Answer to Question #154178 in Classical Mechanics for Max

Question

A 75.0 kg hockey player moving at a speed of + 10.0 m / s collides with a second hockey player, who is motionless. After the collision, the two skaters move together at a speed of + 4.50 m / s. During the collision, what is the impulse received by the 75.0 kg player?

Answer 1

By the definition, the impulse received by the 75.0 kg player equals the change in momentum of the 75.0 kg player:

$$J=m\Delta v=m(v_f-v_i),$$ $$J=75.0\ kg\cdot(4.5\ \dfrac{m}{s}-10.0\ \dfrac{m}{s})=-412.5\ N\cdot s.$$

The sign minus means that the impulse received by the 75.0 kg player directed in the opposite direction to the motion of the 75.0 kg hockey player.

Answer:

$J=412.5\ N\cdot s,$ in the opposite direction to the motion of the 75.0 kg hockey player.