I posted this in another forum, but if anyone's interested, my answers:

1.a)proof, ds/dt = u+at and integrate

b)4.19*10^-10 J

2.a) i) 6.75*10^-3 kg m^2

ii)A)proof, 45rpm = 90pi rev/min = 4.7rad/s

ii)B)3.1rad/s^2

iii)3.6rad/s

iv)further from axis of rotation, increased central force required, friction can't provide this so it slides off

b)angular velocity increases as moment of inertia decreases and angular momentum is conserved

3.a)i)1.98*10^20 N

ii)1020m/s

iii)work done in bringing 1kg mass from infinity to that point

iv)7.6*10^28 J

v)-3.8*10^28 J

b)i)proof, equate kinetic energy and potential energy

ii)2400m/s

4.a)force acting on object is proportional to and in opposite direction to displacement

b)Wasn't entirely sure what they were looking for here. Presumably both y=0.05cos(200pi t) and y=0.05sin(200pi t) are acceptable.

c)1.9*10^4 m/s^2

d)9.5*10^3 N

e)240J

5.a)i)proof

ii)potential at B 92V, so pd 138V

b)bring charged rod close, touch sphere with finger to repel electrons to earth, remove finger then remove charged rod leaving positively charged sphere

c)i)electrostatic force up, weight down

ii)1.92*10^-18 C

iii)12

d)Millikan conjectured that electronic charge is the fundamental charge unit and all charges can be expressed as integer multiples of this. Down quark has charge of e/3 so that's a contradiction.

6.a)Proof. Use v(perp) = v sin69

b)i)2.1*10^-4N

ii)East (I think)

c)Zero, parallel to magnetic field

d)i)0.012m

ii)Cylinder around the wire? Not sure what they were looking for here (I think I got this wrong in the exam, see earlier in thread).

7.a)i)Change in current --> changing magnetic field --> emf across inductor which opposes buildup of current --> time delay in max brightness being reached

ii)0.25A

iii)2.4H

iv)A)presumably resistance stays the same so max current is unchanged

B)takes longer

b)switch off current, changing magnetic field induces large emf for short period of time over inductor hence large pd for short period of time over neon lamp

8.a)i)A)4.75*10^7 m/s

B)8.23*10^7 m/s

ii)perpendicular compenent accelerated into uniform circular motion, parallel component unchanged, recombine and helical motion obtained

b)i)proof, equate central force and magnetic force

ii)1.6*10^-10 s

iii) 7.6*10^-3m

c)circles off in opposite direction, larger radius for circular motion (or greater pitch for helix I think)

9.a)y=0.05sin2pi(3t - x/0.02)

b)y=0.05sin2pi(3t + x/0.02)

c)0.035m

10.a)i)A)pi rad

B)pi rad

ii)optical path difference between the two reflected rays. if optical path difference = (m+0.5)lambda then for this lambda we have destructive interference hence wavelength lambda blocked.

iii)wouldn't get a very pretty picture otherwise

iv)580nm

b)measure space between succesive fringes (with travelling microscope) and measure length of glass plates. Find wavelength of sodium light. Use d=(lambda l)(2 deltax)

11.a)proof. nodes are lambda/2 apart, find lambda

b)(75+/-4)m/s

c)i)Although percentage uncertainty is overall decreased, increasing frequency increases speed of wave and this actually means that the absolute error is larger. Not totally sure about this one though, maybe the wavelength decreases instead.

ii)Measure space between several nodes and scale down.

There's likely to be a few things wrong in there, particularly explanations

## Exam 07

### David_07

Posted 16 May 2007 - 10:49 PM

There's likely to be a few things wrong in there, particularly explanations

I have got very similar answers to you on the questions I was unsure about, such as 5(b), 7, 8, 10 and 11, which has put my mind at rest a little.

I made a stupid mistake on 5 (a) (ii). I was rushing and glanced at the question, I thought that B was 300mm from the charge, not from A. Oh well, it's only one mark.

### ad absurdum

Posted 17 May 2007 - 06:08 PM

Significant figures aren't too big a deal I think, I've probably used more than I should in a few places but in the mark schemes it says you get away with up to two more. For the SHM I put sine in the exam, but I wrote "assuming y=0 at t=0", so maybe I won't get the mark for not noticing it was implied in the question.

What did youse put for 6.d)ii)? I'm still not sure what the answer should be.

By the way, what are the conditions for your Oxford offer paddy?

What did youse put for 6.d)ii)? I'm still not sure what the answer should be.

By the way, what are the conditions for your Oxford offer paddy?

### David_07

Posted 17 May 2007 - 07:53 PM

What did youse put for 6.d)ii)? I'm still not sure what the answer should be.

I put something similar to you, along the lines of 'clockwise around the wire'. I just thought of that left hand thumb rule thing, which would suggest it was a clockwise direction.

### paddyb67

Posted 17 May 2007 - 10:21 PM

i put radial but i know thats wrong, i drew a diagram though of the right answer so i dont know if ill get anything. i suppose something like a cylinder around the wire would have done.

A in maths and A in physics. i hope iv got them, im not going to spend the summer worrying though so we'll just wait and see.

A in maths and A in physics. i hope iv got them, im not going to spend the summer worrying though so we'll just wait and see.

### ad absurdum

Posted 18 May 2007 - 05:44 PM

Yeah, I said a cylinder round the wire. David, you've got the direction of the field right too, I made the same mistake as paddy (i.e., said it was radial).

AA is a pretty generous offer for Oxford. Sure you'll manage that fine.

AA is a pretty generous offer for Oxford. Sure you'll manage that fine.