Question

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 exerted on the nail?​

Answer 1

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