Displacement of a particle moving on the xy axis?

AI Thread Summary
The discussion focuses on calculating the displacement of an electron moving along the x-axis under constant acceleration in both x and y directions. Given the initial velocity and acceleration values, the time taken for the electron to travel a distance of 0.0100 m is determined to be approximately 5.67458 × 10^-10 seconds. The next step involves using this time to find the y-coordinate of the electron, utilizing the known acceleration in the y-direction. Participants suggest using the same kinematic equations applied for the x-direction to solve for the y-coordinate. The conversation emphasizes the importance of correctly applying the equations of motion to find the final displacement.
JessicaJ283782
Messages
41
Reaction score
0
As an example of the manipulation of an electron beam, consider an electron traveling away from the origin along the x-axis in the xy plane with initial velocity vi= vii. As it passes through the region x = 0 to x = d, the electron experiences acceleration a = axi + ayj, where ax and ay are constants. For the case vi = 1.74 ✕ 10^7 m/s, ax = 7.84 ✕ 10^14 m/s2, and ay = 1.51 ✕ 10^15 m/s2, determine the following, at x = d = 0.0100 m.

(a) the position of the electron (yf=?)So, I went through, and I used the equation d=vi*t+.5(at^2)

I plugged in: .01=1.74*10^7t+.5(7.84*10^14)t^2

I got the time to be 5.67458*10^-10

Now, I know you have to plug the time into an equation, along with the acceleration of y, but I am confused on which equation to use since you only know acceleration and time for the y coordinates?(b) the velocity of the electron

So, I used: vf=vi+at

vf(x)=5.67458*10^-10 X 7.84*10^14

and wouldn't you do the same for the y coordinates?

Thank you!
 
Physics news on Phys.org
a)
So at this point in time, you now need to find the y-coordinate of the electron. You know the acceleration in the y-direction. Can't you find an equations, which you could then use? Perhaps the very same equation, you just used to find the time, just with different parameters?
 
Actually you know the velocity of the electron in the y-direction it is 0 m/s, therefore v=1.74*10^7i+0j m/s
and a=7.84*10^14i+1.5*10^15j, you have solved for t and you know vi and ai so you can solve for y. You are almost there.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...

Similar threads

Replies
2
Views
2K
Replies
2
Views
3K
Replies
2
Views
1K
Replies
1
Views
2K
Replies
4
Views
1K
Replies
27
Views
3K
Replies
3
Views
15K
Replies
11
Views
4K
Back
Top