A 62.0 kg ice hockey goalie, originally at rest, catches a 0.150 kg hockey puck

Question

A 62.0 kg ice hockey goalie, originally at rest, catches a 0.150 kg hockey puck slapped at him at a velocity of 33.0 m/s. Suppose the goalie and the ice puck have an elastic collision and the puck is reflected back in the direction from which it came. What would their final velocities (in m/s) be in this case? (Assume the original direction of the ice puck toward the goalie is in the positive direction. Indicate the direction with the sign of your answer.)

puck= ________ m/s

goalie= _________ m/s

Explanation / Answer

for elastic collision,

velocity of approach = velocity of separation

33 = vP + vG

vG = 33 - vP

Applying momentum conservation,

(0.150 x 33) + (62 x 0) = - 0.150 vP + 62 vG  

4.95 = - 0.150 vP + 2046 - 62 vP

vP = 32.84 m/s  

vG = 0.16 m/s

Ans: puck = - 32.84 m/s

goalie = +0.16 m/s

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