Question

A hammer of mass $500g$, moving at $50m{s}^{-1}$, strikes a nail. The nail stops the hammer for a short duration of $0.01s$. What is the force of the nail on the hammer?
[2 Marks]

Answer 1

Given:

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

Initial velocity, $u=50m{s}^{-1}$

Final velcity, $v=0m{s}^{-1}$

Time, $t=0.01s$

Acceleration $a=\frac{(v-u)}{t}=\frac{(0-50)}{0.01}=-5000m{s}^{-2}$
[1 Mark]

Force acting on the nail $\left(F\right)=ma$

$F=\left(0.5kg\right)\times -5000m{s}^{-2}$

$F=-2500N$

According to the third law of motion, every action has an equal and opposite reaction. Therefore, the nail exerts an equal and opposite force on the hammer. Since the force exerted on the nail by the hammer is $-2500N$, the force exerted on the hammer by the nail will be $+2500N$.
[1 Mark]