Solution:

The component of velocity perpendicular to the wall is \(v_n = v sin 60^\circ = \frac{\sqrt{3}}{2} \text{m/s}\). The change in momentum is \(\Delta p = 2 m v_n = 2 \times 1 \times \frac{\sqrt{3}}{2} = \sqrt{3} \text{kg m/s}\). The force is \(F = \frac{\Delta p}{\Delta t} = \frac{\sqrt{3}}{0.1} = 10\sqrt{3} \text{N}\).