A ball with a mass of  3 kg is rolling at 8 m/s and elastically collides with a resting ball with a mass of 4 kg. What are the post-collision velocities of the balls?

Jan 5, 2018

The post collision velocities are $= - 1.14 m {s}^{-} 1$ and $= 6.86 m {s}^{-} 1$

Explanation:

In an elastic collision, there is conservation of momentum and conservation of kinetic energies.

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

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

Here,

The masses are

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

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

The initial velocities are

${u}_{1} = 8 m {s}^{-} 1$

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

$3 \cdot 8 + 4 \cdot 0 = 3 \cdot {v}_{1} + 4 \cdot {v}_{2}$, $\iff$, $3 {v}_{1} + 4 {v}_{2} = 24$........$\left(1\right)$

$\frac{1}{2} \cdot 3 \cdot {8}^{2} + \frac{1}{2} \cdot 4 \cdot {0}^{2} = \frac{1}{2} \cdot 3 \cdot {v}_{1}^{2} + \frac{1}{2} \cdot 4 \cdot {v}_{2}^{2}$

$3 {v}_{1}^{2} + 4 {v}_{2}^{2} = 192$.............................$\left(2\right)$

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

From $\left(1\right)$, ${v}_{2} = \frac{24 - 3 {v}_{1}}{4}$

Substituting is $\left(2\right)$

$3 {v}_{1}^{2} + 4 \cdot {\left(\frac{24 - 3 {v}_{1}}{4}\right)}^{2} = 192$

$12 {v}_{1}^{2} + {\left(24 - 3 {v}_{1}\right)}^{2} = 768$

$12 {v}_{1}^{2} + 576 - 144 {v}_{1} + 9 {v}_{1}^{2} = 768$

$21 {v}_{1}^{2} - 144 v - 192 = 0$

$7 {v}_{1}^{2} - 48 v - 64 = 0$

${v}_{1} = \frac{48 \pm \sqrt{{\left(- 48\right)}^{2} - 4 \cdot 7 \cdot \left(- 64\right)}}{2 \cdot 7}$

$= \frac{48 \pm \sqrt{4096}}{14}$

$= \frac{48 \pm 64}{14}$

So,

${v}_{1} = \left(\frac{48 + 64}{14}\right) = 8 m {s}^{-} 1$, $\implies$, ${v}_{2} = 0 m {s}^{-} 1$

This is the initial conditions.

or

${v}_{2} = \left(\frac{48 - 64}{14}\right) = - 1.14 m {s}^{-} 1$, $\implies$, ${v}_{2} = 6.86 m {s}^{-} 1$