Chapter 4, Problem 77A

(a)

To determine

The acceleration of the baseball.

Answer to Problem 77A

  $-6\times {10}^3{\text{m/s}}^2$

Explanation of Solution

Given:

The initial velocity of the baseball is $v_{\text{i}}=30\text{m/s}$ .

The final velocity of the baseball is $v_{\text{f}}=0\text{m/s}$ .

The time interval for speed up is $\Delta t=0.0050\text{s}$ .

The mass of baseball is $m=0.145\text{kg}$ .

Formula used:

Show the expression for the acceleration $\left(\overline{a}\right)$ of an object as follows:

  $\begin{array}{l}\overline{a}=\frac{\Delta v}{\Delta t}\\ \overline{a}=\frac{v_{\text{f}}-v_{\text{i}}}{\Delta t}\end{array}$

Here, $v_{\text{i}}$ and $v_{\text{f}}$ are the initial and final velocity of the object, and $\Delta t$ is the time interval.

Calculation:

Find the acceleration of baseball as follows:

  $\overline{a}=\frac{v_{\text{f}}-v_{\text{i}}}{\Delta t}$

Apply the values for $v_{\text{i}}$ , $v_{\text{f}}$ , and $\Delta t$ in the above expression,

  $\begin{array}{l}\overline{a}=\frac{0-30}{0.0050}\\ \overline{a}=-6\times {10}^3{\text{m/s}}^2\end{array}$

Conclusion:

Thus, the acceleration of baseball is $-6\times {10}^3{\text{m/s}}^2$ .

(b)

To determine

The magnitude and direction of the force acting on the baseball.

Answer to Problem 77A

The force exerted on baseball is $-8.7\times {10}^2\text{N}$ opposite to the direction of velocity.

Explanation of Solution

Given:

The initial velocity of the baseball is $v_{\text{i}}=0\text{m/s}$ .

The final velocity of the baseball is $v_{\text{f}}=30\text{m/s}$ .

The time interval for speed up is $\Delta t=0.0050\text{s}$ .

The mass of baseball is $m=0.145\text{kg}$ .

Formula used:

Newton’s Second Law:

The object’s acceleration is equivalent to the sum of the forces acting on the object divided by the object’s mass. That is,

  $\begin{array}{l}a=\frac{F_{\text{net}}}{m}\\ F_{\text{net}}=ma\end{array}$

Here, $m$ is the mass of the object and $a$ is the acceleration of the object.

Calculation:

Refer to part (a).

The acceleration of baseball is $-6\times {10}^3{\text{m/s}}^2$ .

Find the force exerted on the baseball as follows:

  $\begin{array}{c}{F}_{\text{net}}=ma\\ =0.145\text{kg}\times \left(-6\times {10}^3\right){\text{m/s}}^2\\ =-8.7\times {10}^2\text{kg}\cdot {\text{m/s}}^2\times \left(\frac{1\text{N}}{1\text{kg}\cdot {\text{m/s}}^2}\right)\\ =-8.7\times {10}^2\text{N}\end{array}$

The negative sign of the force indicates that the direction of the force is opposite to the direction of the velocity of the baseball.

Conclusion:

Thus, the force exerted on the baseball is $-8.7\times {10}^2\text{N}$ in the direction opposite to the direction of the velocity of the ball.

(c)

To determine

Find the magnitude and direction of the force exerted on the player who caught the ball.

Answer to Problem 77A

The force exerted on the player while catching the ball is $8.7\times {10}^2\text{N}$ in the direction of the velocity of the ball.

Explanation of Solution

Given:

The initial velocity of the baseball is $v_{\text{i}}=0\text{m/s}$ .

The final velocity of the baseball is $v_{\text{f}}=30\text{m/s}$ .

The time interval for speed up is $\Delta t=0.0050\text{s}$ .

The mass of baseball is $m=0.145\text{kg}$ .

Formula used:

Newton’s Third Law of Motion:

According to Newton’s Third Law, for every action, there is an equal and opposite reaction. When a body $A$ applies a force on the body $B$ then the body $B$ exerts an equal and opposite force on the body $A$ .

  $F_{\text{A on B}}=-F_{\text{B on A}}$

Calculation:

Refer to part (b).

The force exerted on the baseball is $F_{\text{baseball}}=-8.7\times {10}^2\text{N}$ .

Consider the force exerted on the player while catching the ball is denoted by $F_{\text{player}}$ .

Apply Newton’s Third Law of Motion,

  $\begin{array}{l}{F}_{\text{player}}=-F_{\text{baseball}}\\ F_{\text{player}}=-\left(-8.7\times {10}^2\text{N}\right)\\ F_{\text{player}}=+8.7\times {10}^2\text{N}\end{array}$

The positive sign of the force indicates that the direction of the force is the same as the direction of the velocity of the baseball.

Conclusion:

Thus, the force exerted on the player while catching the ball is $8.7\times {10}^2\text{N}$ in the direction ofthe velocity of the ball.