Particle 1, mass 1 kg, is moving to the right at 1 m/s. Particle 2, mass 3/2 kg, is moving to the left at 3/2 m/s. The particles collide. After the collision, particle 1 is moving to the left at 1 m/s. What is the velocity of particle 2? Was the collision elastic or inelastic?

__Before:__

__After:__

With *x* increasing to the right, the total momentum of the system
is:

Before | After | |

P_{x} |
(1 kg)(1 m/s) - (3/2 kg)(3/2 m/s) | (1 kg)(-1m/s)+(3/2 kg)v_{x} |

Setting the value before equal to the value after, we get:

-5/4 kg· m/s = -1 kg·
m/s + (3/2 kg)*v _{x}*

And thus:

*v _{x}* = -(1/4)(2/3) m/s = -1/6 m/s

Since the particles have different velocities, we can immediately conclude that the collision is NOT totally inelastic. Now let's calculate the kinetic energy:

Before:

*K*_{before}* = *1/2(1 kg)(1 m/s)^{2} + 1/2(3/2
kg)(3/2 m/s)^{2} = (1/2 + 9/16) J = 17/16 J

After:

*K*_{after}* = *1/2(1 kg)(1 m/s)^{2} + 1/2(3/2
kg)(1/6 m/s)^{2} = (1/2 + 1/48) J = 25/48 J

The total kinetic energy is less after the collision, so the collision is inelastic, but not totally inelastic.

We can see this easily if we look in the CM frame:

__Before:__

__After:__

In the CM frame, the speed of each particle has decreased. Thus
the total kinetic energy of the system has decreased.