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2006 Physics Exam How did it go?


Posted 18 May 2006 - 02:33 PM

QUOTE(smit @ May 18 2006, 02:53 PM) View Post

Question 6 the oil drop, how do you get the number of excess? I got how many electrons there were but didnt know what it wanted me to do.

A neutral drop has a charge of 0. So for a drop to have a charge of -1.6e-12 it must have 10,000,000 excess electrons (1.6e-12/1.6e-19).


Posted 18 May 2006 - 02:39 PM

hmm.. was the drop neutral? I thought it needed to have a charge of at least one electron?

Jason Bourne

Posted 18 May 2006 - 02:44 PM



Posted 18 May 2006 - 06:08 PM

For the question on the cyclotron - if you worked out the radius when it enters the right hand dee (7 a ii) i doubled that radius for when it enters the left hand dee because when you look at the diagram (if you have the paper) the diameter of the old path is now the radius of the new path. I cant remember what that gave me but i ll work it out when i have stopped revision for maths for tomorrow


Posted 19 May 2006 - 06:27 PM

That Dee question was uncalled for.

The good news is that almost NOONE got it.

An easy-ish exam.

Jason Bourne

Posted 20 May 2006 - 09:59 AM

Is anyone %100 sure they know how to do the dee question. If so please say how as im sure alot of people would like to know (or maybe not).

Im guessing you had to use  \frac {1} {2} mv^2=QV to get part ai.

Using  Bqv= \frac {mv^2} {r} for part aii.

For part aiii. it was only 2 marks ..............so couldnt have been anything too complicated. Its accelerated by the same potential so i just doubled the velocity, i duno if this is right.

par b:  W=QV=e \times \frac {79e} {4\pi \varepsilon_o r}

part c:  m= \frac {m_o} {\sqrt {1-\frac {v^2} {c^2}}}

Dont know if any of this is right.


Posted 21 May 2006 - 05:46 PM

The new kinetic energy is just equal to the existing kinetic energy plus the work done by the potential to accelerate it.


Posted 23 May 2006 - 09:17 PM

QUOTE(FcukWIT @ May 17 2006, 05:17 PM) View Post
Btw. How many excess electrons were there?

107 or something. You divide the overall charge by e.

Yeah I just doubled the velocity for that question, didn't really know what else to do.


Posted 31 May 2006 - 06:09 PM

It wasn't exactly twice the initial velocity when it went into the left dee because although it was getting the same amount of energy the E(k) isn't directly proportional to the velocity, it's proportional to the velocity^2. I worked it out by saying

Final E(k) = Initial E(k) + QV

But I guess you could have just multiplied that first speed by root 2

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