/*! 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 34 In a \(100-\mathrm{m}\) linear a... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 34

In a \(100-\mathrm{m}\) linear accelerator, an electron is accelerated to \(1.00 \%\) of the speed of light in \(40.0 \mathrm{m}\) before it coasts for \(60.0 \mathrm{m}\) to a target. (a) What is the electron's acceleration during the first \(40.0 \mathrm{m} ?\) (b) How long does the total flight take?

Short Answer

Expert verified
a) The electron's acceleration during the first 40.0 m is \(2.25 \times 10^{14} m/s^2\) b) The total flight takes \(3.33 \times 10^{-8} s.\)

Step by step solution

01

Determine the final velocity

First we need to determine the final velocity of the electron after 40m of acceleration. As the problem statement mentions 1 percent of the speed of light, we use the speed of light (\(c = 3.00 \times 10^{8} m/s\)) and calculate \(1.00 \%\) of it which gives us \(v_f = 0.01c = 3.00 \times 10^{6} m/s\)
02

Calculate the acceleration

Next, to calculate the acceleration, we use the kinematic equation \[ v^2 = u^2 + 2as \] where \(v\) is the final velocity, \(u\) is the initial velocity, \(a\) is the acceleration, \(s\) is the displacement. Here, initial velocity (\(u\)) is 0 as the electron starts from rest, displacement (\(s\)) is 40m and final velocity (\(v\)) is \(3.00 \times 10^{6} m/s\). Substituting these values we get \[ a = \frac{(v_f)^2}{2s} = \frac{(3.00 \times 10^{6})^2}{2*40} = 2.25 \times 10^{14} m/s^2\]
03

Find time taken for first 40m

The time (\(t1\)) taken for the first 40m can be found using the equation \(v = u + at\) where \(v\) is final velocity, \(a\) is acceleration (\(2.25 \times 10^{14} m/s^2\)), \(t\) is time and \(u\) is initial velocity which is 0 here. Rearranging the formula, \(t1 = v / a = 3.00 \times 10^{6} / 2.25 \times 10^{14} = 1.33 \times 10^{-8} s\)
04

Find time taken for last 60m

Since the electron coasts for the last \(60.0 m\), the speed is constant and equals the final velocity \(v_f\) calculated in the first step. Using the equation \(s = v t\) where \(s\) is distance, \(v\) is speed and \(t\) is time, we calculate \(t2 = s / v = 60 / 3.00 \times 10^{6} = 2.00 \times 10^{-8} s\)
05

Calculate Total Flight Time

Finally, we add the times \(t1\) and \(t2\) for the two sections of the trip to find the total time. So, total time \(t = t1 + t2 = 1.33 \times 10^{-8} + 2.00 \times 10^{-8} = 3.33 \times 10^{-8} s\)

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Key Concepts

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

Kinematic Equations
Kinematic equations are fundamental tools used in classical physics to describe the motion of objects under the influence of forces when air resistance is negligible. They connect variables such as initial velocity, final velocity, acceleration, time, and displacement without requiring the detailed forces involved in the motion.

An important kinematic equation used in physics problems, including those involving linear accelerators, is: \[ v^2 = u^2 + 2as \] where:\
  • \( v \) is the final velocity,
  • \( u \) is the initial velocity,
  • \( a \) is acceleration,
  • \( s \) is displacement.
This equation allows us to solve for one variable when the other three are known, providing a base to understand the relationship between an object's motion and the forces acting upon it.
Electron Acceleration
Electron acceleration, particularly in the context of linear accelerators or linacs, involves using electric fields to increase the kinetic energy of electrons. The electrons are accelerated from rest to high speeds, sometimes approaching the speed of light.In our exercise example, an electron in a 100-meter accelerator is being accelerated to a given fraction of the speed of light. This process involves converting potential energy from the electric field into kinetic energy of the electron. The substantial factor in this process is that the acceleration is constant, meaning that the electron increases its velocity at a steady rate while it is within the accelerating component of the linac.
Speed of Light
The speed of light is a crucial constant in physics, denoted as \( c \) and is approximately \( 3.00 \times 10^{8} \) meters per second in a vacuum. It is the highest possible speed at which energy, matter, and information can travel. In theoretical physics and under Einstein's theory of relativity, the speed of light connects space and time and has implications in concepts of simultaneity and the relativity of simultaneity.

In the context of our exercise, the speed of light is used as a reference point for the velocity that the electron is accelerated to within the linear accelerator. Since the electrons can never actually reach the speed of light, they are described in terms of a percentage of \( c \).
Velocity and Acceleration Calculations
Velocity and acceleration calculations are paramount in understanding motion. Velocity is the rate of change of displacement, a vector quantity specifying both magnitude and direction. Acceleration is the rate of change of velocity, which can result in changes in speed or direction or both.

Using the kinematic equations, we can determine an electron's acceleration and the velocities at different points, given some initial conditions and displacements. As shown in the solution, by rearranging kinematic equations, we can extract the acceleration of the electron and use this to find the time taken to reach a certain velocity or to travel a certain displacement without further acceleration鈥攁llowing a comprehensive analysis of the electron's motion during its journey in a linear accelerator.

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Most popular questions from this chapter

The position of a particle moving along the \(x\) axis varies in time according to the expression \(x=3 t^{2},\) where \(x\) is in meters and \(t\) is in seconds. Evaluate its position (a) at \(t=3.00 \mathrm{s}\) and (b) at \(3.00 \mathrm{s}+\Delta t\). (c) Evaluate the limit of \(\Delta x / \Delta t\) as \(\Delta t\) approaches zero, to find the velocity at \(t=3.00 \mathrm{s}\).

A student throws a set of keys vertically upward to her sorority sister, who is in a window \(4.00 \mathrm{m}\) above. The keys are caught 1.50 s later by the sister's outstretched hand. (a) With what initial velocity were the keys thrown? (b) What was the velocity of the keys just before they were caught?

Speedy Sue, driving at \(30.0 \mathrm{m} / \mathrm{s}\), enters a one-lane tunnel. She then observes a slow-moving van 155 m ahead traveling at \(5.00 \mathrm{m} / \mathrm{s} .\) Sue applies her brakes but can accelerate only at \(-2.00 \mathrm{m} / \mathrm{s}^{2}\) because the road is wet. Will there be a collision? If yes, determine how far into the tunnel and at what time the collision occurs. If no, determine the distance of closest approach between Sue's car and the van.

A 745 i BMW car can brake to a stop in a distance of \(121 \mathrm{ft}\). from a speed of \(60.0 \mathrm{mi} / \mathrm{h} .\) To brake to a stop from a speed of \(80.0 \mathrm{mi} / \mathrm{h}\) requires a stopping distance of \(211 \mathrm{ft}\) What is the average braking acceleration for (a) \(60 \mathrm{mi} / \mathrm{h}\) to rest, (b) \(80 \mathrm{mi} / \mathrm{h}\) to rest, \((\mathrm{c}) 80 \mathrm{mi} / \mathrm{h}\) to \(60 \mathrm{mi} / \mathrm{h} ?\) Express the answers in \(\mathrm{mi} / \mathrm{h} / \mathrm{s}\) and in \(\mathrm{m} / \mathrm{s}^{2}\) .

A car is approaching a hill at \(30.0 \mathrm{m} / \mathrm{s}\) when its engine suddenly fails just at the bottom of the hill. The car moves with a constant acceleration of \(-2.00 \mathrm{m} / \mathrm{s}^{2}\) while coasting up the hill. (a) Write equations for the position along the slope and for the velocity as functions of time, taking \(x=0\) at the bottom of the hill, where \(v_{i}=30.0 \mathrm{m} / \mathrm{s} .\) (b) Determine the maximum distance the car rolls up the hill.
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