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

Apr 17, 2016

${u}_{1} = - \frac{24}{7} \frac{m}{s}$
${u}_{2} = \frac{60}{7} \text{ } \frac{m}{s}$
$\text{TESTING}$
$\textcolor{g r e e n}{{E}_{b} = {E}_{a}}$

#### Explanation: $\text{using the equation (1)}$

$5 \cdot 12 + 9 \cdot 0 = 5 \cdot {u}_{1} + 9 \cdot {u}_{2}$
$60 + 0 = 5 \cdot {u}_{1} + 9 \cdot {u}_{2}$

$60 = 5 \cdot {u}_{1} + 9 \cdot {u}_{2} \text{ (a)}$

$\text{using the equation (3)}$

$12 + {u}_{1} = 0 + {u}_{2}$

${u}_{2} = 12 + {u}_{1} \text{ (b)}$

$\text{now use (a)}$

$60 = 5 \cdot {u}_{1} + 9 \cdot \left(12 + {u}_{1}\right)$

$60 = 5 \cdot {u}_{1} + 108 + 9 \cdot {u}_{1}$

$60 - 108 = 14 \cdot {u}_{1}$

$- 48 = 14 \cdot {u}_{1}$

${u}_{1} = - \frac{48}{14} = - \frac{24}{7} \frac{m}{s}$

$\text{finally ;use (b)}$

${u}_{2} = 12 - \frac{24}{7}$

${u}_{2} = \frac{84 - 24}{7}$

${u}_{2} = \frac{60}{7} \text{ } \frac{m}{s}$

$\text{testing}$

$\text{The Total Kinetic Energy before collision:}$
${E}_{b} = \frac{1}{2} \cdot 5 \cdot {12}^{2} + 0$

${E}_{b} = 72 \cdot 5 = 360 J$

$\text{The Total Kinetic Energy after collision:}$

${E}_{a} = \frac{1}{2} \cdot 5 \cdot {\left(- \frac{24}{7}\right)}^{2} + \frac{1}{2} \cdot 9 \cdot {\left(\frac{60}{7}\right)}^{2}$

${E}_{a} = \frac{5 \cdot 576}{2 \cdot 49} + \frac{9 \cdot 3600}{2 \cdot 49}$

${E}_{a} = \frac{2880 + 32400}{98}$

${E}_{a} = \frac{35280}{98}$

${E}_{a} = 360 J$

$\textcolor{g r e e n}{{E}_{b} = {E}_{a}}$