Position of electron motion 2D

In summary: Thank you for the help!In summary, an electron with an initial velocity of vi= viî travels away from the origin along the x-axis in the xy plane. As it passes through the region x = 0 to x = d, it experiences an acceleration of a = axî + ayj, where ax and ay are constants. For the given values of vi = 1.80 ✕ 10^7 m/s, ax = 8.00✕ 10^14 m/s2, and ay = 1.60 ✕ 10^15 m/s2, at x = d = 0.0100 m, we can determine the position, velocity, speed, and direction of travel
  • #1
Ab17
99
2

Homework Statement



consider an electron traveling away from the origin along the x-axis in the xy plane with initial velocity vi= viî. As it passes through the region x = 0 to x = d, the electron experiences acceleration a = axî + ayj, where ax and ay are constants. For the case vi = 1.80 ✕ 10^7 m/s, ax = 8.00✕ 10^14 m/s2, and ay = 1.60 ✕ 10^15 m/s2, determine the following, at x = d = 0.0100 m. (a) the position of the electron, (b) the velocity of the electron, (c) the speed of the electron, and (d) the direction of travel of the electron (i.e., the angle between its velocity and the x axis)

Homework Equations


Xf = Xi + Vt + 1/2at2
Vf2= vi2 + 2ax
Vf= Vi + at

The Attempt at a Solution


Vf2 = vi2 +2ax
Vf = sqrt((1.80x10^7)^2 +2(8.00x10^14)(0.01))
Vf = 18×10^6Time taken to reach this velocity:
Vf = vi + at
t = 225s

Therefore y position after 225s
Yf = yi + viyt +1/2at2
yf = 0 + 1/2(1.60×10^5)(225)^2
Yf = 4.05× 10^9

rf = (0.01î + 4.05x10^9j)m
Magnitude = 4.05×10^9m

(b) vfx = 18x10^6î
Vfy= viy +at
Vfy = 0 + (1.60×10^5)(225)
Vfy = 36×10^6j
Vf = (18× 10^6)î + (36x10^6)j
Magnitude = 40.25 ×10^6
Direction tan@ = 36x10^6/18×10^6
@= 63.43 NE

Are these answers right or did I make a mistake
 
Physics news on Phys.org
  • #2
Without checking your arithmetic, it looks OK.
I think you just need to tidy up your notation and add comments to show your reasoning.
That would also help you (and others) troubleshoot your own work.

ie.
The initial velocity is: ##\vec v(0)= v_i\hat\imath##
Acceleration is: ##\vec a = a_x\hat\imath + a_y\hat\jmath##
Initial and final displacements: ##x_i=x(0)=0,\; x_f=x(T)=d,\; y_i=y(0)=0##
... ie. ##x=d## when ##t=T##.

Position would be: ##\vec r(t) = x(t)\hat\imath + y(t)\hat\jmath##
Also velocity is: ##\vec v(t) = v_x(t)\hat\imath + v_y(t)\hat\jmath##

You need to find the final position: ##\vec r(T)=\vec r_f=d\hat\imath + y(T)\hat\jmath## and the final velocity, ##\vec v_f=\vec v(T)##

See how to keep the variables clear? See how I've distinguished the vectors from their magnitudes?
Also watch how clear my reasoning is when I add comments to show what everything is.

Use suvat equations.

Velocity is the easiest - so do that first:
x-component of final velocity is given by:##v_x^2(T)=v_i^2 + a_xd##
y-component of final velocity is given by: ##v_y^2(T)=a_yy_f##
... that what you had in mind?

##y_f## is given by: ##y_f = \frac{1}{2}a_yT^2## and ##v_x(T)=v_i+a_xT##
So solve the second for ##T## and substitute into the first one. Which I think is what you did.
Gets you something like ##y_f=\frac{1}{2a_x}\Delta v_x^2##

That was the hard work.
From there you can pretty much just read off the answers.
(a) final position: ##\vec r_f = d\hat\imath + y_f\hat\jmath##
(b) the velocity: ##\vec v_f = \sqrt{(v_i^2 + a_xd)}\hat\imath + \sqrt{a_yd}\hat\jmath##
From the velocity, you get (c) the speed: ##v_f = |\vec v_f| = \sqrt{v_i^2 + a_xd + a_yy_f}##
...and also (d) the angle to the x-axis: ##\theta = \tan^{-1}\big( (v_i^2 + a_xd)/(a_yy_f) \big)##

... and we are done.
If I've made a mistake in the above, it will probably be easy to find ;)
 
  • #3
Ok thank you I will setup my solutions like this from now on
 
  • Like
Likes Simon Bridge

1. What is the significance of the position of electron motion in 2D?

The position of electron motion in 2D refers to the location of an electron in a two-dimensional space, such as a plane or surface. This is important because the position of an electron determines its behavior and interactions with other particles, and is a key factor in understanding various physical and chemical phenomena.

2. How is the position of electron motion in 2D measured?

The position of electron motion in 2D can be measured using various techniques, such as scanning tunneling microscopy or electron diffraction. These methods use sophisticated instruments to track the movement of electrons and create images of their positions in a two-dimensional space.

3. Can the position of electron motion in 2D be predicted?

The position of electron motion in 2D can be predicted to some extent using quantum mechanics. However, due to the probabilistic nature of electrons, it is impossible to precisely determine their position at any given time. We can only calculate the probability of an electron being in a certain position at a certain time.

4. How does the position of electron motion in 2D affect the properties of materials?

The position of electron motion in 2D plays a crucial role in determining the properties of materials. For example, in semiconductors, the movement of electrons in a two-dimensional space can be controlled to create specific electrical properties. In metals, the position of electron motion affects their conductivity and thermal properties.

5. Are there any real-world applications of studying the position of electron motion in 2D?

Yes, there are many real-world applications of studying the position of electron motion in 2D. This includes the development of new materials for electronics, understanding the behavior of electrons in nanoscale devices, and improving our understanding of chemical reactions and catalysis. Additionally, studying the position of electron motion in 2D is crucial for advancements in fields such as quantum computing and renewable energy.

Similar threads

  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
720
  • Introductory Physics Homework Help
Replies
2
Views
4K
  • Introductory Physics Homework Help
Replies
3
Views
14K
  • Introductory Physics Homework Help
Replies
4
Views
802
  • Introductory Physics Homework Help
Replies
6
Views
6K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
916
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
1K
Back
Top