Question

A 0.061 kg golf ball is struck by a golf club with a force of 299.4 N. The change in velocity of the ball is 54.3 m/s. How much time is the ball in contact with the club?​

Answer 1

The time the ball is in contact with the club is approximately 0.011063 seconds

Step-by-step explanation:

The question is with regards to Newton's second law of motion which states that a force is equal to the rate of change of momentum produced

The given parameters are;

The mass of the ball, m = 0.061 kg

The force with which the ball is struck with the golf club, F = 299.4 N

The change in velocity of the ball, Δv = 54.3 m/s

By Newton's second law, we have;

`F_(net) = (\Delta P)/(\Delta t) = (m * \Delta v)/(\Delta t)`

Where;

Δt = The time it takes the momentum of the object to change = The time the ball is in contact with the club

Substituting the known values, we get;

`F_(net) = F =299.4 = (0.061 * 54.3)/(\Delta t)`

`\therefore \Delta t = (0.061 * 54.3)/(299.4) \approx 0.011063`

The time the ball is in contact with the club = Δt ≈ 0.011063 seconds.