Competition - post a photo you've taken of nature >>

AQA Mechanics 2 Unofficial Mark Scheme

Scroll to see replies

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

(edited 7 years ago)

Reply 81

Ben :)

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?

Reply 82

C0balt

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

Reply 83

typing

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

\sqrt{5ag}

.

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

v>0

is a necessary and sufficient condition for circular motion. It is not. It is just a necessary condition. What is sufficient is that

v \geq gl

. This is because if

v<gl

, 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

Reply 84

jjsnyder

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.

Posted from TSR Mobile

Reply 85

Cal 1

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?

Reply 86

typing

Original post by jjsnyder

You get what they mean, it's because the radius is often labelled a.

Posted from TSR Mobile

Ah right cheers

Reply 87

IrrationalRoot

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.

Reply 88

naro.n

For question 7 (b) did anyone get

tanx = -μ/4

Reply 89

Ben :)

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!

Does anyone agree?

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...

Reply 92

Ben :)

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!

Reply 93

C0balt

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

Reply 94

Jac_Jac

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 :smile:

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.

Reply 95

C0balt

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 :smile:

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

Reply 96

Jac_Jac

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 :smile:

Reply 97

C0balt

Original post by Jac_Jac

I'm saying that those are the solutions you get when you assume GPE is 0 at R :smile:

Ah okay, sorry I'm not with it rn lol

Reply 98

Jac_Jac

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 :biggrin:

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 :smile:

Sorry for the confusion, I didn't do the exam, was just interested in trying the questions since people were saying it was challenging :biggrin:

(edited 7 years ago)

Reply 99

C0balt

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 :biggrin:

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 :smile: