# A ball with a mass of 2 kg is rolling at 2 m/s and elastically collides with a resting ball with a mass of 4 kg. What are the post-collision velocities of the balls?

Dec 15, 2016

#### Answer:

The 2-kg ball recoils backwards at $\frac{2}{3}$ m/s, while the 4-kg ball rolls in the same direction the 2-kg ball had been moving, at $1 \frac{2}{3}$ m/s

#### Explanation:

This is a lengthy problem to work out!!

The two missing final velocities mean we have two unknowns to determine. This will require two independent equations that include these velocities.

One comes from the law of conservation of momentum, the other from conservation of kinetic energy (not a law, but true in this case because the collision is elastic).

Prior to collision, the total momentum is p = mv = 4 kg m/s

and the total KE is $\frac{1}{2} m {v}^{2}$ = 4 J

For convenience, I will use u for the final velocity of the 2-kg ball, and v for that of the 4-kg ball.

Cons. of momentum tells us:

initial momentum = final momentum

4 = $\left(2\right) u + \left(4\right) v$

which I will write as

2 = $u + 2 v$

Cons of KE tells us:

initial KE = final KE

4 = $\frac{1}{2} \left(2\right) {u}^{2}$ + $\frac{1}{2} \left(4\right) {v}^{2}$

which we write as

4 = ${u}^{2}$ + $2 {v}^{2}$

We now must solve these two equations in two unknowns

Rewrite the momentum equation as $u$ = 2 - $2 v$ and substitute this in place of u in the KE equation:

4 = ${\left(2 - 2 v\right)}^{2}$ + $2 {v}^{2}$

4 = (4 - $8 v$ + $4 {v}^{2}$) + $2 {v}^{2}$

4 = 4 - $8 v$ + $6 {v}^{2}$

0 = $6 {v}^{2}$ - $8 v$

0 = $6 v$ - 8

v = $\frac{8}{6}$ m/s or $1 \frac{1}{3}$ m/s

So, u = $2$ - $2 v$ = - $\frac{2}{3}$ m/s