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Two masses $m$ and $2 m$ are dropped from a certain height $h$ . Then on reaching the ground,

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Kinetic energy of them will be equal

Kinetic energy of the heavier is 6 times the

KE

of the lighter mass

Kinetic energy of the lighter mass will be

4

times the

KE

of the heavier mass

Kinetic energy of the heavier mass is more than the kinetic energy lighter mass

answer is D.

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Detailed Solution

The kinetic energy of the heavier mass is two times the kinetic energy of lighter.
Kinetic Energy for mass

m

, dropped from height

h

we know the fee fall velocity

v = \sqrt{2 gh}

where

v =

free fall velocity

g =

acceleration due to gravity

h =

height of the free fall

Kinetic energy of mass (m) = \frac{1}{2} \times m \times v^{2}

= \frac{1}{2} \times m \times (2 gh)

= mgh

Kinetic energy,

K . E = mgh

Kinetic Energy for mass

2 m

, dropped from height

2 h

We know,
Kinetic energy

= \frac{1}{2} \times m \times v^{2}

= \frac{1}{2} \times (2 m) \times {(\sqrt{(2 g \times 2 h)})}^{2}

= \frac{1}{2} \times 2 m \times 4 gh

= 4 mgh

Kinetic energy,

K . E = 4 mgh

Hence, on reaching the ground, the object of mass

2 m

have more kinetic energy,

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