# 3.3.2 Elasticity of Collisions

Given that 2 masses m1 and m2 travelling at speeds u1 and u2 collide head-on, what are their speeds v1 and v2 after the collision?

Obviously, the outcome must obey PCOM. So \begin{aligned}\sum {{p}_{i}}&=\sum {{p}_{f}}\\{{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}&={{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}\end{aligned}

But with one equation but two unknowns v1 and v2, aren’t there going to be an infinite number of possible solutions? We know that is impossible in practice. Well, it turns out that besides PCOM, the outcome of a collision is also governed by the elasticity of the collision. So what is elasticity?

Unlike total momentum (which must be conserved), some kinetic energy can be “lost” (as heat, deformation energy, sound, etc) during a collision. So the total KE after the collision is usually lower than the total KE before the collision. The less KE lost, the more elastic the collision. Conversely, the more KE lost, the more inelastic the collision. So

• A (perfectly) elastic collision is one which retains 100% of the initial total KE. Note that we usually omit the word “perfectly”. So if someone says “elastic collision”, by default he means “perfectly elastic collision”.
• A perfectly inelastic collision is one which retains the least amount (not necessarily zero) of initial total KE. It can be shown mathematically that this implies that ${{v}_{1}}={{v}_{2}}$. In other words, m1 and m2 will travel at the same speed in the same direction (aka “stuck together”) after a perfectly inelastic collision.
• Anything in between is called an inelastic collision.

What determines the elasticity of a collision? Well, it is determined by the type of material. Elastic collisions tend to occur with hard and stiff objects e.g. billiard balls. Inelastic collisions tend to occur with soft and malleable objects e.g. plasticine balls.

For illustration purpose, let’s look at the “classic” head-on collision between two equal masses m with equal initial speed u. Keep in mind that the total momentum before the collision is $mu+m(-u)=0$.

Depicted above are just 5 of the infinitely many possible outcomes for such a collision, from the completely elastic “bouncy balls” collision on top, to the perfectly inelastic “sticky balls” collision at the bottom. Notice that momentum-wise, all of them have the total momentum of zero after the collision, as dictated by PCOM. But KE-wise, depending on how elastic the collision is, the amount of total KE remaining after the collision range from 100% (for the elastic “billiard balls” collision), down to 0% (for the perfectly inelastic “plasticine balls” collision).

For another example, let’s look at the classic “sitting duck” collision: a mass m, with initial speed of u, colliding into another mass m initially at rest. Make a mental note that total initial momentum of the system is mu.

Depicted above are just 5 of the infinitely many PCOM-obeying outcomes for such a collision, from the completely elastic “bouncy balls” collision on top, to the perfectly inelastic “sticky balls” collision at the bottom.

Notice that momentum-wise, all of them have the total momentum of mu after the collision, as dictated by PCOM. But KE-wise, the (perfectly) elastic “billiard balls” retain 100% of the initial KE, whereas the perfectly inelastic “sticky balls” collision retains 50%. Why is it not zero? Because of PCOM! Losing all the KE also means losing all the momentum, which is an impossible outcome for this collision since we started with initial total momentum of mu.

Video Explanation

Elastic vs Inelastic Collisions