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CHEM1STRY
That wasn't too bad. I didn't get K but got the right working (i hate calculators :P) so I shouldnt lose too many. And also I am not sure about the tension... did you put 250N or 125N


I put 250N and then was like hangggg on and drew a lovely diagram and then did 2T = 250N so 125N, i think other people out 125N too, so HOPEFULLY.
What did you get for the displacement graph thing?
I felt so silly losing 6 marks over that graph..:mad:
judd789
really oh crap i dont remember the question but i definitely didn't covert any units on q1

Q1 was something about dropping something 0.75cm above the ground and it taking 3s, im not sure if it was 3 seconds, but you get the idea. Dropping a particle downwards type question.
I used a=9.8m/s, I think thats ok :o:
Reply 62
LisaWilliams
i wrote a very angry note about the last question involving theta! It was riduclous to lose SIXTEEN marks just because one couldn't find out theta!!

haha lmao same, you had to find theta by solving the 250sin40 and 250sin)theta to find theta by making them equal because it was in the i direction....i messed it up too, i hope my value of 87.9' that i got could be used, cos all my working for parts 3 and 4 was right, just i messed up the part 2 to find theta, resulting in parts 4 and 5 being wrong....

GAY!!! i could have got 96%
Reply 63
sinesquared.
I put 250N and then was like hangggg on and drew a lovely diagram and then did 2T = 250N so 125N, i think other people out 125N too, so HOPEFULLY.
What did you get for the displacement graph thing?
I felt so silly losing 6 marks over that graph..:mad:


I wasnt sure, I put 125 then put 250. If it was two separate strings then it is 125 but if it is one string then I wasn't sure :s-smilie:
Reply 64
I thought it went pretty good. However for some stupid reason when having to integrate for finding C I worked out what the value was meant to be and just put that when I should of put the difference, oh well should get a few working marks.

Also wasn’t really sure if it was 205N or 125N but went with 250N.

Also for when v=4 I only got 0.5s?
Reply 65
CHEM1STRY
I wasnt sure, I put 125 then put 250. If it was two separate strings then it is 125 but if it is one string then I wasn't sure :s-smilie:

i put 125, it was a pretty good paper(apart from taht last stupid question...if only i remembered to solve the sideways components were equal.. what did you get for that last part Q.6 for 3<t<4 velocity equations, i think i got y=(x^2-4)+3.5...
Reply 66
Mr.Singh
i put 125, it was a pretty good paper(apart from taht last stupid question...if only i remembered to solve the sideways components were equal.. what did you get for that last part Q.6 for 3<t<4 velocity equations, i think i got y=(x^2-4)+3.5...

Which one? I dont remember that
Reply 67
Jederrs
I thought it went pretty good. However for some stupid reason when having to integrate for finding C I worked out what the value was meant to be and just put that when I should of put the difference, oh well should get a few working marks.

Also wasn’t really sure if it was 205N or 125N but went with 250N.

Also for when v=4 I only got 0.5s?


That's correct but you need to put in -4 aswell because it said the speed was 4ms-1, I missed that :P

I think I got 66-68/72
Jederrs
I thought it went pretty good. However for some stupid reason when having to integrate for finding C I worked out what the value was meant to be and just put that when I should of put the difference, oh well should get a few working marks.

Also wasn’t really sure if it was 205N or 125N but went with 250N.

Also for when v=4 I only got 0.5s?


:eek3: I didnt integrate once on the whole paper! What question was this? OH DEAR.
Reply 69
sinesquared.
:eek3: I didnt integrate once on the whole paper! What question was this? OH DEAR.


You didn't need to. it was the one where you put (t-3) into the SUVAT eqtn
Reply 70
CHEM1STRY
Which one? I dont remember that

Section B, First Question, Last part, i think it was 5 marks to state the equation of the displacement for 3<t<4
it was a X^2 graph shape, so i just geussed, it went through points (3.5,-4)...
Reply 71
just incase someone has the paper, it'll be great if you post it.
Reply 72
When Finding K in that vector equation, did anyone find it =2(rhs) so that it was equal
CHEM1STRY
You didn't need to. it was the one where you put (t-3) into the SUVAT eqtn

:frown: I dont remember doing that either...haha.
Reply 74
Mr.Singh
Section B, First Question, Last part, i think it was 5 marks to state the equation of the displacement for 3<t<4
it was a X^2 graph shape, so i just geussed, it went through points (3.5,-4)...


Yeah that's wrong. You find velocity at t=3 and you get -16ms-1... that's the initial velocity.

Then you have

S = ut + 1/2 at^2 but we are three seconds later so we need (t-3) translation (to the right).

s = -16(t-3) + 1/2 x 32 x (t-3)^2

Then expand that
Mr.Singh
Section B, First Question, Last part, i think it was 5 marks to state the equation of the displacement for 3<t<4
it was a X^2 graph shape, so i just geussed, it went through points (3.5,-4)...

Hmmm.
I used SUVAT.
Worked out the velocity by finding the gradient, but realised i forgot to add the +12 to the S= equation, as its displacement right?
Damn.
CHEM1STRY
Yeah that's wrong. You find velocity at t=3 and you get -16ms-1... that's the initial velocity.

Then you have

S = ut + 1/2 at^2 but we are three seconds later so we need (t-3) translation (to the right).

s = -16(t-3) + 1/2 x 32 x (t-3)^2

Then expand that


Oh (t-3) here huh?
Hmm I missed that.
Didnt you need to add 12 as thats the starting position? (i forgot to do that too)
Reply 77
CHEM1STRY
That's correct but you need to put in -4 aswell because it said the speed was 4ms-1, I missed that :P


Ahh yeah I saw it said times not time but didn’t really bother thinking about it too long. If I remember it was only 3 marks so probs only lost one.

Hopefully I only lost 5 or 6 marks all together, fingers crossed.
Reply 78
sinesquared.
Oh (t-3) here huh?
Hmm I missed that.
Didnt you need to add 12 as thats the starting position? (i forgot to do that too)


No because its from 3 --> 4 which is a separate curve
Reply 79
CHEM1STRY
Yeah that's wrong. You find velocity at t=3 and you get -16ms-1... that's the initial velocity.

Then you have

S = ut + 1/2 at^2 but we are three seconds later so we need (t-3) translation (to the right).

s = -16(t-3) + 1/2 x 32 x (t-3)^2

Then expand that

this is cool, but i think it'll have been easier to just look at the symmetry of the graph between those points. just doing s=(t-3)(t-4) gives the same answer without the expansion hassle

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