A ball with a mass of  6 kg is rolling at 2 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 30, 2018

The post collision velocities of the balls are $= - \frac{2}{7} m {s}^{-} 1$ and $= \frac{12}{7} m {s}^{-} 1$

Explanation:

As the collision is elastic, there is conservation of momentum and conservation of kinetic energy

${m}_{1} {u}_{1} + {m}_{2} {u}_{2} = {m}_{1} {v}_{1} + {m}_{2} {v}_{2}$

$\frac{1}{2} {m}_{1} {u}_{1}^{2} + \frac{1}{2} {m}_{2} {u}_{2}^{2} = \frac{1}{2} {m}_{1} {v}_{1}^{2} + \frac{1}{2} {m}_{2} {v}_{2}^{2}$

The mass the first ball is ${m}_{1} = 6 k g$

The velocity of the first ball before the collision is ${u}_{1} = 2 m {s}^{-} 1$

The mass of the second ball is ${m}_{2} = 8 k g$

The velocity of the second ball before the collision is ${u}_{2} = 0 m {s}^{-} 1$

The velocity of the first ball after the collision is $= {v}_{1}$

The velocity of the second ball after the collision is $= {v}_{2}$

$6 \cdot 2 + 8 \cdot 0 = 6 {v}_{1} + 8 {v}_{2}$

$3 {v}_{1} + 4 {v}_{2} = 6$

${v}_{1} = \frac{6 - 4 {v}_{2}}{3}$.....................$\left(1\right)$

$\frac{1}{2} \cdot 6 \cdot {2}^{2} + \frac{1}{2} \cdot 8 \cdot 0 = \frac{1}{2} \cdot 6 \cdot {v}_{1}^{2} + \frac{1}{2} \cdot 8 \cdot {v}_{2}^{2}$

$6 {v}_{1}^{2} + 8 {v}_{2}^{2} = 24$

$3 {v}_{1}^{2} + 4 {v}_{2}^{2} = 12$...........................$\left(2\right)$

From equations $\left(1\right)$ and $\left(2\right)$, calculate $v 1$ and ${v}_{2}$

$3 \cdot {\left(\frac{6 - 4 {v}_{2}}{3}\right)}^{2} + 4 {v}_{2}^{2} = 12$

$36 - 48 {v}_{2} + 16 {v}_{2}^{2} + 12 {v}_{2}^{2} = 36$

$28 {v}_{2}^{2} - 48 {v}_{2} = 0$

$4 {v}_{2} \left(7 {v}_{2} - 12\right) = 0$

${v}_{2} = 0$ or ${v}_{2} = \frac{12}{7}$

${v}_{1} = 2$ or ${v}_{1} = - \frac{2}{7}$