Question

The changes in a momentum of a particle,when the kinetic energy is increased by 0.1% will be:-

• A. 0.05%
• B. 0.1%
• C. 1.0%
• D. 10%

Answer 1

The correct option is A 0.05%

$$\begin{array}{c}ThekineticenergyisE=\frac{1}{2}m{v}^2\\ Themomentumisp=mv\\ v=\sqrt{\frac{2E}{m}}\\ therefore,\\ \Rightarrow p=mv=m\sqrt{\frac{2E}{m}}=\sqrt{2mE}\\ taking\log s,\\ \Rightarrow \log p=\frac{1}{2}\left(\log 2+\log m+\log E\right)\\ differentiating:\\ \Rightarrow \frac{\Delta p}{p}=\frac{1}{2}\left(\frac{\Delta m}{m}\right)+\frac{1}{2}\left(\frac{\Delta E}{E}\right)\\ but,\\ \Rightarrow \frac{\Delta E}{E}=0.1\\ so.therefore,\\ \Rightarrow \frac{\Delta p}{p}=\frac{1}{2}\times 0.1=0.05\\ sothecorrectoptionisA.\end{array}$$