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Chapter 6: Problem 13

The mass of a proton is 1836 times the mass of an electron. (a) A proton is traveling at speed \(V\) . At what speed (in terms of \(V )\) would an electron have the same kinetic energy as the proton? (b) An electron has kinetic energy \(K\) . If a proton has the same speed as the electron, what is its kinetic energy (in terms of \(K ) ?\)

Short Answer

Expert verified
(a) Electron's speed is \( V\sqrt{1836} \). (b) Proton's kinetic energy is \( 1836K \).

Step by step solution

01

Understanding the Relationship between Kinetic Energies

Kinetic energy is expressed by the formula \( KE = \frac{1}{2}mv^2 \). For both the proton and the electron to have the same kinetic energy, the kinetic energy of the proton must equal that of the electron. Thus, we have the equation \( \frac{1}{2}m_pV^2 = \frac{1}{2}m_eu^2 \), where \( m_p \) and \( m_e \) are the masses of the proton and electron, respectively, \( V \) is the proton's speed, and \( u \) is the electron's speed.
02

Expressing Proton Mass in Terms of Electron Mass

Given that the mass of a proton is 1836 times the mass of an electron, we can express the proton mass as \( m_p = 1836m_e \). Substituting this into the kinetic energy equation gives \( \frac{1}{2}(1836m_e)V^2 = \frac{1}{2}m_eu^2 \).
03

Solving for Electron Speed

Cancel out \( \frac{1}{2}m_e \) on both sides of the equation to simplify it to \( 1836V^2 = u^2 \). Taking the square root of both sides, we find \( u = V\sqrt{1836} \). Thus, the speed of the electron is \( V\sqrt{1836} \).
04

Understanding Kinetic Energy Expression for Electron's Speed

If the electron has kinetic energy \( K \), its kinetic energy expression is \( K = \frac{1}{2}m_ev^2 \), where \( v \) is the electron's speed. We need to determine the kinetic energy of a proton moving at the same speed \( v \).
05

Relating Electron's Speed to Proton's Kinetic Energy

With the proton having the same speed \( v \), its kinetic energy is \( KE_p = \frac{1}{2}m_pv^2 = \frac{1}{2}(1836m_e)v^2 \).
06

Expressing Proton's Kinetic Energy in Terms of Electron's Energy

Using the expression for electron's kinetic energy \( K = \frac{1}{2}m_ev^2 \), replace \( v^2 \) in the proton's energy equation: \( KE_p = 1836 \cdot K \). Thus, the proton's kinetic energy at that speed is \( 1836K \).

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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 a crucial concept in physics that describes the energy an object possesses due to its motion. The formula for kinetic energy is given by \( KE = \frac{1}{2}mv^2 \), where \( m \) is the mass of the object and \( v \) is its velocity. Understanding this formula is fundamental as it helps us calculate how much energy is needed for an object to move or how much it possesses while in motion. When comparing the kinetic energy of different particles, like a proton and an electron, we want them to have the same kinetic energy to determine equivalency or specific relationships between their speeds. For example, the kinetic energy of a proton can be set equal to that of an electron to derive their speed relationship. This principle is essential in the calculation of their relative energies in particle physics.
Mass Ratio of Proton and Electron
The mass ratio between a proton and an electron is a significant factor that influences their kinetic energy comparison. A proton is 1836 times heavier than an electron. This means that when both particles have the same kinetic energy, the intensity of speed they each require vastly differs due to their mass difference. This discrepancy plays a vital role when using the kinetic energy equation to solve problems involving particles of different masses. The larger mass of the proton means that it needs much higher velocity compensation to equate its kinetic energy with a much lighter electron. Therefore, the equation \( 1836V^2 = u^2 \) is derived from substituting this mass relationship into the kinetic energy formula, which is then used to solve for the speed ratio between the two.
Speed Calculation
Speed calculation for particles such as protons and electrons is a practical application of kinetic energy and mass ratios. To find out at what speed an electron must travel to have the same kinetic energy as a proton traveling at speed \( V \), the equation \( 1836V^2 = u^2 \) is solved. This results in the electron's speed being \( u = V\sqrt{1836} \). Breaking this down, the equation is derived by equating the kinetic energies and substituting the mass ratio. By simplifying the equation, we cancel terms to isolate the electron's speed \( u \), and then take the square root of both sides. Understanding this calculation process is important for accurately predicting particle behavior at different energies and velocities. It further illustrates how, due to their mass difference, the speed requirements differ drastically while maintaining the same kinetic energy.

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Most popular questions from this chapter

A 12 -pack of Omni-Cola (mass 4.30 \(\mathrm{kg} )\) is initially at rest on a horizontal floor. It is then pushed in a straight line for 1.20 \(\mathrm{m}\) by a trained dog that exerts a horizontal force with magnitude 36.0 \(\mathrm{N}\). Use the work-energy theorem to find the final speed of the 12 -pack if (a) there is no friction between the 12 -pack and the floor, and (b) the coefficient of kinetic friction between the 12 -pack and the floor is 0.30.

The aircraft carrier John \(F .\) Kennedy has mass \(7.4 \times 10^{7} \mathrm{kg}\). When its engines are developing their full power of \(280,000 \mathrm{hp}\), the John \(F\) . Kennedy travels at its top speed of 35 knots \((65 \mathrm{km} / \mathrm{h})\). If 70\(\%\) of the power output of the engines is applied to pushing the ship through the water, what is the magnitude of the force of water resistance that opposes the carrier's motion at this speed?

A mass \(m\) slides down a smooth inclined plane from an initial vertical height \(h,\) making an angle \(\alpha\) with the horizontal. (a) The work done by a force is the sum of the work done by the components of the force. Consider the components of gravity parallel and perpendicular to the surface of the plane. Calculate the work done on the mass by each of the components, and use these results to show that the work done by gravity is exactly the same as if the mass had fallen straight down through the air from a height \(h\) . (b) Use the work-energy theorem to prove that the speed of the mass at the bottom of the incline is the same as if it had been dropped from height \(h\) , independent of the angle \(\alpha\) of the incline. Explain how this speed can be independent of the slope angle. (c) Use the results of part (b) to find the speed of a rock that slides down an icy frictionless hill, starting from rest 15.0 \(\mathrm{m}\) above the bottom.

The gravitational pull of the earth on an object is inversely proportional to the square of the distance of the object from the center of the earth. At the earth's surface this force is equal to the object's normal weight \(m g,\) where \(g=9.8 \mathrm{m} / \mathrm{s}^{2},\) and at large distances, the force is zero. If a \(20,000-k g\) asteroid falls to earth from a very great distance away, what will be its minimum speed as it strikes the earth's surface, and how much kinetic energy will it impart to our planet? You can ignore the effects of the earth's atmosphere.

A force of 160 \(\mathrm{N}\) stretches a spring 0.050 \(\mathrm{m}\) beyond its unstretched length. (a) What magnitude of force is required to stretch the spring 0.015 \(\mathrm{m}\) beyond its unstretched length? To compress the spring 0.020 \(\mathrm{m} ?\) (b) How much work must be done to stretch the spring 0.015 \(\mathrm{m}\) beyond its unstretched length? To compress the spring 0.020 \(\mathrm{m}\) from its unstretched length?
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