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Question 3
A hammer of mass 500 g, moving at 50 ms−1, strikes a nail. The nail stops the hammer in a very short time of 0.01 s. What is the force of the nail on the hammer?
A hammer of mass 500 g, moving at 50 ms−1, strikes a nail. The nail stops the hammer in a very short time of 0.01 s. What is the force of the nail on the hammer?
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Solution
Mass of the hammer, m = 500 g = 0.5 kg
Initial velocity of the hammer, u = 50 ms−1
Time taken by the nail to the stop the hammer, t = 0.01 s
Velocity of the hammer, v = 0 (since the hammer finally comes to rest)
From Newton’s second law of motion:
Force, f = m(v−u)t = 0.5(0−50)0.01 = -2500 N
The hammer strikes the nail with a force of −2500 N.
Hence, from Newton’s third law of motion, the force of the nail on the hammer is equal and opposite, i.e., +2500 N.
Initial velocity of the hammer, u = 50 ms−1
Time taken by the nail to the stop the hammer, t = 0.01 s
Velocity of the hammer, v = 0 (since the hammer finally comes to rest)
From Newton’s second law of motion:
Force, f = m(v−u)t = 0.5(0−50)0.01 = -2500 N
The hammer strikes the nail with a force of −2500 N.
Hence, from Newton’s third law of motion, the force of the nail on the hammer is equal and opposite, i.e., +2500 N.
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