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Question

A pendulum bob of mass 250 gm is raised to a height of 4 cm and then released. At the bottom-most position of its swing, an insect of mass 25 gm sits on the bob. Find the maximum height to which the combined mass will rise. [Take g=10 m/s2]

A
5.6 cm
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B
2.7 cm
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C
3.3 cm
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D
1.5 cm
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Solution

The correct option is C 3.3 cm
The velocity of the bob at the bottom of its swing after falling through a vertical height h=4 cm,
v1=2gh=2×10×0.04
v1=0.8 m/s
Initial velocity of the insect is v2=0 m/s
Let v be the velocity of the combined mass of (insect+ bob) system.
m1 mass of the bob
m2 mass of the insect
Applying the conservation of momentum on system of (bob+insect) when insect sits on the bob
Pi=Pf
m1v1+m2v2=(m1+m2)v
(0.25×0.8)+(0.025×0)=(0.25+0.025)v
v=0.25×0.80.275=(0.909×0.8) m/s

Maximum height (final velocity=0) of the combined mass is given by,
v2=2ghmax
hmax=v22g=(0.909×0.8)22×10
hmax=0.6620=0.033 m=3.3 cm

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