/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 75 A particle of unknown mass \(M\)... [FREE SOLUTION] | 91Ó°ÊÓ

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A particle of unknown mass \(M\) decays into two particles of known masses \(m_{a}=0.5 \mathrm{GeV} / c^{2}\) and \(m_{b}=1.0 \mathrm{GeV} / c^{2},\) whose momenta are measured to be \(\mathbf{p}_{a}=2.0 \mathrm{GeV} / \mathrm{c}\) along the \(x_{2}\) axis and \(\mathbf{p}_{b}=1.5 \mathrm{GeV} / c\) along the \(x_{1}\) axis. \(\left(1 \mathrm{GeV}=10^{9} \mathrm{eV} .\right)\) Find the unknown mass \(M\) and its speed.

Short Answer

Expert verified
\(M \approx 2.94 \, \text{GeV}/c^2\), \(v \approx 0.660c\).

Step by step solution

01

Calculate Energy of Particle a

Use the energy-momentum relation for particle \(a\):\[ E_a = \sqrt{(m_a c^2)^2 + (p_a c)^2} \]Given \(m_a = 0.5 \, \text{GeV}/c^2\) and \(p_a = 2.0 \, \text{GeV}/c\), calculate \(E_a\):\[ E_a = \sqrt{(0.5)^2 + (2.0)^2} = \sqrt{0.25 + 4} = \sqrt{4.25} \, \text{GeV} \]\[ E_a \approx 2.06 \, \text{GeV} \]
02

Calculate Energy of Particle b

Use the energy-momentum relation for particle \(b\):\[ E_b = \sqrt{(m_b c^2)^2 + (p_b c)^2} \]Given \(m_b = 1.0 \, \text{GeV}/c^2\) and \(p_b = 1.5 \, \text{GeV}/c\), calculate \(E_b\):\[ E_b = \sqrt{(1.0)^2 + (1.5)^2} = \sqrt{1 + 2.25} = \sqrt{3.25} \, \text{GeV} \]\[ E_b \approx 1.80 \, \text{GeV} \]
03

Calculate Total Energy and Momentum of Initial Particle

The total energy \(E\) of the original particle is the sum of the energies of particles \(a\) and \(b\):\[ E = E_a + E_b = 2.06 + 1.80 = 3.86 \, \text{GeV} \]Calculate the total momentum vector \(\mathbf{p}\):\[ \mathbf{p} = (p_{b(x_1)}, p_{a(x_2)}) = (1.5 \, \text{GeV}/c, 2.0 \, \text{GeV}/c) \]Magnitude of \(\mathbf{p}\):\[ p = \sqrt{1.5^2 + 2.0^2} = \sqrt{2.25 + 4} = \sqrt{6.25} = 2.5 \, \text{GeV}/c \]
04

Find Unknown Mass M of Initial Particle

Use the invariant mass equation:\[ M c^2 = \sqrt{E^2 - (pc)^2} \]Substitute values:\[ M = \sqrt{(3.86)^2 - (2.5)^2} = \sqrt{14.8996 - 6.25} = \sqrt{8.6496} \, \text{GeV}/c^2 \]\[ M \approx 2.94 \, \text{GeV}/c^2 \]
05

Calculate Speed of Initial Particle

Use the relation \( E = \gamma M c^2 \) and \( p = \gamma M v \):First, find \( \gamma \):\[ \gamma = \frac{E}{M c^2} = \frac{3.86}{2.94} \approx 1.31 \]Find \( v \) using \( p = \gamma M v \):\[ v = \frac{p}{\gamma M} = \frac{2.5}{1.31 \times 2.94} c \]\[ v \approx 0.660 c \]
06

Conclusion: Unknown Mass and Speed

The unknown mass \(M\) is approximately \(2.94 \, \text{GeV}/c^2\), and its speed is approximately \(0.660c\).

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Key Concepts

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

Energy-Momentum Relation
The energy-momentum relation is a crucial concept in relativistic mechanics. It links the energy of a particle with its momentum and mass, allowing us to derive quantities like energy when momentum and mass are known. Given as:
  • \( E = \sqrt{(m c^2)^2 + (pc)^2} \)
This equation stems from Einstein’s theory of relativity. It shows how energy is not merely dependent on mass but also momentum, especially significant at high speeds.
For our case with particle \( a \) and \( b \), using this relation helps us find their respective energies knowing their masses and momenta.
Knowing the energy of the final state particles is key as it helps find the initial state quantities such as total energy and momentum, later used to find the invariant mass.
Invariant Mass
Invariant mass, often denoted as \( M \), is a scalar quantity that remains the same, irrespective of the reference frame. It's sometimes called rest mass and provides a direct measure of a particle system's "weight".
The invariant mass can be determined using the total energy and momentum of the system through:
  • \( M c^2 = \sqrt{E^2 - (pc)^2} \)
Here, \( E \) is the total energy, and \( p \) is the total momentum.
This formula tells us that even though the energy and momentum can vary based on the observer's frame of reference, the invariant mass is constant.
In our particle decay example, knowing the decay products' energies and momenta helps compute the original particle's mass.
Particle Decay
Particle decay refers to a process where a particle transforms into other particles. It's a common phenomenon in high-energy physics and is governed by conservation laws.
In the decay process considered in the problem, a particle of unknown mass decays into two particles with known masses and measurable momenta.
Important points to remember about particle decay:
  • Conservation of Energy: Total energy before decay equals total energy after.
  • Conservation of Momentum: Total momentum before decay equals total momentum after.
By applying these conservation laws, you can find the energy and momentum of the decayed particles and use them to compute unknown quantities, such as the original particle's invariant mass and its speed, using relativistic equations.
Relativistic Speed
Relativistic speed occurs when a particle moves at speeds close to the speed of light, \( c \). At these speeds, traditional velocity calculations using Newtonian mechanics break down, as relativistic effects become significant.
The speed of a particle can be found using the relation:
  • \( E = \gamma M c^2 \)
  • \( p = \gamma M v \)
Here, \( \gamma \) is the Lorentz factor given by \( \gamma = \frac{1}{\sqrt{1 - (v/c)^2}} \), reflecting how time and length distort at high velocities.
In our calculation, the unknown particle's speed approximates to 0.660c, indicating significant relativistic effects due to its high velocity.
Understanding relativistic speed ensures accurate analysis of a particle's behavior at speeds nearing that of light, crucial in high-energy physics experiments.

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Most popular questions from this chapter

A low-flying earth satellite travels at about 8000 m/s. What is the factor \(\gamma\) for this speed? As observed from the ground, by how much would a clock traveling at this speed differ from a ground-based clock after one hour (as measured by the latter)? What is the percent difference?

The muons created by cosmic rays in the upper atmosphere rain down more-or- less uniformly on the earth's surface, although some of them decay on the way down, with a half-life of about \(1.5 \mu\) s (measured in their rest frame). A muon detector is carried in a balloon to an altitude of \(2000 \mathrm{m}\), and in the course of an hour detects 650 muons traveling at \(0.99 c\) toward the earth. If an identical detector remains at sea level, how many muons should it register in one hour? Calculate the answer taking account of the relativistic time dilation and also classically. (Remember that after \(n\) half-lives, \(2^{-n}\) of the original particles survive.) Needless to say, the relativistic answer agrees with experiment.

(a) Show that if a body has speed \(v < c\) in one inertial frame, then \(v < c\) in all frames. [Hint: Consider the displacement four-vector \(d x=(d \mathbf{x}, c d t),\) where \(d \mathbf{x}\) is the three-dimensional displacement in a short time \(d t .]\) (b) Show similarly that if a signal (such as a pulse of light) has speed \(c\) in one frame, its speed is \(c\) in all frames.

An excited state \(X^{*}\) of an atom at rest drops to its ground state \(X\) by emitting a photon. In atomic physics it is usual to assume that the energy \(E_{\gamma}\) of the photon is equal to the difference in energies of the two atomic states, \(\Delta E=\left(M^{*}-M\right) c^{2},\) where \(M\) and \(M^{*}\) are the rest masses of the ground and excited states of the atom. This cannot be exactly true, since the recoiling atom X must carry away some of the energy \(\Delta E .\) Show that in fact \(E_{\gamma}=\Delta E\left[1-\Delta E /\left(2 M^{*} c^{2}\right)\right] .\) Given that \(\Delta E\) is of order a few ev, while the lightest atom has \(M\) of order \(1 \mathrm{GeV} / c^{2},\) discuss the validity of the assumption that \(E_{\gamma}=\Delta E\)

Frame \(\mathcal{S}^{\prime}\) travels at speed \(V_{1}\) along the \(x\) axis of frame \(\mathcal{S}\) (in the standard configuration). Frame \(\mathcal{S}^{\prime \prime}\) travels at speed \(V_{2}\) along the \(x^{\prime}\) axis of frame \(\mathcal{S}^{\prime}\) (also in the standard configuration). By applying the standard Lorentz transformation twice find the coordinates \(x^{\prime \prime}, y^{\prime \prime}, z^{\prime \prime}, t^{\prime \prime}\) of any event in terms of \(x, y, z, t .\) Show that this transformation is in fact the standard Lorentz transformation with velocity \(V\) given by the relativistic "sum" of \(V_{1}\) and \(V_{2}\)

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