Question

If the momentum of a body increases by 50% its kinetic energy increases by ?

Answer 1

Solution:

Let initial momentum be $\text{p}$ p and kinetic energy be$\text{K}$

From definition of momentum we can say, $\text{p=mv}$

If $\text{p}$ increases by 50%, the new momentum

$p^{\prime }=p+\frac{p}{2}=\frac{3p}{2}$
Relation between kinetic energy and momentum is given by;

$K=\frac{1}{2}{mv}^2$

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

The new kinetic energy

$K^{\prime }=\frac{p^{\prime 2}}{2m}$

So,

$\frac{K^{\prime }}{K}=\frac{\frac{p^{\prime 2}}{2m}}{\frac{p^2}{2m}}=\frac{p^{\prime 2}}{p^2}=\frac{\frac{{\left(3p\right)}^2}{2}}{p^2}=\frac{9}{4}$

$K^{\prime }=\frac{9}{4}K$

$\frac{K^{\prime }-K}{K}\times 100⟹\frac{\frac{9}{4}K-K}{K}\times 100⟹125\%$

Hence, its kinetic energy increases by: 125%