Vertical deflection of an electron in a uniform force field

[Matternot]

If we instead consider the position of a particle I mentioned earlier. If we keep the x co-ordinate constant, but changed the y co-ordinate, the direction we would need to travel in order to reach the particle has changed. It doesn't mean the x co-ordinate has.

Do you see how we can relate this to the velocities?

 

[J-dizzal]

If we instead consider the position of a particle I mentioned earlier. If we keep the x co-ordinate constant, but changed the y co-ordinate, the direction we would need to travel in order to reach the particle has changed. It doesn't mean the x co-ordinate has.

Do you see how we can relate this to the velocities?

i see know

 

[Matternot]

maybe not, if there are no forces acting in the -x direction, then the x component of its velocity doesn't change. its only the y component increasing right?

Exactly! The y component is increasing, thus changing the direction. The x component does not need to change for this to happen

 

[Matternot]

So the x velocity is constant. What equation can we therefore use to calculate the time the particle is in motion for?

 

[J-dizzal]

If we instead consider the position of a particle I mentioned earlier. If we keep the x co-ordinate constant, but changed the y co-ordinate, the direction we would need to travel in order to reach the particle has changed. It doesn't mean the x co-ordinate has.

Do you see how we can relate this to the velocities?

so

Exactly! The y component is increasing, thus changing the direction. The x component does not need to change for this to happen

So the x velocity is constant. What equation can we therefore use to calculate the time the particle is in motion for?

t=dist/v = 3.091x10^-9s

 

[Matternot]

That seems correct to me.

We now need to use that time to calculate the vertical distance travelled.

 

[J-dizzal]

That seems correct to me.

We now need to use that time to calculate the vertical distance travelled.

yes just distance now. so Δy = v₀t - 1/2 at² this look like a good formula of motion to use?

 

[Matternot]

This looks good.

What values of v, a and t are you going to use?

 

[J-dizzal]

This looks good.

What values of v, a and t are you going to use?

v₀=0, a= 3.5126x10¹⁴, and t=3.091x10^-9

 

[Matternot]

Great! v₀ should be 0 because in the y direction it is initially at rest .

You may want to check your equation again though. Are you sure about the - sign?

Let me know what you get as your final answer.

 

[J-dizzal]

Great! v₀ should be 0 because in the y direction it is initially at rest .

You may want to check your equation again though. Are you sure about the - sign?

Let me know what you get as your final answer.

542872.33 m that seems huge. yea i rechecked formula thanks its + because v is the unknown not v₀

 

[J-dizzal]

Great! v₀ should be 0 because in the y direction it is initially at rest .

You may want to check your equation again though. Are you sure about the - sign?

Let me know what you get as your final answer.

542872.33 m seems large

 

[J-dizzal]

542872.33 m seems large

forgot to sqare the time one sec

 

[J-dizzal]

1.678 m

 

[Matternot]

Almost. Check your units.

 

[J-dizzal]

Almost. Check your units.

1.67810^-3

 

[J-dizzal]

lol thanks

 

[Matternot]

Great! Glad I could help. It seems like you learned a lot doing that question

 

[J-dizzal]

Almost. Check your units.

Thanks stephen very helpful!
have time to look at my circular motion problem? https://www.physicsforums.com/threads/uniform-circular-motion-problem.820429/#post-5149618

 

[J-dizzal]

Great! Glad I could help. It seems like you learned a lot doing that question

yea i learned some fundamental concepts that i know will help on the exam

 

[J-dizzal]

main thing being that force changes the velocity of a particles seems so obvious now

 

[Matternot]

Yeah. The other important thing is to know that you can simply treat the velocities separately. It is often easier to do this than try to work with vectors.

 

[Matternot]

I just had a look at the circular motion problem. Which part are you having problems with?

 

[J-dizzal]

I just had a look at the circular motion problem. Which part are you having problems with?

i can't seem to find any unknowns. i know that 3/4 of the distance of the circle is traveled by the particle in 9.90s-3.30s but when i try to derive an equation for total time(T) i cant. to solve for T i have so far 6.6T = 3pi/2 but plugging that solution into r =vt/2pi its to get the radius its not working

 

[Matternot]

So is the y co-ordinate obvious from your diagram?

To calculate T, a bit of common sense needs to be used. If 3/4 of the circle is completed in 6.6s, how long does it take to complete a full circle?

 

[J-dizzal]

So is the y co-ordinate obvious from your diagram?

To calculate T, a bit of common sense needs to be used. If 3/4 of the circle is completed in 6.6s, how long does it take to complete a full circle?

8.8s would be the whole right?
the y coordinate of?

 

[J-dizzal]

So is the y co-ordinate obvious from your diagram?

To calculate T, a bit of common sense needs to be used. If 3/4 of the circle is completed in 6.6s, how long does it take to complete a full circle?

ok 8.8 works, it must of made a mistake in the math last time i tried it.

 

[Matternot]

So that question is now solved?

 

[J-dizzal]

So that question is now solved?

just working on the y coordinate now

 

[Matternot]

Take a look at your diagram.