91影视

The momentum of a particle: The momentum$\stackrel{\mathbf{鈫�}}{\mathbf{P}}$ of the particle鈥檚 mass mand velocity $\stackrel{\mathbf{鈫�}}{\mathbf{V}}$.

$\stackrel{\mathbf{鈫�}}{\mathbf{P}}\mathbf{=}\mathit{m}\stackrel{\mathbf{鈫�}}{\mathbf{V}}$

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

$\underset{\mathbf{}}{\mathbf{鈭�}}\mathbf{F}\mathbf{=}\frac{\mathbf{dp}}{\mathbf{dt}}$

Impulse and momentum theorem

$\stackrel{\mathbf{鈫�}}{\mathbf{J}}\mathbf{=}{\stackrel{\mathbf{鈫�}}{\mathbf{P}}}_{\mathbf{2}}\mathbf{-}\stackrel{\mathbf{鈫�}}{{\mathbf{P}}_{\mathbf{1}}}$

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$v_f=55\mathrm{m}/\mathrm{s}$

$鈭�t=2ms$

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

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

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

From impulse-momentum theorem.

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

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

From Newton鈥檚 second law

$$\begin{gathered} \underset{}{鈭�}F=\frac{dp}{dt} \\ \underset{}{鈭�}F=\frac{p_2-p_1}{鈭�t} \\ =\frac{\left(7.975\right)-\left(-6.525\right)}{2脳{10}^{-3}} \\ =7250\mathrm{N} \end{gathered}$$

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