91Ó°ÊÓ

According to Newton’s second law, the rate of change of momentum of an object is equal to the net force applied on it, i.e.,

\(F = \frac{{\Delta p}}{{\Delta t}}\).

Here,\(\Delta p\)is the change in momentum of the object in\(\Delta t\)time.

In this problem,the average force exerted on the ball by the golf club is equal to the rate of change of momentum of the golf ball.

Mass of the golf ball,\(m = 0.045\;{\rm{kg}}\).

Contact time between the golf club and the ball is\(\Delta t = 3.5 \times {10^{ - 3}}\;{\rm{s}}\).

Ifthe direction of motion of the golf ball from the golf club to the pitch is considered as the positive direction, then

The initial velocity of the golf ball is\(u = 0\;{\rm{m/s}}\).

The final velocity of the golf ball with which it hits off the tee is \(v = 38\;{\rm{m/s}}\).

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

\(\begin{array}{c}{\rm{Impulse}} = \Delta p\\ = m\left( {v - u} \right)\\ = \left( {0.045\;{\rm{kg}}} \right)\left[ {38\;{\rm{m/s}} - 0\;{\rm{m/s}}} \right]\\ = 1.71\;{\rm{m/s}}\\ = 1.7\;{\rm{m/s}}\end{array}\)

Thus, the impulse imparted to the golf ball is 1.7 m/s.

From Newton’s second law, theaverage force exerted on the ball is calculated as follows:

\(\begin{array}{c}\bar F = \frac{{\Delta p}}{{\Delta t}}\\ = \frac{{1.7\;{\rm{m/s}}}}{{\left( {3.5 \times {{10}^{ - 3}}\;{\rm{s}}} \right)}}\\ = \;0.49 \times {10^3}\;{\rm{N}}\\ = 490\;{\rm{N}}\end{array}\)

Thus, the average force exerted on the ball by the golf club is 490 N.