The bob of a simple pendulum (mass $m$ and length $l$ ) dropped from a horizontal position strikes a block of the same mass elastically placed on a horizontally frictionless table. The $K . E$ . of the block will be:

2 m g l

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m g l / 2

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m g l

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Solution

The correct option is C $m g l$
When the bob reaches its lowest point it has kinetic energy which is equal to the work done by gravity while coming down.
$K . E = m \times g \times l$
When it strikes the block of same mass elastically the law of conservation of momentum gives that the velocity will be exchanged
$m_{1} \times u_{1} + m_{2} \times u_{2} = m_{1} \times v_{1} + m_{2} \times v_{2}$
Now, as their masses are same and also velocity is exchanged , the kinetic energy of the block will be:
$K E = m \times g \times l$

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The bob of a simple pendulum (mass m and length l) dropped from a horizontal position strikes a block of the same mass elastically placed on a horizontally frictionless table. The K.E. of the block will be:

The bob of a simple pendulum (mass $m$ and length $l$ ) dropped from a horizontal position strikes a block of the same mass elastically placed on a horizontally frictionless table. The $K . E$ . of the block will be: