Question

If the kinetic energy of a particle becomes four times, then % change in momentum will be?

• A. $200$
• B. $100$
• C. $150$
• D. $50$

Answer 1

The correct option is B

$100$

Explanation for the correct option:

Option(B):$100$

Step 1:Formula for kinetic energy:

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

The new kinetic energy is given equally to 4 times the initial kinetic energy.

Hence, when the KE is taken as ‘KE’ and the new kinetic energy is taken as KE’, which can be given as:
$K{E}^{\prime }=4KE\to (2)$

Also,the kinetic energy formula is given as:
$K.E.={\displaystyle \frac{1}{2}}m{v}^2$
Where, m= mass of the given particle

v= velocity of the given particle
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,
$\rho =$the momentum of the given particle

Step 3: Calculation of the change in momentum:
Putting the value of the equation (1) in equation (2) it becomes,
$\Rightarrow K{E}^{\prime }={\displaystyle \frac{\rho {}^{\prime 2}}{2m}}$

( Where ,$\rho =$the momentum of the given particle)
$\Rightarrow \frac{\rho {}^{\text{'}2}}{2m}=\frac{4{\rho }^2}{2m}$
$\Rightarrow \rho {}^{\prime 2}=4{\rho }^2$
$\Rightarrow p^{\prime }=\sqrt{4p^2}$
$\Rightarrow p^{\prime }=2p\to (3)$
Step 4: Calculation of the % increase in momentum:
The % increase in momentum can be calculated as:

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