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# AQA Mechanics 2 Unofficial Mark Scheme watch

1. Guys, for the last question, i did forces rather than energy... So the Tension of the string was 100N (extrension was 5m). The tension in the other string past A = 40x

Then 100 + Wsin30 - 40x - muR = 0

I actually put muRx accidentally and i solved for x in this case and go 2.5. So my answer was 6.5

How many marks out of 8 would this grant? Is it even a correct approach?

Also: wasn't question 1+ 2mu^2? If i got +, how many method marks would i get? I took the moments and subbed R for F
2. (Original post by C0balt)
I got 7.58
I got x=0 and x=3.58
For last q obv
Surely x can't have been 0 because that would have meant that there was no displacement so when it was released then surely it shouldn't have moved?
3. (Original post by Ben :))
Surely x can't have been 0 because that would have meant that there was no displacement so when it was released then surely it shouldn't have moved?
Well x was 0 when it was released and it was at rest
4. (Original post by IrrationalRoot)
Nope. Alright I'll explain this because it seems half the people here don't understand it.

This question has come up before. And the answer was definitely .

Something similar to this has come up before, with a BEAD so that there is a REACTION force instead of tension. Now a reaction force can act upwards unlike tension, hence the difference.

For the tension problem, what you are saying is that is a necessary and sufficient condition for circular motion. It is not. It is just a necessary condition. What is sufficient is that . This is because if , then the particle is going so slow that the centripetal force is so small that we would actually need the tension to go upwards for this to happen. But this is impossible.
Why are people saying its root of 5ga?
It's not its root of 5gl it didn't mention a in the question
5. (Original post by typing)
Why are people saying its root of 5ga?
It's not its root of 5gl it didn't mention a in the question
You get what they mean, it's because the radius is often labelled a.

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6. 255 UMS last year, roughly 59/75 in this exam, 55/75 for C3 and 60/75 for C4. Reckon that's enough for an A overall?
7. (Original post by jjsnyder)
You get what they mean, it's because the radius is often labelled a.

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Ah right cheers
8. (Original post by typing)
Why are people saying its root of 5ga?
It's not its root of 5gl it didn't mention a in the question
I don't know why other people are mentioning a, but I was referring to a different question in which they used a for the radius, hence the answer has an a in it. Notice that I wasn't referring to the question in our paper.
9. For question 7 (b) did anyone get

tanx = -μ/4
10. (Original post by C0balt)
Well x was 0 when it was released and it was at rest
Yeah it was at rest but it was being held at rest so wouldn't naturally have been at rest! So the equation shouldn't have produced x=0 as a root!

11. Does anyone agree?
12. (Original post by Ben :))
Yeah it was at rest but it was being held at rest so wouldn't naturally have been at rest! So the equation shouldn't have produced x=0 as a root!
Well i got the right answer anyway...
13. (Original post by C0balt)
Well i got the right answer anyway...
Unless it wasn't the right answer, which is the point i'm trying to make! I think the unofficial markscheme is wrong on this answer even though a lot of people agree, I haven't seen any proof to it!
14. (Original post by Ben :))
Unless it wasn't the right answer, which is the point i'm trying to make! I think the unofficial markscheme is wrong on this answer even though a lot of people agree, I haven't seen any proof to it!
I can't see anything wrong with the method
15. (Original post by C0balt)
I can't see anything wrong with the method
If you assume GPE to be 0 at the point R then your solutions would be x = 4m and x = 6.53m This is convincing since 4m is the original distance the particle is from Q when it is first at rest. Thus, the other solution for x would be the distance the particle is next at rest.
16. (Original post by Jac_Jac)
If you assume GPE to be 0 at the point R then your solutions would be x = 4m and x = 6.53m This is convincing since 4m is the original distance the particle is from Q when it is first at rest. Thus, the other solution for x would be the distance the particle is next at rest.
I don't get what you're saying... I didn't assume gpe to be 0 at R but when the particle is next at rest
17. (Original post by C0balt)
I don't get what you're saying... I didn't assume gpe to be 0 at R but when the particle is next at rest
I'm saying that those are the solutions you get when you assume GPE is 0 at R
18. (Original post by Jac_Jac)
I'm saying that those are the solutions you get when you assume GPE is 0 at R
Ah okay, sorry I'm not with it rn lol
19. (Original post by C0balt)
Ah okay, sorry I'm not with it rn lol
Lol I've done it again and I recon your answer is correct since I got your answer with both approaches now When I did it initially I read on this forum that the frictional force was 0.4. In your working you used 0.4 as your value for mu I assume, so I did that and got your result Sorry for the confusion, I didn't do the exam, was just interested in trying the questions since people were saying it was challenging
20. (Original post by Jac_Jac)
Lol I've done it again and I recon your answer is correct since I got your answer with both approaches now When I did it initially I read on this forum that the frictional force was 0.4. In your working you used 0.4 as your value for mu I assume, so I did that and got your result Sorry for the confusion, I didn't do the exam, was just interested in trying the questions since people were saying it was challenging
Ah right, no worries

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