# A ball with a mass of 2 kg is rolling at 6 m/s and elastically collides with a resting ball with a mass of 1 kg. What are the post-collision velocities of the balls?

Mar 1, 2016

${v}_{1}^{'} = 2 \frac{m}{s}$
${v}_{2}^{'} = 8 \frac{m}{s}$

#### Explanation:

${\vec{P}}_{1} + {\vec{P}}_{2} = {\vec{P}}_{2}^{'} + {\vec{P}}_{2}^{'} \text{ the conservation of momentum}$
${m}_{1} \cdot {v}_{1} + {m}_{2} \cdot {v}_{2} = {m}_{1} \cdot {v}_{1}^{'} + {m}_{2} \cdot {v}_{2}^{'}$
$2 \cdot 6 + 0 = 2 \cdot {v}_{1}^{'} + 1 \cdot {v}_{2}^{'}$
$12 = 2 \cdot {v}_{1}^{'} + {v}_{2}^{'} \text{ (1)}$
v_1+v_1^'=v_2+v_2^'" (2)
$6 + {v}_{1}^{'} = 0 + {v}_{2}^{'}$
${v}_{2}^{'} = 6 + {v}_{1}^{'} \text{ (3)}$
$\text{use equation (1)}$
$12 = 2 \cdot {v}_{1}^{'} + 6 + {v}_{1}^{'}$
$12 - 6 = 3 \cdot {v}_{1}^{2}$
$6 = 3 \cdot {v}_{1}^{'} \text{ } {v}_{1}^{'} = 2 \frac{m}{s}$
$\text{use equation (3)}$
${v}_{2}^{'} = 6 + 2$
${v}_{2}^{'} = 8 \frac{m}{s}$