Question

In a golf ball game, a person hits the golf ball with a club. The club is in contact with the ball, which is initially at rest, for 2.45 ms. The ball has a mass of 0.0550 kg and leaves the club with a speed of 1.80 X 102 m/s. What is the average force exerted on the ball by the club

Answer 1

F=4040.81 N

Step-by-step explanation:

Given that

Time ,t= 2.45 ms

Mass ,m= 0.055 kg

v= 1.8 x 10² m/s = 180 m/s

We know that rate of change in the linear momentum is known as force.

Momentum P = m v

`F=(dP)/(dt)`

Therefore force F

`F=(\Delta P)/(\Delta t)`

`F=(0.055* 180)/(2.45* 10^(-3))\ N`

F=4040.81 N

Therefore force on the ball will be 4040.81 N