Suppose the force acting on a tennis ball (mass 0.060 kg) points in the \({\bf{ + x}}\) direction and is given by the graph of Fig. 7–33 as a function of time.

(a) Use graphical methods (count squares) to estimate the total impulse given the ball.

(b) Estimate the velocity of the ball after being struck; assuming the ball is being served so it is nearly at rest initially. [Hint: See Section 6–2.]

FIGURE 7-33 Problem 23.

The impulse of a force is the product of force and the time interval for which it works on the particle. It is also equal to the difference in the momentum produced in the body.

The mass of the ball is\(m = 0.060\;{\rm{kg}}\).

The initial velocity of the ball is\({v_i} = 0\).

(a)

From the graph, the total impulse given to the ball is equal to the area of 10 rectangles.

\(I = 10A\)

The above expression can be rewritten as\(I = 10wh\).

Here,\(w\)is the width and\(h\)is the height.

Substitute\(0.01\;{\rm{s}}\)for\(w\)and\(50\;{\rm{N}}\)for\(h\)in the above equation.

\(\begin{array}{l}I = 10\left( {0.01\;{\rm{s}}} \right)\left( {50\;{\rm{N}}} \right)\\I = 5\;{\rm{N}} \cdot {\rm{s}}\end{array}\)

Thus, the impulse given to the ball is\(5\;{\rm{N}} \cdot {\rm{s}}\).

(b)

The final velocity of the ball can be calculated using the formula of impulse.

\(\begin{array}{c}I = m\left( {{v_f} - {v_i}} \right)\\{v_f} - {v_i} = \frac{m}{I}\\{v_f} - 0 = \frac{{\left( {5\;{\rm{N}} \cdot {\rm{s}}} \right)}}{{\left( {0.060\;{\rm{kg}}} \right)}}\\{v_f} = 83.3\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.\\} {\rm{s}}}\end{array}\)

Thus, the velocity of the ball after being struck is \(83.3\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.\\} {\rm{s}}}\).