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

Question

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)  (mgl/2)  (D)  2mgl

Tardigrade Admin · Accepted Answer

The given condition can be shown as
Potential energy of bob at point

A = m g l

The total energy is converted into kinetic energy.

∴

Kinetic energy of bob at point B

= m g l

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

Q. 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

zero

$m g l$

$\frac{m g l}{2}$

$2 m g l$

Q. 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

zero

mgl

2mgl​

2mgl

$m g l$

$\frac{m g l}{2}$

$2 m g l$