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A meter stick moves past you at great speed. Its motion relative to you is parallel to its long axis. If you measure the length of the moving meter stick to be 1.00 ft 11 ft = 0.3048 m2-for example, by comparing it to a 1-foot ruler that is at rest relative to you-at what speed is the meter stick moving relative to you?

Short Answer

Expert verified
The meter stick is moving at approximately \(2.85 \times 10^8\) m/s relative to you.

Step by step solution

01

Identify the Concept

We are dealing with a problem that involves length contraction, a concept in special relativity. When an object moves at high speeds relative to an observer, its length appears contracted along the direction of motion.
02

Set Up the Length Contraction Formula

The length contraction formula is given by \[ L = L_0 \sqrt{1 - \frac{v^2}{c^2}} \]where \( L \) is the observed contracted length, \( L_0 \) is the proper length, \( v \) is the velocity of the object, and \( c \) is the speed of light. Given \( L_0 = 1 \text{ m} \) and \( L = 0.3048 \text{ m} \).
03

Rearrange the Formula to Solve for Speed

Rearrange the formula to solve for velocity \( v \). First, square both sides of the equation to get:\[ \left(\frac{L}{L_0}\right)^2 = 1 - \frac{v^2}{c^2} \]Then solve for \( \frac{v^2}{c^2} \):\[ \frac{v^2}{c^2} = 1 - \left(\frac{L}{L_0}\right)^2 \]
04

Substitute the Known Values

Substitute \( L = 0.3048 \text{ m} \), \( L_0 = 1 \text{ m} \), and \( c = 3 \times 10^8 \text{ m/s} \) into the equation:\[ \frac{v^2}{c^2} = 1 - \left(\frac{0.3048}{1}\right)^2 \] Calculate \[ \frac{v^2}{c^2} = 1 - 0.0927 = 0.9073 \]
05

Solve for v

Finally, solve for \( v \):\[ v = c \sqrt{0.9073} \]Plug in the known value of \( c \):\[ v = 3 \times 10^8 \text{ m/s} \times \sqrt{0.9073} \]\[ v \approx 2.85 \times 10^8 \text{ m/s} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Length Contraction
Length contraction is a fascinating concept from Einstein's theory of special relativity. It comes into play when objects move at a significant fraction of the speed of light relative to an observer. In simple terms, when you observe a fast-moving object, it appears shorter along the direction of motion, compared to when it is at rest.
This isn't an illusion but a real effect that Einstein's theories predict. The effect becomes noticeable only at speeds close to the speed of light.The formula for length contraction is \[ L = L_0 \sqrt{1 - \frac{v^2}{c^2}} \]- **L**: contracted length (what the observer measures)- **L_0**: proper length (the object's length at rest)- **v**: velocity of the object - **c**: speed of light When you apply this formula, it allows you to calculate how much the length contracts based on its speed. This concept helps us understand that measurements of space and time are not absolute but depend on the observer's relative motion.
Calculating Velocity in Special Relativity
In special relativity, calculating the velocity of a moving object involves more than just dividing distance by time. This is due to relativistic effects, including length contraction and time dilation. For the given exercise, we're interested in finding the velocity of a moving meter stick observed to be shorter than its rest length. Using the length contraction formula, we rearrange it to solve for velocity.Start by squaring both sides. This leads to: \[ \left(\frac{L}{L_0}\right)^2 = 1 - \frac{v^2}{c^2} \]Next, isolate \( \frac{v^2}{c^2} \) by moving terms around: \[ \frac{v^2}{c^2} = 1 - \left(\frac{L}{L_0}\right)^2 \]From this equation, you can find the square of velocity as a ratio involving the speed of light. Finally, take the square root and multiply by the speed of light to solve for \( v \). This process highlights how dramatically different relativistic velocity calculations are from classical ones.
The Role of the Speed of Light
The speed of light, denoted by \( c \), is a fundamental constant in physics and plays a crucial role in special relativity. It's about \( 3 \times 10^8 \text{ m/s} \), and it serves as a universal speed limit. No object with mass can reach or exceed this speed.In the realm of special relativity, \( c \) not only limits speed but also affects time and space. For instance, as objects approach the speed of light, time dilation and length contraction become significant.This constant appears in the mathematical expressions for relativity, including the length contraction formula:\[ L = L_0 \sqrt{1 - \frac{v^2}{c^2}} \]Here, \( c \) ensures that velocity \( v \) is compared to a scale that defines maximum possible speed. It also emphasizes Einstein’s insight that space and time are intertwined in the fabric of the universe. Understanding how \( c \) interacts with other variables helps illustrate why it's impossible to observe objects at or above this speed. It serves as a reminder of the unique and counterintuitive aspects of our universe.

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Most popular questions from this chapter

Space pilot Mavis zips past Stanley at a constant speed relative to him of 0.800c. Mavis and Stanley start timers at zero when the front of Mavis's ship is directly above Stanley. When Mavis reads 5.00 s on her timer, she turns on a bright light under the front of her spaceship. (a) Use the Lorentz coordinate transformation derived in Example 37.6 to calculate x and t as measured by Stanley for the event of turning on the light. (b) Use the time dilation formula, Eq. (37.6), to calculate the time interval between the two events (the front of the spaceship passing overhead and turning on the light) as measured by Stanley. Compare to the value of \(t\) you calculated in part (a). (c) Multiply the time interval by Mavis's speed, both as measured by Stanley, to calculate the distance she has traveled as measured by him when the light turns on. Compare to the value of \(x\) you calculated in part (a).

(a) Consider the Galilean transformation along the \(x\)-direction : \(x' = x - vt\) and \(t' = t\). In frame \(S\) the wave equation for electromagnetic waves in a vacuum is $${\partial^2E(x, t)\over \partial x^2} - {1 \over c^2} {\partial^2E(x, t)\over \partial t^2} = 0$$ where \(E\) represents the electric field in the wave. Show that by using the Galilean transformation the wave equation in frame \(S'\) is found to be $$(1 - {v^2 \over c^2} ) {\partial^2E(x', t') \over \partial{x'^2}} + {2v \over c^2} {\partial^2E(x', t') \over \partial{x'} \partial{t'}} - {1 \over c^2} {\partial^2E(x', t') \over \partial{t'^2}} = 0$$ This has a different form than the wave equation in \(S\). Hence the Galilean transformation \(violates\) the first relativity postulate that all physical laws have the same form in all inertial reference frames. (\(Hint\): Express the derivatives \(\partial/\partial{x}\) and \(\partial/\partial{t}\) in terms of \(\partial/\partial{x'}\) and \(\partial/\partial{t'}\) by use of the chain rule.) (b) Repeat the analysis of part (a), but use the Lorentz coordinate transformations, Eqs. (37.21), and show that in frame \(S'\) the wave equation has the same form as in frame \(S\): $${\partial^2E(x', t') \over \partial{x'}^2} - {1 \over c^2} {\partial^2E(x', t') \over \partial{t'}^2} = 0$$ Explain why this shows that the speed of light in vacuum is c in both frames \(S\) and \(S'\).

An electron is acted upon by a force of 5.00 \(\times\) 10\(^{-15}\) N due to an electric field. Find the acceleration this force produces in each case: (a) The electron's speed is 1.00 km/s. (b) The electron's speed is 2.50 \(\times\) 10\(^8\) m/s and the force is parallel to the velocity.

Two particles are created in a high-energy accelerator and move off in opposite directions. The speed of one particle, as measured in the laboratory, is 0.650c, and the speed of each particle relative to the other is 0.950c. What is the speed of the second particle, as measured in the laboratory?

An alien spacecraft is flying overhead at a great distance as you stand in your backyard. You see its searchlight blink on for 0.150 s. The first officer on the spacecraft measures that the searchlight is on for 12.0 ms. (a) Which of these two measured times is the proper time? (b) What is the speed of the spacecraft relative to the earth, expressed as a fraction of the speed of light \(c\)?

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