If the momentum of a body of mass ' $m$ ' is increased by $20 \%$ then its kinetic energy increases by

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Aicharya · Accepted Answer

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. $\begin{aligned} \Delta \mathrm{KE} & =\left(\mathrm{K} \cdot \mathrm{E}^{\prime}-\mathrm{K} \cdot \mathrm{E}\right) \\ & =(1.44 \mathrm{~K} \cdot \mathrm{E}-\mathrm{K} \cdot \mathrm{E}) \\ & =0.44 \mathrm{~K} . \mathrm{E} \\ & \Rightarrow 44 \% \end{aligned}$

If the momentum of a body of mass ' $m$ ' is increased by $20 \%$ then its kinetic energy increases by

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