A hockey goalie is standing on ice. Another player fires a puck (m = 0.170 kg) at the goalie with a velocity of +55.1 m/s

goalie-with-velocity

#1

A hockey goalie is standing on ice. Another player fires a puck (m = 0.170 kg) at the goalie with a velocity of +55.1 m/s. (a) If the goalie catches the puck with his glove in a time of 3.98 x 10-3 s, what is the magnitude of the average force exerted on the goalie by the puck? (b) Instead of catching the puck, the goalie slaps it with his stick and returns the puck straight back to the player with a velocity of -55.1 m/s. The puck and stick are in contact for a time of 3.98 x 10-3 s

Answer:

For this problem, I think you must use the equations: momentum= mass*velocity , impulse =change in momentum, and impulse=force * time.

For part a.
(p1) initial momentum= 0.170 [kg] * 55.1 [m/s] = 9.37 [kg * m/s]
(p2) final momentum= 0, because final velocity is 0 [m/s]
impulse = p1-p2= 9.37 [kg m/s]
Finally impulse= Force* time,
Force= 9.37/ 3.98 x 10^-3 s = 2354.3 [N]

for part b, it is the same process but the final momentum is not 0.
p1=9.37
p2=- 9.37 (negative because opposite direction)
impulse= p1-p2= + 18.74 [kg m/s]

Then Force= 18.74/ (3.98 x 10^-3 s) = 4708.5 [N]