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Question: The linear momentum of a body is increased by \( 10\% \) . What is the percentage change in its kine...

The linear momentum of a body is increased by 10%10\% . What is the percentage change in its kinetic energy?
(A) 15%15\%
(B) 17%17\%
(C) 21%21\%
(D) 27%27\%

Explanation

Solution

Hint
Percentage change of a value is determined by taking the fraction of the difference in values and the original value. The kinetic energy of a body depends directly on the square of its angular momentum, while the percentage change in kinetic energy depends on the factor by which linear momentum changes.
Formula used: E=p22mE = \dfrac{{{p^2}}}{{2m}} , where EE is the kinetic energy of the body, pp is the linear momentum and mm is its mass. The SI unit of kinetic energy is Joules (J).

Complete step by step answer
In this question, we are not provided with any values and only the percentage increase in the linear momentum of a body. Let us suppose the following:
Initial momentum is pp
Increased momentum is pp'
Initial kinetic energy is EE
Changed kinetic energy is EE'
We are asked to determine the percentage change in the kinetic energy. For that let’s calculate the increased momentum first:
p=p+10% of pp' = p + 10\% {\text{ of }}p [As the momentum increases]
p=p+10100 pp' = p + \dfrac{{10}}{{100}}{\text{ }}p
This gives us the new value of momentum as:
p=p+0.1 p=1.1pp' = p + 0.1{\text{ }}p = 1.1p
We know that the kinetic energy depends on the linear momentum with the following relation:
E=p22mE = \dfrac{{{p^2}}}{{2m}} [Eq. 1]
With the increased momentum, kinetic energy would become:
E=p22m=(1.1p)22mE' = \dfrac{{p{'^2}}}{{2m}} = \dfrac{{{{(1.1p)}^2}}}{{2m}} [As the mass remains constant] [Eq. 2]
Dividing Eq. 2 by Eq. 1, we get:
EE=(1.1p)2p2\dfrac{{E'}}{E} = \dfrac{{{{(1.1p)}^2}}}{{{p^2}}} [2m mass cancels out]
Upon solving further, we get:
EE=(1.1)21=1.21\dfrac{{E'}}{E} = \dfrac{{{{(1.1)}^2}}}{1} = 1.21
Cross multiplying gives us:
E=1.21EE' = 1.21E
To calculate the percentage increase in the kinetic energy, we divide the change in energy by the original value as
EEE×100\dfrac{{E' - E}}{E} \times 100 [Multiply by 100 to get the percentage]
1.21EEE×100\Rightarrow \dfrac{{1.21E - E}}{E} \times 100
0.21×100=21%\Rightarrow 0.21 \times 100 = 21\% .
Since this value is positive, there is an increase in the percentage and the correct option is (C).

Note
Both the linear momentum and kinetic energy are related to the mass and velocity of the body. But the difference between the two is that linear momentum is a vector while kinetic energy is not. It tells about the energy possessed by the body as a whole due to its motion.