A 1.2 kg hammer slams down on a nail at 5.0 m/s and bounces off at 1.0 m/s. If the impact lasts 1.0 ms, what average force is ex
erted on the nail?
1 answer:
Answer:
Explanation:
Impulse results in a change of momentum
FΔt = mΔV
F = mΔV/Δt
The impulse acting on the hammer will equal the impulse acting on the nail
If we assume upward is the positive direction
F = m(vf - vi)/t
F = 1.2(1.0 - (-1.5)) / 0.001
F = 3000 N
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