A 7.30-kg bowling ball strikes a 1.60-kg pin at rest head-on. Before the collision, the velocity of the ball is +6.00 m/s. After the collision, the velocity of the ball is +5.40 m/s. What is the velocity of the pin after the collision?
A)+0.6 m/s
B)+5.4 m/s
C)+1.2 m/s
D)+2.7 m/s
E)+3.2 m/s

Question

Quizplus · Accepted Answer

Using the conservation of momentum: (7.30 kg * 6.00 m/s) = (7.30 kg * 5.40 m/s) + (1.60 kg * v_pin). Solving for v_pin gives v_pin = 2.7 m/s.

A 730-Kg Bowling Ball Strikes a 1