If kinetic energy of a body is increased by 300%, then percentage change in momentum will be

The correct option is B: 100%

We know,

Kinetic energy is given as

$K.E=\frac{1}{2}m{v}^2$

From momentum, $p=mv$

$K.E=\frac{1}{2}\frac{m^2{v}^2}{m}=\frac{1p^2}{2m}$

So,

$p=\sqrt{2\times K.E.\times m}$

Initial kinetic energy, $=K_1$

Initial momentum, $=p_1$

Final momentum, $=p_2$

After increment kinetic energy will be, $=K_1+300\%of{K}_1=4K_1$

So, ratio of momentum in both condition will be,

$\frac{p_1}{p_2}=\sqrt{\frac{K_1}{4K_1}}=\sqrt{\frac{1}{4}}=\frac{1}{2}$

$p_2=2p_1$

Now putting all the values and finding percentage change in momentum,

$\frac{\Delta p}{p_1}=\frac{p_2-p_1}{p_1}\times 100⟹\frac{2p_1-p_1}{p_1}\times 100=100\%$