# A ball with a mass of  3 kg is rolling at 6 m/s and elastically collides with a resting ball with a mass of  8 kg. What are the post-collision velocities of the balls?

Apr 7, 2018

$- \text{2.72 m/s}$ and $\text{3.27 m/s}$

#### Explanation:

$\text{Initial total momentum = Final total momentum}$

${m}_{1} {u}_{1} + {m}_{2} {u}_{2} = {m}_{1} {v}_{1} + {m}_{2} {v}_{2}$

("3 kg × 6 m/s") + ("8 kg × 0 m/s") = ("3 kg" × v_1) + ("8 kg" × v_2)

18 = 3v_1 + 8v_2 color(white)(...)……(1)

Coefficient of restitution $\left(e\right)$ is given by

$e = \frac{{v}_{2} - {v}_{1}}{{u}_{1} - {u}_{2}}$

For perfectly elastic collision $e = 1$

$1 = \frac{{v}_{2} - {v}_{1}}{6 - 0}$

${v}_{2} - {v}_{1} = 6$

Multiply both sides by $3$

3v_2 - 3v_1 = 18 color(white)(...) ……(2)

Add equations $\left(1\right)$ and $\left(2\right)$

$18 + 18 = 3 {v}_{1} + 8 {v}_{2} + 3 {v}_{2} - 3 {v}_{1}$

$36 = 11 {v}_{2}$

${v}_{2} = \frac{36}{11} = \textcolor{b l u e}{\text{3.27 m/s}}$

Substitute ${v}_{2} = 3.27$ in equation $\left(1\right)$

18 = 3v_1 + (8 × 3.27)

v_1 = (18 - (8 × 3.27))/3 = color(blue)(-"2.72 m/s")