A 12-kg hammer strikes a nail at a velocity of 7.5 m/s and comes to rest in a time interval of 8.0 ms.

(a) What is the impulse given to the nail?

(b) What is the average force acting on the nail?

Impulse imparted to an object is the product of force (\(\vec F\)) and the time (\(\Delta t\)) the force acts on the object. Also,theimpulse is equal to the change in momentum of the object, i.e.,

\({\rm{Impulse}} = \Delta \vec p = \vec F\Delta t\).

In this problem,the impulse imparted to the hammer is equal to the change in momentum of the hammer.

Mass of the hammer,\(m = 12\;{\rm{kg}}\).

Time taken by the hammer to strike the nail is\(\Delta t = 8.0\;{\rm{ms}} = 8.0 \times {10^{ - 3}}\;{\rm{s}}\).

Ifthe direction of motion of the hammer from the initial point toward the nail is taken as the positive direction, then

The initial velocity of the hammer is\(u = 7.5\;{\rm{m/s}}\).

The final velocity of the hammer is \(v = 0\;{\rm{m/s}}\).

The impulse imparted to the hammer is equal to the total change in the momentum of the hammer, i.e.,

\(\begin{array}{c}{\rm{Impulse}} = \Delta p\\ = m\left( {v - u} \right)\\ = \left( {12\;{\rm{kg}}} \right)\left[ {0\;{\rm{m/s}} - 7.5\;{\rm{m/s}}} \right]\\ = - 90\;{\rm{kg}} \cdot {\rm{m/s}}\\ = - 9.0 \times {\rm{1}}{{\rm{0}}^1}\;{\rm{kg}} \cdot {\rm{m/s}}\end{array}\)

Since the impulse imparted to the hammer is equal and opposite to the impulse imparted to the nail, the impulse given to the nail is \(9.0 \times {\rm{1}}{{\rm{0}}^1}\;{\rm{kg}} \cdot {\rm{m/s}}\).

The average force acting on the nail is equal to the impulse imparted to it divided by the time over which the force acts, i.e.,

\(\begin{array}{c}\bar F = \frac{{{\rm{Impulse}}}}{{\Delta t}}\\ = \frac{{9.0 \times {\rm{1}}{{\rm{0}}^1}\;{\rm{kg}} \cdot {\rm{m/s}}}}{{\left( {8.0 \times {{10}^{ - 3}}\;{\rm{s}}} \right)}}\\ = \;1.1 \times {10^4}\;{\rm{N}}\end{array}\)

Thus, the average force acting on the nail is \(1.1 \times {10^4}\;{\rm{N}}\).