Question:
If the Kinetic energy of a moving body becomes four times its initial Kinetic energy, then the percentage change in its momentum will be :
Correct Option: 1
Solution:
$\mathrm{K}_{2}=4 \mathrm{~K}_{1}$
$\frac{1}{2} \mathrm{mv}_{2}^{2}=4 \frac{1}{2} \mathrm{mv}_{1}^{2}$
$\mathrm{v}_{2}=2 \mathrm{v}_{1}$
$\mathrm{P}=\mathrm{mv}$
$\mathrm{P}_{2}=\mathrm{mv}_{2}=2 m \mathrm{v}_{1}$
$\mathrm{P}_{1}=\mathrm{mv}_{1}$
$\%$ change $=\frac{\Delta \mathrm{P}}{\mathrm{P}_{1}} \times 100=\frac{2 \mathrm{mv}_{1}-\mathrm{mv}_{1}}{\mathrm{mv}_{1}} \times 100=100 \%$
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