Applied Maths aftermath

DarraghF197 wrote: »

Found that the exam was really tough. Started off with Relative velocity at the start and I thought it would have been a damage limitation exam. However, as I progressed, it was better and better! Went back to my mistakes, fixed them, and hopefully an A1.

Question 1 was hard enough. It's usually my weakest so I was happy to see I got it out.
Question 2 was difficult for me. I always delayed equations until they're either directly East or North of eachother. Having no distances meant I had to improvise.
Question 3 was actually pretty fair. I completely forgot about the three metres at the start. Then redid it and worked out then (hopefully!).
I messed up a bit with question 4 part two of B. Forgot about the tension for forces up equal down.
Question 5 and 10 were really easy.

For Q2 (rel vel) the secod part was easy enough, A was 2km east of B when B was at the intersection. What did you end up doing? Probably over complicated it

Chickennuggets wrote: »

Was it not 1km east of b

Nope, when B got to the intersection A had been travelling east at 60kmh for 2 minutes
2 minutes is 1/30th of an hour
1/30th of 60km is 2km

Could not be happier with that exam. I was so anxious about how it was going to go since I had put off studying it for the last week's to focus on other exams, and when I opened the paper I was like oh hell no this is horrible, and was beginning to settle for a C overall.

Started with Q4 as it's my best, which ironically turned out to be the hardest! It went well though, then moved onto Q5 which was piss, as was Q2, bar maybe a few mistakes, Q3a probably posed the greatest challenge but I figured it out and got it all out, bar maybe some minor slips. 3b was piss. Q10 seemed strange at first but turned out to be very simple and then Q1 went perfectly.

Was so worried that it was not going to go well and that I'd be on a serious downer finishing the LC, but now I couldn't be happier. B1 minimum or maybe an A1 if my slips weren't too serious.
Delighted!

Could a kind soul post a scan of the paper?

I hope you all did well!

AlfaJack wrote: »

hopefully i'll get the marks for the method :P

You absolutely will.

Best to put it all behind yerselves now and just remember that it's graded on a curve so it'll be fine.

PotmBottom wrote: »

I was going to ask, is it similar to the old maths course with the blunder minus 3 and then it's marked normally, like if an incorrect acceleration causes an incorrect tension is it still marked correct?

Yes. Slips lose one mark, blunders lose 3 and a trivial misreading loses 1.

Everything after the mistake is marked as if you were correct so you only lose marks once per mistake.

It's a very fairly marked exam tbh.

DarraghF197 wrote: »

Oh wow, yeah I did overcomplicated it. I notice that I'm never up to speed with things at the start of an exam, and I've missed a pretty obvious thing there! I wasted a bit of time doing unnecessary work there, hopefully I still got the right answer (0.655km?)

Yup, would've said I got about 0.64km but really anything arouns there is a matter of rounding :P out of curiosity, how did you go about working out working out that part?

Doctorhopeful wrote: »

Yup, would've said I got about 0.64km but really anything arouns there is a matter of rounding :P out of curiosity, how did you go about working out working out that part?

I don't fully remember but it was something like this:

I used s=ut and let the distance from A to the junction equal x. Then s=ut for B and got the distance in terms of X.

I then drew a relative velocity line from B to wherever it crossed A (this was a bit confusing as I thought it would intersect ahead of A rather than behind, which made it all that bit more messy!).

I used the triangle [point intersects A, junction, and B]. Was able to use the sine rule to find the point it intersects A line in terms of X.

Then subtracted that from x to get the distance the relative velocity line was from A. This gave me another triangle and I was able to find the shortest distance with the Sine of the angle between Relative Velocity line and A line.

I wonder if that makes any sense at all lol.

qweerty wrote: »

It's not clear what your problem with it is. If you take away the picture (which makes it look like Project Maths) it's similar in style to the 2009 question.

My problem has nothing to do with the diagram - it is to do with the numbers. All we seem to do is use our calculator all the time. Surely, it is not beyond the wit of someone setting the question to choose numbers that give nice answers. This is what one does when you set questions - it is part of the challenge of setting a really nice question.

Or if the question setter insists on going ahead with their awful numbers then perhaps give some indication as to what the answer will be like (two decimal places, less than a degree below the horizontal...).

There is also the ridiculous situation of using the other answer. The server would have to be able to exert the same force to launch the tennis ball almost vertically into the air and wait around 10 s for the ball to land.

To me it is all about being able to write good questions that will equally test the skills of the students - that's all.

qweerty wrote: »

I didn't think your problem with the question was the diagram. Just mentioned it in case you had taken issue with the style.

The majority of angles in applied maths aren't clean. I guess you could gripe that they didn't ask for it to a particular number of significant figures, but I expect they'll allow for rounding in that case. I highly doubt that the reason it wasn't an integer was because they were lazy or incompetent.

The only complexity in that question was the need to use two trig identities in succession; everything else was standard. I might agree with you if it were a monster pulleys question.

As for near-vertical serves, underarm is legal!

Not from 3 m above the ground.

My answers to the last 5 questions are as follows:

Q6 (a) (i) mg/d
(ii) assume test-tube is pushed down a distance x. Therefore the upward force on the tube is now F = k(d+x) while the downward force remains mg. The net force acting on the tube at this point is F-mg = k(d+x) - mg = ma => kd + kx - mg = ma . But mg = kd from part (i) => kd + kx -kd = ma => kx = ma => (mg/d)x = ma => a = (g/d)x. As acceleration is directly proportional to displacement x from equilibrium, this means the tube will oscillate with SHM.

The period of the motion is 2Pi*root(d/g) seconds

(b) alpha = 21.7044 degrees

Q7 (a) (i) T = W*root(c^2 + l^2)/c Newtons
(ii) R = (W/2C)*(9l^2 + c^2) Newtons

(b) (i)Rx = 7W/4 N, Ry = 5W/4 N
(ii) Take moments about z for the rod zy. This will yield an equation allowing us to relate the frictional force at Y to the angle theta. We get Fy = (3W/4)*tan(theta/2). This must be the same as the frictional force at X also. Slipping occurs at X when Fx > uRx i.e. when (3W/4)*tan(theta/2) > uRx i.e. > u*(7W/4) i.e when tan(theta/2) becomes > 7u/3. Similarly slipping occurs at Y when Fy > uRy i.e. when (3W/4)*tan(theta/2)> u*(5W/4) i.e. when tan(theta/2) becomes > 5u/3. As theta increases tan(theta/2) will also increase and it will exceed 5u/3 before it exceeds 7u/3. Therefore slippage will occur at Y before X.

(iii) 3/5

Q8 (a) text-book proof via integration
(b) (i) 1/12x + x metres
(ii) x = 1/root(2) m
(iii) 1.82298 seconds

Q9 (a) 8676.0736 kg/m3
(b) (i) 482.8125 tonnes
(ii) It will sink to an overall depth of 0.38625m i.e. a further 0.01125 m from its current depth.

Q10 (a) (i) Va = 16 m/s and Vb = 16m/s
(ii) 16m
(iii) draw the 2 curves on the graph using the equations given and the distance between the cars after 4 seconds is the area of the region between the two curves on the graph

(b) (i) C(x) = 74x +0.55x^2 + 0.01x^3 + F
(ii) 32800
(iii) 2695

Again, just to reiterate, I can't be 100% sure about the answers (bar the proofs) as I could easily have made a slip or blunder along the way

japester wrote: »

(iii) funny question, I'm inclined to believe the answer here is also 18g/10 m/s i.e. when P begins to move it is "jerked" with an initial speed equal to the speed Q has after travelling the 3m since at this time the string joining them suddenly becomes taut

You need to employ conservation of momentum here.