# A ball with a mass of  5 kg is rolling at 7 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 6, 2016

I found: $0.78 \mathmr{and} 7.78 \frac{m}{s}$ but check my maths!

#### Explanation:

We can use conservation of momentum $\vec{p} = m \vec{v}$ in one dimension (along $x$). So we get:

${p}_{\text{before")=p_("after}}$

So:

$\left(5 \cdot 7\right) + \left(4 \cdot 0\right) = \left(5 \cdot {v}_{1}\right) + \left(4 \cdot {v}_{2}\right)$

Being an elastic collision also kinetic energy $K = \frac{1}{2} m {v}^{2}$ is conserved:

${K}_{\text{before")=K_("after}}$

So:
$\frac{1}{2} \cdot 5 \cdot {7}^{2} + \frac{1}{2} \cdot 4 \cdot {0}^{2} = \frac{1}{2} \cdot 5 \cdot {v}_{1}^{2} + \frac{1}{2} \cdot 4 \cdot {v}_{2}^{2}$

We can use the two equations together and we get:

$\left\{\begin{matrix}35 = 5 {v}_{1} + 4 {v}_{2} \\ 245 = 5 {v}_{1}^{2} + 4 {v}_{2}^{2}\end{matrix}\right.$

From the first equation:
${v}_{1} = \frac{35 - 4 {v}_{2}}{5}$
Substitute into the second and get:
$245 = \cancel{5} {\left(35 - 4 {v}_{2}\right)}^{2} / {\cancel{25}}^{5} + 4 {v}_{2}^{2}$
$\cancel{1225} = \cancel{1225} - 280 {v}_{2} + 16 {v}_{2}^{2} + 20 {v}_{2}^{2}$
$36 {v}_{2}^{2} - 280 {v}_{2} = 0$
${v}_{2} = 0 \frac{m}{s}$ (not) and ${v}_{2} = \frac{280}{36} = \frac{70}{9} = 7.78 \frac{m}{s}$ yes.
Corresponding to:
${v}_{1} = 7 \frac{m}{s}$ and ${v}_{1} = \frac{7}{9} = 0.78 \frac{m}{s}$