# A ball with a mass of  5 kg is rolling at 14 m/s and elastically collides with a resting ball with a mass of 4 kg. What are the post-collision velocities of the balls?

May 26, 2016

${v}_{1}^{'} = \frac{14}{9} \text{ } \frac{m}{s}$

${v}_{2}^{'} = \frac{140}{9} \text{ } \frac{m}{s}$

#### Explanation:

${m}_{1} = 5 \text{ } k g$
${v}_{1} = 14 \text{ } \frac{m}{s}$
$\text{Momentum before collision for the object with mass of 5 kg}$
${P}_{1} = {m}_{1} \cdot {v}_{1} = 5 \cdot 15 = 70 \text{ } k g \cdot \frac{m}{s}$

${m}_{2} = 4 \text{ } k g$
${v}_{2} = 0$
$\text{Momentum before collision for the object with mass of 4 kg}$

${P}_{2} = {m}_{2} \cdot {v}_{2}$
${P}_{2} = 4 \cdot 0 = 0$

$\text{The total momentum of objects before collision}$
$\Sigma {P}_{b} = {P}_{1} + {P}_{2}$
$\Sigma {P}_{b} = 70 + 0 = 70$

$\text{Post collision velocities :}$

${v}_{1}^{'} = \frac{2 \cdot \Sigma {p}_{b}}{{m}_{1} + {m}_{2}} - {v}_{1}$

${v}_{1}^{'} = \frac{2 \cdot 70}{5 + 4} - 14$

${v}_{1}^{'} = \frac{140}{9} - 14$

${v}_{1}^{'} = \frac{140 - 126}{9}$

${v}_{1}^{'} = \frac{14}{9} \text{ } \frac{m}{s}$

${v}_{2}^{'} = \frac{2 \cdot \Sigma {p}_{b}}{{m}_{1} + {m}_{2}} - {v}_{2}$

${v}_{2}^{'} = \frac{2 \cdot 70}{5 + 4} - 0$

${v}_{2}^{'} = \frac{140}{9} \text{ } \frac{m}{s}$