Question

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

• A. 2 mgl
• B. $\frac{mgl}{2}$
• C. mgl
• D. 0

Answer 1

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

and as the collision between bob and block (of same mass) is elastic, after collision bob will come to rest and total Kinetic energy will be transferred to block. So kinetic energy of block = mgl