Question

A particle is moving in translatory motion. If momentum of the particle decreases by $10\%$, kinetic energy will decrease by

• A. $20\%$
• B. $19\%$
• C. $10\%$
• D. $5\%$

Answer 1

Answer: (B)

$19\%$

Answer 2

According to question, Momentum of the particle decrease by $10 \%$ As we know, momentum of particle $p=\sqrt{2 m K}$ $P= \sqrt{2}{mK}$ $\Rightarrow p \propto \sqrt{K}$ and $p_{1} \propto \sqrt{K_{1}}$ Final momentum, $p_{1}=p-\frac{10 p}{100}$ $Rightarrow p_{1}=\frac{90 p}{100}$ So, $\frac{p}{p_{1}}=\sqrt{\frac{K}{K_{1}}}$ $\Rightarrow\left(\frac{100}{90}\right)^{2}=\frac{K}{K_{1}}$ $\Rightarrow \frac{10000}{8100}=\frac{K}{K_{1}}$ $\Rightarrow \frac{81}{100}=\frac{K_{1}}{K}$ Decreasing both sides from 1 , $1-\frac{81}{100} =1-\frac{K_{1}}{K}$ $\Rightarrow \frac{100-81}{100} =\frac{K-K_{1}}{K}$ $\frac{K-K_{1}}{K} =\frac{19}{100}$ Change in kinetic energy $\Rightarrow \left(\frac{K-K_{1}}{K}\right) \times 100=\left(\frac{19}{100}\right) \times 100=19 \%$