Question

If the K.E of a body is increased by $300\%$, its momentum will increase by?

• A. $100\%$
• B. $1560\%$
• C. $\surd 300\%$
• D. $175\%$

Answer 1

The correct option is A

$100\%$

Explanation for the correct option:

Step 1: Given data:

The kinetic energy of the particle is given to be increased by$=300\%$

Step 2: Relation between kinetic energy and momentum:

The relation between kinetic energy and momentum can be given as:
$K.E.={\displaystyle \frac{{\rho }^2}{2m}}\to (1)$

Where, $\mathrm{\rho }=$momentum, $\mathrm{m}=$ mass of the given particle, and $\mathrm{K}.\mathrm{E}=$the kinetic energy of the particle

Step 3: kinetic energy formula:

The kinetic energy of a particle is defined as the energy that it possesses due to its motion.

kinetic energy formula can be given as:
$K.E.={\displaystyle \frac{1}{2}}m{v}^2$
When the KE is increased by $300\%$ is taken as $\mathrm{KE}$ and the new kinetic energy is taken as $\mathrm{KE}’$
$K{E}^{\text{'}}=KE+\frac{300}{100}KE$
$$K{E}^{\prime }=KE+3KE$$
$$\begin{gathered} K{E}^{\text{'}}=4KE \\ K{E}^{\text{'}}=\frac{4p^2}{2m}\to \left(2\right) \end{gathered}$$

This refers that the new kinetic energy being equal to 4 times the initial kinetic energy.
Step 4: Calculation of increase in momentum:
By substituting the value of equation (2) in equation (1) it becomes,
$\Rightarrow K{E}^{\prime }={\displaystyle \frac{\rho {}^{\prime 2}}{2m}}$ (Where ${\mathrm{\rho }}^{\text{'}}=$increases momentum)
$\Rightarrow \frac{\rho {}^{\text{'}2}}{2m}=\frac{4{\rho }^2}{2m}$
$\Rightarrow \rho {}^{\prime 2}=4{\rho }^2$
$\Rightarrow p^{\prime }=\sqrt{4p^2}$
$\Rightarrow p^{\prime }=2p\to (3)$

Step 5: Calculation of % increase in momentum:
The % increase in momentum can be calculated as:

=${\displaystyle \frac{{\rho }^{\prime }-\rho }{\rho }}\times 100\mathrm{\%}$
$={\displaystyle \frac{2\rho -\rho }{\rho }}\times 100\mathrm{\%}$
$={\displaystyle \frac{\rho }{\rho }}\times 100\mathrm{\%}$
$=100\mathrm{\%}$
Thus, the increase in momentum is calculated to be $100\%.$

Hence, option A is correct.