Question

A ball of mass m is dropped from a cliff of height H. The ratio of its kinetic energy to the potential energy when it is fallen through a height $\frac { 3 } { 4 } H$is:

• A. 3 : 4
• B. 4 : 3
• C. 1 : 3
• D. 3 : 1

Answer 1

Answer: (C)

1 : 3

Answer 2

Total mechanical energy at height, H

Let be velocity of the ball at height h $\left( = \frac { 3 } { 4 } \mathrm { H } \right)$

total mechanical energy at height h,

$\mathrm { E } _ { \mathrm { h } } = \mathrm { mgh } + \frac { 1 } { 2 } \mathrm { mv } _ { \mathrm { h } } ^ { 2 }$

According to law of conservation of mechanical energy.

$E _ { H } = E _ { h }$

$\mathrm { mgH } = \mathrm { mgh } + \frac { 1 } { 2 } \mathrm { mv } _ { \mathrm { h } } ^ { 2 }$

Required ratio of kinetic energy to potential energy at height h is

$\frac { \mathrm { K } _ { \mathrm { h } } } { \mathrm { V } _ { \mathrm { h } } } = \frac { \frac { 1 } { 2 } \mathrm { mv } _ { \mathrm { h } } ^ { 2 } } { \mathrm { mgh } } = \frac { \frac { 1 } { 2 } \mathrm {~m} 2 \mathrm {~g} ( \mathrm { H } - \mathrm { h } ) } { \mathrm { mgh } }$

$= \frac { \mathrm { H } - \mathrm { h } } { \mathrm { h } } = \left( \frac { \mathrm { H } } { \mathrm { h } } - 1 \right)$

$= \left( \frac { 4 } { 3 } - 1 \right)$ $\left( \therefore \mathrm { h } = \frac { 3 } { 4 } \mathrm { H } \right)$

$= \frac { 1 } { 3 }$