A golf ball of mass 0.050 kg is at rest on the tee. Just after being struck, it has a velocity of 102 m/s. If the club and ball were in contact for 0.81 ms, what is the average force exerted on the ball by the club?
A) 5.5 kN
B) 4.9 kN
C) 7.1 kN
D) 6.3 kN

Question

A golf ball of mass  0.050 kg is at rest on the tee. Just after being struck, it has a velocity of  102 m/s. If the club and ball were in contact for  0.81 ms, what is the average force exerted on the ball by the club?
A)  5.5 kN
B)  4.9 kN
C)  7.1 kN
D)  6.3 kN

Quizplus · Accepted Answer

The average force exerted on the ball can be calculated using the formula $F = \frac{\Delta p}{\Delta t}$, where $\Delta p$ is the change in momentum and $\Delta t$ is the time interval. The change in momentum $\Delta p = m \Delta v = 0.050 \times 102 = 5.1 \, \text{kg m/s}$. The time interval $\Delta t = 0.81 \, \text{ms} = 0.81 \times 10^{-3} \, \text{s}$. Thus, $F = \frac{5.1}{0.81 \times 10^{-3}} = 6296.3 \, \text{N}$, which rounds to $6.3 \, \text{kN}$.

A Golf Ball of Mass 0