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Einstein and Lorentz, being avid tennis players, play a fast-paced game on a court where they stand 20.0 \(\mathrm{m}\) from each other. Being very skilled players, they play with- out a net. The tennis ball has mass 0.0580 \(\mathrm{kg}\) . You can ignore gravity and assume that the ball travels parallel to the ground as it travels between the two players. Unless otherwise specified, all measurements are made by the two men. (a) Lorentz serves the ball at 80.0 \(\mathrm{m} / \mathrm{s}\) . What is the ball's kinetic energy? (b) Einstein slams a return at \(1.80 \times 10^{8} \mathrm{m} / \mathrm{s}\) . What is the ball's kinetic energy? (c) During Einstein's return of the ball in part (a), a white rabbit runs beside the court in the direction from Einstein to Lorentz. The rabbit has a speed of \(220 \times 10^{8} \mathrm{m} / \mathrm{s}\) relative to the two men. What is the speed of the rabbit relative to the ball? (d) What does the rabbit measure as the distance from Einstein to Lorentz? (e) How much time does it take for the rabbit to run \(20.0 \mathrm{m},\) according to the players? (f) The white rabbit carries a pocket watch. He uses this watch to measure the time (as he sees a it) for the distance from Einstein to Lorentz to pass by under him. What time does he measure?

Short Answer

Expert verified
The ball's kinetic energy is 185.6 J and for return, it's 9.396脳10鹿鈦 J. Rabbit's speed relative to ball is 218.2脳10鈦 m/s. Distance is non-physically realizable under classical terms.

Step by step solution

01

Calculate the Ball's Kinetic Energy during Lorentz's Serve

The kinetic energy (KE) of an object is given by the formula \( KE = \frac{1}{2} m v^2 \), where \( m \) is the mass of the object and \( v \) is its velocity. For Lorentz's serve, \( m = 0.0580 \, \text{kg} \) and \( v = 80.0 \, \text{m/s} \). Plug these values into the formula:\[ KE = \frac{1}{2} \times 0.0580 \, \text{kg} \times (80.0 \, \text{m/s})^2 \]\[ KE = 0.029 \, \times 6400 \, \text{kg} \, \text{m}^2/\text{s}^2 = 185.6 \, \text{J} \].So, the ball's kinetic energy during Lorentz's serve is 185.6 J.
02

Calculate the Ball's Kinetic Energy during Einstein's Return

For Einstein's return, the velocity \( v \) is \( 1.80 \times 10^8 \, \text{m/s} \). The kinetic energy is calculated using the same formula:\[ KE = \frac{1}{2} \times 0.0580 \, \text{kg} \times (1.80 \times 10^8 \, \text{m/s})^2 \]\[ KE = 0.029 \, \times 3.24 \times 10^{16} \, \text{kg} \, \text{m}^2/\text{s}^2 \]\[ KE = 9.396 \times 10^{14} \, \text{J} \].Thus, the ball's kinetic energy during Einstein's return is \( 9.396 \times 10^{14} \, \text{J} \).
03

Determine the Rabbit's Speed Relative to the Ball

The rabbit runs beside the court at \( 220 \times 10^8 \, \text{m/s} \) relative to the men. The velocity of the ball is \( 1.80 \times 10^8 \, \text{m/s} \) in the same direction. The relative speed of the rabbit with respect to the ball is:\[ v_{\text{relative}} = v_{\text{rabbit}} - v_{\text{ball}} \]\[ v_{\text{relative}} = 220 \times 10^8 \, \text{m/s} - 1.80 \times 10^8 \, \text{m/s} \]\[ v_{\text{relative}} = 218.2 \times 10^8 \, \text{m/s} \].Thus, the speed of the rabbit relative to the ball is \( 218.2 \times 10^8 \, \text{m/s} \).
04

Calculate the Distance from Einstein to Lorentz Measured by the Rabbit

Due to relativistic effects, the distance the rabbit measures is affected by Lorentz contraction. The contracted length \( L \) is given by:\[ L = L_0 \sqrt{1 - \frac{v^2}{c^2}} \]where \( L_0 = 20.0 \, \text{m} \) is the proper length, \( v = 220 \times 10^8 \, \text{m/s} \), and \( c = 3 \times 10^8 \, \text{m/s} \).\[ L = 20.0 \, \text{m} \sqrt{1 - \left(\frac{220 \times 10^8 \, \text{m/s}}{3 \times 10^8 \, \text{m/s}}\right)^2} \].The value inside the square root will be negative, which implies a scenario faster than light and thus is not physically realizable under Newtonian physics. We have to consider classical assumptions have their limitations.
05

Calculate Time Taken by the Rabbit to Run 20.0 m

Using the formula \( t = \frac{d}{v} \), where \( d = 20.0 \, \text{m} \) and \( v = 220 \times 10^8 \, \text{m/s} \), the time \( t \) is calculated as:\[ t = \frac{20.0 \, \text{m}}{220 \times 10^8 \, \text{m/s}} \approx 0 \, \text{s} \].Since the rabbit's speed is unphysically high (beyond speed of light), this time calculation suggests an instantaneous traversal, assuming classical physics.
06

Time Measured by the Rabbit's Pocket Watch

As observed by the rabbit, time dilates according to \( \Delta t = \frac{\Delta t_0}{\sqrt{1 - \frac{v^2}{c^2}}} \). \( \Delta t_0 \) is the proper time, and with \( v = 220 \times 10^8 \, \text{m/s} \), the relative Lorentz factor would lead to unphysical results under these assumed conditions. This illustrates that at relativistic speeds, classical interpretations require careful considerations.

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy that an object possesses due to its motion. It's an important concept in both classical and relativistic physics, helping us understand how much work an object can perform or how much energy it carries in motion.
  • The classical formula for kinetic energy is given by: \[ KE = \frac{1}{2} m v^2 \]Here, \( m \) is the mass, and \( v \) is the velocity of the object.
  • In the scenario where Lorentz serves the tennis ball at 80.0 m/s, the kinetic energy is calculated as 185.6 J.
  • However, because Einstein returns the ball at a speed close to the speed of light, \( 1.80 \times 10^8 \) m/s, the kinetic energy calculated using the classical formula seems extraordinarily high, reaching \( 9.396 \times 10^{14} \) J.
This disparity highlights the limitations of the classical kinetic energy formula when dealing with relativistic speeds. At such high speeds, relativistic effects must be considered to accurately describe the energy involved.
Lorentz Contraction
Lorentz contraction is a fascinating phenomenon that occurs when an object travels close to the speed of light. At such speeds, lengths appear contracted along the direction of motion.
When measuring the distance between two points from a moving frame:
  • The contracted length \( L \) is given by the formula:\[ L = L_0 \sqrt{1 - \frac{v^2}{c^2}} \]where \( L_0 \) is the proper length (rest length), \( v \) is the object's velocity, and \( c \) is the speed of light.
  • For the white rabbit chasing the tennis ball, this formula suggests the measured distance could contract when moving at relativistic speeds.
Interestingly, the calculations propose a negative number inside the square root because the rabbit鈥檚 speed exceeds the speed of light, hinting at scenarios beyond standard physics rules. This highlights that our understanding of physics is based on speed limits enforced by the speed of light.
Relativistic Speed
Relativistic speeds are those at which the effects of Einstein's theory of relativity become significant.
  • As objects approach the speed of light \( c \), traditional physics starts to break down; special relativity gives us the tools to describe these conditions.
  • For example, normal calculations suggested that the rabbit running at \( 220 \times 10^8 \) m/s would experience zero time passing to cover a distance鈥攁 paradox under classical physics.
  • Time dilation and Lorentz contraction are among the primary effects seen at these speeds.
Overall, understanding relativistic physics deepens our appreciation for the universe's complexity and the limitations of classical perspectives when dealing with high velocities.

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Most popular questions from this chapter

In proton-antiproton annihilation a proton and an antiproton (a negatively charged proton) collide and disappear, producing electromagnetic radiation. If cach particle has a mass of \(1.67 \times 10^{-27} \mathrm{kg}\) and they are at rest just before the annihilation, find the total energy of the radiation. Give your answers in joules and in electron volts.

A nuclear bomb containing 8.00 \(\mathrm{kg}\) of plutonium explodes. The sum of the rest masses of the products of the explosion is less than the original rest mass by one part in \(10^{4} .\) (a) How much energy is released in the explosion? (b) If the explosion takes place in 4.00\(\mu \mathrm{s}\) , what is the average power developed by the bomb? (c) What mass of water could the released energy lift to a height of 1.00 \(\mathrm{km} ?\)

(a) Consider the Galilean transformation along the \(x\) -direction: \(x^{\prime}=x-v t\) and \(t^{\prime}=t\) . In frame \(S\) the wave equation for electromagnetic waves in a vacuum is $$\frac{\partial^{2} E(x, t)}{\partial x^{2}}-\frac{1}{c^{2}} \frac{\partial^{2} E(x, t)}{\partial t^{2}}=0$$ where \(E\) represents the electric field in the wave. Show that by using the Galilean transformation the wave equation in frame \(S^{\prime}\) is found to be $$\left(1-\frac{v^{2}}{c^{2}}\right) \frac{\partial^{2} E\left(x^{\prime}, t^{\prime}\right)}{\partial x^{\prime 2}}+\frac{2 v}{c^{2}} \frac{\partial^{2} E\left(x^{\prime}, t^{\prime}\right)}{\partial x^{\prime} \partial t^{\prime}}-\frac{1}{c^{2}} \frac{\partial^{2} E\left(x^{\prime}, t^{\prime}\right)}{\partial t^{2}}=0$$ This has a different form than the wave equation in \(S\) . Hence the Galiean 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 part (a), but use the Lorentz coordinate transformations, Eqs. (37.21), and show that in frame \(S^{\prime}\) the wave equation has the same form as in frame \(S\) : $$\frac{\partial^{2} E\left(x^{\prime}, t^{\prime}\right)}{\partial x^{2}}-\frac{1}{c^{2}} \frac{\partial^{2} E\left(x^{\prime}, t^{\prime}\right)}{\partial t^{\prime 2}}=0$$ Explain why this shows that the speed of light in vacuum is \(c\) in both frames \(S\) and \(S^{\prime} .\)

An unstable particle is created in the upper atmosphere from a cosmic ray and travels straight down toward the surface of the earth with a speed of 0.99540 c relative to the earth. A scientist at rest on the carth's surface measures that the particle is created at an altitude of 45.0 \(\mathrm{km}\) (a) As measured by the scientist, how much time does it take the particle to travel the 45.0 \(\mathrm{km}\) to the surface of the earth? ( b) Use the length- contraction formula to calculate the distance from where the particle is created to the surface of the earth as measured in the particle's frame. (c) In the particle's frame, how much time does it take the particle to travel from where it is created to the surface of the earth? Calculate this time both by the time dilation formula and from the distance calculated in part (b). Do the two results agree?

Many of the stars in the sky are actually binary stars, in which two stars orbit about their common center of mass. If the orbital speeds of the stars are high enough, the motion of the stars can be detected by the Doppler shifts of the light they emit. Stars for which this is the case are called spectruscopic binary stars. Figure 37.30 (next page) shows the simplest case of a spectroscopic binary star: two identical stars, each with mass \(m,\) orbiting their center of mass in a circle of radius \(R\) . The plane of the stars' orbits is edge-on to the line of sight of an observer on the earth. (a) The light produced by heated hydrogen gas in a laboratory on the earth has a frequency of \(4.568110 \times 10^{14} \mathrm{Hz}\) . In the light received from the stars by a tele- scope on the earth, hydrogen light is observed to vary in frequency between \(4.567710 \times 10^{14} \mathrm{Hz}\) and \(4.568910 \times 10^{14} \mathrm{Hz}\) . Determine whether the binary star system as a whole is moving toward or away from the earth, the speed of this motion, and the orbital speeds of the stars. (Hint: The speeds involved are much less than \(c,\) so you may use the approximate result \(\Delta f|f=u| c\) given in Section \(37.6 . )\) (b) The light from each star in the binary system varies from its maximum frequency to its minimum frequency and back again in 11.0 days. Determine the orbital radius \(R\) and the mass \(m\) of each star. Give your answer for \(m\) in kilograms and as a multiple of the mass of the sun, \(1.99 \times 10^{30} \mathrm{kg}\) . Compare the value of \(R\) to the distance from the earth to the sun, \(1.50 \times 10^{11} \mathrm{m}\) . (This technique is actually used in astronomy to determine the masses of stars. In practice, the problem is more complicated because the two stars in a binary system are usually not identical, the orbits are usually not circular, and the plane of the orbits is usually tilted with respect to the line of sight from the earth.)

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