A ball of mass $200\ \text{g}$ rests on a vertical post of height $20\ \text{m}$. A bullet of mass $10\ \text{g}$, travelling in horizontal direction, hits the centre of the ball. After collision both travel independently. The ball hits the ground at a distance $30\ \text{m}$ and the bullet at a distance of $120\ \text{m}$ from the foot of the post. The value of initial velocity of the bullet will be (if $g=10\ \text{m s}^{-2}$)
Answer: (D) $360\ \text{m s}^{-1}$
Both fall $20\ \text{m}$: $t=\sqrt{\dfrac{2\times20}{10}}=2\ \text{s}$.
Ball: $v_1=\dfrac{30}{2}=15\ \text{m s}^{-1}$; bullet: $v_2=\dfrac{120}{2}=60\ \text{m s}^{-1}$.
Momentum conservation: $0.01u=0.2\times15+0.01\times60\Rightarrow u=300+60=360\ \text{m s}^{-1}$.
Solution by Sreeraj P, M.Sc Physics