Question

If the moment of an object increases by $20\% $ , what will be the percent increase in its kinetic energy?

Answer 1

Hint: A body's momentum is proportional to its velocity, with the mass being constant. The square of the velocity of a body is precisely proportional to its kinetic energy. As a result, kinetic energy and momentum are related to one another via velocity. As a result, we'll use this to calculate the % increase in kinetic energy.

Formula used:
$P = mv$
Where, $P$ is the momentum, $m$ is the mass and $v$ is the velocity.
And,
$K.E. = \dfrac{1}{2}m \times {v^2}$
Where, $K.E$ is the kinetic energy, $m$ is the mass and $v$ is the velocity.
And, using both the above formula we can find that,
$K.E = \dfrac{{{p^2}}}{{2m}}$

Complete step by step answer:
Here, as per the question it is given that the momentum of a body is increased by $20\% $. The momentum changes kinetic energy also changes.
$K.E = \dfrac{{{p^2}}}{{2m}}$
So, after the increase in kinetic energy, we get,
$K.E' = \dfrac{{{{\left( {p + 20\% p} \right)}^2}}}{{2m}} \\
\Rightarrow K.E' = \dfrac{{{p^2} + 0.04{p^2} + 2\left( {0.2{p^2}} \right)}}{{2m}} \\
\Rightarrow K.E' = \dfrac{{1.44{p^2}}}{{2m}} \\ $
As a result, we can state that the kinetic energy increase will be equal to the total sum of kinetic energy initial and kinetic energy initial of a percent increase equals $K.E'$.
${\text{K}}{\text{.E initial of a percent increase = }}K.E' - K.E$
And, increase in $K.E = K.E \times \dfrac{a}{{100}}$
$\left( {K.E} \right) \times \dfrac{a}{{100}} = \dfrac{{1.44{p^2}}}{{2m}} - \dfrac{{{p^2}}}{{2m}} \\
\Rightarrow \dfrac{{{p^2}}}{{2m}} \times \dfrac{a}{{100}} = \dfrac{{0.44{p^2}}}{{2m}} \\$
Now, cross multiplying and solving,
$\Rightarrow \dfrac{a}{{100}} = \dfrac{{2m}}{{{p^2}}} \times \dfrac{{0.44{p^2}}}{{2m}} \\
\Rightarrow \dfrac{a}{{100}} = 0.44 \\
\therefore a = 44\% \\ $
Therefore, the kinetic energy will increase by $44\% $.

Note: We must enter the values carefully in the calculations, keeping in mind that the initial and ultimate masses will be the same. As a result, when expressing the beginning and final kinetic energy in terms of momentum, only momentum will be used as the initial and final values, rather than mass. The mass does not change.