Question

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

• A. 2 mgl
• B. mgl/2
• C. mgl
• D. 0

Answer 1

The correct option is C mgl

Initial potential energy of the bob when its string is horizontal $=mgl$. This is also the kinetic energy of bob just before it strikes the block. Since mass of the block is same as that of the bob and $u=0$ this becomes the case of exchange of velocities, i.e. after the collision, the velocity of bob becomes zero and block starts moving with the velocity of the both just before the collision. Thus, kinetic energy of the bob just before the collision = kinetic energy of the bob after the collision $=mgl$