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 horizontal frictionless table. The K.E. of the block will be

2 mgl

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mgl/2

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mgl

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0

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Solution

The correct option is C mgl
P.E. of bob at point A = mgl
This amount of energy will be converted into kinetic energy
$∴$ K.E. of bob at point B = mgl

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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 horizontal frictionless table. The K.E. of the block will be

The correct option is C mglP.E. of bob at point A = mgl This amount of energy will be converted into kinetic energy ∴ K.E. of bob at point B = mgl

The correct option is C mgl
P.E. of bob at point A = mgl
This amount of energy will be converted into kinetic energy
$∴$ K.E. of bob at point B = mgl