Q105 / 150 · MHT CET 2024 (10 May Shift 1) · Physics
If the momentum of a body of mass ' $m$ ' is increased by $20 \%$ then its kinetic energy increases by
- Correct answerA$44 \%$
- B$55 \%$
- C$66 \%$
- D$77 \%$
Solution
K.E. $=\frac{\mathrm{p}^2}{2 \mathrm{~m}}$
When momentum increases by $20 \%$, new momentum is $\mathrm{p}^{\prime}$
$\begin{aligned}
\mathrm{p}^{\prime} & =\mathrm{p}+20 \% \mathrm{p} \\
& =\mathrm{p}+0.2 \mathrm{p}=1.2 \mathrm{p}
\end{aligned}$
New kinetic energy, $\begin{aligned} \mathrm{K} \cdot \mathrm{E}^{\prime} & =\frac{\mathrm{p}^{\prime 2}}{2 \mathrm{~m}} \\ & =\frac{(1.2 \mathrm{p})^2}{2 \mathrm{~m}}=\frac{1.44 \mathrm{p}^2}{2 \mathrm{~m}} \end{aligned}$ $\therefore \quad K \cdot E^{\prime}=1.44 \mathrm{~K} . \mathrm{E}$. Increase in kinetic energy is calculated by change in kinetic energy. $\begin{aligned} \Delta \mathrm{KE} & =\left(\mathrm{K} \cdot \mathrm{E}^{\prime}-\mathrm{K} \cdot \mathrm{E}\right) \\ & =(1.44 \mathrm{~K} \cdot \mathrm{E}-\mathrm{K} \cdot \mathrm{E}) \\ & =0.44 \mathrm{~K} . \mathrm{E} \\ & \Rightarrow 44 \% \end{aligned}$
New kinetic energy, $\begin{aligned} \mathrm{K} \cdot \mathrm{E}^{\prime} & =\frac{\mathrm{p}^{\prime 2}}{2 \mathrm{~m}} \\ & =\frac{(1.2 \mathrm{p})^2}{2 \mathrm{~m}}=\frac{1.44 \mathrm{p}^2}{2 \mathrm{~m}} \end{aligned}$ $\therefore \quad K \cdot E^{\prime}=1.44 \mathrm{~K} . \mathrm{E}$. Increase in kinetic energy is calculated by change in kinetic energy. $\begin{aligned} \Delta \mathrm{KE} & =\left(\mathrm{K} \cdot \mathrm{E}^{\prime}-\mathrm{K} \cdot \mathrm{E}\right) \\ & =(1.44 \mathrm{~K} \cdot \mathrm{E}-\mathrm{K} \cdot \mathrm{E}) \\ & =0.44 \mathrm{~K} . \mathrm{E} \\ & \Rightarrow 44 \% \end{aligned}$
Hence, the correct answer is option: 'A'.