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

[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.

 

[Matternot]

as long as i remember thesedoes'nt the effect of gravity depend on mass a =f/m?

or is this out of the rhealm of Newtons laws?

But what is F in this case?

 

[Nathanael]

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

No because "f" in the case of gravity is proportional to mass, so it cancels out.
This is the reason that the acceleration of gravity near Earth (9.8 m/s/s) is the same for all objects.

i don't know if I am cut out for physics

In my opinion, whether you're "cut out for physics" is more dependent on your work ethic than on your natural ability. You can lack natural ability and still make it by hard work, and if you have natural ability you still won't make it without hard work.

 

[J-dizzal]

But what is F in this case?

ok, g=g is the bottom line

 

[Matternot]

There is also the idea that despite finding the problems previously difficult, you now find them easy. All Physicists, good or bad find problems they can't solve. what separates the good from the bad is their ability to learn and persist. You seem to have done both here.

 

[J-dizzal]

No because "f" in the case of gravity is proportional to mass, so it cancels out.
This is the reason that the acceleration of gravity near Earth (9.8 m/s/s) is the same for all objects.In my opinion, whether you're "cut out for physics" is more dependent on your work ethic than on your natural ability. You can lack natural ability and still make it by hard work, and if you have natural ability you still won't make it without hard work.

Yea, I am just trying get quicker at solving physics problems. I will fail an exam if i take as long as i do to solve these homework problems

 

[J-dizzal]

There is also the idea that despite finding the problems previously difficult, you now find them easy. All Physicists, good or bad find problems they can't solve. what separates the good from the bad is their ability to learn and persist. You seem to have done both here.

what do you think is the best way to solve physics problems quickly? i think doing a bunch of practice problems would be the only way.

 

[Matternot]

Drawing a diagram is almost always useful but you seem to already do that well. Understanding e.g. formulae is always more important that simply memorising. Practice problems are helpful. Make sure you try doing questions without looking at the answer sheets.

 

[J-dizzal]

what do you think is the best way to solve physics problems quickly? i think doing a bunch of practice problems would be the only way.

and the difficult part about these problems is that i can be asked a problem just like the one previously solved but it will have

Drawing a diagram is almost always useful but you seem to already do that well. Understanding e.g. formulae is always more important that simply memorising. Practice problems are helpful. Make sure you try doing questions without looking at the answer sheets.

Drawing a diagram is almost always useful but you seem to already do that well. Understanding e.g. formulae is always more important that simply memorising. Practice problems are helpful. Make sure you try doing questions without looking at the answer sheets.

to me physics problems are difficult in that the technique used to solve the problem is not given, as opposed to a math class where you'd be solving a set of problem with a common technique that works for each problem.

 

[Matternot]

A skill which should come with practice and understanding.