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 0.0013 s. If the ball has a mass of 55 g and leaves the head of the club with a speed of 1.2X10^2 ft/s, find the average force exerted on the ball by the club.

To find the average force exerted on the ball by the club, we can use Newton's second law of motion, which states that force is equal to mass multiplied by acceleration. In this case, the acceleration is the change in velocity of the ball.

First, we need to convert the speed of the ball from feet per second to meters per second since the mass of the ball is given in grams.

1 foot = 0.3048 meters
1 second = 1000 milliseconds

So, the speed of the ball is:
1.2 x 10^2 ft/s =
1.2 x 10^2 ft/s x 0.3048 m/ft x 1 s/1000 ms =
36.576 m/s

Now, we can calculate the acceleration using the equation:
Acceleration = (final velocity - initial velocity) / time

The initial velocity of the ball is 0 m/s since it was initially at rest. The final velocity is 36.576 m/s. The time of contact between the club and the ball is given as 0.0013 s.

Acceleration = (36.576 m/s - 0 m/s) / 0.0013 s =
28135.38 m/s^2

Next, we can calculate the force using Newton's second law:
Force = mass x acceleration

The mass of the ball is given as 55 g, which is equivalent to 0.055 kg.

Force = 0.055 kg x 28135.38 m/s^2 =
1547.44 N

Therefore, the average force exerted on the ball by the club is approximately 1547.44 Newtons.

m•v=F•Δt

F= m•v/Δt