A 75-kg man is riding in a 30-kg cart at 2.0 m/s. He jumps off in such a way as to land on the ground with no horizontal velocity. The resulting change in speed of the cart is:
A) zero
B) 2.0 m/s
C) 3.0 m/s
D) 5.0 m/s
E) 7.0 m/s

Question

Quizplus · Accepted Answer

Using conservation of momentum, the initial momentum of the system is (75 kg + 30 kg) * 2.0 m/s = 210 kg*m/s. After the man jumps off, the cart's momentum is 30 kg * v. Since the man's horizontal velocity is zero, the cart's velocity must be 7.0 m/s to conserve momentum, resulting in a change of 5.0 m/s.

A 75-Kg Man Is Riding in a 30-Kg Cart at 2.0