Question

The momentum of a body increases by 20%. Then the percentage increase in its kinetic energy is

• A. $36$
• B. $44$
• C. $52$
• D. $60$

Answer 1

The correct option is B

$44$

Step 1: Given:

The momentum of a body increases by 20%

Step 2: Finding percentage increase in KE

We use the formula $K.E.=\frac{p^2}{2m}$

Here, $K.E.=kineticenergy,p=momentum,m=mass$

Let initial momentum is $p_i$.

Initial kinetic energy -

$$\begin{gathered} K,E_i=\frac{1}{2}{\left(m{v}^{}\right)}^2/m \\ =\frac{1}{2}{p}^2/m \end{gathered}$$

Now, $p$increases by $20\%$, so new momentum $\left(p_f\right)$-

$p_f=p+20\%ofp=p+\frac{p}{5}=\frac{6p}{5}.$

New kinetic energy-

$$\begin{gathered} K.E_f=\frac{1}{2}{\left(\frac{6p}{5}\right)}^2/m \\ =\frac{1}{2}\frac{p^2}{m}\frac{36}{25} \end{gathered}$$

$$\begin{gathered} \%increaseinK.E.=\frac{changeinK.E}{initialK.E}·100 \\ =\frac{\left({\displaystyle \frac{36}{25}}·\left({\displaystyle \frac{1}{2}}{p}^2\right)/m\right)-\left({\displaystyle \frac{1}{2}}{p}^2\right)/m}{\left({\displaystyle \frac{1}{2}}{p}^2\right)/m} \\ =\frac{11}{25}·100 \\ =44\% \end{gathered}$$

Hence, option (B) is the correct answer.