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8. FORCE AND LAWS OF MOTION
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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
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