A bullet with mass 200gm and velocity vb m/s hits a ballistic pendulum bob of mass 800gm and rises along with pendulum to a height h=0.2m. Determine the value of vb. Take g=10m/s2.
A
100m/s
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B
10m/s
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C
8m/s
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D
6m/s
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Solution
The correct option is B10m/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 ⇒(0.2vb)2=2×10×0.2 ∴vb=√2×10×0.20.04=10m/s