A ball with a mass of  3 kg is rolling at 1 m/s and elastically collides with a resting ball with a mass of 4 kg. What are the post-collision velocities of the balls?

Mar 9, 2016

${v}_{1}^{'} = - \frac{1}{7} \frac{m}{s}$
${v}_{2}^{'} = \frac{6}{7} \frac{m}{s}$

Explanation:

${m}_{1} \cdot {v}_{1} + {m}_{2} {v}_{2} = {m}_{1} {v}_{1}^{'} + {m}_{2} {v}_{2}^{'} \text{ } \left(1\right)$
$\text{total momentum before collision=total momentum after collision}$
$\frac{1}{2} \cdot {m}_{1} \cdot {v}_{1}^{2} + \frac{1}{2} \cdot {m}_{2} \cdot {v}_{2}^{2} = \frac{1}{2.} {m}_{1} \cdot {v}_{1} {'}^{2} + \frac{1}{2} \cdot {m}_{2} \cdot {v}_{2} {'}^{2}$

${m}_{1} \cdot {v}_{1}^{2} + {m}_{2} \cdot {v}_{2}^{2} = {m}_{1} \cdot {v}_{1} {'}^{2} + {m}_{2} \cdot {v}_{2} {'}^{2} \text{ } \left(2\right)$
$\text{we obtain the equation of (3),if we arrange (1) and (2)}$
${v}_{1} + {v}_{1}^{'} = {v}_{2} + {v}_{2}^{'} \text{ } \left(3\right)$
$3 \cdot 1 + 4 \cdot 0 = 3 \cdot {v}_{1}^{'} + 4 \cdot {v}_{2}^{'}$
$3 = 3 \cdot {v}_{1}^{'} + 4 \cdot {v}_{2}^{'} \text{ } \left(4\right)$
$\text{use (3)}$
$1 + {v}_{1}^{'} = 0 + {v}_{2}^{'}$
${v}_{2}^{'} = 1 + {v}_{1}^{'} \text{ } \left(5\right)$
$\text{use (4)}$
$3 = 3 \cdot {v}_{1}^{'} + 4 \left(1 + {v}_{1}^{'}\right)$
$3 = 3 \cdot {v}_{1}^{'} + 4 + 4 {v}_{1}^{'}$
$- 1 = 7 {v}_{1}^{'} \text{ } {v}_{1}^{'} = - \frac{1}{7} \frac{m}{s}$
$\text{use (5)}$
${v}_{2}^{'} = 1 - \frac{1}{7}$
${v}_{2}^{'} = \frac{6}{7} \frac{m}{s}$