# A ball with a mass of 2 kg is rolling at 5 m/s and elastically collides with a resting ball with a mass of 1 kg. What are the post-collision velocities of the balls?

May 18, 2016

#### Answer:

${v}_{1}^{'} = \frac{5}{3} \frac{m}{s}$

${v}_{2}^{'} = \frac{20}{3} \frac{m}{s}$

#### Explanation:

$\text{Before Collision :}$
${m}_{1} = 2 k g$
${v}_{1} = 5 \frac{m}{s}$
${P}_{1} = {m}_{1} \cdot {v}_{1} \text{ } {P}_{1} = 2.5 = 10 k g \cdot \frac{m}{s}$

${m}_{2} = 1 k g$
${v}_{2} = 0$
${P}_{2} = {m}_{2} \cdot {v}_{2} \text{ } {P}_{2} = 1 \cdot 0 = 0$

$\text{Total Momentum before Collision :}$

$\Sigma {P}_{b} = {P}_{1} + {P}_{2} \text{ "Sigma P_b=10+0 " } \Sigma {P}_{b} = 10 k g \cdot \frac{m}{s}$

$\text{Post collision velocities :}$

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

${v}_{1}^{'} = \frac{2 \cdot 10}{2 + 1} - 5$

${v}_{1}^{'} = \frac{20}{3} - 5 \text{ "v_1^'=(20-15)/3" } {v}_{1}^{'} = \frac{5}{3} \frac{m}{s}$

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

${v}_{2} = \frac{2.10}{2 + 1} - 0$

${v}_{2}^{'} = \frac{20}{3} \frac{m}{s}$