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Question

# As a meterstick moves past you, your measurements show its momentum to be twice its classical momentum and its length to be $1 \mathrm{~m}$. In what direction is the stick pointing?

Solution

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Explanation: It is given that the momentum of the stick to be twice its classical momentum. i.e.

\begin{aligned} mv&=2m_{0}v\\ m&=2m_{0}----\left( 1 \right)\\ \text{where,}\ m&=\text{the mass of the moving stick}\\ &=\frac{m_{0}}{\sqrt{1-\frac{v^{2}}{c^{2}} }} \\ m_{0}&=\text{the rest mass of the stick}\\ v&=\text{the speed of the moving stick}\\ \end{aligned}

after putting the value in the equation 1,

\begin{aligned} \frac{m_{0}}{\sqrt{1-\frac{v^{2}}{c^{2}} }} &=2m_{0}\\ 1&=2\sqrt{1-\frac{v^{2}}{c^{2}} }\\ \textbf{after taking square on the both sides}\\ 1&=4\left( 1-\frac{v^{2}}{c^{2}}\right)\\ \frac{v^{2}}{c^{2}}&=1-\frac{1}{4}\\ &=\frac{3}{4}\\ \frac{v}{c}&=\sqrt{\frac{3}{4}\\}\\ &=\frac{1.738}{2}\\ &=0.869\\ v&=\bf{0.869c} \end{aligned}

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