91Ó°ÊÓ

Impulse imparted to an object is the product of force (\(\vec F\)) and time (\(\Delta t\)) over which 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 given to the ball is equal to the total change in the momentum of the ball.

Mass of the tennis ball,\(m = 0.060\;{\rm{kg}}\).

The tennis ball strikes the wall at an angle,\(\theta = 45^\circ \).

The initial velocity of the tennis ball at an angle\(\theta \)to the wall is\(u = 28\;{\rm{m/s}}\).

The final velocity of the tennis ball at an angle \(\theta \) from the wall is \(v = u = 28\;{\rm{m/s}}\).

Theimpulse given to the ball along the vertical direction, i.e., along the plane of the wall, will be zero as the momentum of the ball does not change along the vertical direction.

Thus, the impulse imparted to the ball due to the change in momentum along the horizontal direction, i.e., along the horizontal line drawn perpendicular to the wall at the point of contact.

Considerthe direction of motion of the tennis ball away from the wall along a line perpendicular to the wall to be the positive direction.

The horizontal component of the initial velocity is\( - u\cos \theta \), directed towards the wall.The horizontal component of the final velocity is\(v\cos \theta \), directed away from the wall.

So, the impulse given to the ball is as follows:

\(\begin{array}{c}{\rm{Impulse}} = \Delta p\\ = m\left[ {v\sin \theta - \left( { - u\sin \theta } \right)} \right]\\ = m\left[ {v\sin \theta + v\sin \theta } \right]\\ = 2mv\sin \theta \\ = 2 \times \left( {0.060\;{\rm{kg}}} \right) \times \left( {28\;{\rm{m/s}}} \right)\sin 45^\circ \\ = \frac{{3.36}}{{\sqrt 2 }}\;{\rm{kg}} \cdot {\rm{m/s}}\\ = 2.4\;{\rm{kg}} \cdot {\rm{m/s}}\end{array}\)

Thus, the magnitude of the impulse given to the ball is \(2.4\;{\rm{kg}} \cdot {\rm{m/s}}\), which is directed away from the wall.