An electron is shot through a box containing a constant electric field, getting accelerated in the process. The acceleration inside the box is \(a=2000 \mathrm{~m} / \mathrm{s}^{2}\). The width of the box is \(1 \mathrm{~m}\) and the electron enters the box with a velocity of \(100 \mathrm{~m} / \mathrm{s}\). (a) What is the velocity of the electron when it exits the box?

Short Answer

Expert verified
The final velocity of the electron when it exits the box is approximately 118.32 m/s.

Step by step solution

01

Identify the known values

Determine the given variables: The acceleration inside the box is specified as \(a = 2000 \text{ m/s}^2\), the width of the box is \(d = 1 \text{ m}\), and the initial velocity is \(v_i = 100 \text{ m/s}\).
02

Use the kinematic equation

Apply the kinematic equation that relates initial velocity, acceleration, distance, and final velocity: \[ v_f^2 = v_i^2 + 2ad \] where \(v_f\) is the final velocity.
03

Substitute the known values

Plug the known values into the equation: \[ v_f^2 = (100 \text{ m/s})^2 + 2 \times 2000 \text{ m/s}^2 \times 1 \text{ m} \]
04

Simplify the equation

Simplify the right-hand side of the equation: \[ v_f^2 = 10000 + 4000 = 14000 \]
05

Solve for the final velocity

Take the square root of both sides to find the final velocity: \[ v_f = \sqrt{14000} \text{ m/s} \approx 118.32 \text{ m/s} \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Kinematics
Kinematics is the study of motion without considering the forces causing it. When dealing with kinematics, we often use equations that describe the relationships between displacement, velocity, acceleration, and time. In our given problem, we used one of these kinematic equations to find out how an electron's velocity changes as it moves through a box with a constant acceleration. This equation helps us predict the final velocity of the electron based on its initial velocity, the acceleration it experiences, and the distance it travels. Such problems can be solved by carefully identifying the known values and choosing the appropriate kinematic equation.
Electric Field
An electric field is a region around a charged particle where a force would be exerted on other charged particles. The strength and direction of the electric field are determined by the nature of the charges and the distribution of those charges around the space. In our example, the electron moves through a box containing a constant electric field. This electric field applies a force on the electron, causing it to accelerate. This acceleration is a crucial factor that determines how the electron's velocity changes as it travels through the box. By understanding the role of the electric field, we can better analyze and predict the motion of charged particles within that field.
Acceleration
Acceleration is the rate of change of velocity of an object. It's a vector quantity, which means it has both magnitude and direction. In the exercise, the electron is subjected to a constant acceleration within the box. This acceleration is provided by the electric field and it tells us how much the electron's velocity increases each second. We're given an acceleration of 2000 m/s². By using the kinematic equation and the known values, we determine how the acceleration impacts the electron's final velocity when it exits the box. The concept of acceleration is vital for solving motion problems as it links the force exerted on an object to the change in its velocity over time.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Your roommate sets off early to school, walking leisurely at \(0.5 \mathrm{~m} / \mathrm{s}\). Thirty minutes after she left, you realize that she forgot her lecture notes. You decide to run after her to give her the notes. You run at a healthy \(3 \mathrm{~m} / \mathrm{s}\). (a) What is her position when you start running? (b) What is your position when \(tt_{1}\). (k) How can you use this result to find where you catch up with your roommate? (1) Where do you catch up with your roommate? (m) What parts of your solution strategy are general, that is, what parts of your strategy do not change if we change how either person moves?

In the film "The World's Fastest Indian" Anthony Hopkins plays Burt Munro who reaches a velocity of 201 mph in his 1920 Indian motorcycle. (a) At this velocity, how far does the Indian travel in 10 s? (b) How long time does the Indian need to travel \(1 \mathrm{~km}\) ?

In this project we address the motion of an object sliding on a slippery surface-such as a ski sliding in a snowy track. You will learn how to find the equation of motion for sliding systems both analytically and numerically, and to interpret the results. We start by studying a simplified situation called frictional motion: A block is sliding on a surface. moving with a velocity \(v\) in the positive \(x\)-direction. The forces from the interactions with the surface results in an acceleration: $$ a=\left\\{\begin{array}{cc} -\mu(|v|) g & v>0 \\ 0 & v=0 \\ \mu(|v|) g & v<0 \end{array}\right. $$ where \(g=9.8 \mathrm{~m} / \mathrm{s}^{2}\) is the acceleration of gravity. Let us first assume that \(\mu(v)=\) \(\mu=0.1\) for the surface. That is, we assume that the coefficient of friction does not depend on the velocity of the block. We give the block a push and release it with a velocity of \(5 \mathrm{~m} / \mathrm{s}\). (a) Find the velocity, \(v(t)\), of the block. (b) How long time does it take until the block stops? (c) Write a program where you find \(v(t)\) numerically using Euler's or Euler- Cromer's method. (Hint: You can find a program example in the textbook.) Use the program to plot \(v(t)\) and compare with your analytical solution. Use a timestep of \(\Delta t=0.01 .\) The description of friction provided above is too simplified. The coefficient of friction is generally not independent of velocity. For dry friction, the coefficient of friction can in some cases be approximated by the following formula: $$ \mu(v)=\mu_{d}+\frac{\mu_{s}-\mu_{d}}{1+v / v^{*}} $$ where \(\mu_{d}=0.1\) often is called the dynamic coefficient of friction, \(\mu_{s}=0.2\) is called the static coefficient of friction, and \(v^{*}=0.5 \mathrm{~m} / \mathrm{s}\) is a characteristic velocity for the contact between the block and the surface. (d) Show that the acceleration of the block is: $$ a(v)=-\mu_{d g}-g \frac{\mu_{s}-\mu_{d}}{1+v / v^{*}} $$ for \(v>0\). (e) Use your program to find \(v(t)\) for the more realistic model, with the same starting velocity, and compare with your previous results. Are your results reasonable? Explain. The model we have presented so far is only relevant at small velocities. At higher velocities the snow or ice melts, and the coefficient of friction displays a different dependency on velocity: $$ \mu(v)=\mu_{m}\left(\frac{v}{v_{m}}\right)^{-\frac{1}{2}} \text { when } v>v_{m} $$ where \(v_{m}\) is the velocity where melting becomes important. For lower velocities the model presented above with static and dynamic friction is still valid. (f) Show that $$ \mu_{m}=\mu_{d}+\frac{\mu_{s}-\mu_{d}}{1+v_{m} / v^{*}}, $$ in order for the coefficient of friction to be continuous at \(v=v_{m}\). (g) Modify your program to find the time development of \(v\) for the block when \(v_{m}=1.5 \mathrm{~m} / \mathrm{s}\). Compare with the two other models above: The model without velocity dependence and the model for dry friction. Comment on the results. (h) The process may be clearer if you plot the acceleration for all the three models in the same plot. Modify your program to plot \(a(t)\), plot the results, and comment on the results. What would happen if the initial velocity was much higher or much lower than \(5 \mathrm{~m} / \mathrm{s}\) ?

Is is possible to envision a motion where you for a period have no displacement, but non-zero velocity? (You may use an \(x\) ( \(t\) ) plot for illustration).

A car travelling at \(36 \mathrm{~km} / \mathrm{h}\) crashes into a mountainside. The crunchzone of the car deforms in the collision, so that the car effectively stops over a distance of \(1 \mathrm{~m}\). (a) Let us assume that the acceleration is constant during the collision, what is the acceleration of the car during the collision? (b) Compare with the acceleration of gravity, which is \(g=9.8 \mathrm{~m} / \mathrm{s}^{2}\).

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free