Center of Mass, Momentum & Collision

JEE Main level, numerical answer, medium

92 of 130

Question 92 of 130

A pendulum of length 2m2m consists of a wooden bob of mass 50g50g. A bullet of mass 75g75g is fired towards the stationary bob with a speed vv. The bullet emerges out of the bob with a speed v3\dfrac{v}{3} and the bob just completes the vertical circle. The value of vv is _____ ms−1m{s}^{-1}
(if g=10ms−2g=10m{s}^{-2}).

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Answer: 10

For the bob to completer the vertical circle, vb=5gl{v}_{b}=\sqrt{5gl}.
From the conservation of momentum,
m1v=m2vb+m1v3⇒75v=505gl+75(v3){m}_{1}v={m}_{2}{v}_{b}+{m}_{1}\dfrac{v}{3} \Rightarrow 75v=50\sqrt{5gl}+75\left(\dfrac{v}{3}\right)
⇒75(2v3)=505×10×2\Rightarrow 75\left(\dfrac{2v}{3}\right)=50\sqrt{5\times 10\times 2}
⇒75(2v3)=50×10\Rightarrow 75\left(\dfrac{2v}{3}\right)=50\times 10
⇒v=500×3150=50050=10ms−1\Rightarrow v=\dfrac{500\times 3}{150}=\dfrac{500}{50}=10m{s}^{-1}

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