**Applied Maths 2012 Before/After**

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