question_answer

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