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Question

The string of a pendulum is attached to a bob and is released from its initial rest position. A fixed peg is present in the path of motion of the pendulum. What are the velocities of the bob at positions B and C respectively? The distances are as shown in the figure.


A
3.16 m/s and 7.07 m/s
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B
7.07 m/s and 3.16 m/s
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C
5.34 m/s and 10.78 m/s
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D
7.07 m/s and 6.32 m/s
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Solution

The correct option is D 7.07 m/s and 6.32 m/s
As the bob is released, it swings downwards, Hence its potential energy is converted completely into kinetic energy at point B. Taking reference level at point B

KEB=PEA12mvB2=mghvB2=2ghvB=2gh=2×10×2.5vB=7.07 m/s
Therefore velocity of the bob at point B is 7.07 m/s.

From the point B to C, only the thread below the peg can swing. Therefore, applying conservation of energy at points B and C

KEC + PEC = KEB=PEA12mvC2 + mghC=mgh12mvC2=mg(hhC)vC=2g(hhC)vC=6.32 m/s

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