Question

A hammer of mass 0.5 kg moving at 50 ms-¹ , hits a nail. The nail stops the hammer in a very short time of 0.02 seconds. Calculate the force of the nail on the hammer .​

Answer 1

Answer:

The initial velocity of the hammer (u) = 50 m/s

Time is taken by the nail to the stop the hammer (t) = 0.01 s

The velocity of the hammer (v) = 0 (since the hammer finally comes to rest)

From Newton’s second law of motion

f = m (v – u) / t

Where f = force

f = 0.5 × (0 – 50) / 0.01

f = -2500 N

-ve sign indicates the force of nail on the hammer moving in the opposite direction to that of hammer movement.

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