# A ball with a mass of 2 kg is rolling at 1 m/s and elastically collides with a resting ball with a mass of 3 kg. What are the post-collision velocities of the balls?

Apr 25, 2016

${u}_{1} = - \frac{1}{5} \frac{m}{s}$
${u}_{2} = \frac{4}{5} \frac{m}{s}$

#### Explanation:

$\text{given data:}$
${m}_{1} = 2 k g$
${v}_{1} = 1 \frac{m}{s}$
$\text{Momentum before collision for the first ball:}$
${\vec{P}}_{1} = {m}_{1} \cdot {v}_{1} = 2 \cdot 1 = 2 k g \cdot \frac{m}{s}$

${m}_{2} = 3 k g$
${v}_{2} = 0$
$\text{momentum before collision for the second ball:}$
${\vec{P}}_{2} = 3 \cdot 0 = 0$

$\text{Total momentum before collision:}$
${\vec{P}}_{b} = {\vec{P}}_{1} + {\vec{P}}_{2}$
${\vec{P}}_{b} = 2 + 0 = 2 k g \cdot \frac{m}{s}$

$\text{Velocities for post collision :}$

${u}_{1} : \text{Post collision velocity of the first ball}$

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

${u}_{1} = \frac{2 \cdot 2}{2 + 3} - 1$

${u}_{1} = \frac{4}{5} - 1$

${u}_{1} = - \frac{1}{5} \frac{m}{s}$

${u}_{2} : \text{Post collision velocity of the second ball}$

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

${u}_{2} = \frac{2 \cdot 2}{2 + 3} - 0$

${u}_{2} = \frac{4}{5} \frac{m}{s}$