Impulse & Momentum Conservation question from NEET 2000 (AIPMT)

Answer: (a)

1. The path makes $60^\circ$ with the wall, so the velocity component normal to the wall is $v\sin 60^\circ = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3}$ m/s.
2. The component along the wall is unchanged; the normal component reverses.
3. $\Delta p = 2m v\sin 60^\circ = 2 \times 3 \times 5\sqrt{3} = 30\sqrt{3}$ kg m/s
4. $F = \frac{\Delta p}{\Delta t} = \frac{30\sqrt{3}}{0.2} = 150\sqrt{3}$ N
5. So the answer is (a) $150\sqrt{3}$ N.