8. FORCE AND LAWS OF MOTION
medium

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(in $N$) of the nail on the hammer?

A

$5000$

B

$2500$

C

$3500$

D

$4500$

Solution

Mass of the hammer, $ m= 500\, g = 0.5 \,kg$

Initial velocity of the hammer, $u= 50 \,m/s$

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, $F=\frac{m(v-u)}{t}=\frac{0.5(0-50)}{0.01}=-2500 \,N$

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., $+2500\, N$.

Standard 9
Science

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