Answer:
The time the ball is in contact with the club is approximately 0.011063 seconds
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;
[tex]F_{net} = \dfrac{\Delta P}{\Delta t} = \dfrac{m \times \Delta v}{\Delta t}[/tex]
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;
[tex]F_{net} = F =299.4 = \dfrac{0.061 \times 54.3}{\Delta t}[/tex]
[tex]\therefore \Delta t = \dfrac{0.061 \times 54.3}{299.4} \approx 0.011063[/tex]
The time the ball is in contact with the club = Δt ≈ 0.011063 seconds.
