Question

a 77 kg man is riding on a 32 kg cart traveling at a speed of 3.8 m/s. he jumps off with zero horizontal speed relative to the ground. what is the resulting change in the cart's speed, including sign?

Answer 1

See below

Step-by-step explanation:

Using conservation of momentum:

Initial momentum = (77+32 kg)* 3.8 m/s = 414.2 kg m/s

Jumping off with zero speed relative to ground (I think this is what zero horizontal speed MEANS) means all of his momentum is given to the cart ( think about jumping off of a skateboard)

final momentum of the cart

mv = 414.2

v = 414.2/32 = 12.94 m/s for a + 9.13 m/s change