# A ball with a mass of  5 kg is rolling at 8 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 23, 2016

${v}_{1}^{'} = \frac{8}{9} \text{ } \frac{m}{s}$

${v}_{2}^{'} = \frac{80}{9} \text{ } \frac{m}{s}$

#### Explanation:

${m}_{1} = 5 \text{ } k g$
${v}_{1} = 8 \text{ } \frac{m}{s}$
${P}_{1} = {m}_{1} \cdot {v}_{1} = 5 \cdot 8 = 40 \text{ } k g \cdot \frac{m}{s}$

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

$\Sigma {P}_{b} = {P}_{1} + {P}_{2} \text{ total momentum}$
$\Sigma {P}_{b} = 40 + 0 = 40 \text{ } k g \cdot \frac{m}{s}$

$\text{post collision velocities :}$

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

${v}_{1}^{'} = \frac{2 \cdot 40}{5 + 4} - 8$

${v}_{1}^{'} = \frac{80}{9} - 8$

${v}_{1}^{'} = \frac{80 - 72}{9}$

${v}_{1}^{'} = \frac{8}{9} \text{ } \frac{m}{s}$

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

${v}_{2}^{'} = \frac{2 \cdot 40}{5 + 4} - 0$

${v}_{2}^{'} = \frac{80}{9} \text{ } \frac{m}{s}$