# A ball with a mass of  3 kg is rolling at 2 m/s and elastically collides with a resting ball with a mass of 4 kg. What are the post-collision velocities of the balls?

Jun 26, 2017

#### Answer:

The velocity of the first ball is $= - \frac{2}{7} m {s}^{-} 1$ and the second ball is $= \frac{12}{7} m {s}^{-} 1$

#### Explanation:

Since the collision is elastic, there is conservation of linear momentum and conservation of kinetic energy.

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

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

${m}_{1} = 3 k g$

${m}_{2} = 4 k g$

${u}_{1} = 2 m {s}^{-} 1$

${u}_{2} = 0 m {s}^{-} 1$

So,

$3 \cdot 2 + 4 \cdot 0 = 3 \cdot {v}_{1} + 4 \cdot {v}_{2}$

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

$\frac{1}{2} \cdot 3 \cdot {2}^{2} + \frac{1}{2} \cdot 4 \cdot 0 = \frac{1}{2} \cdot 3 \cdot {v}_{1}^{2} + \frac{1}{2} \cdot 4 \cdot {v}_{2}^{2}$

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

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

${v}_{2} = \frac{6 - 3 {v}_{1}}{4}$

$3 {v}_{1}^{2} + 4 \cdot {\left(\frac{6 - 3 {v}_{1}}{4}\right)}^{2} = 12$

$12 {v}_{1}^{2} + 36 - 36 {v}_{1} + 9 {v}_{1}^{2} = 48$

$21 {v}_{1}^{2} - 36 {v}_{1} - 12 = 0$

$7 {v}_{1}^{2} - 12 {v}_{1} - 4 = 0$

${v}_{1} = \frac{12 \pm \sqrt{{12}^{2} + 4 \cdot 4 \cdot 7}}{14}$

$= \frac{12 \pm 16}{14}$

${v}_{1} = 2 m {s}^{-} 1$ or ${v}_{1} = - \frac{2}{7} = - 0.28 m {s}^{-} 1$

We keep the second result ${v}_{1} = 0.28 m {s}^{-} 1$

${v}_{2} = \frac{6 + 3 \cdot 0.28}{4} = \frac{12}{7} = 1.71$

Verification

$4 \cdot \frac{12}{7} - 3 \cdot \frac{2}{7} = 6$