91Ó°ÊÓ

The momentum of a particle: The momentum $\stackrel{\mathbf{→}}{\mathbf{P}}$of a particle is a vector quantity equal to the product of the particle’s mass mand velocity $\stackrel{\mathbf{→}}{\mathbf{V}}$.

$\stackrel{\mathbf{→}}{\mathbf{P}}\mathbf{=}\mathit{m}\stackrel{\mathbf{â‡\P Ä}}{\mathbf{v}}$

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

$\underset{\mathbf{}}{\mathbf{∑}}\mathit{F}\mathbf{=}\frac{\mathbf{d}\mathbf{p}}{\mathbf{d}\mathbf{t}}$

Given in the question

$$\begin{gathered} m=0.0450\mathrm{kg} \\ {v}_2=25\mathrm{m}/\mathrm{s} \\ ∆t=2.00\mathrm{ms} \end{gathered}$$

Since the ball is initially at rest so initial momentum is zero.

$p_1=0$

Final momentum

$$\begin{gathered} p_2=m{v}_2 \\ =\left(0.045\right)\left(25\right) \\ =1.125kg.\mathrm{m}/\mathrm{s} \end{gathered}$$

From Newton’s second law

role="math" localid="1665122942949" $$\begin{gathered} \underset{}{∑}F=\frac{dp}{dt} \\ \underset{}{∑}F=\frac{p_2-p_1}{∆t}=\frac{1.125-0}{2×{10}^{-3}}=562.5\mathrm{N} \end{gathered}$$

The average force acting on the ball is

The weight of the ball can be given as

$$\begin{gathered} W=mg \\ W=0.045×9.8 \\ W=0.441N \end{gathered}$$

Since the weight of the ball is very small in comparison to average forces acting on the ball, therefore, we can neglect its effect during the time of contact.