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# Official OCR A G481 mechanics 20th May 2013 Watch

1. (Original post by JustFacts)
Did anybody else work out the youngs modulas by finding the gradient of the stress /strain graph? I got 1.875X10-12 doing it that way. I didn't attempt the deriving k.e question and the moments question I think I got wrong. Apart from that it was ok, although i'm sure I dropped marks here and there.
1.875^-11 is the general consensus, I believe. We think that if you got 1.875^(something), then you should get 2 marks.
2. (Original post by OllieGCSEs)
The Stress was in GPa but the strain was in %, so the strain was 0.8/100, i.e 0.008 or something
Yay! I did the exact same thing, my friends were like "It doesnt matter if it's in %" but if it was just 0.8 it would give a low Youngs Modulus which didn't seem right. Overall I thought the exam was pretty good. I really liked the deriving 1/2mv^2, question infact I liked it so much I showed two different ways of deriving it ^^
3. how do you change from cm3 to m3? cause didnt we have to convert that for the density question?
4. (Original post by Lucas96)
Initial acceleration for the trolley? 14?
I got 3.5 from F=kx the substituting that in F=ma
5. (Original post by skitz241)
I got 3.5 from F=kx the substituting that in F=ma
3.5 here too
6. For the density I got 2000 and something, I got 14 for the acceleration of the spring , the time for the ball to drop in the oil was something like 0.4 secs, and then 13.3 for the tangent thing.
7. (Original post by teenajohny777)
how do you change from cm3 to m3? cause didnt we have to convert that for the density question?
You needed to divide by 10^6, as it's cm^3, you need to divide by 100, 3 times to get it into m^3
8. (Original post by sohailkm96)
Yay! I did the exact same thing, my friends were like "It doesnt matter if it's in %" but if it was just 0.8 it would give a low Youngs Modulus which didn't seem right. Overall I thought the exam was pretty good. I really liked the deriving 1/2mv^2, question infact I liked it so much I showed two different ways of deriving it ^^
What was the second way? I got f = ma / w = fx --> w = max --> SUVAT for ax --> ax = 1/2v^2 so w = m1/2v^2 (or 1/2mv^2)
9. (Original post by sohailkm96)
You needed to divide by 10^6, as it's cm^3, you need to divide by 100, 3 times to get it into m^3
Whooo thats good, people were saying to me i was wrong cause i did that
10. (Original post by Joeyoh9292)
What was the second way? I got f = ma / w = fx --> w = max --> SUVAT for ax --> ax = 1/2v^2 so w = m1/2v^2 (or 1/2mv^2)
I did F = ma, W = Fx therefore, W = max, Then v^2 = u^2 + 2as, Then v^2 -u^2 (0) / 2a = s(x) (rearrange again to get a) Put that back into the equation and you can get either

W = ma(v^2)/2a (cancel out a to get W = mv^2 /2)

or

W = mx(v^2)/2x (cancel out x to get W = mv^2/2)
11. (Original post by yodawg321)
What did people say about the forces of graph.

For the first point I said about the drag force being greater than the weight force which means it deaccelerates.

And for the second point I put about the weight and drag forces being equal which makes constant velocity.

anyone say anything like this?
Instead of saying drag force I said the resistance of the oil. Do you think I will get the mark(s)?
12. (Original post by skitz241)
Whooo thats good, people were saying to me i was wrong cause i did that
Nahhh I'm pretty sure that's right, I did it wrong first and did 7^3 x 10^-6, then I realised what an idiot I was and I had to ask for more paper XD
13. (Original post by skitz241)
I got 3.5 from F=kx the substituting that in F=ma
But there were 2 springs, so 14 x 0.3 + 14x 0.5 gives 11.2/0.8= 14
14. (Original post by Ferrari_1996)
Instead of saying drag force I said the resistance of the oil. Do you think I will get the mark(s)?
I'm sure you should, but it's more correct to say drag force.
15. (Original post by Lucas96)
But there were 2 springs, so 14 x 0.3 + 14x 0.5 gives 11.2/0.8= 14

Yeah but the forces are acting in opposite directions, so it's (14 x 0.5) - (14 x 0.3)/0.8 = 3.5
16. Can someone tell me what they did on the component question?
17. (Original post by Joeyoh9292)
I'm sure you should, but it's more correct to say drag force.
Ok thanks
18. (Original post by Hasslol)
Can someone tell me what they did on the component question?
For the perpendicular component you need to use the cos of the angle, so it was 8x10^-5 (if im right) x cos30

For the parallel component you need to use the sin of the angle, so it was 8 x 10 ^-5 x sin30
19. (Original post by Lucas96)
But there were 2 springs, so 14 x 0.3 + 14x 0.5 gives 11.2/0.8= 14
The question was weird. It said that they were both being stretched, so I just kind of assumed that the first one was negligible (as it wouldn't be pushing, but instead pulling, so the maths just either ignored that or it was included in a previous part). I usually read too much into questions, but I hope not. Edit: looks like I forgot the whole question, but I was right. Spring A is pulling on Spring B, so it's B-A.
20. (Original post by Hasslol)
Can someone tell me what they did on the component question?
Which question was that?
EDIT: Nevermind.

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