**Applied Maths 2012 Before/After**

BrianC2095

Wesc. wrote: »

But isn't theta the difference in the temperature of a body and the surroundings? So at the start theta is 60 because that's 80 - 20..

Aw yea i misread the question damn i thought theta was the temperature of the body...... Still got 42 though somehow Ah well it looks like i was never gonna get my a1 :rolleyes:

Plonker!

Anyone confirm or deny these?

Q1 (a) 44.1m
(b) (i) 40secs
(ii) 352m

Q3 (a) 53.13 or 36.87
(b) (i) 60 degrees
(ii) 70m/s

Q4 (a) (i) 1.23m/s
(ii) 1/11 m
(b) (i) f*cked this up, I think (8/5)(mew)mg
for tension
Q5 proofs, lovely!

Q6 (a) (i)1.973 ( it was (1/5)pi^2 )
(ii) 5.9 N
(b) (i) 1.4 m/s

Q8 (a) Proof.
(b) 2m

Q10(a) (i) k=(1/600)ln(4/3)
(ii) 38.something can't remember

(b) proofs, handy 'nuff.

paddyhes

paddyhes wrote: »

My answers:

Q1
(a) 44.1m
(b) (i) 40s (ii) 352m

Q2
(a)87 degrees (probably wrong)
(b) (ii) 13:07(12 seconds) (ii) 96 minutes

Q3
(a) (i) roughly 36 and 53 (ii) 4 seconds
(b) (i) 60 degrees (ii) 70m/s

Q4
(a) (i) root(2g/13) (ii) 1/11 m
(b) (i) (4mg(1+''mew''))/5 (ii) (3f-2)/6

Q5
(a) showed V1<V3<V4
(b) proof

Q10
(a) (i) k=roughly 6.75x10^-4 (used seconds) (ii) 41.77 degrees
(b) proof

plonker ^

ganon

Woow, loads of the same answers as everyone, for 1,2,3 definitely, 4 was a bit dodgy but think my a is right, 10 i seemed to get a different k to everyone but got the next part as 41.77 so i dunno

For 5a, did anyone find they were getting e<thingy rather than e>thingy
I did so i just reversed the inequality at the end hoping that i'd forgot some rule that you're supposed to do that

Tankosaur

For q1 I wrote d = 24.5 (which it is I believe) then when testing if this is correct I worked out the total distance the first particle moved in that time which was 44.1 m.

But I circled the d=24.5 bit as I assumed it was what we were looking for, but the actuall value we were asked for was 44.1m.


I still worked out the total distance from P when the collided was 44.1m so will I still get full marks?

Dicksboro_man

made some stupid mistakes today, q10a was a nightmare for me, had no idea what to do with it :-/

I tried to study q6 outside school and did it as an extra question today but didnt even attempt the last part of, b (ii). how many marks ye reckon is gonna be going for 10a or 6b(ii)?

RedTexan

Did anyone try 6(b)(ii) i came out with k=1.56? it was 1.560000000 on the calculator and it was asked to 2 decimal places, made me confident at the time! Got 1.4 for the speed!

RedTexan

Dicksboro_man wrote: »

made some stupid mistakes today, q10a was a nightmare for me, had no idea what to do with it :-/

I tried to study q6 outside school and did it as an extra question today but didnt even attempt the last part of, b (ii). how many marks ye reckon is gonna be going for 10a or 6b(ii)?

Well for q6 (ii) I did a lot of work, had to get, used projectile vectors and all that and thought it was pretty clever the way it worked out, would be delighted if it was right! Well I would say that it was this year's trickiest, probably 10 marks I'd say.

Liveforrugby

That exam made me orgasm uncontrollably

ganon

Did anyone notice how much it was like the pres?
Q.1a was the same qustion as the pre questino that had root8ga in it (forget which company)
And 4.b was the same as one of the pres, except they gave you 7kg and 5kg (again forget which company)

paddyhes

Messed up Q2 a.
Took velocity of car to be 20m/s instead of -20m/s.
How many marks dya think ill lose

Tankosaur

paddyhes wrote: »

Messed up Q2 a.
Took velocity of car to be 20m/s instead of -20m/s.
How many marks dya think ill lose

I think mis-reads are minus 3,and as long as you make no other mistakes in the rest of the question all you'll lose is that -3.

I made like 3/4toal misreads in this paper. Hoping I still get my a1 though

paddyhes

Tankosaur wrote: »

I think mis-reads are minus 3,and as long as you make no other mistakes in the rest of the question all you'll lose is that -3.

I made like 3/4toal misreads in this paper. Hoping I still get my a1 though

yeah my method was perfect otherwise.
that was the only mistake i made (that i know of :L)
so here's hopin.

Comrade C

Osric wrote: »

Thought the paper was fair, although some parts really hard

Here are my answers:

(1)
(a) 44.1m
(b) 40 seconds
352m

(3)
(a) 36.87 and 53.13 (roughly)
4 seconds
(b) 60 degrees
70m/s

(4)
(a) 1.23m/s
1/11m
(b) (4mg(1+u))/5
(2g-f)/3

(5)
(a) proof
(b) proof

(9)
(a) s=7.79
(b) 1.2m
Theta = 60

(10)
(a) k = ln(1.5) / 600 (in seconds)
Temp = 41.77 degrees (roughly)
(b) proof
Proof

Very happy with a lot of those answers! Forgot to add 20 onto 21.77, was raging! Just regarding the hydrostatics! I'm fairly good at hydrostatics, but was wondering did you use BwSl=Bl ? My friends got 1.2m, but they didn't use that. I got 1.29, and I'm thinking that's approx the difference between using water and that liquid of relative density 0.9

Osric

Comrade C wrote: »

Very happy with a lot of those answers! Forgot to add 20 onto 21.77, was raging! Just regarding the hydrostatics! I'm fairly good at hydrostatics, but was wondering did you use BwSl=Bl ? My friends got 1.2m, but they didn't use that. I got 1.29, and I'm thinking that's approx the difference between using water and that liquid of relative density 0.9

BwSl=Bl?

Comrade C

When you want to find the buoyant force acting on a rod or any object in water, you find the fraction of the weight under water and divide it by the relative density of the rod. However, when finding the buoyant force exerted by a liquid other than water, you must multiply the force exerted by water on the rod (Bw) and multiply it by the relative density of the liquid (Sl). This will give you Bl the force exerted by that liquid on the rod

Cathalog

Tankosaur wrote: »

paddyhes wrote: »

Messed up Q2 a.
Took velocity of car to be 20m/s instead of -20m/s.
How many marks dya think ill lose

I think mis-reads are minus 3,and as long as you make no other mistakes in the rest of the question all you'll lose is that -3.

I made like 3/4toal misreads in this paper. Hoping I still get my a1 though

Misreads are actually -1. So all is good

Anyways, I'm finished! Exam was fine. In fact, all exams were fine. Just going to forget about it now until August.

ROLE ON THE SUMMER!!!

japester

Hi folks, my solutions for the first 3 questions as follows:

Q1 (a) 44.1m (b) (i) 40 s (ii) 352 m
Q2 (a) 82.542 degrees (b) (i) 73.739 degrees North of East (ii) 1:20:55 pm (iii) 1 hr 8 min 33.36 s
Q3 (a) 36.869 degrees and 53.131 degrees (ii) 4 s (b) (i) 60 degrees (ii) 70m/s

more answers to come shortly

japester

My solutions for questions 4-6 are:

Q4: (a) (i) sqrt(2g/13) m/s (ii) 1/11 m (b) (i) 4mg(1+mew)/5 (ii) 3f-2g
Q5: (a) proof involving finding the velocity of A after its collision with B and the velocity of B after its collision with C and then saying that no further collision can occur as long as velocity of A exceeds velocity of B (b) proof using NEL, PCM along the line of impact and the fact that the j-component of velocity for each sphere is the same before and after impact, meaning that u1Sin(alpha) = v1Sin(theta + alpha). After this some trigonometric manipulation is needed to get an expression for tan(theta).
Q6: (a) (i) 1.974 m/s/s (ii) 5.9N (b) (i) 1.4m/s (ii) 0.93

more solutions to come shortly

japester

My solutions for questions 7-10 are as follows:

Q7: (a) 1.316 (b) the proof here involves taking forces left and right and up and down at either of the rough rings and also taking moments about the chosen rough ring. This will give 3 equations. Then use the fact that Sin(alpha) = sqrt(l*l-(d*d/4))/l to get the expression involving d , mew and l

Q8: (a) text-book proof (b) (i) mg-(r*r*m*m*g/(I+m*r*r)) (ii) 8m

Q9: (a) 7.79 (b) (i) 1.2m (ii) 60 degrees

Q10: (a) (i) 0.000675775 or -(1/600)ln(2/3) (ii) 41.773 degrees (b) (i) to prove this take m(dv/dt) = -mk*v*v and integrate to get an expression involving v, k, t and and the constant of integration C. Use known values at time zero and time = 0.01 seconds to get values for C and k respectively (ii) to prove this take mv(dv/ds) = -mk*v*v and integrate to get an expression involving v, k, s and C. Use the fact that v= 1000/(9900t+1) and substitute and the known values at time zero and time = 0.01 s to get C and then s, the length of the block of gel.

hope these help