Chapter 26, Problem 13P

(a)

To determine

The length and height of the spacecraft when it is at rest.

Answer to Problem 13P

The rest length is $6.39m$ .

The height is $1.25m$ .

Explanation of Solution

Given info:

Length is $4.80m$

Speed is $0.660c$

Formula used:

  $L=L_0{\left[1-{\left(\frac{v}{c}\right)}^2\right]}^{\frac{1}{2}}$

Where,

  $L_0$ is rest length

  $L$ is length (contracted)

  $v$ is velocity

  $c$ is speed of light, $3.00\times {10}^{8}m/s$

Calculation:

Substitute the given values in the formula

  $L=L_0{\left[1-{\left(\frac{v}{c}\right)}^2\right]}^{\frac{1}{2}}$

  $4.80m=L_0{\left[1-{\left(0.660\right)}^2\right]}^{\frac{1}{2}}$

  $L_0=6.39m$

It is known that distances perpendicular to the motion do not change.

So, the rest height will be $1.25m$ .

(b)

To determine

The dilated time in the spacecraft.

Answer to Problem 13P

  $15.0s$

Explanation of Solution

Given info:

Time is $20.0s$ .

Speed is $0.660c$ .

Formula used:

  $\Delta t=\frac{\Delta t_0}{{\left[1-\left(\frac{v^2}{c^2}\right)\right]}^{\frac{1}{2}}}$

Where,

  $t_0$ is rest time

  $t$ is time

  $v$ is velocity

  $c$ is a factor, $3.00\times {10}^{8}m/s$

Calculation:

Substitute the given values in the formula $\Delta t=\frac{\Delta t_0}{{\left[1-\left(\frac{v^2}{c^2}\right)\right]}^{\frac{1}{2}}}$

  $20.0s=\frac{\Delta t_0}{{\left[1-{\left(0.660\right)}^2\right]}^{\frac{1}{2}}}$

  $\Delta t_0=15.0s$

(c)

To determine

The relative speed.

Answer to Problem 13P

  $0.660c$ .

Explanation of Solution

Introduction:

Relativistic effects are significant only at high speeds, close to speed of light, which itself is ultimate speed in universe.

Relativistic effects are relative. Means that speed relative (a person) to friend will be the same as friend's speed relative to that person. According to the question, to friend, a person moved at the same relative speed that is $0.660c$ .

(d)

To determine

The elapsed time.

Answer to Problem 13P

  $15.0s$

Explanation of Solution

Introduction: Relativistic effects are relative. Means that speed relative to friend will be the same as friend's speed relative.

According to the question, as anyone saw her watch slow down, she will see the watch slow down by the same amount. So, she would measure the same time dilation: $15.0s$ .