The bob of a simple pendulum of mass m and length ; is droped from the horizontal position strikes a block of the same mass elastically placed on a horizontal frictionless table. The kinetic energy of the block will be
A) zero
B) \[mgl\]
C) \[\frac{mgl}{2}\]
D) \[2mgl\]
Correct Answer:
B
Solution :
The given condition can be shown as Potential energy of bob at point \[A=mgl\] The total energy is converted into kinetic energy. \[\therefore \] Kinetic energy of bob at point B \[=mgl\] As collision between bob and block is elastic so after collision bob will come to rest and total kinetic energy will be transferred to block. So, kinetic energy of the block = mgl