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Q31, compare the maximum height reached by a vertically launched projectile under the two different assumptions. Assumption A - constant (uniform) gravitational field. g=9.8 m/s^2 and potential energy U = mgh. Assumption B - (downward) gravitational acceleration is \frac{GM}{r^2} and...

That's a fair criticism, 20-B is certainly a correct statement. It just doesn't give as good a comparison of time dilatation effects for all three clocks as does 20-C. So 20-C will definitely be the accepted answer. That's taking the inertial reference frame approximation way too far...

I just took a quick look at this test today. Some interesting question, and parts of Q14 are definitely pretty difficult. Q14d has an interesting kink that I don't think any of the current solutions have captured. The question doesn't state that 0 \le \theta \le \pi/2, though most of the...

Yes you're correct, you need to show a rhombus specifically for the angle bisect property. Sorry, I deleted that quoted post just before reading your reply and was about to acknowlege the correction.

Note that the lines from O to 1, from O to z and from z to z+1 form a parallelogram rhombus. The result follows immediately from the parellelogram rhombus property that the diagonals bisect the corner angles.

I think that the problem with the equation given in Q37 is that the SO3^(2-) just wont exist in aqueous equilibrium with H+, not without also forming HSO3-. I'm starting to think that maybe the correct answer to the last question is that an equilibrium cannot be established with only the...

Guys, I've got a query about the equilibrium constant (approx 1 x 10^-9) in the last question. I agree that it was very easy to algebraically come up with that solution, but I'm trying to understand the utility (usefulness) of that overall equilibrium equation. I'm probably doing something dumb...

It looks dirty Agreed that it looks dirty, but it's probably still valid. The point is that it's not about proving that it is the correct answer, it's about proving that the other surd is not the correct answer. It would of course be invalid to try to use a calculator to prove that it was...

You can also verify it's C by reading off the exact values of the concs, and then calculating the equilb constant ratios with both the initial and final values. You'll find A just doesn't work numerically. The soln given by Beanie above is a lot quicker though.

Yeah, it definitely should be like the blue one, and it kind of needs some sloppy working to get that red soln. Look for example at the column of slope lines immediately to the right of P (at x about -0.8). There you can see that TL's red solution had to take a gradient that was significantly...

Ok I see, I hate it when that happens. MC can be annoying like that, with so little time it's very easy to know your stuff but still make some really minor mistakes that will lose you the marks. Meanwhile someone who knows close to nothing can often jag about 50%.

With const acceleration from rest you get: x = \frac{1}{2} a_x t^2 y = \frac{1}{2} a_y t^2 Eliminate "t" to get a linear relation between x and y. - KE increases from (2 phi - phi) to (4 phi - phi), so KE is tripled. - v is proportional to sqrt(KE), so v is increase by factor sqrt(3) - r...

Thanks notime, you're absolutely right. I only skimmed through this late last night and didn't notice that mass loss was per minute (yes I calculated per second). The other question 14 again you are correct. Equal an opposite forces, not acceleration. Thanks for pointing that out as I had a...

I only got to look at the paper late tonight so have just skimmed through, but my prelim MC solns are 1 ........... 5 B C D B A 6 ...........10 D C D C B 11 ........ 15 A D B A C 16 ....... 20 B C A A D Edited 14 and 16 thanks to notime123

Yep, ans to Q18 is (a). With accelerating motion (from rest) in both x and y directions then the trajectory is linear. (Even if the accelerations are unequal.) After the x accel is removed (E field removed) it follows the usual parabolic trajectory.

Also, neither the Itute nor TL solns provided a satisfactory explanation for why the negative \sqrt{5} solution was invalid in Q14c(ii). \cos^2{(\pi/10)} = \sqrt{\frac{5 \pm \sqrt{5}}{8} } Itute gave no reasoning for selection of the +ive inner surd, while TL reasoned that cos^2() must be...

Ok cool. So it could be anyone then, like even a teacher or other staff member. I think it's a good thing to get the papers circulating asap, so as to facilitate discussion of the questions while they're still fresh in people's minds. And as you say, they're going to be officially released soon...

Hopefully you'll get an answer straight from the horses mouth. But I'm assuming that either someone manages to sneak a copy of the paper out of the exam, or they manage to photo/scan it during the examination.

Thanks.

Does anyone have a copy of the exam they can share?