Centre Of Mass Ques 36
- A ball of mass $0.2 kg$ rests on a $vm / s$ 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
(2011)
(a) $250 m / s$
(b) $250 \sqrt{2} m / s$
(c) $400 m / s$
(d) $500 m / s$
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Answer:
Correct Answer: 36.(d)
Solution:
Formula:
Motion of Centre of Mass and Conservation of Momentum:
$$ \begin{array}{rlrl} R & =u \sqrt{\frac{2 h}{g}} \Rightarrow 20=v _1 \sqrt{\frac{2 \times 5}{10}} \\ \text { and } & & 100=v _2 \sqrt{\frac{2 \times 5}{10}} \\ \Rightarrow \quad v _1 & =20 m / s, v _2=100 m / s . \end{array} $$
Applying momentum conservation just before and just after the collision
$(0.01)(v)=(0.2)(20)+(0.01)(100) v=500 m / s$
BETA
