Posted 16 May 2007 - 09:51 PM
1.a)proof, ds/dt = u+at and integrate
2.a) i) 6.75*10^-3 kg m^2
ii)A)proof, 45rpm = 90pi rev/min = 4.7rad/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
iii)work done in bringing 1kg mass from infinity to that point
b)i)proof, equate kinetic energy and potential energy
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.
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
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
ii)East (I think)
c)Zero, parallel to magnetic field
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
iv)A)presumably resistance stays the same so max current is unchanged
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
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
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)
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
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
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
Posted 16 May 2007 - 10:49 PM
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.
Posted 17 May 2007 - 01:52 PM
Posted 17 May 2007 - 02:32 PM
I thought that so I used sin.
Posted 17 May 2007 - 06:08 PM
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?
Posted 17 May 2007 - 07:53 PM
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.
Posted 17 May 2007 - 10:21 PM
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.
Posted 18 May 2007 - 05:44 PM
AA is a pretty generous offer for Oxford. Sure you'll manage that fine.
1 user(s) are reading this topic
0 members, 1 guests, 0 anonymous users