Question

If the kinetic energy of the body increases by $400\% $ , what would be its percentage increase in momentum?
A. $20$
B. $50$
C. $10$
D. $100$

Answer 1

Hint From the formula for finding momentum using kinetic energy, we observe that the momentum of the body is directly proportional to the kinetic energy of the body. By using this relation we can find the percentage increase in momentum
Formula used:
$P = \sqrt {2m(K.E)} $ , where P is the momentum and K.E is the kinetic energy

Step By Step Solution
Percentage increase in kinetic energy = $400\% $
Let ${P_1}$, ${P_2}$ be the initial and final momentum of the body and ${E_1}$ , ${E_2}$ be the initial and final kinetic energies of the body. According to the given data, the final kinetic energy of the body will be
$ \Rightarrow {E_2} = {E_1} + \dfrac{{400}}{{100}}({E_1})$
$ \Rightarrow {E_2} = {E_1} + 4{E_1}$
$ \Rightarrow {E_2} = 5{E_1}$ ….. $(1)$
We know that,
Percentage increase in momentum = $\dfrac{{\Delta P}}{P} \times 100$
Where $\Delta P$ = change in momentum = ${P_2} - {P_1}$
                 $P$ = Original momentum = ${P_1}$
$ \Rightarrow $ Percentage increase in momentum = $\dfrac{{{P_2} - {P_1}}}{{{P_1}}} \times 100 = \left[ {\dfrac{{{P_2}}}{{{P_1}}} - 1} \right] \times 100$ ….. $(2)$
The formula for finding momentum from kinetic energy is
$P = \sqrt {2m(K.E)} $
From this formula, we can observe that, $P \propto \sqrt {K.E} $
$ \Rightarrow \dfrac{{{P_2}}}{{{P_1}}} = \sqrt {\dfrac{{{E_2}}}{{{E_1}}}} $
By using this relation in the equation $(2)$ , we get
$ \Rightarrow $ Percentage increase in momentum = $$\left( {\sqrt {\dfrac{{{E_2}}}{{{E_1}}}} - 1} \right) \times 100$$
By using the value of ${E_2}$ from the equation $(1)$ , we get
$ \Rightarrow $ Percentage increase in momentum = $$\left( {\sqrt {\dfrac{{5{E_1}}}{{{E_1}}}} - 1} \right) \times 100$$
$ \Rightarrow $ Percentage increase in momentum = $$\left( {\sqrt 5 - 1} \right) \times 100$$
$ \Rightarrow $ Percentage increase in momentum = $$\left( {2.2 - 1} \right) \times 100 = 1.2 \times 100 = 120\% $$
$\therefore $ Percentage increase in momentum is $120\% $
There is no correct option in the given options
Additional Information:
The relation between kinetic energy and momentum is
$ \Rightarrow K.E = \dfrac{{{P^2}}}{{2m}}$
By rearranging this equation, we get
$ \Rightarrow {P^2} = 2m \times K.E$
$ \Rightarrow P = \sqrt {2m(K.E)} $

Note
1. In the question, the percentage increase in momentum is given, so it should be added to the initial momentum to obtain final momentum
2. Take the approximate value of $\sqrt 5 $ as $2.2$ while substituting in the calculation, as the value of $\root {} \of 5 $ has many decimal places.