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Chapter 35: Problem 6

A proton with a momentum of \(3.0 \mathrm{GeV} / \mathrm{c}\) is moving with what velocity relative to the observer? a) \(0.31 c\) c) \(0.91 c\) e) \(3.2 c\) b) \(0.33 c\) d) \(0.95 c\)

Short Answer

Expert verified
Answer: The velocity of the proton relative to the observer is approximately 0.955c, which is closest to choice (d) 0.95c.

Step by step solution

01

Recall the relativistic momentum formula

The relativistic momentum formula is given by: \(p = \frac{m_0 v}{\sqrt{1 -\frac{v^2}{c^2}}}\), where \(p\) is the momentum, \(m_0\) is the rest mass of the particle, \(v\) is the velocity, and \(c\) is the speed of light. In this exercise, the value \(p = 3.0 \,\text{GeV}/\text{c}\) is given.
02

Substitute the known values

The mass of a proton is \(m_0 = 1.67 \times 10^{-27} \,\text{kg}\), and \(p = 3.0 \,\text{GeV}/\text{c}\) (which is \(3.0 \times 10^9 \,\text{eV}/\text{c}\)). We can now substitute these values in the relativistic momentum formula: \(\frac{3.0 \times 10^9 \,\text{eV}/\text{c}}{c} = \frac{1.67 \times 10^{-27} \,\text{kg} \cdot v}{\sqrt{1 -\frac{v^2}{c^2}}}\).
03

Convert mass-energy units

To properly solve the equation, let's convert the momentum from eV to kg.m/s. Using the conversion factor \(1 \,\text{eV}/\text{c}^2 = 1.783\times 10^{-36} \,\text{kg}\), we can write the momentum as: \(p = \frac{3.0\times 10^9 \,\text{eV}}{c} \times 1.783\times 10^{-36} \,\text{kg}/(\text{eV}/\text{c}^2) = 5.349\times 10^{-27} \,\text{kg.m/s}\).
04

Solve for the velocity

Now, we will input the momentum using SI units in the equation we derived in Step 2: \(\frac{5.349 \times 10^{-27} \,\text{kg.m/s}}{c} = \frac{1.67 \times 10^{-27} \,\text{kg} \cdot v}{\sqrt{1 -\frac{v^2}{c^2}}}\). The next step is to isolate \(v\) and solve for it using algebra. Multiplying both sides by \(c\) and the denominator, we get: \(5.349 \times 10^{-27} \,\text{kg.m/s} = 1.67 \times 10^{-27} \,\text{kg} \cdot v\sqrt{1-\frac{v^2}{c^2}}\). To simplify, let \(x = \frac{v}{c}\), which gives us: \(5.349 \times 10^{-27} \text{kg.m/s} = 1.67 \times 10^{-27} \,\text{kg} \cdot xc\sqrt{1-x^2}\). Solving for \(x\) (using numerical methods or online calculator) gives \(x \approx 0.955\).
05

Find the velocity

Now, we can find the velocity by multiplying \(x\) by the speed of light \(c\): \(v = x\cdot c = 0.955c\). Therefore, the velocity of the proton relative to the observer is approximately \(0.955c\). The correct answer is closest to choice (d) \(0.95c\).

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Key Concepts

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

Proton Velocity
Understanding the velocity of a proton moving at relativistic speeds requires a bit of insight into physics. When a proton moves at a significant fraction of the speed of light, we cannot use the classical momentum formula. Instead, we use the relativistic momentum equation, which accounts for the increase in inertia of particles as they approach the speed of light. The equation given was \[ p = \frac{m_0 v}{\sqrt{1 -\frac{v^2}{c^2}}} \] where \( m_0 \) is the rest mass, \( v \) is velocity, and \( c \) is the speed of light. To solve for the velocity \( v \), we rearrange the equation and isolate \( v \). This exercise demonstrates how these calculations are not straightforward and require numerical methods or tools for solving complex equations. By discovering that \( v \approx 0.955c \), students can appreciate both the complexity and beauty of relativistic dynamics.
Rest Mass
The rest mass \( m_0 \) of a proton plays a critical role in calculating its relativistic momentum. Unlike classical physics, where mass is invariant, in relativity, the effective mass of an object seems to increase as it moves faster. Rest mass is the mass of a particle when it is at rest, which means it is not moving relative to an observer. For protons, the rest mass is a constant 1.67 \( \times 10^{-27} \) kg. This quantity is crucial because it serves as the reference mass in the relativistic momentum equation. Without knowing the rest mass, it's impossible to calculate how much the mass effectively 'increases' as velocity increases.
  • Rest mass refers to the mass of an object when it is not in motion.
  • Critical in calculations involving particles moving near the speed of light.
Momentum Conversion
Momentum conversion involves transforming given quantities from one unit of measure to another, which is often necessary when working with relativistic equations. In the problem, momentum was provided in units of eV/c, a common energy unit in particle physics. To convert this to the more familiar SI unit of kg.m/s, we used the equivalency 1 eV/c\(^2\) equals 1.783 \( \times 10^{-36} \) kg. This conversion made it easier to use the relativistic momentum equation. Converting units is a vital skill because it ensures accuracy and consistency across various scientific calculations. Correct conversion steps ensure all values used in equations match in dimensions. By converting units, the problem can be analyzed and solved correctly.
Speed of Light
The speed of light \( c \) plays a pivotal role in the theory of relativity and is fundamental to understanding relativistic momentum. Considered a universal constant, \( c \approx 299,792,458 \) m/s in a vacuum. In all advanced physics calculations, it sets a speed limit that no object with mass can exceed. This constant features in many physics formulas, such as the relativistic momentum formula. As objects approach speeds near \( c \), they demonstrate significantly altered behaviors due to relativistic effects, such as increases in relativistic mass and time dilation.
  • Central to the effects detailed in Einstein's theory of relativity.
  • Influences how objects move and behave at high velocities.
Comprehending \( c \) is key for students to understand the bounds of high-speed physics and why certain results appear as they do in relativistic calculations.

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Most popular questions from this chapter

Consider motion in one spatial dimension. For any velocity \(v,\) define parameter \(\theta\) via the relation \(v=c \tanh \theta\) where \(c\) is the vacuum speed of light. This quantity is variously called the velocity parameter or the rapidity corresponding to velocity \(v\). a) Prove that for two velocities, which add according to the Lorentzian rule, the corresponding velocity parameters simply add algebraically, that is, like Galilean velocities. b) Consider two reference frames in motion at speed \(v\) in the \(x\) -direction relative to one another, with axes parallel and origins coinciding when clocks at the origin in both frames read zero. Write the Lorentz transformation between the two coordinate systems entirely in terms of the velocity parameter corresponding to \(v\), and the coordinates.

Consider two clocks carried by observers in a reference frame moving at speed \(v\) in the positive \(x\) -direction relative to ours. Assume that the two reference frames have parallel axes, and that their origins coincide when clocks at that point in both frames read zero. Suppose the clocks are separated by distance \(l\) in the \(x^{\prime}-\) direction in their own reference frame; for instance, \(x^{\prime}=0\) for one clock and \(x^{\prime}=I\) for the other, with \(y^{\prime}=z^{\prime}=0\) for both. Determine the readings \(t^{\prime}\) on both clocks as functions of the time coordinate \(t\) in our reference frame.

Radar-based speed detection works by sending an electromagnetic wave out from a source and examining the Doppler shift of the reflected wave. Suppose a wave of frequency \(10.6 \mathrm{GHz}\) is sent toward a car moving away at a speed of \(32.0 \mathrm{~km} / \mathrm{h}\). What is the difference between the frequency of the wave emitted by the source and the frequency of the wave an observer in the car would detect?

The most important fact we learned about aether is that: a) It was experimentally proven not to exist. b) Its existence was proven experimentally. c) It transmits light in all directions equally. d) It transmits light faster in longitudinal direction. e) It transmits light slower in longitudinal direction.

A square of area \(100 \mathrm{~m}^{2}\) that is at rest in the reference frame is moving with a speed \((\sqrt{3} / 2) c\). Which of the following statements is incorrect? a) \(\beta=\sqrt{3} / 2\) b) \(\gamma=2\) c) To an observer at rest, it looks like another square with an area less than \(100 \mathrm{~m}^{2}\) d) The length along the moving direction is contracted by a factor of \(\frac{1}{2}\)
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