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Found in: Page 1147

Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

Bullwinkle in reference frame S' passes you in reference frame S along the common direction of the x' and x axes, as in Fig. 37-9. He carries three meter sticks: meter stick 1 is parallel to the x' axis, meter stick 2 is parallel to the y' axis, and meter stick 3 is parallel to the z' axis. On his wristwatch he counts off 15.0 s, which takes 30.0 s according to you. Two events occur during his passage. According to you, event 1 occurs at ${{\mathbf{x}}}_{{\mathbf{1}}}{\mathbf{=}}{\mathbf{33}}{\mathbf{.}}{\mathbf{0}}{\mathbf{}}{\mathbf{m}}$ and ${\mathbf{t}}_{{\mathbf{1}}}{\mathbf{=}}{\mathbf{22}}{\mathbf{.}}{\mathbf{0}}{\mathbf{}}{\mathbf{ns}}$, and event 2 occurs at ${{\mathbf{x}}}_{\mathbf{2}}{\mathbf{=}}{\mathbf{53}}{\mathbf{.}}{\mathbf{0}}{\mathbf{}}{\mathbf{m}}$ and ${\mathbf{t}}_{\mathbf{2}}{\mathbf{=}}{\mathbf{62}}{\mathbf{.}}{\mathbf{0}}{\mathbf{}}{\mathbf{ns}}$. According to your measurements, what is the length of (a) meter stick 1, (b) meter stick 2, and (c) meter stick 3? According to Bullwinkle, what are (d) the spatial separation and (e) the temporal separation between events 1 and 2, and (f) which event occurs first?

1. The length of meter stick 1 is 0.5 m.
2. The length of meter stick 2 is 1 m.
3. The length of meter stick 3 is 1 m.
4. The spatial separation is 19.216 m.
5. The temporal separation is -35.46 ns.
6. The event 2 is occurred first.
See the step by step solution

Step 1: Describe the expression for dilation time

The expression of time dilation is given by,

${\mathbf{∆}}{\mathbf{t}}{\mathbf{=}}{\mathbf{\gamma }}{\mathbf{∆}}{{\mathbf{t}}}_{{\mathbf{o}}}$ …… (1)

Step 2: Find the length of meter stick 1 (a)

Here, the time according to watch is, $∆\mathrm{t}=30\mathrm{s}$, and the time according to Bull winkle is, $∆{\mathrm{t}}_{\mathrm{o}}=15\mathrm{s}$.

From equation (1),

$\begin{array}{rcl}\mathrm{\gamma }& =& \frac{30\mathrm{s}}{15\mathrm{s}}\\ & =& 2\end{array}$

It is known that then,

$\begin{array}{rcl}\frac{{\mathrm{v}}^{2}}{{\mathrm{c}}^{2}}& =& 1-\frac{1}{{\mathrm{\gamma }}^{2}}\\ \frac{\mathrm{v}}{\mathrm{c}}& =& \sqrt{1-\frac{1}{{\mathrm{\gamma }}^{2}}}\\ & =& \sqrt{1-\frac{1}{{2}^{2}}}\\ \mathrm{v}& =& 0.866\mathrm{c}\end{array}$

The expression to calculate the length is given by,

$\mathrm{L}=\frac{{\mathrm{L}}_{\mathrm{o}}}{\mathrm{\gamma }}$ ……. (2)

Substitute 1 m for Lo, and 2 for $\mathrm{\gamma }$ in equation (2).

$\begin{array}{rcl}\mathrm{L}& =& \frac{1\mathrm{m}}{2}\\ & =& 0.5\mathrm{m}\end{array}$

Therefore, the length of meter stick 1 is 0.5 m.

Step 3: Find the length of meter stick 2 (b)

The meter stick 2 is along y-axis, and the frame S' is moving along positive x-axis. Therefore, there is no length contraction along y' direction. So, the length of the meter stick parallel to y' axis will remain same.

Hence, according to the measurement of length of the meter stick parallel to y' axis is 1 m.

Step 4: Find the length of meter stick 3 (c)

There is no length contraction for the meter stick placed parallel to z axis. Thus, according to the measurement of length of the meter stick parallel to z' axis is 1 m.

Step 5: Find the spatial separation (d)

The expression to calculate the spatial separation is given by,

$∆{\mathrm{x}}^{\text{'}}=\mathrm{\gamma }\left(∆\mathrm{x}-\mathrm{v}∆\mathrm{t}\right)$ ……. (3)

Find $∆\mathrm{x}$ as follows.

$\begin{array}{rcl}∆\mathrm{x}& =& {\mathrm{x}}_{2}-{\mathrm{x}}_{1}\\ & =& 53\mathrm{m}-33\mathrm{m}\\ & =& 20\mathrm{m}\end{array}$

Find $∆\mathrm{t}$ as follows.

$\begin{array}{rcl}∆\mathrm{t}& =& {\mathrm{t}}_{2}-{\mathrm{t}}_{1}\\ & =& 62\mathrm{ns}-22\mathrm{ns}\\ & =& \left(40\mathrm{ns}\right)\left(\frac{{10}^{-9}\mathrm{s}}{1\mathrm{ns}}\right)\\ & =& 40×{10}^{-9}\mathrm{s}\end{array}$

Find v as follows.

$\begin{array}{rcl}\mathrm{v}& =& \left(0.866\right)\left(3×{10}^{8}\mathrm{m}/\mathrm{s}\right)\\ & =& 2.598×{10}^{8}\mathrm{m}/\mathrm{s}\end{array}$

Substitute 2 for $\mathrm{\gamma }$, 20 m for $∆\mathrm{x}$, $40×{10}^{-9}\mathrm{s}$ for $∆\mathrm{t}$, and $2.598×{10}^{8}\mathrm{m}/\mathrm{s}$ for v in equation (3).

$\begin{array}{rcl}∆{\mathrm{x}}^{\text{'}}& =& 2\left[20\mathrm{m}-\left(2.598×{10}^{8}\mathrm{m}/\mathrm{s}\right)\left(40×{10}^{-9}\mathrm{s}\right)\right]\\ & =& 2\left(20\mathrm{m}-10.392\mathrm{m}\right)\\ & =& 19.216\mathrm{m}\end{array}$

Therefore, the spatial separation is 19.216 m.

Step 6: Find the temporal separation between events 1 and 2 (e)

The expression to calculate temporal separation is given by,

$∆{\mathrm{t}}^{\text{'}}=\mathrm{\gamma }\left(∆\mathrm{t}-\mathrm{v}\frac{∆\mathrm{x}}{{\mathrm{c}}^{2}}\right)$ ……. (4)

Substitute 2 for $\mathrm{\gamma }$, 20 m for $∆\mathrm{x}$, $40×{10}^{-9}\mathrm{s}$ for $∆\mathrm{t}$, $3×{10}^{8}\mathrm{m}/\mathrm{s}$ for c, and $2.598×{10}^{8}\mathrm{m}/\mathrm{s}$ for v in equation (4).

$\begin{array}{rcl}∆{\mathrm{t}}^{\text{'}}& =& 2\left[40×{10}^{-9}\mathrm{s}-\frac{\left(0.866\right)\left(3×{10}^{8}\mathrm{m}/\mathrm{s}\right)\left(20\mathrm{m}\right)}{{\left(3×{10}^{8}\mathrm{m}/\mathrm{s}\right)}^{2}}\right]\\ & =& 2\left[40×{10}^{-9}\mathrm{s}-5.77×{10}^{-8}\mathrm{s}\right]\\ & =& -35.46×{10}^{-9}\mathrm{s}\\ & =& -35.46\mathrm{ns}\end{array}$

Therefore, the temporal separation is -35.46 ns.

Step 7: Find the event that occur first (f)

The value of temporal separation is negative. Therefore, the event 2 is occurred before event 1 in S'.

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