Chapter 9, Problem 43A

To determine

The speed of the ball at $4.0\text{s}$ .

Answer to Problem 43A

  $25.33\text{m/s}$

Explanation of Solution

Given:

The mass of the ball, $m=0.150\text{kg}$

Initial velocity of the ball, $v_i=12\text{m/s}$ in the positive direction.

The force versus time graph of the considered ball.

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:

Now, using the above formula,

  $\begin{array}{c}F\Delta t=p_f-p_i\\ \text{Area under the given curve}=m\left(v_f-v_i\right)\end{array}$

Then, using the given values in above,

  $\begin{array}{c}F\Delta t=p_f-p_i\\ \frac{1}{2}\left(2\text{}\text{N}\right)\left(2\text{s}\right)=0.150\left(v_f-12\right)\\ v_f=25.33\text{}\text{m/s}\end{array}$

Conclusion:

Thus, the speedof the ball at $4.0\text{s}$ is $25.33\text{m/s}$ .