Question

If momentum of an object is increased by \[10\;\% \] then its kinetic energy will increase by
1. \[20\;\% \]
2. \[21\;\% \]
3. \[40\;\% \]
4. \[19\;\% \]

Answer 1

Hint:The above problem can be resolved using the concept and fundamentals of kinetic energy and other forms of energy. The kinetic energy is calculated by taking the value of the subject's speed and the mass of the subject. Then by applying a suitable relation, we can obtain the final result. In this problem, we are quite handy with the variables like percentage increase in the momentum, such that the modified form of the kinetic energy is applied. This modified formula of kinetic energy involves the momentum and the mass in the denominator. Then we have to substitute the values in the formula to get the percentage energy change.

Complete step by step answer:
Given:
The percentage increased in momentum is, \[{P_1} = \left( {1 + \dfrac{{10}}{{100}}} \right)\].
The expression for the kinetic energy is,
\[\Rightarrow KE = \dfrac{{{P^2}}}{{2m}}\]
 As, the momentum is increased by \[10\;\% \].
Then substitute the above value in the expression of kinetic energy as,
\[
\Rightarrow K{E_1} = \dfrac{{{{\left( {{P_1}} \right)}^2}}}{{2m}}\\
\Rightarrow K{E_1} = \dfrac{{{{\left( {1 + \dfrac{{10}}{{100}}} \right)}^2}}}{{2m}}\\
\Rightarrow K{E_1} = \dfrac{{121P}}{{100 \times 2m}}\\
\Rightarrow K{E_1} = \dfrac{{121}}{{100}}\left( {KE} \right)
\]
Then the change in kinetic energy is,
\[
\Rightarrow \Delta KE = K{E_2} - KE\\
\Rightarrow \Delta KE = \dfrac{{121}}{{100}}\left( {KE} \right) - KE\\
\Rightarrow \Delta KE = \dfrac{{21}}{{100}}\;KE
\]
Further solving as,
\[
\Rightarrow \dfrac{{\Delta KE}}{{KE}} \times 100 = 21\\
\Rightarrow \dfrac{{\Delta KE}}{{KE}} = 21\;\%
\]
Therefore, the kinetic energy will increase by \[21\;\% \] and option (B) is correct.


Note: To solve the given problem, we must understand the concept behind several forms of energy. The energy in kinematics is of two forms namely kinetic energy and the potential energy. The kinetic energy is associated with any object's speed, while the potential energy is associated with the height at which the object is kept. Moreover, there are several applications of the kinetic as well as potential energy in resolving the typical problems.