# A ball with a mass of  3 kg is rolling at 16 m/s and elastically collides with a resting ball with a mass of  8 kg. What are the post-collision velocities of the balls?

Jun 29, 2016

${v}_{r}^{'} = - 7.27 \text{ "m/s" The red ball's velocity after collision}$
${v}_{b}^{'} = 8.73 \text{ "m/s" The blue ball's velocity after collision}$

#### Explanation: $\text{BEFORE COLLISION}$

${m}_{r} = 3 \text{ kg The red ball's mass}$
${v}_{r} = 16 \text{ "m/s " The velocity of red ball before collision}$

${P}_{r} = {m}_{r} \cdot {v}_{r} = 16 \cdot 3 = 48 k g \cdot \frac{m}{s}$
${P}_{r} : \text{(The red ball's momentum before collision)}$

${m}_{b} = 8 \text{ kg The blue ball's mass}$
${v}_{r} = 0 \text{ "m/s " The velocity of blue ball before collision}$

${P}_{b} = {m}_{b} \cdot {v}_{b} = 8 \cdot 0 = 0$
${P}_{b} = 0$

$P = {P}_{r} + {P}_{b} = 48 + 0 = 48 k g \cdot \frac{m}{s}$

$P : \text{(The Total momentum before collision)}$

$\text{AFTER COLLISION}$

${P}_{r}^{'} = 3 \cdot {v}_{r}^{'} \text{ The red ball's momentum after collision}$
${P}_{b}^{'} = 8 \cdot {v}_{b}^{'} \text{ The blue ball's momentum after collision}$
${P}_{^} ' = {P}_{r}^{'} + {P}_{b}^{'}$
${P}^{'} = 3 \cdot {v}_{r}^{'} + 8 \cdot {v}_{b}^{'}$

${P}^{'} : \text{The total momentum after collision}$

$P = {P}^{'} \text{ The conservation of momentum}$

$48 = 3 \cdot {v}_{r}^{'} + 8 \cdot {v}_{b}^{'} \text{ equation 1}$

${v}_{r} + {v}_{r}^{'} = {v}_{b} + {v}_{b}^{'} \text{ equation 2}$
$16 + {v}_{r}^{'} = 0 + {v}_{b}^{'}$
${v}_{b}^{'} = 16 + {v}_{r}^{'} \text{ plug in equation 1}$

$48 = 3 \cdot {v}_{r}^{'} + 8 \cdot \left(16 + {v}_{r}^{'}\right)$

$48 = 3 {v}_{r}^{'} + 128 + 8 {v}_{r}^{'}$

$11 {v}_{r}^{'} = 48 - 128$

$11 {v}_{r}^{'} = - 80$

${v}_{r}^{'} = - \frac{80}{11}$

${v}_{r}^{'} = - 7.27 \text{ "m/s" The red ball's velocity after collision}$

$\text{So;}$
${v}_{b}^{'} = 16 + {v}_{r}^{'}$

${v}_{b}^{'} = 16 - 7.27$

${v}_{b}^{'} = 8.73 \text{ "m/s" The blue ball's velocity after collision}$