Chapter 9, Problem 48A

To determine

The average force applied by T on the ball if the interaction time between her hands and the ball was $0.025\text{s}$ .

Answer to Problem 48A

  $59.52\text{N}$

Explanation of Solution

Given:

The mass of the volleyball, $m=0.24\text{kg}$

Initial velocity of the ball with which it approaches T, $v_i=3.8\text{}\text{m/s}$

Final velocity of the ball after T bumps it, $v_f=2.4\text{m/s}$ in the opposite direction

The interaction time between her hands and the ball, $\Delta t=0.025\text{s}$ .

Formula used:

Momentum of an object of mass m and moving with velocity v is given by,

  $p=mv$

Impulse on an object is equal to the change in momentum of an object,

  $\text{Impulse = }F\Delta t=p_f-p_i$

Where $F$ is the applied force for time $\Delta t$ , $p_f$ is the final momentum, $p_i$ is the initial momentum.

Calculation:

Consider the direction of initial motion of the ball as negative then the opposite direction will be positive. Thus, the velocities can be written as,

  $\begin{array}{c}{v}_i=-3.8\text{}\text{m/s}\\ v_f=+2.4\text{m/s}\end{array}$

Then, using all the given values in the formula mentioned, the average force applied by T on the ball will be,

  $\begin{array}{c}F\Delta t=m\left(v_f-v_i\right)\\ F\left(0.025\right)=0.24\left(2.4-\left(-3.8\right)\right)\\ F=59.52\text{N}\end{array}$

Conclusion:

Thus, the average force applied by T on the ball is $59.52\text{N}$ .