A hockey goalie is standing on ice. Another player fires a puck ( m = 0.11 kg) a

Question

A hockey goalie is standing on ice. Another player fires a puck (m = 0.11 kg) at the goalie with a velocity of +67 m/s

A hockey goalie is standing on ice. Another player fires a puck (m = 0.11 kg) at the goalie with a velocity of +67 m/s (a) If the goalie catches the puck with his glove in a time of 3.0 x 10^-3 s. Now, what is the average force exerted on the goalie by the puck? x 10^-3 s, what is the average what is the average force (magnitude and direction) 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 ?67 m/s. The puck and stick are in contact for a time of 3.0

Explanation / Answer

a.) calculate the momentum before the puck hits the goalie: momentum = mass*velocity
p = 0.11 * 67
p = 7.37 kg.m/s

Because the final momentum is zero, the change in momentum is also 7.37, which is also called impulse.

impulse = Force*time
7.37 = 3.0*10-3 * F
F = 2.456*103 N
Therefore the average force exerted is 2456 N

b.) As we said from the previous question the original momentum is 7.37 kg.m/s. However, this time the final momentum is not zero: apply momentum = mass*velocity:

p = 0.11* -67
p = -7.37 kg.m/s

So this time the change in momentum is 14.74 kg.m/s, this is also the impulse.

impulse = Force*time
14.74 = 3.0*10-3 * F
F = 14.74 / 3.0 *10-3
F = 4.913*103 N

So now the average force is 4913 N

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