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5.Work, Energy, Power and Collision
normal
In the non-relativistic regime, if the momentum, is increased by $100\%$, the percentage increase in kinetic energy is
A
$100$
B
$200$
C
$300$
D
$400$
Solution
Kinetic energy, $K=\frac{p^{2}}{2 m}$
since $m$ remains constant $\therefore K \propto p^{2}$
$\frac{K_{1}}{K_{2}}=\left(\frac{p_{1}}{p_{2}}\right)^{2}=\left(\frac{p_{1}}{2 p_{1}}\right)^{2}=\frac{1}{4}$ or $K_{2}=4 K_{1}$
Percentage increase in kinetic energy
$=\frac{K_{2}-K_{1}}{K_{1}} \times 100=\frac{4 K_{1}-K_{1}}{K_{1}} \times 100=300 \%$
Standard 11
Physics
