Force of a Baseball Swing. A baseball has mass 0.145 kg.

(a) If the velocity of a pitched ball has a magnitude of

and the batted ball’s velocity is in the opposite direction,

find the magnitude of the change in momentum of the ball and

of the impulse applied to it by the bat. (b) If the ball remains in

contact with the bat for 2.00 ms, find the magnitude of the average

force applied by the bat.

The momentum of a particle: The momentum$\overrightarrow{\mathbf{P}}$ of the particle’s mass mand velocity $\overrightarrow{\mathbf{V}}$.

$\overrightarrow{\mathbf{P}}\mathbf{=}\mathit{m}\overrightarrow{\mathbf{V}}$

Newton’s second law says that the net force on a particle is equal to the rate of change of the particle’s momentum.

$\mathbf{\sum }_{\mathbf{}}\mathbf{F}\mathbf{=}\frac{\mathbf{dp}}{\mathbf{dt}}$

Impulse and momentum theorem

$\overrightarrow{\mathbf{J}}\mathbf{=}{\overrightarrow{\mathbf{P}}}_{\mathbf{2}}\mathbf{-}\overrightarrow{{\mathbf{P}}_{\mathbf{1}}}$

Given in the question

Mass of the baseball,$m_b=0.145\mathrm{kg}$

Considering the direction of the batted ball as the positive direction

The velocity of the pitched ball $v_i=-45m/s$

The velocity of the batted ball$v_f=55\mathrm{m}/\mathrm{s}$

$∆t=2ms$

Initial momentum

$$\begin{gathered} p_1=m{v}_i \\ =\left(0.145\mathrm{kg}\right)\left(-45\mathrm{m}/\mathrm{s}\right) \\ =-6.525\mathrm{kg}.\mathrm{m}/\mathrm{s} \end{gathered}$$

Final momentum

$$\begin{gathered} p_2=m{v}_f \\ =\left(0.145\mathrm{kg}\right)\left(55\mathrm{m}/\mathrm{s}\right) \\ =7.975\mathrm{kg}.\mathrm{m}/\mathrm{s} \end{gathered}$$

Change in momentum

$$\begin{gathered} ∆\overrightarrow{P}=\overrightarrow{P_2}-\overrightarrow{P_1} \\ =\left(7.975\mathrm{kg}.\mathrm{m}/\mathrm{s}\right)-\left(-6.525\mathrm{kg}.\mathrm{m}/\mathrm{s}\right) \\ =14.5\mathrm{kg}.\mathrm{m}/\mathrm{s} \end{gathered}$$

The net change in the momentum of the ball is 14.5 kg.m/s.

From impulse-momentum theorem.

$$\begin{gathered} \overrightarrow{J}=\overrightarrow{P_2}-\overrightarrow{P_1} \\ =\left(7.975\mathrm{kg}.\mathrm{m}/\mathrm{s}\right)-\left(-6.525\mathrm{kg}.\mathrm{m}/\mathrm{s}\right) \\ =14.5\mathrm{kg}.\mathrm{m}/\mathrm{s} \end{gathered}$$

The impulse applied by the bat on the ball is 14.5 kg.m/s .

From Newton’s second law

$$\begin{gathered} \sum _{}F=\frac{dp}{dt} \\ \sum _{}F=\frac{p_2-p_1}{∆t} \\ =\frac{\left(7.975\right)-\left(-6.525\right)}{2\times {10}^{-3}} \\ =7250\mathrm{N} \end{gathered}$$

The average force applied by the bat on the ball is 7250 N.