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