Question

A high-speed photograph of a club hitting a golf ball is shown in the figure below. The club was in contact with a ball, initially at rest, for about 0.0052 s. If the ball has a mass of 55 g and leaves the head of the club with a speed of 1.6 102 ft/s, find the average force exerted on the ball by the club.

Answer 1

Given:

The time duration for which the club was in contact with the ball, t=0.0052 s

The mass of the ball, m=55 g=55×10⁻³ kg

The initial speed of the ball, u=0 m/s

The final speed with which the ball leaves the club, v=1.6×10² ft/s=48.77 m/s

To find:

The force exerted on the ball, F.

Step-by-step explanation:

The acceleration of an object is the time rate of change of velocity.

Thus the acceleration of the ball is given by,

`a=(v-u)/(t)`

On substituting the known values,

`\begin{gathered} a=(48.77-0)/(0.0052) \\ =9378.8\text{ m/s}^2 \end{gathered}`

From Newton's second law of motion,

`F=ma`

On substituting the known values,

`\begin{gathered} F=55*10^(-3)*9378.8 \\ =515.8\text{ N} \end{gathered}`

Final answer:

Thus the average force exerted on the ball is 515.8 N