Question

Consider a golf club hitting a golf ball that results in the following graph of force versus time on a 45 gram golf ball. If the final velocity of the ball has a magnitude of 41 m/s, determine the value of Fmax

Answer 1

The question is missing the graph. So, the graph is attached below.

Answer:

The value of $F_{max}$ is 7380 N.

Explanation:

Given:

Mass of the ball is, $m=45\ g=0.045\ kg$

Initial velocity of the ball is, $u=0\ m/s(Assuming)$

Final velocity of the ball is, $v=41\ m/s$

From the graph,

Time interval for which the force acts, $\Delta t=0.5\ ms=0.5\times 10^{-3}\ s$

Height of the triangle is equal to the maximum force acting on the ball = $F_{max}$

Now, we know that, impulse acting on the ball is equal to the area under the curve of force and time graph.

So, impulse is equal to the area of the triangle and is given as:

Impulse = $\frac{1}{2}\times base\times height$

Here, base is time interval $\Delta t$ and height is $F_{max}$.

Impulse = $\frac{1}{2}\times (0.5\times 10^{-3})\times F_{max}$

Impulse = $0.25\times 10^{-3}F_{max}$

Now, we also know that, impulse is equal to the change in momentum of a body.

Therefore, change in momentum (Δp) is given as:

$\Delta p=m(v-u)\\\Delta p = 0.045(41-0)=1.845\ kgms^{-1}$

Now, change in momentum is equal to impulse acting on the ball. Thus,

$0.25\times 10^{-3}F_{max}=1.845\\\\F_{max}=\frac{1.845}{0.25\times 10^{-3}}\\\\F_{max}=7380\ N$

Therefore, the value of $F_{max}$ is 7380 N.