Question

A golf club exerts an average force of 500N on a 0.1kg golf ball and the contact time is 0.01s.

What is the change in velocity of the golf ball?​

Answer 1

`50\; {\rm m\cdot s^(-1)}`, assuming that the force on the golf ball was constant.

Step-by-step explanation:

Assume that the force the golf club exerted on the golf ball a constant force of magnitude
`F`. Let
`\Delta t` denote the duration the club was in contact with the golf ball. The impulse of this force would be:

`J = F\, \Delta t`.

Let
`\Delta p` denote the change in the momentum of this golf ball. By the impulse-momentum theorem, this
`\Delta p\!` would be equal to the impulse on the golf ball.

`\Delta p = J = F\, \Delta t`.

The momentum
`p` of this golf ball is the scalar product between the mass
`m` of this golf ball and the velocity
`v` of this golf ball. The mass of this golf ball stays the same. Thus, when the momentum of this golf ball changes by
`\Delta p`, the velocity of this golf ball would change by
`(\Delta p) / (m)`.

The change in the velocity of this golf ball would thus be:

`\begin{aligned}\Delta v &= (\Delta p)/(m) & \genfrac{}{}{0em}{}{(\text{change in momentum})}{}\\ &= (J)/(m) & \genfrac{}{}{0em}{}{(\text{impulse on the golf ball})}{}\\ &= (F\, \Delta t)/(m) \\ &= \frac{500\; {\rm N} * 0.01\; {\rm s}}{0.1\; {\rm kg}} \\ &\approx 50\; {\rm m\cdot s^(-1)}\end{aligned}`.

(Note that
`1\; {\rm N} = 1\; {\rm kg \cdot m^(2)\cdot s^(-2)}`.)