Chapter 9, Problem 42A

a.

To determine

The change in momentum of the ball.

Answer to Problem 42A

  $2.32\text{kg}\cdot \text{m/s}$

Explanation of Solution

Given:

The mass of the baseball, $m=0.145\text{kg}$

Initial velocity of the baseball, $v_i=42\text{}\text{m/s}$

Final velocity of the baseball, $v_f=58\text{}\text{m/s}$

Formula used:

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

  $p=mv$

Calculation:

Using the given values in above formula, the change in momentum will be

  $\begin{array}{c}{p}_f-p_i=m{v}_f-m{v}_i\\ \Delta p=m\left(v_f-v_i\right)\\ =0.145\left(58-42\right)\\ =2.32\text{kg}\cdot \text{m/s}\end{array}$

Conclusion:

Thus, the change in momentum of the ball is $2.32\text{kg}\cdot \text{m/s}$ .

b.

To determine

The average force during the contact.

Answer to Problem 42A

  $5043.47\text{N}$

Explanation of Solution

Given:

The mass of the baseball, $m=0.145\text{kg}$

Initial velocity of the baseball, $v_i=42\text{}\text{m/s}$

Final velocity of the baseball, $v_f=58\text{}\text{m/s}$

From part (a), the change in momentum of the ball, $\Delta p=2.32\text{kg}\cdot \text{m/s}$

Time of contact of between bat and the ball is $\Delta t=4.6\times {10}^{-4}\text{s}$

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

Calculation:

Then, using the given and calculated values in above formula, the force during contact will be,

  $\begin{array}{c}F\Delta t=\Delta p\\ F\left(4.6\times {10}^{-4}\right)=2.32\text{kg}\cdot \text{m/s}\\ F=5043.47\text{N}\end{array}$

Conclusion:

Thus, the average force during contact is $5043.47\text{N}$ .