Question

two stones one twice the mass of the other are dropped from a cliff. Just before hitting the ground what is the Kinetic energy of the heavy stone compared to light one?

Answer 1

Velocity of both stone is same
So the body of twice mass has double KE .

Answer 2

Given:

Mass of one stone is twice that of other.

To find:

Kinetic energy of the heavy stone with respect to the light stone.

Solution:

Kinetic energy of a body is the energy possessed by a body by virtue of its velocity .

Mathematically,    $KE=\frac{1}{2}mv^{2}$

The acceleration of the bodies while falling down is same for both the masses as this acceleration is equal to the acceleration due to gravity (g).

All of the kinetic energy possessed by the body at the top of he cliff is converted to kinetic energy of the body just before falling on the ground.

Hence,          $mgh=\frac{1}{2}mv^{2}$

                        $v^{2} =2gh$

$v$ is the velocity of the body just before touching the ground and wee see that the velocity is purely depended on height from which the masses are released.

According to the question, both the masses are released from the same height that is the cliff. Hence, both masses will possess the same velocity.

For the first body, kinetic energy before touching the ground

$KE_{1} =\frac{1}{2}mv^{2}$

Similarly for the second body,kinetic energy will be

$KE_{2} =\frac{1}{2}(2m)v^{2}$

        $=2(\frac{1}{2} mv^{2})$

$KE_{2}$  $=2KE_{1}$  

Or

$\frac{KE_{1} }{KE_{2} } =\frac{1}{2}$

Final answer:

For the stone, which is twice the mass of the first stone, will have kinetic energy twice the first stone having lower mass.