Answer:
[tex]50\; {\rm m\cdot s^{-1}}[/tex], assuming that the force on the golf ball was constant.
Explanation:
Assume that the force the golf club exerted on the golf ball a constant force of magnitude [tex]F[/tex]. Let [tex]\Delta t[/tex] denote the duration the club was in contact with the golf ball. The impulse of this force would be:
[tex]J = F\, \Delta t[/tex].
Let [tex]\Delta p[/tex] denote the change in the momentum of this golf ball. By the impulse-momentum theorem, this [tex]\Delta p\![/tex] would be equal to the impulse on the golf ball.
[tex]\Delta p = J = F\, \Delta t[/tex].
The momentum [tex]p[/tex] of this golf ball is the scalar product between the mass [tex]m[/tex] of this golf ball and the velocity [tex]v[/tex] of this golf ball. The mass of this golf ball stays the same. Thus, when the momentum of this golf ball changes by [tex]\Delta p[/tex], the velocity of this golf ball would change by [tex](\Delta p) / (m)[/tex].
The change in the velocity of this golf ball would thus be:
[tex]\begin{aligned}\Delta v &= \frac{\Delta p}{m} & \genfrac{}{}{0em}{}{(\text{change in momentum})}{}\\ &= \frac{J}{m} & \genfrac{}{}{0em}{}{(\text{impulse on the golf ball})}{}\\ &= \frac{F\, \Delta t}{m} \\ &= \frac{500\; {\rm N} \times 0.01\; {\rm s}}{0.1\; {\rm kg}} \\ &\approx 50\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].
(Note that [tex]1\; {\rm N} = 1\; {\rm kg \cdot m^{2}\cdot s^{-2}}[/tex].)
