Question

A hammer of mass 500 g, moving at 50m/s, strikes a nail. The nail stops the hammer in a very short time of 0.01s. What is the force of the nail on the hammer?

Answer 1

Hint: We are given the mass of a hammer and the speed with which it is moving. It is said that the hammer is made to strike on a nail and it stops within a very short time. From the second law of motion we know that the force applied will be directly proportional to the rate of change of momentum. Using this principle we can solve the question.

Formula used:
$F\propto \dfrac{m\left( v-u \right)}{t}$

Complete step-by-step answer:
In the question it is said that a hammer moving with a speed of 50m/s strikes a nail and the nail stops the hammer in 0.01 seconds.
The mass of the hammer is given to us as,
$m=500g$
Converting this into standard unit of mass, kilogram we get,
$m=\dfrac{500}{1000}=0.5kg$
From the second law of motion by Newton, we know that the force applied is directly proportional to the rate of change in momentum, i.e.
$\Rightarrow F=\dfrac{m\left( v-u \right)}{t}$, were ‘F’ is the force applied, ‘m’ is the mass, ‘v’ is the final velocity, ‘u’ is the initial velocity and ‘t’ is the time.
In this case we are given the initial velocity of the hammer as 50 m/s, i.e.
$u=50m/s$
We know that the hammer stops when it strikes the nail.
Therefore the final velocity of the hammer is zero, i.e.
$v=0m/s$
We are also given the time taken by the nail to stop the hammer,
$t=0.01s$
Therefore the force on the hammer will be,
$\Rightarrow F=\dfrac{0.5\left( 0-50 \right)}{0.01}$
By solving this we get,
$\Rightarrow F=\dfrac{-25}{0.01}$
$\Rightarrow F=-2500N$
Therefore the force applied by the nail on the hammer is -2500N.
The negative sign in the value of force indicates that the force applied is in the opposite direction of the movement of the hammer.

Note: Second law of motion by Newton simply states that the force applied on a system with constant mass is directly proportional to the change in its momentum with respect to time.
This can be mathematically written as,
$F\propto \dfrac{m\left( v-u \right)}{t}$, were ‘F’ is the force applied, ‘m’ is the mass, ‘v’ is the final velocity, ‘u’ is the initial velocity and ‘t’ is the time.
This proportionality in the equation is equated using a constant ‘k’ which is known as force constant.
$\Rightarrow F=k\dfrac{m\left( v-u \right)}{t}$
When the value of force constant, k is 1 we get the force applied as,
$\Rightarrow F=\dfrac{m\left( v-u \right)}{t}$