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Question

# A bullet with mass 200 gm and velocity vb m/s hits a ballistic pendulum bob of mass 800 gm and rises along with pendulum to a height h=0.2 m. Determine the value of vb. Take g=10 m/s2.

A
100 m/s
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B
10 m/s
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C
8 m/s
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D
6 m/s
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Solution

## The correct option is B 10 m/s Since, net external force on the system of (pendulum bob+ bullet) is zero in horizontal direction. vp→initial velocity of pendulum bob vb→initial velocity of bullet bfore collision vs is the velocity of system of (bullet +pendulum bob) just after collision Applying conservation of linear momentum in horizontal direction: Pi=Pf mbvb+mpvp=(mb+mp)vs ⇒vs=mbvb+mpvpmb+mp ∴vs=0.2×vb+01=0.2vb ....(i) Applying mechanical energy conservation from position, just after collision till system reaches maximum height i.e (vs=0): Loss in K.E of system=Gain in P.E of system 12(mb+mp)v2s=(mb+mp)gh ...(ii) ⇒v2s=2gh ⇒(0.2 vb)2=2×10×0.2 ∴vb=√2×10×0.20.04=10 m/s

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