Question

If Momentum of A body increases By 50 Percentage, By what percentage will the energy increases ?

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)
$E=\frac{1}{2}m{v}^2$
$\Rightarrow E=\frac{m}{2m}m{v}^2$
$\Rightarrow E=\frac{1}{2m}\left(m{v}^2\right)$
$\Rightarrow E=\frac{p^2}{2m}$ (this is valid when mass is constant)
Also,
(Cosidering 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,
$p^{\prime }=p+\frac{50}{100}p=\frac{3}{2}p$
If the mass of the system is constant then,
The new kinetic energy is,
$E^{\prime }=\frac{(p^{\prime }{)}^2}{2m}=\frac{(\frac{3}{2}p{)}^2}{2m}=\frac{9}{4}\times \frac{p^2}{2m}$
$\Rightarrow E^{\prime }=\frac{9}{4}\times E$
Therefore, $\%$ increase in kinetic enrgy is,
$\frac{E^{\prime }-E}{E}\times$
$=\frac{\frac{9}{4}\times E-E}{E}\times 100=\frac{5}{4}\times 100=125\%$
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, $\%$ increse in kinetic energy is,
$\frac{E^{\prime }-E}{E}\times =\frac{\frac{3}{2}\times E-E}{E}\times 100=\frac{1}{2}\times 100=50\%$