Question

The linear momentum of a body increased by 10% What is the persentage charge in kinetic energy?

Answer 1

Let $p$ be the linear momentum

Given, Percentage increase in momentum$=\frac{\Delta p}{p}\times 100=10$

$\frac{\Delta p}{p}=0.1$

Also we know that ,

kinetic energy , $K=\frac{p^2}{2m}$

Using error analysis

$\frac{\Delta K}{K}=2\frac{\Delta p}{p}=0.2$

So, Percentage increase in kinetic energy $=0.2\times 100=20\text{\%}$