# A ball with a mass of  5 kg is rolling at 3 m/s and elastically collides with a resting ball with a mass of 4 kg. What are the post-collision velocities of the balls?

Feb 4, 2016

${v}_{1}^{'} = \frac{1}{3} \frac{m}{s}$
${v}_{2}^{'} = \frac{10}{3} \frac{m}{s}$

#### Explanation:

$\text{momentum before collision=momentum after collision}$
m_1=3 kg; v_1=3 m/s; m_2=4 kg; v_2=0
${P}_{b} = {m}_{1} \cdot {v}_{1} + {m}_{2} \cdot {v}_{2} = 5 \cdot 3 + 4 \cdot 0$
${P}_{b} = 15 \text{(momentum before collision)}$
${P}_{a} = {m}_{1} \cdot {V}_{1}^{'} + {m}_{2} \cdot {V}_{2} ' \text{(momentum after collision)}$
${P}_{a} = 5 \cdot {V}_{1}^{'} + 4 \cdot {V}_{2}^{'}$
${P}_{b} = {P}_{a}$
$15 = 5 \cdot {V}_{1}^{'} + 4 \cdot {V}_{2}^{'}$; $e q u a t i o n 1$
${v}_{1} + {v}_{1}^{'} = {v}_{2} + {v}_{2}^{'}$
$3 + {v}_{1}^{'} = 0 + {v}_{2}^{'}$;$e q u a t i o n 2$
write $3 + {v}_{1}^{'}$ instead ${v}_{2}^{'}$ in equation 1
$15 = 5 \cdot {v}_{1}^{'} + 4 \cdot \left(3 + {v}_{1}^{'}\right)$
$15 = 5 \cdot {v}_{1}^{'} + 12 + 4 \cdot {v}_{1}^{'}$
$15 - 12 = 9 \cdot {v}_{1}^{'}$
$3 = 9 \cdot {v}_{1}^{'}$
${v}_{1}^{'} = \frac{1}{3} \frac{m}{s}$
$\text{now use equation 2}$
$3 + {v}_{1}^{'} = {v}_{2}^{'}$
$3 + \frac{1}{3} = {v}_{2}^{'}$
${v}_{2}^{'} = \frac{10}{3} \frac{m}{s}$