# Linear Momentum and Collisions

James D. Jones,
M. Casco Associates, 81 Turkey Ridge Road, East Stroudsburg, PA 18301, USA

Consider two spherical objects (balls) one of mass m1 and the other of mass m2. Let's set things up so these balls are approaching each other along the line joining their centers, a recipe for a head on collision. Let these balls not be rotating or vibrating. The motion is purely translation. Under these conditions we may choose a reference frame which is one dimensional, with the objects on the x-axis.

Applying the conservation of linear momentum to this situation we have,

m1v1i + m2v2i = m1v1 + m2v2

where vi is the initial velocity of each object and v is the final velocity of each object. The conservation of kinetic energy gives us the equation

m1v1i2 / 2 + m2v2i2 / 2 = m1v12 / 2 + m2v22 / 2

These may be rewritten as follows:

m1 (v1i - v1) = m2 (v2 - v2i)

for the momentum equation and

m1 (v1i2 - v12) = m2 (v22 - v2i2)

for the energy.

As long as the difference between final and initial velocities is not zero for either object (meaning a collision actually happens), we may divide the second equation by the first one which yields

v1i + v1 = v2 + v2i

or

v1i - v2i = v2 - v1

In other words in a one dimensional elastic collision, the relative velocity of approach before the collision equals the relative velocity of separation after collision.

To get the final velocities in terms of the initial velocities and the masses, you would solve the last equation above for v2 and plug that into the momentum equation and solve to get

v1 = v1i (m1 - m2) / (m1 + m2) + v2i (2 m2) / (m1 + m2)

Likewise

v2 = v1i (2 m1) / (m1 + m2) + v2i (m2 - m1) / (m2 + m1)

So we have the basis for a model of a one dimensional elastic collision. For initial conditions v1i and v2i, if a collision happens, the final velocities depend on the masses as above.

Now if the two masses, m1 and m2 are equal we can replace both with simply m, like this.

v1 = v1i (m - m) / (m + m) + v2i (2 m) / (m + m)

Likewise

v2 = v1i (2 m) / (m + m) + v2i (m - m) / (m + m)

This reduces to the following expressions for final velocities v1 and v2.

v1 = v2i and v2 = v1i

In the case of equal mass collisions, the two objects just exchange velocities. Of course if the initial velocity of one of the balls were zero, then the one colliding with it would just stop and the hit ball would take off at the original speed of the incoming ball.

Now consider the action of the series of balls illustrated here. It looks like there must be a series of two ball collisions of the sort described above so that the last ball in the line leaves the group with approximately the same velocity as the ball which struck the group originally. This should tell you something about the nature of the contact between adjacent balls in the line. If the contact between the balls were rigid, the incoming ball would see a single massive object comprised of several balls. It would then rebound from the collision in accordance with the rules developed above for collision between different masses. Our reasoning leads us to conclude that even though the balls in the line appear to be in contact, there must be initially a very soft reaction between adjacent balls.

Authors with theoretical explanations, interesting photos
and videos, GIF animations, JAVA applets are welcomed.