# A ball with a mass of  5 kg is rolling at 18 m/s and elastically collides with a resting ball with a mass of 2 kg. What are the post-collision velocities of the balls?

Jan 6, 2017

#### Answer:

After collision, the 5-kg ball moves in the same direction at 7.71m/s while the 2-kg ball also moves in this direction at 25.7 m/s.

#### Explanation:

This is a lengthy problem that asks us to solve for two unknowns (the two final velocities). To do this, we must generate two equations involving these two unknowns, and solve them simultaneously.

One equation will come from conservation of momentum, the other will come from kinetic energy conservation.

First, cons. of momentum:

${m}_{1} {v}_{1 i} + {m}_{2} {v}_{2 i} = {m}_{1} {v}_{1 f} + {m}_{2} {v}_{2 f}$

Inserting the values we know:

$5 \left(18\right) + 2 \left(0\right) = 5 {v}_{1 f} + 2 {v}_{2 f}$

$90 = 5 {v}_{1 f} + 2 {v}_{2 f}$

Now, kinetic energy conservation.

$\frac{1}{2} {m}_{1} {v}_{1 i}^{2} + \frac{1}{2} {m}_{2} {v}_{2 i}^{2} = \frac{1}{2} {m}_{1} {v}_{1 f}^{2} + \frac{1}{2} {m}_{2} {v}_{2 f}^{2}$

$\frac{1}{2} \left(5\right) {\left(18\right)}^{2} + \frac{1}{2} \left(2\right) {\left(0\right)}^{2} = \frac{1}{2} \left(5\right) {v}_{1 f}^{2} + \frac{1}{2} \left(2\right) {v}_{2 f}^{2}$

$810 = 2.5 {v}_{1 f}^{2} + \left(1\right) {v}_{2 f}^{2}$

With all that complete, our two equations are:

$90 = 5 {v}_{1 f} + 2 {v}_{2 f}$

$810 = 2.5 {v}_{1 f}^{2} + \left(1\right) {v}_{2 f}^{2}$

Rewrite the first equation as ${v}_{2 f} = \frac{90 - 5 {v}_{1 f}}{2}$

Substitute this value into the second equation:

$2.5 {v}_{1 f}^{2} + {\left(\frac{90 - 5 {v}_{1 f}}{2}\right)}^{2} = 810$

Now, we must solve this equation. It will be simpler if we multiply every term by 4, to eliminate the denominators:

$10 {v}_{1 f}^{2} + {\left(90 - 5 {v}_{1 f}\right)}^{2} = 3240$

$10 {v}_{1 f}^{2} + \left(8100 - 900 {v}_{1 f} + 25 {v}_{1 f}^{2}\right) = 3240$

$35 {v}_{1 f}^{2} - 900 {v}_{1 f} + 4860 = 0$

Use the quadratic formula to solve for v_(1f)

${v}_{1 f} = \frac{900 \pm \sqrt{{\left(- 900\right)}^{2} - 4 \left(35\right) \left(4860\right)}}{2 \left(35\right)}$

$= \frac{900 \pm 360}{70}$

There are two answers:

${v}_{1 f} = \frac{900 + 360}{70} = 18 \frac{m}{s}$ and

${v}_{1 f} = \frac{900 - 360}{70} = 7.71 \frac{m}{s}$

Substituting each answer back into ${v}_{2 f} = \frac{90 - 5 {v}_{1 f}}{2}$

we get ${v}_{2 f} = \frac{90 - 5 \left(18\right)}{2} = 0 \frac{m}{s}$

and ${v}_{2 f} = \frac{90 - 5 \left(7.71\right)}{2} = 25.7 \frac{m}{s}$

We must reject the first answer in each case, as this would result in both balls continuing in their original directions. This could only happen if there was no collision.

So, the only acceptable answers are

${v}_{1 f} = - 7.71 \frac{m}{s}$

${v}_{2 f} = 25.7 \frac{m}{s}$