Question

Force of a Golf Swing. A 0.0450-kg golf ball initially at rest is given a speed of 25 m/s when a club strikes it. If the club and ball are in contact for 2.00 ms, what average force acts on the ball? Is the effect of the ball’s weight during the time of contact significant? Why or why not?

Short answer

The average force acting on the ball is 562.5 N.

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.

Step-by-step solution

Formula used

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

$\overrightarrow{\mathbf{P}}\mathbf{=}\mathit{m}\stackrel{\mathbf{\rightharpoonup }}{\mathbf{v}}$

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

$\mathbf{\sum }_{\mathbf{}}\mathit{F}\mathbf{=}\frac{\mathbf{d}\mathbf{p}}{\mathbf{d}\mathbf{t}}$

Finding the average force.

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" $$\begin{gathered} \sum _{}F=\frac{dp}{dt} \\ \sum _{}F=\frac{p_2-p_1}{∆t}=\frac{1.125-0}{2\times {10}^{-3}}=562.5\mathrm{N} \end{gathered}$$

The average force acting on the ball is

The weight of the ball can be given as

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

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.