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Chapter 35: Problem 3

A particle of rest mass \(m_{0}\) travels at a speed \(v=0.20 c\) How fast must the particle travel in order for its momentum to increase to twice its original momentum? a) \(0.40 c\) c) \(0.38 c\) e) \(0.99 c\) b) \(0.10 c\) d) \(0.42 c\)

Short Answer

Expert verified
Answer: (c) \(0.38c\)

Step by step solution

01

Relativistic momentum formula

Using the formula for relativistic momentum, we have: \(p = \frac{m_0 v}{\sqrt{1 - \frac{v^2}{c^2}}}\) Where - \(p\) is the momentum of the particle - \(m_0\) is the rest mass of the particle - \(v\) is the speed of the particle - \(c\) is the speed of light
02

Calculate initial momentum

We are given the initial speed of the particle, \(v = 0.20c\). Let's calculate its initial momentum, \(p_0\): \(p_0 = \frac{m_0 (0.20c)}{\sqrt{1 - \frac{(0.20c)^2}{c^2}}} = \frac{0.20m_0c}{\sqrt{1 - 0.04}} = \frac{0.20m_0c}{\sqrt{0.96}}\)
03

Set up the equation to find the final speed

We want to find the speed at which the momentum of the particle becomes twice its initial momentum. So, the equation we need to solve is: \(2p_0 = \frac{m_0v_f}{\sqrt{1 - \frac{v_f^2}{c^2}}}\) Where \(v_f\) is the final speed of the particle.
04

Solve the equation

Now, we can substitute the initial momentum (\(p_0\)) into the equation: \(2\left(\frac{0.20m_0c}{\sqrt{0.96}}\right) = \frac{m_0v_f}{\sqrt{1 - \frac{v_f^2}{c^2}}}\) We can simplify and solve for \(v_f\): \(\frac{0.40m_0c}{\sqrt{0.96}} = \frac{m_0v_f}{\sqrt{1 - \frac{v_f^2}{c^2}}}\) Notice that \(m_0c\) can be canceled from both sides: \(\frac{0.40}{\sqrt{0.96}} = \frac{v_f}{\sqrt{1 - \frac{v_f^2}{c^2}}}\) Now, we can square both sides: \(0.40^2 \cdot 0.96 = (1 - \frac{v_f^2}{c^2})v_f^2\) Simplify and solve for \(v_f\): \(0.1536c^2 = v_f^2 - \frac{v_f^4}{c^2}\) \(v_f^4 - 0.8464c^2v_f^2 + 0.9536c^4 = 0\) This is a quadratic equation in terms of \(v_f^2\). We can use any method to solve it (e.g., factoring, completing the square, or quadratic formula). The solution is: \(v_f^2 = 0.146c^2\) \(v_f = \sqrt{0.146c^2} = 0.382c\)
05

Select the correct answer

From the given choices, the speed at which the particle's momentum becomes twice its original value is closest to \(0.38c\). The correct answer is (c) \(0.38c\).

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Key Concepts

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

Special Relativity
Special relativity is a fundamental theory in physics introduced by Albert Einstein in 1905. It revolutionized our understanding of space, time, and energy. Unlike classical mechanics, which operate under the assumption that speeds are much less than the speed of light, special relativity accounts for phenomena that occur at or near the speed of light.
Key points of special relativity include:
  • Time Dilation: Time appears to move slower for an object in motion relative to a stationary observer.
  • Length Contraction: An object in motion is measured to be shorter in the direction of motion as observed from a stationary frame.
  • Relativity of Simultaneity: Two events that occur simultaneously in one frame of reference may not be simultaneous in another moving frame of reference.
These effects are particularly pronounced and must be taken into account when dealing with speeds close to the speed of light, such as in the calculations of relativistic momentum.
Lorentz Factor
The Lorentz factor, commonly denoted by the Greek letter gamma (\(\gamma\)), is a key component in the equations of special relativity. It represents how much time, length, and relativistic mass increase by a factor as the object鈥檚 speed approaches the speed of light.

Formula:

The Lorentz factor is calculated as:\[\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}\]Here,
  • \(v\) is the velocity of the object.
  • \(c\) is the speed of light.
As velocity \(v\) increases towards the speed of light \(c\), the Lorentz factor grows significantly, affecting time dilation and relativistic mass.Considerations involving the Lorentz factor are crucial for describing and analyzing phenomena when speeds approach \(c\), especially for calculating relativistic momentum as done in the original exercise.
Speed of Light
The speed of light, denoted as \(c\), is a fundamental constant in physics, valued at approximately 299,792,458 meters per second. It is the maximum speed at which all energy, matter, and information in the universe can travel. The exact value of the speed of light is crucial in calculations concerning special relativity.
Significance includes:
  • As an upper limit, it ensures that no object with mass can ever reach or exceed this speed.
  • It serves as a foundational element in Einstein's theory of special relativity, affecting the perception of time and space.
  • The speed of light is used in the determination of the Lorentz factor and thus influences calculations like momentum in relativistic speeds.
Understanding the speed of light is essential for solving problems involving relativistic momentum because it defines both the framework and limitations for particle velocities and interactions at high speeds.

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Most popular questions from this chapter

Sam sees two events as simultaneous: (i) Event \(A\) occurs at the point (0,0,0) at the instant 0: 00: 00 universal time; (ii) Event \(B\) occurs at the point \((500, \mathrm{~m}, 0,0)\) at the same moment. Tim, moving past Sam with a velocity of \(0.999 c \hat{x}\), also observes the two events. a) Which event occurred first in Tim's reference frame? b) How long after the first event does the second event happen in Tim's reference frame?

A wedge-shaped spaceship has a width of \(20.0 \mathrm{~m}\) a length of \(50.0 \mathrm{~m},\) and is shaped like an isosceles triangle. What is the angle between the base of the ship and the side of the ship as measured by a stationary observer if the ship is traveling by at a speed of \(0.400 c\) ? Plot this angle as a function of the speed of the ship.

Prove that in all cases, two sub-light-speed velocities "added" relativistically will always yield a sub-light-speed velocity. Consider motion in one spatial dimension only.

An astronaut in a spaceship flying toward Earth's Equator at half the speed of light observes Earth to be an oblong solid, wider and taller than it appears deep, rotating around its long axis. A second astronaut flying toward Earth's North Pole at half the speed of light observes Earth to be a similar shape but rotating about its short axis. Why does this not present a contradiction?

Although it deals with inertial reference frames, the special theory of relativity describes accelerating objects without difficulty. Of course, uniform acceleration no longer means \(d v / d t=g,\) where \(g\) is a constant, since that would have \(v\) exceeding \(c\) in a finite time. Rather, it means that the acceleration experienced by the moving body is constant: In each increment of the body's own proper time \(d \tau,\) the body acquires velocity increment \(d v=g d \tau\) as measured in the inertial frame in which the body is momentarily at rest. (As it accelerates, the body encounters a sequence of such frames, each moving with respect to the others.) Given this interpretation: a) Write a differential equation for the velocity \(v\) of the body, moving in one spatial dimension, as measured in the inertial frame in which the body was initially at rest (the "ground frame"). You can simplify your equation, remembering that squares and higher powers of differentials can be neglected. b) Solve this equation for \(v(t),\) where both \(v\) and \(t\) are measured in the ground frame. c) Verify that your solution behaves appropriately for small and large values of \(t\). d) Calculate the position of the body \(x(t),\) as measured in the ground frame. For convenience, assume that the body is at rest at ground-frame time \(t=0,\) at ground-frame position \(x=c^{2} / g\) e) Identify the trajectory of the body on a space-time diagram (Minkowski diagram, for Hermann Minkowski) with coordinates \(x\) and \(c t,\) as measured in the ground frame. f) For \(g=9.81 \mathrm{~m} / \mathrm{s}^{2},\) calculate how much time it takes the body to accelerate from rest to \(70.7 \%\) of \(c,\) measured in the ground frame, and how much ground-frame distance the body covers in this time.
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