

the F(t) versus t curve. This integral is called the impulse or impulse

integral. If the curve of F(t) versus t were a parabola, (reasonable for an

impulse), then the right side of the equation would be the area under the curve.**Example 5**

A 1.2 kg croquet ball moving at 2.0 m/s is struck from behind by an impulse

force, what is the final velocity of the ball?

Solution:

The initial momentum is \(p_{i}=1.2 kg(2.0)m/s=2.4 kg*m/s\)

The impulse integral, the difference in momentum, can be calculated by

inspection. The area under the curve(1/2 base times height) is:

\[\Delta (mv)=\int F(t) dt=(1/2)(0.20s)100N=10kg*m/s\]

Th final momentum then is \[

p_{i}=12.4kg*m/s\]

Final velocity is:

\[v_{f}=\frac{12.4kg*m/s}{1.2kg}=10.3m/s\]