Question

The linear momentum of a body is increased by $50\%$ its kinetic energy will increase by

• A. $150\%$
• B. $125\%$
• C. $100\%$
• D. $50\%$

Answer 1

Answer: (B)

$125\%$

Let, the initial momentum of the body is $p$.

The change in momentum is

$p^{\prime }=p+\frac{50}{100}p$

$p^{\prime }=p+0.5p=1.5p$

The kinetic energy of the body is

$E=\frac{1}{2}m{v}^2$

$E=\frac{(mv{)}^2}{2m}=\frac{p^2}{2m}$ .......(1)

Since, $p=mv$

where $m$ is the mass of the body

Now, the kinetic energy of the body due to changed momentum is

$E^{\prime }=\frac{(m{v}^{\prime }{)}^2}{2m}=\frac{p^{\prime 2}}{2m}$ ..........(2)$($since, $p^{{}^{\prime }}=m{v}^{{}^{\prime }})$

From (1) and (2)

$\frac{E^{\prime }}{E}=\frac{p^{\prime 2}}{p^2}$

$\frac{E^{\prime }}{E}=\frac{(1.5p{)}^2}{p^2}$

$\frac{E^{\prime }}{E}=2.25$

$E^{\prime }=2.25E$

The percent change in kinetic energy is

$\frac{E^{{}^{\prime }}-E}{E}\times 100=\frac{2.25E-E}{E}\times 100$

$\Rightarrow 1.25\times 100=125\%$