A ball of mass $0.2\ kg$ rests on a vertical post of height $5\ m$. A bullet of mass $0.01\ kg$, travelling with a velocity $V\ m/s$ in a horizontal direction, hits the centre of the ball. After the collision, the ball and bullet travel independently. The ball hits the ground at a distance of $20\ m$ and the bullet at a distance of $100\ m$ from the foot of the post. The initial velocity $V$ of the bullet is [image:6_image36.emf]

Question

Dr. Abubacker · Accepted Answer

(d) $R = u\sqrt{\frac{2h}{g}} \Rightarrow 20 = V_{1}\sqrt{\frac{2 	imes 5}{10}}$ and $100 = V_{2}\sqrt{\frac{2 	imes 5}{10}}$
$\Rightarrow V_{1} = 20\ m/s,\ V_{2} = 100\ m/s$
Applying momentum conservation just before and just after the collision $(0.01)(V) = (0.2)(20) + (0.01)(100)$
$V = 500\ m/s$

Explanation

Reviewed by Dr. Abubacker

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