You are Here: Home >< Maths

# AQA Mechanics 2 Unofficial Mark Scheme

Announcements Posted on
Why bother with a post grad course - waste of time? 17-10-2016
1. (Original post by Jm098)
Did anyone get 2 rootlg?

I did for 5d
2. Guys, it's since we have and we require at the top of the circle, hence .
3. How can so many people not see that it's just 2 root lg? I'm confused
4. (Original post by aoxa)
No, tension was root(4gl) which simplified, was 2root(gl). Which gives non negative tension.

Posted from TSR Mobile
i think you're forgetting about the gravitational force of the weight

at top (mv^2)/r = T + mg

for complete rotation T>>0 so

(mv^2)/r = mg

this gave v= root(5gl)
5. (Original post by jjsnyder)
Here are my answers from todays exam, not all of them will necessarily be correct so please correct me if you think they are wrong!
1. a) 9.6
b) i) 24.3
ii) 12.7
iii) No Air Resistance
2.
a)
b) i)
ii) 8.20
c)
3. a) By symmetry
b)10.4 cm
c) 16.5 degrees
d) Explanation Q
4. a)78.4
b) 41.4 degrees
c) r=2.89
5. Lots of very long answers, will upload later if I have time. Part d) was
6. a) Show that Q
b)
7. a) Diagram
b)
8. a) 17.0
b) I got 6.05m but many others got 7.58m so this seems like the correct answer, can anyone confirm this?
Question 5) was just root 5gl
Question 8) I got 6.87 I think but forgot to include the potential energy still in the bottom string I think!
6. (Original post by IrrationalRoot)
Guys, it's since we have and we require at the top of the circle, hence .
Mate this exact question has come up before word for word and the answer is 2rootlg
7. (Original post by Jm098)
How can so many people not see that it's just 2 root lg? I'm confused
Because it's not.
8. (Original post by CoIIins)
Mate this exact question has come up before word for word and the answer is 2rootlg
Did you even read my method...
9. (Original post by IrrationalRoot)
Guys, it's since we have and we require at the top of the circle, hence .
Yes that is right you the mvp
10. (Original post by IrrationalRoot)
Guys, it's since we have and we require at the top of the circle, hence .

Surely v>0 for compete circular motion rather than v>gl, I'm sorry I'm confused to see where you got that from
11. (Original post by jjsnyder)
Here are my answers from todays exam, not all of them will necessarily be correct so please correct me if you think they are wrong!
1. a) 9.6
b) i) 24.3
ii) 12.7
iii) No Air Resistance
2.
a)
b) i)
ii) 8.20
c)
3. a) By symmetry
b)10.4 cm
c) 16.5 degrees
d) Explanation Q
4. a)78.4
b) 41.4 degrees
c) r=2.89
5. Lots of very long answers, will upload later if I have time. Part d) was
6. a) Show that Q
b)
7. a) Diagram
b)
8. a) 17.0
b) I got 6.05m but many others got 7.58m so this seems like the correct answer, can anyone confirm this?
Agree with all of these except 8 wasn't split in to a) and b) was it? Thought it was just one 8 marker? Also got 7.5m for 8 BUT forgot the cos30 when I was finding friction, but cos30 is about 0.86 so my term for friction wouldn't have been too far out, so I think 7.58 is probably right.

Also got 2root(ag) for for 5, surely you've got 0.5mv^2 > 0.5mu^2 - 2 ag, so if v > 0 for complete revolution, u must be > root(4ag) i.e. 2root(ag)?
12. Q4a was 10.4 im pretty sure
13. (Original post by benjammy)
Surely v>0 for compete circular motion rather than v>gl, I'm sorry I'm confused to see where you got that from
I'm with you, the condition for continued circular motion is v>0 at the top isn't it?
14. (Original post by benjammy)
Surely v>0 for compete circular motion rather than v>gl, I'm sorry I'm confused to see where you got that from
Because if then the tension is negative which is impossible.
15. (Original post by IrrationalRoot)
Guys, it's since we have and we require at the top of the circle, hence .
Yeah, but if you go off the basis that to make revolutions, v>0 this gives min u = root (4gl). This question has come up before, and the answer was definitely 2root(gl) then.

Posted from TSR Mobile
16. (Original post by benjammy)
Surely v>0 for compete circular motion rather than v>gl, I'm sorry I'm confused to see where you got that from
The question you are talking about had no tension so u need the velocity to be greater than 0 but in the exam there was a string so need to the tension greater than 0
17. 7.58 isn't right for the last question because there is work done against friction
18. (Original post by CoIIins)
this question dosen't account for tension in the wire
19. Got all the same, except 7.58 for Q8 and 2root(gl) for the other one, I may be wrong though

## Register

Thanks for posting! You just need to create an account in order to submit the post
1. this can't be left blank
2. this can't be left blank
3. this can't be left blank

6 characters or longer with both numbers and letters is safer

4. this can't be left empty
1. Oops, you need to agree to our Ts&Cs to register

Updated: June 28, 2016
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Today on TSR

### A level maths

Which modules should I take?

Poll
Useful resources