Question 89 of 130

A ball of mass $200g$ rests on a vertical post of height $20m$. A bullet of mass $10g$, travelling in horizontal direction, hits the centre of the ball. After collision both travels independently. The ball hits the ground at a distance $30m$ and the bullet at a distance of $120m$ from the foot of the post. The value of initial velocity of the bullet will be (if $g=10m{s}^{-2}$) :

A$120m{s}^{-1}$B$60m{s}^{-1}$C$400m{s}^{-1}$D$360m{s}^{-1}$

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Answer: (D) $360m{s}^{-1}$

Let initial velocity of bullet be $u$.
Given, mass of bullet $m=10g$ and mass of ball $M=200g$,
Initially ball is at height $5m$ and at rest, the only acceleration acting is due to gravity.
Time of flight of each ball and bullet is $t=\sqrt{\dfrac{2h}{g}}$.
Now, velocity of ball and bullet as ${v}_{1}=\dfrac{30}{\sqrt{\dfrac{2h}{g}}}and{v}_{2}=\dfrac{120}{\sqrt{\dfrac{2h}{g}}}$
As the collision is elastic, so applying conservation of linear momentum.
$mu+M\left(0\right)=M{v}_{1}+m{v}_{2}$
$\Rightarrow \left(0.01\right)u=\left(0.2\right)\dfrac{30\sqrt{g}}{\sqrt{2h}}+\left(0.01\right)\dfrac{120\sqrt{g}}{\sqrt{2h}}$
$\Rightarrow u=300+60=360m{s}^{-1}$