AQA A2 MM2B Mechanics 2 – 27th June 2016 [Exam Discussion Thread]

Yes there is the correct answer is

\sqrt{5gl}

and that is certain.

I certainly hope so since that's what I put!

Reply 221

jjsnyder

OP

10

Original post by IrrationalRoot

I needed 0UMS too XD. Still really disappointing that I got 80s in this paper, worst mark I've got in an M2 paper by far and that's just from two tiny mistakes.

Me too! Made a couple silly errors, but I don't mind as I had only done a couple past papers and was never taught the module, actually just read the M2 book on a train XD

Reply 222

MrCahooner

1

Original post by LiesandAlibis

Also, is there a general consensus on 2(root)gl versus (root)5gl?

In my opinion:

The answer is (root)5gl because this is the answer found with the assumption that the tension in the string must be greater than or equal to zero at the top.

An answer of 2(root)gl is yielded when assuming that the velocity is greater than or equal to zero at the top.

This is true for a constrained circular motion, e.g. a bead on a wire where it cannot go anywhere else other than along the wire, but for situations where the particle is unconstrained, the tension must also be greater than or equal to zero. This is an additional condition which stops the particle from taking a decay path before reaching the top.

Reply 223

IrrationalRoot

20

Original post by jjsnyder

Me too! Made a couple silly errors, but I don't mind as I had only done a couple past papers and was never taught the module, actually just read the M2 book on a train XD

Wow fair play to you then, I was taught the whole module. Also I did quite a few past papers (10 or so). So for me, no excuses for stupid mistakes lol.

Reply 224

IrrationalRoot

20

Original post by MrCahooner

In my opinion:

The answer is (root)5gl because this is the answer found with the assumption that the tension in the string must be greater than or equal to zero at the top.

An answer of 2(root)gl is yielded when assuming that the velocity is greater than or equal to zero at the top.

This is true for a constrained circular motion, e.g. a bead on a wire where it cannot go anywhere else other than along the wire, but for situations where the particle is unconstrained, the tension must also be greater than or equal to zero. This is an additional condition which stops the particle from taking a decay path before reaching the top.

Yep this is correct reasoning. Bead on a wire can have a reaction force upwards but tension will not act upwards.

Reply 225

tanyapotter

20

do people think this paper was comparable to june 2014, or was it easier/harder?

Reply 226

Hjyu1

2

Original post by tanyapotter

do people think this paper was comparable to june 2014, or was it easier/harder?

Really similar but I think there were little more things u could trip up on so I think slightly Lower gb that 2014

Reply 227

mcride98

1

How many marks was the last question?

Reply 228

scenes28

1

1-u/2 for the ladder question? 7.15m last question?

Reply 229

scenes28

1

Original post by mcride98

How many marks was the last question?

8 i think

Reply 230

higginsy7

for the last question, i have a strong opinion that the final answer is 5.46m. Both extensions where taken from the distance from q where the next point of rest is. one of the extensions was 4+x and the other i think was 11-x.

Reply 231

Andrew Brockbank

Working for question 8.
6.78m being final answer.

Reply 232

higginsy7

Original post by Andrew Brockbank

Working for question 8.
6.78m being final answer.

you have used the conservation of energy, but i used forces, the next point of rest will mean that the forces are in equilibrium from each string.

Reply 233

Andrew Brockbank

Original post by higginsy7

you have used the conservation of energy, but i used forces, the next point of rest will mean that the forces are in equilibrium from each string.

The forces dont need to be in equilibrium. Where the forces are in equilibrium will be the final place the mass comes to rest, it simply asks for the next postition the mass comes to rest. It will only be breifly at 6.78m before going up again towards Q.