Vertical Deflection of Electron: 1.1×107 m/s & 3.2×10-16 N

[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.

 

[J-dizzal]

Take a look at your diagram.

how is the y coordinate of center of the circle also 6.8 i don't see it

 

[J-dizzal]

how is the y coordinate of center of the circle also 6.8 i don't see it

ok i see it know that i draw it on a proper sized coordinate system

 

[Matternot]

where is the centre in relation to $(5.1, 6.8)$

 

[J-dizzal]

where is the centre in relation to $(5.1, 6.8)$

point 5.10,6.80 is horizontal to the center therefore the same y component. lol

 

[Matternot]

Cool :D Glad that's settled.

 

[J-dizzal]

where is the centre in relation to $(5.1, 6.8)$

ive spent over 2 days on these questions... they seem so easy now. i don't know if I am cut out for physics

 

[Nathanael]

Gravity will be negligible in this question as m<<1.

Are you implying that the effect (acceleration) of gravity depends on mass?
The real reason it would be negligible is because 9.8<<10^14 (≈F/m)
(But the problem didn't even mention that it was taking place near Earth.)

P.S.
Doesn't it make more sense to discuss the circular motion problem in the circular motion thread?

 

[Matternot]

Are you implying that the effect (acceleration) of gravity depends on mass?
The real reason it would be negligible is because 9.8<<10^14 (≈F/m)
(But the problem didn't even mention that it was taking place near Earth.)

P.S.
Doesn't it make more sense to discuss the circular motion problem in the circular motion thread?

Sorry, yeah, you're right about that. so used to thinking about acceleration of gravity being far less that of electrostatic. Slip of the mind.

Re P.S. probably.

 

[J-dizzal]

ive spent over 2 days on these questions... they seem so easy now. i don't know if I am cut out for physics

as long as i remember these

Are you implying that the effect (acceleration) of gravity depends on mass?
The real reason it would be negligible is because 9.8<<10^14 (≈F/m)
(But the problem didn't even mention that it was taking place near Earth.)

does'nt the effect of gravity depend on mass a =f/m?

or is this out of the rhealm of Newtons laws?

 

[Matternot]

But in practice my automatic method of thinking was correct. For a larger mass the weight would be comparable. Looking back, not the best way of explaining why it's negligable. Comparing the forces leads to the same result.