Q. 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) $mgl/2$
C) $mgl$
D) 0
Explanation

(c) P.E. of bob at point $A = mgl$
This amount of energy will be converted into kinetic energy
$\therefore$ K.E. of bob at point $B = mgl$
And 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 $= mgl$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.