# A ball with a mass of  6 kg is rolling at 6 m/s and elastically collides with a resting ball with a mass of  8 kg. What are the post-collision velocities of the balls?

Mar 28, 2016

${v}_{1}^{'} = - \frac{6}{7} \text{ } \frac{m}{s}$

${v}_{2}^{'} = \frac{36}{7} \text{ } \frac{m}{s}$

#### Explanation:

${m}_{1} \cdot {v}_{1} + {m}_{2} \cdot {v}_{2} = {m}_{1} \cdot {v}_{1}^{'} + {m}_{2} \cdot {v}_{2}^{'}$
$6 \cdot 6 + 8 \cdot 0 = 6 \cdot {v}_{1}^{'} + 8 \cdot {v}_{2}^{'}$
$36 + 0 = 6 \cdot {v}_{1}^{'} + 8 \cdot {v}_{2}^{'}$
$36 = 6 \cdot {v}_{1}^{'} + 8 \cdot {v}_{2}^{'} \text{ } \left(1\right)$

$\text{ we use the equation } {v}_{1} + {v}_{1}^{'} = {v}_{2} + {v}_{2}^{'}$
$\text{ for calculating velocities after collision}$

$6 + {v}_{1}^{'} = 0 + {v}_{2}^{'}$
$6 + {v}_{1}^{'} = {v}_{2}^{'} \text{ } \left(2\right)$

$36 = 6 \cdot {v}_{1}^{'} + 8 \left(6 + {v}_{1}^{'}\right)$
$36 = 6 \cdot {v}_{1}^{'} + 48 + 8 \cdot {v}_{1}^{'}$
$36 - 48 = 14 \cdot {v}_{1}^{'}$
$- 12 = 14 \cdot {v}_{1}^{'}$
${v}_{1}^{'} = - \frac{12}{14} = - \frac{6}{7} \text{ } \frac{m}{s}$

$\text{use (2)}$
$6 - \frac{6}{7} = {v}_{2}^{'}$
$\frac{42 - 6}{7} = {v}_{2}^{'}$

${v}_{2}^{'} = \frac{36}{7} \text{ } \frac{m}{s}$