# 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 9 kg. What are the post-collision velocities of the balls?

Sep 1, 2017

The velocity of the first ball is $= - 3.82 m {s}^{-} 1$
The velocity of the second ball is $= 2.18 m {s}^{-} 1$

#### Explanation:

As the collision is elastic, we have conservation of linear momentum and conservation of kinetic energy.

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

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

so,

$2 \cdot 6 + 9 \cdot 0 = 2 {v}_{1} + 9 {v}_{2}$,

$\implies$, ${v}_{1} = \frac{12 - 9 {v}_{2}}{2}$........................$\left(1\right)$

$\frac{1}{2} \cdot 2 \cdot {6}^{2} + \frac{1}{2} \cdot 9 \cdot 0 = \frac{1}{2} \cdot 2 \cdot {v}_{1}^{2} + \frac{1}{2} \cdot 9 \cdot {v}_{2}^{2}$

$2 {v}_{1}^{2} + 9 {v}_{2}^{2} = 72$..........................$\left(2\right)$

Solving for ${v}_{1}$ and ${v}_{2}$ from equations $\left(1\right)$ and $\left(2\right)$

$2 \cdot {\left(\frac{12 - 9 {v}_{2}}{2}\right)}^{2} + 9 {v}_{2}^{2} = 72$

${\left(12 - 9 {v}_{2}\right)}^{2} + 18 {v}_{2}^{2} = 144$

$144 - 216 {v}_{2} + 81 {v}_{2}^{2} + 18 {v}_{2}^{2} = 144$

${v}_{2} \left(99 {v}_{2} - 216\right) = 0$

${v}_{2} = 0$ or ${v}_{2} = \frac{216}{99} = 2.18 m {s}^{-} 1$

${v}_{1} = \frac{1}{2} \left(12 - 9 \cdot 2.18\right) = - 3.82 m {s}^{-} 1$