Question

If the K.E. of a particle is doubled, then its momentum will

• A. Remain unchanged
• B. Be doubled
• C. Be quadrupled
• D. Increase $\sqrt{2}$ times

Answer 1

The correct option is D

Increase $\sqrt{2}$ times

Explanation for correct answer:

Option (D) is correct

Step 1 Given data:

The kinetic energy of a particle is doubled.

Step 2 Formula used:

Kinetic energy, $\mathrm{E}=\frac{1}{2}{\mathrm{mv}}^2$

Momentum, $\mathrm{P}=\mathrm{mv}$

Where,

$\mathrm{m}$ is Mass

$\mathrm{v}$ is Volume

Step 3 Kinetic energy:

$\mathrm{E}=\frac{1}{2}{\mathrm{mv}}^2$

Rearranging this equation,

$\mathrm{E}=\frac{{\mathrm{mv}}^2}{2}....\left(1\right)$

Multiplying both numerator and denominator of equation(1) by $\mathrm{m}$

$$\begin{gathered} \mathrm{E}=\frac{{\mathrm{mv}}^2\times \mathrm{m}}{2\times \mathrm{m}} \\ \mathrm{E}=\frac{{\mathrm{m}}^2{\mathrm{v}}^2}{2\mathrm{m}}...\left(2\right) \end{gathered}$$

The equation to find momentum is $\mathrm{P}=\mathrm{mv}$

Substituting this in equation (2)

$$\begin{gathered} \mathrm{E}=\frac{{\left(\mathrm{mv}\right)}^2}{2\mathrm{m}} \\ \mathrm{E}=\frac{{\mathrm{P}}^2}{2\mathrm{m}}.....\left(3\right) \end{gathered}$$

Rearranging this equation (3)

$$\begin{gathered} {\mathrm{P}}^2=2\mathrm{mE} \\ \mathrm{P}=\sqrt{2\mathrm{mE}}....\left(4\right) \end{gathered}$$

From this equation (4)

$\mathrm{P}\propto \sqrt{\mathrm{E}}$

Here, the kinetic energy of a particle is doubled($2\mathrm{E}$),

That is,

$\mathrm{P}\propto \sqrt{2\mathrm{E}}$

Therefore, Kinetic energy increase $\sqrt{2}$ times.