Question

If the momentum of an object is increased by $10\%$ then its kinetic energy will increase by

• A. $20\%$
• B. $21\%$
• C. $40\%$
• D. $19\%$

Answer 1

The correct option is B

$21\%$

Step 1: Given Data

Let the initial momentum be $p$.

Given, the final momentum $p\text{'}=p\left(1+10\%\right)=\frac{11p}{10}$

Let the initial kinetic energy be $KE$.

Let the mass be $m$.

Let the velocity be $v$.

Step 2: Formula Used

$$\begin{gathered} KE=\frac{1}{2}m{v}^2 \\ p=mv \end{gathered}$$

Step 3: Calculate the Kinetic Energy

We know that kinetic energy,

$$\begin{gathered} KE=\frac{1}{2}m{v}^2 \\ \Rightarrow KE=\frac{1}{2m}.m^2{v}^2 \\ \Rightarrow KE=\frac{1}{2m}{p}^2\left(\because p=mv\right) \end{gathered}$$

Now, the final kinetic energy,

$$\begin{gathered} KE\text{'}=\frac{1}{2m}p{\text{'}}^2 \\ =\frac{1}{2m}\frac{121p^2}{100} \end{gathered}$$

Step 4: Calculate the Percentage Increase

Therefore, change in kinetic energy,

$$\begin{gathered} △KE=\frac{1}{2m}\frac{121p^2}{100}-\frac{1}{2m}{p}^2 \\ =\frac{1}{2m}{p}^2\left(\frac{121}{100}-1\right) \\ =\frac{1}{2m}{p}^2\left(\frac{21}{100}\right) \\ =KE\left(\frac{21}{100}\right) \end{gathered}$$

Therefore, the percentage increase in kinetic energy,

$$\begin{gathered} =\frac{21}{100}\times 100 \\ =21\% \end{gathered}$$

Hence, the correct answer is option (B).