Correct option is (4) Increase √2 times
Given,
The kinetic energy of a particle is doubled.
Kinetic energy, \(\mathrm{E}=\frac{1}{2} \mathrm{mv}^2\)
Momentum, \({P}=\mathrm{mv}\)
Where,
m is Mass
v is Volume
Kinetic energy:
\(\mathrm{E}=\frac{1}{2} \mathrm{mv}^2\)
Rearranging this equation,
\(\mathrm{E}=\frac{\mathrm{mv}^2}{2}\) ......(1)
Multiplying both numerator and denominator of equation (1) by m
\(\mathrm{E}=\frac{\mathrm{mv}^2 \times \mathrm{m}}{2 \times \mathrm{m}} \)
\( \mathrm{E}=\frac{\mathrm{m}^2 \mathrm{v}^2}{2 \mathrm{~m}} \) ......(2)
The equation to find momentum is \(P=\mathrm{mv}\)
Substituting this in equation (2)
\( \mathrm{E}=\frac{(\mathrm{mv})^2}{2 \mathrm{~m}}\)
\(\mathrm{E}=\frac{\mathrm{p}^2}{2 \mathrm{~m}} \) ......(3)
Rearranging this equation (3)
\(\mathrm{P}^2=2 \mathrm{mE} \)
\(\mathrm{P}=\sqrt{2 \mathrm{mE}}\) .....(4)
From this equation (4)
\(P \propto \sqrt{E}\)
Here, the kinetic energy of a particle is doubled (2E),
That is,
\(\mathrm{P} \propto \sqrt{2 \mathrm{E}}\)
Therefore, Kinetic energy increase \(\sqrt{2}\) times.
