Discussion
A hockey player hits a ball with an impulse of 15 Ns. If time of hit is 0.2 s, the average force exerted by the player on the ball is
Answer: A
The average force is 75 N.
Impulse is average force times the time of contact, \(J = F\Delta t\):
\[ F = \frac{J}{\Delta t} = \frac{15}{0.2} = 75\ \text{N} \]
15 N treats the impulse as the force, and 50 N or 25 N come from slips in dividing by 0.2.
Impulse equals the change in momentum, so the ball's momentum changes by 15 kg m s-1 whatever its mass; a shorter contact time would mean a larger force for the same impulse.
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