Home Page Textbooks Physics Page 958: Lesson Check

Physics

1st Edition

Walker

ISBN: 9780133256925

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Page 958: Lesson Check

Exercise 12

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The time between ticks of a moving clock is greater than the time between ticks of a clock at rest because these two intervals are connected by the time dilation relation

$$
Delta t_{moving} = frac{Delta t_{rest}}{sqrt{1-frac{v^2}{c^2}}}.
$$

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Exercise 13

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This is not correct because the time dilation doesn’t have anything specific to do with clocks but it is related to all physical processes. Any physical process will appear to last longer to you (even a heartbeat of another person) if it is happening in a frame moving in respect to you that if it’s at rest.

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Exercise 14

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According to length contraction formula

$$
l_{moving} = l_{rest}sqrt{1-frac{v^2}{c^2}}
$$

the length of a moving object appears to you to be less than that of an identical object that rests in respect to you.

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Exercise 15

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Due to the phenomenons of time dilation and length contraction, length and time are not the same for every observer, They are different for different observers, ie. their frames of reference.

These phenomenons are not really observable in everyday life; for example, if the moving observer is in a moving train, he will not experience time dilation and length contraction, because the train is moving at nonrelativistic speeds. However, if he was to find himself in a spaceship, he would experience these phenomenons, because the speed of the spaceship can be compared to the speed of light in vacuum $c$.

Time and length do not have absolute values; those values depend on the frame of reference. But, they can only be experienced when an observer is moving with speeds close to the speed of light.

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Time dilation and length contraction are significant only when an observer is moving with a speed that is close to the speed of light in vacuum $c$; in everyday life, these phenomenons are insignificant.

Exercise 16

Solution 1

Solution 2

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$textbf{Given:}$

$t_o = 25$ s

$v = 0.750 c$

$textbf{Find:}$ time, $Delta t$ it takes for the unstable isotope to decay at speed, $v$

Using the time dilation formula we can compute for $Delta t$ as follows:

$$
begin{align*}
Delta t &= dfrac{Delta t_o}{sqrt{1 – dfrac{v^2}{c^2}}} \
&= dfrac{25 text{ s}}{sqrt{1 – dfrac{(0.750 c)^2}{c^2}}} \
&= boxed{37.8 text{ s}}
end{align*}
$$

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$Delta t = 37.8$ s

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Measured from the resting frame of reference it takes

$$
Delta t = frac{25text{ s}}{sqrt{1-0.75^2}} =37.8text{ s}
$$

according to time dilation formula.

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Exercise 17

Solution 1

Solution 2

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$textbf{Given:}$

$L_o = 1.00$ m

$v = 0.910 c$

$textbf{Find:}$ Length, $L$ of the meter stick at speed, $v$

Using the length contraction formula we can compute for $L$ as follows:

$$
begin{align*}
L &= L_o times sqrt{1 – dfrac{v^2}{c^2}} \
&= 1.00 text{ m} times sqrt{1 – dfrac{(0.910 c)^2}{c^2}} \
&= boxed{0.415 text{ m}}
end{align*}
$$

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$L = 0.415$ m

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Measured from the resting frame this length appears to be

$$
l=1text{ m}sqrt{1-0.91^2} =0.41text{ m}.
$$

Result

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Exercise 18

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$textbf{Given:}$

$t = 13.2$ s

$v = 0.885 c$

$textbf{Find:}$ time, $Delta t_o$ elapsed for an observer at rest

Using the time dilation formula we can compute for $Delta t_o$ as follows:

$$
begin{align*}
Delta t_o &= Delta t times sqrt{1 – dfrac{v^2}{c^2}} \
&= 13.2 text{ s} times sqrt{1 – dfrac{(0.885 c)^2}{c^2}} \
&= boxed{6.15 text{ s}}
end{align*}
$$

Result

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$Delta t_o = 6.15$ s

Exercise 19

Solution 1

Solution 2

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$textbf{Given:}$

$L_o = 1.00$ m

$L = 0.950$ m

$textbf{Find:}$ speed of the spaceship, v

We use the length contraction formula to obtain an equation that will help us calculate v, such that:

$$
begin{align*}
L &= L_o sqrt{1 – dfrac{v^2}{c^2}} \
dfrac{L}{L_o} &= sqrt{1 – dfrac{v^2}{c^2}} \
dfrac{L^2}{L_{o}^2} &= 1 – dfrac{v^2}{c^2} \
dfrac{v^2}{c^2} &= 1 – dfrac{L^2}{L_{o}^{2}} \
v &= c left(1 – dfrac{(0.950 text{ m})^2}{(1.00 text{ m})^2}right) \
v &= 0.0975 c
end{align*}
$$

The speed of the spaceship is $boxed{text{0.0975 c}}$

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$$
0.0975 c
$$

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According to time dilation formula we have

$$
0.95text{ m}= 1text{ m}sqrt{1-frac{v^2}{c^2}}.
$$

Solving for $v$ and using known value for $c$ we get

$$
v=9.37times10^7text{ m/s}.
$$

Result

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