In the non-relativistic regime, if the moment | vedclass

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Question

Kinetic energy, $K=\frac{p^{2}}{2 m}$

since $m$ remains constant $	herefore K \propto p^{2}$

$\frac{K_{1}}{K_{2}}=\left(\frac{p_{1}}{p_{2}}ig

Vedclass · Accepted Answer

$300$

Vedclass · Answer

Kinetic energy, $K=\frac{p^{2}}{2 m}$

since $m$ remains constant $	herefore K \propto p^{2}$

$\frac{K_{1}}{K_{2}}=\left(\frac{p_{1}}{p_{2}}ight)^{2}=\left(\frac{p_{1}}{2 p_{1}}ight)^{2}=\frac{1}{4}$ or $K_{2}=4 K_{1}$

Percentage increase in kinetic energy

$=\frac{K_{2}-K_{1}}{K_{1}} 	imes 100=\frac{4 K_{1}-K_{1}}{K_{1}} 	imes 100=300 \%$

In the non-relativistic regime, if the momentum, is increased by $100\%$, the percentage increase in kinetic energy is

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