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
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a
2 mgl
b
mgl/2
c
mgl
d
0
answer is C.
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Detailed Solution
P.E. of bob at point A = mglThis amount of energy will be converted into kinetic energy ∴ K.E. of bob at point B = mgland as the collision between bob and block (of same mass) is elastic so after collision bob will come to rest and total Kinetic energy will be transferred to block. So kinetic energy of block = mgl