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If momentum is increased by $20 \%$, then kinetic energy increases by
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Verified Answer
The correct answer is:
$44 \%$
Hints: $\mathrm{K}=\frac{\mathrm{P}^2}{2 m}$
Here $\mathrm{P}^{\prime}=1.2 \mathrm{P}$
Hence, $K^{\prime}=\frac{(1.2 \mathrm{P})^2}{2 m}$
$$
\mathrm{K}^{\prime}=1.44 \frac{\mathrm{P}^2}{2 m}
$$
$\mathrm{K}^{\prime}=1.44 \mathrm{~K}$ or Percentage increase in $\mathrm{K}=44 \%$
Here $\mathrm{P}^{\prime}=1.2 \mathrm{P}$
Hence, $K^{\prime}=\frac{(1.2 \mathrm{P})^2}{2 m}$
$$
\mathrm{K}^{\prime}=1.44 \frac{\mathrm{P}^2}{2 m}
$$
$\mathrm{K}^{\prime}=1.44 \mathrm{~K}$ or Percentage increase in $\mathrm{K}=44 \%$
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