Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

Answers without the blur.

Just sign up for free and you're in.

Short Answer

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 $x_{1} = 33.0 m$ and $t_{1} = 22.0 ns$ , and event 2 occurs at $x_{2} = 53.0 m$ and $t_{2} = 62.0 ns$ . 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?

The length of meter stick 1 is 0.5 m.
The length of meter stick 2 is 1 m.
The length of meter stick 3 is 1 m.
The spatial separation is 19.216 m.
The temporal separation is -35.46 ns.
The event 2 is occurred first.

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Describe the expression for dilation time

The expression of time dilation is given by,

$∆ t = γ ∆ t_{o}$ …… (1)

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

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

From equation (1),

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

It is known that $γ = \frac{1}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}$ then,

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

The expression to calculate the length is given by,

$L = \frac{L_{o}}{γ}$ ……. (2)

Substitute 1 m for L_o, and 2 for $γ$ in equation (2).

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

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,

$∆ x^{'} = γ (∆ x - v ∆ t)$ ……. (3)

Find $∆ x$ as follows.

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

Find $∆ t$ as follows.

$\begin{array}{rcl} ∆ t & = & t_{2} - t_{1} \\ = & 62 ns - 22 ns \\ = & (40 ns) (\frac{10^{- 9} s}{1 ns}) \\ = & 40 \times 10^{- 9} s \end{array}$

Find v as follows.

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

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

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

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,

$∆ t^{'} = γ (∆ t - v \frac{∆ x}{c^{2}})$ ……. (4)

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

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

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^'.

Find Study Materials for

Create Study Materials

Select your language

Fundamentals Of Physics

Answers without the blur.

Short Answer

Step by Step Solution

Step 1: Describe the expression for dilation time

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

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

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

Step 5: Find the spatial separation (d)

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

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

Most popular questions for Physics Textbooks

Want to see more solutions like these?

Recommended explanations on Physics Textbooks

Select your language

Fundamentals Of Physics

Answers without the blur.

Short Answer

Step by Step Solution

Step 1: Describe the expression for dilation time

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

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

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

Step 5: Find the spatial separation (d)

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

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

Most popular questions for Physics Textbooks

Want to see more solutions like these?

Recommended explanations on Physics Textbooks

Work Energy and Power

Radiation

Astrophysics

Physics of Motion

Scientific Method Physics

Fluids

Torque and Rotational Motion

Nuclear Physics

Fields in Physics

Measurements

Space Physics

Electricity

Physical Quantities and Units

Medical Physics

Electrostatics

Further Mechanics and Thermal Physics

Mechanics and Materials

Engineering Physics

Energy Physics

Force

Particle Model of Matter

Waves Physics

Magnetism

Modern Physics

Atoms and Radioactivity

Dynamics

Electricity and Magnetism

Conservation of Energy and Momentum

Fundamentals of Physics

Kinematics Physics

Circular Motion and Gravitation

Linear Momentum

Rotational Dynamics

Oscillations

Translational Dynamics

Geometrical and Physical Optics

Thermodynamics

Magnetism and Electromagnetic Induction

Electromagnetism

Electric Charge Field and Potential

Turning Points in Physics