Question

If K.E. of a body increases by 0.1% the percentage increase in its momentum will be :

Answer 1

Given,

% Increase in kinetic Energy is $\frac{\Delta K.E}{K.E}=0.1\%$

Kinetic Energy, $K.E=\frac{1}{2}m{v}^2\Rightarrow v=\sqrt{\frac{2\times K.E}{m}}$

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

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

Taking ln on both sides then differentiate

$\ln p=\frac{1}{2}(\ln 2+\ln m+\ln K.E)$

Differentiating

$\frac{\Delta p}{p}=\frac{1}{2}\left(\frac{\Delta m}{m}\right)+\frac{1}{2}\left(\frac{\Delta K.E}{E}\right)$

$\frac{\Delta p}{p}=\frac{1}{2}\left(\frac{0}{m}\right)+\frac{1}{2}\times 0.1$

$\frac{\Delta p}{p}=0.05$

Percentage change in momentum is $\text{0}\text{.05}\%$