## Related questions with answers

In the preceding exercise, if the stick is moving in a direction along its length (like a properly thrown spear), how long will you measure its length to be?

Solutions

VerifiedIn the problem it is assumed that the stick has $p = 2 p_{0}$ where we have denoted classical momentum $p_{0} = mv$. The relativistic momentum is given by

$p = \gamma mv$

so we can conclude that $\gamma = 2$. The length contraction is given by the formula

$L = \frac{L_{0}}{\gamma}$

and we are given that the length of the stick was $L_{0} = 1m$. Thus we conclude that we would measure the stick to be

$L = 0.5m$

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

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

**after putting the value in the equation 1,**

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

## Create a free account to view solutions

## Create a free account to view solutions

## Recommended textbook solutions

#### Physics for Scientists and Engineers: A Strategic Approach with Modern Physics

4th Edition•ISBN: 9780133942651 (8 more)Randall D. Knight#### Mathematical Methods in the Physical Sciences

3rd Edition•ISBN: 9780471198260 (1 more)Mary L. Boas#### Fundamentals of Physics

10th Edition•ISBN: 9781118230718 (3 more)David Halliday, Jearl Walker, Robert Resnick## More related questions

1/4

1/7