Question

If the Kinetic energy of a moving body becomes four times its initial Kinetic energy, then the percentage change in its momentum will be :

• A. $100\text{\%}$
• B. $200\text{\%}$
• C. $300\text{\%}$
• D. $400\text{\%}$

Answer 1

The correct option is A $100\text{\%}$

The K.E of a moving body has relation with momentum as
$K.E,K=\frac{P^2}{2m}$
$\Rightarrow P\propto \sqrt{K}$
Given, the Kinetic energy of a moving body becomes four times its initial Kinetic energy.
Let initial K.E of the body be $K$
$\Rightarrow$ final K.E of the body will be $4K$
$\therefore \frac{P_2}{P_1}=\sqrt{\frac{4K}{K}}$, where, $P_2$ and $P_1$ are the final momentum and initial momentum of the body respectively.
$\Rightarrow \frac{P_2}{P_1}=2$
Percentage change in momentum is given by,
$\left(\frac{P_2-P_1}{P_1}\right)\text{\%}=(\frac{P_2}{P_1}-1)\times 100=(2-1)\times 100=100\text{\%}$

$\Rightarrow \left(\frac{\Delta P}{P_1}\right)\text{\%}=100\text{\%}$
Hence, option (a) is correct.