Question

If linear momentum is increased by 50% then how much will the kinetic energy increase by, and why?

Answer 1

Let us consider a body of mass ‘m’ be moving with velocity ‘v’.
The momentum of the body is, p = mv.
The kinetic energy of the body is, $E=\frac{1}{2}m{v}^2$
Now, considering the mass to be constant.
$E=\frac{1}{2}m{v}^2$
$\Rightarrow E=\frac{m}{2m}m{v}^2$
$\Rightarrow E=\frac{1}{2m}(mv{)}^2$
$\Rightarrow E=\frac{p^2}{2m}$ (this is valid when mass is constant )
Also, considering the velocity to be constant.
$P=mv$
$\Rightarrow p=\frac{2v}{2v}\times mv$
$\Rightarrow p=\frac{2}{v}\times \frac{1}{2}m{v}^2$
$\Rightarrow p=\frac{2}{v}\times E$ (this is valid when velocity is constant).
Now, when momentum is increased by $50\%$ then the new momentum is,
$p=p+\frac{50}{100}p=\frac{3}{2}p$
If the mass of the system is constant then,
The new kinetic energy is,
$E=\frac{(p{)}^2}{2m}=\frac{{\left(\frac{3}{2}p\right)}^2}{2m}=\frac{9}{4}\times \frac{p^2}{2m}$
$\Rightarrow E=\frac{9}{4}\times E$
Therefore, $\%$ increase in kinetic energy is,
$\frac{E-E}{E}\times 100$
$=\frac{\frac{9}{4}\times E-E}{E}\times 100$
$=\frac{5}{4}\times 100$
$=125\%$
If the velocity of the system is constant then,
The new kinetic energy E can be found using,
$P^{{}^{\prime }}=\frac{2}{v}\times E^{{}^{\prime }}$
$\Rightarrow E^{{}^{\prime }}=\frac{p^{{}^{\prime }}v}{2}$
$\Rightarrow E^{{}^{\prime }}=\frac{\frac{3}{2}pv}{2}=\frac{3}{2}E$
Therefore, $\%$ increase in kinetic energy is,
$\frac{E^{{}^{\prime }}-E}{E}\times 100$
$=\frac{\frac{3}{2}\times E-E}{E}\times 100$
$=\frac{1}{2}\times 100$
$=50\%$