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A proton (rest mass \(1.67 \times 10^{-27} \mathrm{~kg}\) ) has total energy that is 4.00 times its rest energy. What are (a) the kinetic energy of the proton; (b) the magnitude of the momentum of the proton; (c) the proton's speed?

Short Answer

Expert verified
The kinetic energy of the proton is \(4.50 \times 10^{-10} \mathrm{J}\), the magnitude of the momentum is \(1.8859 \times 10^{-19} \mathrm{kg \cdot m/s}\), and the speed of the proton is \(2.6788 \times 10^{8} \mathrm{~m/s}\).

Step by step solution

01

Find the Rest Energy

The rest energy of the proton can be calculated using Einstein's equation for rest energy: \(E_{0} = mc^{2}\) where \(m\) is the rest mass of the proton, \(1.67 \times 10^{-27} \mathrm{~kg}\) , and \(c\) is the speed of light, \(2.998 \times 10^{8} \mathrm{~m/s}\). This gives \(E_{0} = 1.67 \times 10^{-27} \mathrm{~kg} \times (2.998 \times 10^{8} \mathrm{~m/s})^{2} = 1.50 \times 10^{-10} \mathrm{J}\).
02

Calculate the Total and Kinetic Energy

Given the total energy is 4 times the rest energy, so it is \(4 \times 1.50 \times 10^{-10} \mathrm{J} = 6.00 \times 10^{-10} \mathrm{J}\). And the kinetic energy is the difference between the total energy and the rest energy, \(6.00 \times 10^{-10} \mathrm{J} - 1.50 \times 10^{-10} \mathrm{J} = 4.50 \times 10^{-10} \mathrm{J}\).
03

Calculate the Momentum

The magnitude of the momentum is given by the equation \(p = \sqrt{E^{2} - E_{0}^{2}}/c\), where \(E\) is the total energy we found in step 2. The result is \(p = \sqrt{(6.00 \times 10^{-10} \mathrm{J})^{2} - (1.50 \times 10^{-10} \mathrm{J})^{2}}/(2.998 \times 10^{8} \mathrm{~m/s}) = 1.8859 \times 10^{-19} \mathrm{kg \cdot m/s}\).
04

Calculate the Speed

The equation relating speed \(v\), light speed \(c\), rest mass \(m\) and the total mass \(m'\) is \(v = c \sqrt{1 - (m/m')^{2}}\). Here, rest mass \(m\) is rest energy divided by \(c^{2}\), and the total mass \(m'\) is total energy divided by \(c^{2}\). Substituting these along with \(c\) value gives \(v = 2.998 \times 10^{8} \mathrm{~m/s} \cdot \sqrt{1 - (1.50 \times 10^{-10} \mathrm{J}/6.00 \times 10^{-10} \mathrm{J})} = 2.6788 \times 10^{8} \mathrm{~m/s}\).

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Key Concepts

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

Rest Energy
Rest energy is a fundamental concept in physics that describes the energy possessed by an object due to its mass, even when it is at rest. According to Einstein's famous equation, rest energy is calculated as \( E_0 = mc^2 \), where \( m \) represents the mass of the object and \( c \) is the speed of light, approximately \( 2.998 \times 10^8 \ \mathrm{m/s} \).

For a proton, with a rest mass of \( 1.67 \times 10^{-27} \ \mathrm{kg} \), its rest energy is \( 1.50 \times 10^{-10} \ \mathrm{J} \). This energy is a baseline level and it does not change due to the state of motion, underscoring the intrinsic nature of mass-energy equivalence.

Rest energy provides a platform to understand how energy in a high-speed particle like a proton behaves under the principles of relativity.
Kinetic Energy
Kinetic energy is the additional energy an object possesses due to its motion relative to a chosen frame of reference. In a relativistic context, the kinetic energy is not merely \( \frac{1}{2}mv^2 \) as used in classical mechanics. Instead, it is derived from the difference between the total energy and rest energy.

For the proton whose exercise we are analyzing, the total energy is four times the rest energy. Therefore, the kinetic energy, \( KE \), calculated as \( E_{total} - E_0 \), translates to \( 6.00 \times 10^{-10} \ \mathrm{J} - 1.50 \times 10^{-10} \ \mathrm{J} = 4.50 \times 10^{-10} \ \mathrm{J} \).

This kinetic energy accounts for how energy scales as speed approaches that of light, providing insight into energy transformations in high-speed physics.
Relativistic Momentum
Relativistic momentum adjusts the classical definition of momentum to pinpoint the behavior of objects moving at significant fractions of the speed of light. Classical momentum is defined as \( p=mv \), but in relativistic terms, the relation \( p = \sqrt{E^2 - E_0^2}/c \) is used.

This equation shows that momentum is influenced more significantly by energy in a system where velocities approach the speed of light, \( c \). In this instance, for the proton, substituting the known values yields a momentum magnitude of \( 1.8859 \times 10^{-19} \ \mathrm{kg \cdot m/s} \).

Understanding relativistic momentum helps comprehend changes in mass and energy with respect to speed, showcasing the intricacies of moving particles.
Einstein's Equation
Einstein's Equation, often signified by \( E=mc^2 \), lays the groundwork for understanding how mass can be converted into energy and vice versa. It is a central equation in the realm of special relativity, expressing the interconvertability between mass and energy.

Applying this equation, the total energy of a moving proton is evaluated by its product of mass and the square of the speed of light, illustrating how significant energy values rise beyond rest energy in rapid motion, a concept vital to nuclear physics and particle science.

This equation not only calculates rest energy but guides us in analyzing scenarios such as nuclear reactions, where mass defect is transformed into enormous amounts of energy.
Proton Speed
Proton speed involves examining how fast protons travel, especially under conditions where their kinetic and total energy rise to substantial levels. To determine the proton's speed, we use the equation \( v = c \sqrt{1 - (m/m')^2} \). This takes into account the reduction of speed relative to light as the mass increases with energy.

With the given data, the proton’s speed calculates to approximately \( 2.6788 \times 10^8 \ \mathrm{m/s} \), a noteworthy percentage of the speed of light.

Probing proton speed helps illustrate the effects of special relativity, highlighting how objects appear to "slow down" relative to their energy, despite accelerating continuously within the confines of relativistic physics.

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Most popular questions from this chapter

Physicists and engineers from around the world came together to build the largest accelerator in the world, the Large Hadron Collider (LHC) at the CERN Laboratory in Geneva, Switzerland. The machine accelerates protons to high kinetic energies in an underground ring \(27 \mathrm{~km}\) in circumference. (a) What is the speed \(v\) of a proton in the \(\mathrm{LHC}\) if the proton's kinetic energy is \(7.0 \mathrm{TeV} ?\) (Because \(v\) is very close to \(c,\) write \(v=(1-\Delta) c\) and give your answer in terms of \(\Delta .\) ) (b) Find the relativistic mass, \(m_{\text {rel }}\), of the accelerated proton in terms of its rest mass.

A cube of metal with sides of length \(a\) sits at rest in a frame \(S\) with one edge parallel to the \(x\) -axis. Therefore, in \(S\) the cube has volume \(a^{3}\). Frame \(S^{\prime}\) moves along the \(x\) -axis with a speed \(u .\) As measured by an observer in frame \(S^{\prime},\) what is the volume of the metal cube?

As a research scientist at a linear accelerator, you are studying an unstable particle. You measure its mean lifetime \(\Delta t\) as a function of the particle's speed relative to your laboratory equipment. You record the speed of the particle \(u\) as a fraction of the speed of light in vacuum \(c .\) The table gives the results of your measurements. $$ \begin{array}{l|lllllll} u / c & 0.70 & 0.80 & 0.85 & 0.88 & 0.90 & 0.92 & 0.94 \\ \hline \Delta t\left(10^{-8} \mathrm{~s}\right) & 3.57 & 4.41 & 5.02 & 5.47 & 6.05 & 6.58 & 7.62 \end{array} $$ (a) Your team leader suggests that if you plot your data as \((\Delta t)^{2}\) versus \(\left(1-u^{2} / c^{2}\right)^{-1},\) the data points will be fit well by a straight line. Construct this graph and verify the team leader's prediction. Use the best-fit straight line to your data to calculate the mean lifetime of the particle in its rest frame. (b) What is the speed of the particle relative to your lab equipment (expressed as \(u / c\) ) if the lifetime that you measure is four times its rest-frame lifetime?

The net force \(\vec{F}\) on a particle of mass \(m\) is directed at \(30.0^{\circ}\) counterclockwise from the \(+x\) -axis. At one instant of time, the particle is traveling in the \(+x\) -direction with a speed (measured relative to the earth) of \(0.700 c .\) At this instant, what is the direction of the particle's acceleration?

Two events are observed in a frame of reference \(S\) to occur at the same space point, the second occurring \(1.80 \mathrm{~s}\) after the first. In a frame \(S^{\prime}\) moving relative to \(S\), the second event is observed to occur \(2.15 \mathrm{~s}\) after the first. What is the difference between the positions of the two events as measured in \(S^{\prime} ?\)

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