Question

DQuestion 2 1 pts A 0.28 kg volleyball is spiked and knocks a stationary player in the head. If the 41 kg player hit moves backward at 0.5 m/s and the volleyball bounces in the opposite direction at 40 m/s. What was the speed (in m/s) of the spiked volleyball? ROUND ANSWER TO ONE DECIMAL PLACE.

Answer 1

$\text{Solution :}\\ \text{This problem is based on the law of conservation of momentum.}\\ \text{We have :}\\ \text{Mass of the volleyball } m_1= 0.28kg \\ \ \ \ \ \ \ \ \ \ \ \text{Initial speed of the ball } u_1 = ??\\ \ \ \ \ \ \ \ \ \ \ \text{Final speed of the ball (Bounces back) } u_2 = -40m/s \\ \boldsymbol {And}\\ \text{Mass of the Player } \ m_2= 41kg\\ \text{Initial speed of the player } v_1 = 0\\ \text{Final speed of the player } \ v_2 = 0.5m/s\\ \ \\ \text{Momentum Before collision = Momentum after collision }\\ m_1u_1+m_2v_1= m_1u_2 +m_2v_2\\ 0.28kg\times u_1 + 41kg\times0 = 0.28kg\times(-40m/s)+41kg\times 0.5m/s\\ u_1 = 33.2\ m/s$