Question

A 10kg ball moving at 9.0 m/s strikes a stationary 130kg man on the chest after which the ball bounces back at 5.0 m/s What is the final velocity of the person after the ball bounces off him

Answer 1

We know that

• The mass of the ball is 10kg.

,

• The initial velocity of the ball is 9m/s.

,

• The mass of the man is 130kg.

,

• The initial velocity of the man is null (stationary).

,

• The final velocity of the ball is 5m/s.

To find the final velocity of the person, we have to use the law of conservation of momentum.

$p_i=p_f$

Where p = mv. Using all the given information, we have

$\begin{gathered} m_{^{}_{ball}}v_{i_{ball}}+m_{_{man}}v_{i_{man}}=m_{\text{ball}}v_{f_{ball}}+m_{man}v_{f_{man}}_{} \\ 10\operatorname{kg}\cdot9m/s+130\operatorname{kg}\cdot0=10\operatorname{kg}\cdot5m/s_{}_{}+130\operatorname{kg}\cdot v_{f_{man}} \end{gathered}$

Then, we solve for v

$\begin{gathered} 90\operatorname{kg}\cdot m/s=50\operatorname{kg}\cdot m/s+130\operatorname{kg}\cdot v_{f_{man}} \\ \frac{90\operatorname{kg}\cdot m/s-50\operatorname{kg}\cdot m/s}{130\operatorname{kg}}=v_{f_{man}} \\ v_{f_{man}}=\frac{40\operatorname{kg}\cdot m/s}{130\operatorname{kg}}\approx0.31m/s \end{gathered}$

Therefore, the velocity of the person after the ball bounces off him is 0.31 meters per second, approximately.