Q88 / 160 · JIPMER 2014 · Physics

If we increase kinetic energy of a body $300 \%$, then per cent increase in its momentum is

  • A
    $50 \%$
  • B
    $300 \%$
  • Correct answer
    C
    $100 \%$
  • D
    $150 \%$

Solution

$\begin{aligned} & p^2=2 M E \\ & \therefore \quad p \propto \sqrt{E} \\ & \therefore \quad \frac{p_1}{p_2}=\sqrt{\frac{E_1}{E_2}}=\sqrt{\frac{E}{E+3 E}}=\frac{1}{2} \\ & \end{aligned}$ $\begin{aligned} \Rightarrow \quad p_2 & =2 p_1 \\ \% \text { Increase } & =\frac{p_2-p_1}{p_1} \times 100\end{aligned}$ $\Rightarrow \frac{2 p_1-p_1}{p_1} \times 100=100 \%$

Hence, the correct answer is option: 'C'.

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