A ball with a mass of 2 kg is rolling at 3 m/s and elastically collides with a resting ball with a mass of 4 kg. What are the post-collision velocities of the balls?

Mar 11, 2018

Speed of moving ball after collision $\text{= 1 m/s}$ (in the direction opposite to it’s initial direction of motion)
Speed of resting ball after collision $\text{= 2 m/s}$

Explanation:

Apply conservation of linear momentum

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

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

(2 × 3) + (4 × 0) = 2v_1 + 4v_2

$6 = 2 {v}_{1} + 4 {v}_{2}$

Didvide both sides by $2$

$3 = {v}_{1} + 2 {v}_{2} \textcolor{w h i t e}{\times \times \times \times \times \times \times \times} \left(1\right)$

Now, use formula of coefficient of restitution ($e$)

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

For an elastic collision $e = 1$

$1 = \frac{{v}_{2} - {v}_{1}}{3 - 0}$

${v}_{2} - {v}_{1} = 3 \textcolor{w h i t e}{\times \times \times \times \times \times \times \times x} \left(2\right)$

Add equations $\text{(1) & (2)}$

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

$6 = 3 {v}_{2}$

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

Substitute ${v}_{2} = \text{2 m/s}$ in equation $\left(1\right)$

3 = v_1 + (2 × "2 m/s")

$\textcolor{b l u e}{{v}_{1} = - \text{1 m/s}}$