Question

When momentum of a body increases by 200%, its KE increases by
A. $200\% $
B. $300\% $
C. $400\% $
D. $800\% $

Answer 1

Hint: The momentum of a body is directly proportional to its velocity, the mass being constant. The kinetic energy of a body is directly proportional to the square of the velocity of that body. So, the kinetic energy and the momentum are related to each other through velocity. So, using this, we will compute the percentage of increase in kinetic energy.

Formula used:
$p = mv$
Where,
$p$is the momentum of an object,
$m$is the mass of the object and
$v$ is the velocity of the object.
And,
$K.E = \dfrac{1}{2}m \times {v^2}$
$K.E$is the kinetic energy of an object,
$m$is the mass of the object and
$v$ is the velocity of the object.
Using both the formula we can get,
$K.E = \dfrac{{{p^2}}}{{2m}}$

Complete step by step solution:
According to the question, the momentum of a body increases by $200\% $ .
The momentum changes kinetic energy also changes.
The kinetic energy is given by the formula given below:
$K.E = \dfrac{{{p^2}}}{{2m}}$
After increase in kinetic energy, we get,
\[
  K.E' = \dfrac{{{{(p + 200\% p)}^2}}}{{2m}} \\
   \Rightarrow K.E' = \dfrac{{{p^2} + 4{p^2} + 2(2{p^2})}}{{2m}} \\
   \Rightarrow K.E' = \dfrac{{9{p^2}}}{{2m}} \\
 \]
Therefore, we can say that the increase in kinetic energy will be the total sum of kinetic energy initial and kinetic energy initial of a percent increase equals $K.E'$ .
Therefore, ${\text{K}}{\text{.E initial of a percent increase}} = K.E' - K.E$
And, increase in $K.E = K.E \times \dfrac{a}{{100}}$
\[
  (K.E) \times \dfrac{a}{{100}} = \dfrac{{9{p^2}}}{{2m}} - \dfrac{{{p^2}}}{{2m}} \\
   \Rightarrow \dfrac{{{p^2}}}{{2m}} \times \dfrac{a}{{100}} = \dfrac{{8{p^2}}}{{2m}} \\
   \Rightarrow \dfrac{a}{{100}} = \dfrac{{8{p^2}}}{{2m}} \times \dfrac{{2m}}{{{p^2}}} \\
   \Rightarrow \dfrac{a}{{100}} = 8 \\
   \Rightarrow a = 800\% \\
 \]
Therefore, the kinetic energy will increase by \[800\% \] .
Hence, the correct option is D.

Note:
We should carefully put the values in the equations, we should keep in mind that the initial and the final masses will be the same. Therefore, when expressing the initial and the final kinetic energy in terms of momentum, only momentum will be taken as initial and final and not the mass. Mass remains constant.