A ball with a mass of  5 kg is rolling at 5 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 28, 2016

${v}_{1}^{'} = \frac{5}{9} \text{ } \frac{m}{s}$

${v}_{2}^{'} = \frac{50}{9} \text{ } \frac{m}{s}$

Explanation:

${m}_{1} = 5 \text{ } k g$
${\vec{v}}_{1} = 5 \text{ } \frac{m}{s}$

${\vec{P}}_{1} = {m}_{1} \cdot {\vec{v}}_{1}$
${\vec{P}}_{1} = 5 \cdot 5 = 25 \text{ "kg*m/s" momentum before collision for object } {m}_{1}$

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

${\vec{P}}_{2} = {m}_{2} \cdot {\vec{v}}_{2} = 4 \cdot 0 = 0 \text{ the object with " m_2= 4 kg" has no momentum}$

$\Sigma {\vec{P}}_{b} = 25 + 0 = 25 \text{ "kg*m/s" total momentum before collision}$

$\text{post collision velocity of object with "m_1="5 kg}$

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

${v}_{1}^{'} = \frac{2 \cdot 25}{5 + 4} - 5$

${v}_{1}^{'} = \frac{50}{9} - 5$

${v}_{1}^{'} = 50 - \frac{45}{9}$

${v}_{1}^{'} = \frac{5}{9} \text{ } \frac{m}{s}$

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

${v}_{2}^{'} = \frac{2 \cdot 25}{5 + 4} - 0$

${v}_{2}^{'} = \frac{50}{9} \text{ } \frac{m}{s}$