# A ball with a mass of  5 kg is rolling at 12 m/s and elastically collides with a resting ball with a mass of 4 kg. What are the post-collision velocities of the balls?

Nov 23, 2017

#### Answer:

The velocities are $\frac{4}{3} m {s}^{-} 1$ and $\frac{40}{3} m {s}^{-} 1$

#### Explanation:

Here it's an elastic collision with no loss of kinetic energy.

We have the conservation of momentum.

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

${m}_{1} = 5 k g$

${u}_{1} = 12 m {s}^{- 1}$

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

${u}_{2} = 0$

v_1=?

v_2=?

$5 \cdot 12 + 4 \cdot 0 = 5 \cdot {v}_{1} + 4 \cdot {v}_{2}$

$5 {v}_{1} + 4 {v}_{2} = 60$............$\left(1\right)$

There is also conservation of kinetic energies

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

$5 \cdot {12}^{2} + 4 \cdot 0 = 5 \cdot {v}_{1}^{2} + 4 \cdot {v}_{2}^{2}$

$5 {v}_{1}^{2} + 4 {v}_{2}^{2} = 720$.........$\left(2\right)$

From $\left(1\right)$, we get

${v}_{1} = \frac{60 - 4 {v}_{2}}{5}$

Plugging this value in $\left(2\right)$

$5 \cdot {\left(\frac{60 - 4 {v}_{2}}{5}\right)}^{2} + 4 {v}_{2}^{2} = 720$

${\left(\left(60 - 4 {v}_{2}\right)\right)}^{2} / 5 + 4 {v}_{2}^{2} = 720$

$16 {\left(15 - {v}_{2}\right)}^{2} + 20 {v}_{2}^{2} = 720 \cdot 5 = 3600$

$4 {\left(15 - {v}_{2}\right)}^{2} + 5 {v}_{2}^{2} = 900$

$4 \left(225 - 30 {v}_{2} + {v}^{2}\right) + 5 {v}_{2}^{2} = 900$

$900 - 120 {v}_{2} + 4 {v}_{2}^{2} + 5 {v}_{2}^{2} = 900$

$9 {v}_{2}^{2} - 120 {v}_{2} = 0$

${v}_{2} \left(9 {v}_{2} - 120\right) = 0$

${v}_{2} = 0$ or ${v}_{2} = \frac{120}{9} = \frac{40}{3} = 13.3 m {s}^{-} 1$

${v}_{1} = \frac{60 - 4 \cdot \frac{40}{3}}{5} = \frac{4}{3} m {s}^{-} 1$