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Standard XII

Physics

Law of Conservation of Mechanical Energy

The length of...

Question

The length of a ballistic pendulum is $1 m$ and mass of its block is $0.98 k g$ . A bullet of mass $20 g$ strikes the block along horizontal direction and gets embedded in the block. If block + bullet completes vertical circle of radius $1 m$ , the striking velocity of bullet is:

280 m / s

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350 m / s

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420 m / s

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490 m / s

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Solution

The correct option is C $350 m / s$
Using conservation of momentum:
Initial momentum of bullet = Final momentum of bullet and pendulum
$0.02 \times v_{i} = (0.02 + 0.98) v_{f}$
So, $v_{i} = 50 \times v_{f}$
Kinetic Energy after the collision=potential energy
$\frac{1}{2} \times 1 \times v_{f}^{2} = 1 \times g \times 2$
$v_{f} = 2 \sqrt{g}$
Thus, $v_{i} =$ $50 \times 2 \sqrt{g}$

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The length of a ballistic pendulum is 1 m and mass of its block is 0.98 kg. A bullet of mass 20 g strikes the block along horizontal direction and gets embedded in the block. If block + bullet completes vertical circle of radius 1 m, the striking velocity of bullet is:

The correct option is C 350 m/sUsing conservation of momentum:Initial momentum of bullet = Final momentum of bullet and pendulum0.02×vi=(0.02+0.98)vfSo, vi=50×vfKinetic Energy after the collision=potential energy12×1×v2f=1×g×2vf=2√gThus, vi=50×2√g

The length of a ballistic pendulum is $1 m$ and mass of its block is $0.98 k g$ . A bullet of mass $20 g$ strikes the block along horizontal direction and gets embedded in the block. If block + bullet completes vertical circle of radius $1 m$ , the striking velocity of bullet is: