Solution:

The angle with the normal is \(90^\circ - 60^\circ = 30^\circ\). The change in momentum perpendicular to the wall is \(\Delta p = 2mv\cos(30^\circ) = 2(1)(1)\left(\frac{\sqrt{3}}{2}\right) = \sqrt{3}\text{ kg m s}^{-1}\). Average force \(F = \frac{\Delta p}{\Delta t} = \frac{\sqrt{3}}{0.1} = 10\sqrt{3}\text{ N}\).