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If momentum of an object is increased by 10 % then its kinetic energy will increase by:
A) 20 %.
B) 21 %.
C) 40 %.
D) 19 %.

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Last updated date: 13th Jun 2024
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Answer
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Hint:The momentum is defined as the product of mass and velocity. The kinetic energy is defined as the energy which is possessed by the body due to the virtue of motion of that body. If the momentum of the body is increased then it means that the body's velocity is increased because the mass of the body is not increased.

Formula used:The formula of the momentum is given by,
$ \Rightarrow P = m \times v$
Where momentum is P. the mass is m and the velocity is v.
The formula of the kinetic energy is given by,
$ \Rightarrow K \cdot E = \dfrac{1}{2} \times m \times {v^2}$
Where K.E is the kinetic energy, the mass is m and the velocity is v.

Step by step solution:
It is asked in the problem if the momentum of an object is increased by 10 % then we need to find the increase in kinetic energy of the body.
The formula of the momentum is given by,
$ \Rightarrow P = m \times v$
Where momentum is P. the mass is m and the velocity is v.
$ \Rightarrow v = \dfrac{P}{m}$………eq. (1)
The formula of the kinetic energy is given by,
$ \Rightarrow K \cdot E = \dfrac{1}{2} \times m \times {v^2}$………eq. (2)
Where K.E is the kinetic energy, the mass is m and the velocity is v.
Replacing the value of the velocity from the equation (1) to equation (2) we get.
$ \Rightarrow K \cdot E = \dfrac{1}{2} \times m \times {v^2}$
$ \Rightarrow K \cdot E = \dfrac{1}{2} \times m \times {\left( {\dfrac{P}{m}} \right)^2}$
$ \Rightarrow K \cdot E = \dfrac{{{P^2}}}{{2m}}$.
Now it is given that the momentum is increased by the 10% we get,
$ \Rightarrow \dfrac{{10}}{{100}} \times P = \dfrac{P}{{10}}$
The final momentum will be,
$ \Rightarrow {P_f} = P + \dfrac{P}{{10}}$
$ \Rightarrow {P_f} = \dfrac{{11P}}{{10}}$
$ \Rightarrow {P_f} = \dfrac{{11P}}{{10}}$
The final kinetic energy,
$ \Rightarrow K \cdot E = \dfrac{{{P^2}}}{{2m}}$
$ \Rightarrow K \cdot {E_f} = \dfrac{{{{\left( {\dfrac{{11P}}{{10}}} \right)}^2}}}{{2m}}$
$ \Rightarrow K \cdot {E_f} = \dfrac{{121{P^2}}}{{200m}}$
Percentage increase in the kinetic energy,
$ \Rightarrow \dfrac{{\Delta K \cdot E}}{{K \cdot E}} = \dfrac{{\dfrac{{121{P^2}}}{{200m}} - \dfrac{{{P^2}}}{{2m}}}}{{\dfrac{{{P^2}}}{{2m}}}} \times 100$
$ \Rightarrow \dfrac{{\Delta K \cdot E}}{{K \cdot E}} = \dfrac{{\dfrac{{121}}{{200}} - \dfrac{1}{2}}}{{\dfrac{1}{2}}} \times 100$
$ \Rightarrow \dfrac{{\Delta K \cdot E}}{{K \cdot E}} = \dfrac{{\dfrac{{121 - 100}}{{200}}}}{{\dfrac{1}{2}}} \times 100$
$ \Rightarrow \dfrac{{\Delta K \cdot E}}{{K \cdot E}} = \dfrac{{\dfrac{{121 - 100}}{{200}}}}{{\dfrac{1}{2}}} \times 100$
$ \Rightarrow \dfrac{{\Delta K \cdot E}}{{K \cdot E}} = \dfrac{{121 - 100}}{{100}} \times 100$
$ \Rightarrow \dfrac{{\Delta K \cdot E}}{{K \cdot E}} = 21\% $.
The percentage increase in the kinetic energy is equal to$\dfrac{{\Delta K \cdot E}}{{K \cdot E}} = 21\% $.

The correct answer for this problem is option B.

Note:The students are advised to understand and remember the formula of the momentum and also the formula of the kinetic energy. It is necessary to make a relation between the momentum and the kinetic energy to solve this problem.