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 $0.81 \mathrm {~ms} \text {, }$ what is the average force exerted on the ball by the club?
A)6.3 kN
B)7.1 kN
C)5.5 kN
D)4.9 kN

Question

Quizplus · Accepted Answer

The average force exerted on the ball by the club 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 0.81 ms, 0.81 \mathrm {~ms} \text {, }0.81 ms,

A Golf Ball of Mass 0 $0.81 \mathrm {~ms} \text {, }$