OCR MEI Mechanics 2 (M2) 19th May 2014

Going by this, I think I got 68/72 too. I messed up the question on 2) iii). I knew I should have done 1-y coordinate, but I took very long on question 1 and 2 and was stressed out, so when I did the final arctan on my calculator, I used the wrong saved value.
I didn't have time to check it and because I used a long and stupid method with no diagram, I'll definitely lose all marks on this question.

it was 4 marks right?

Also Q 2i) didn't it just ask you to work out the Z coordinate? or did it ask for all of them? I might have misread the question if its all of them as I only worked out the Z coordinate. that might be 2 more marks gone if so.

right so that exam went okay for me, hopefully something like 68/72!

my answers are as follows:
1.a) i) I know i got the diagram right
ii) i got 1/12 u towards the left ( same direction as before) (edit: i also got 5/16 for coefficient of restitution)
b) i) v=8 u=9
ii) 87J

2. i) by symmetry x co-ord = 0 & y co-ord = 0.5a. worked out z co-ord to be 0.4a.
ii) (0, 9/16 a, 5/8 a)
iii) 35 degrees
iv) i got all the term correct for this but somehow miscalculated the answer -.- should have been 0.342 and achieved by (4mg-16m)/(4root3 *mg)

3. a) i) y=20 x=0 r=80
ii)ac is thrust, ab tension, bc thrust plus all external forces.
iii) Tbc = 40 root 3
Tba = 20 root 3
Tac = 40
b) i) P = 148
ii) Q = 168
then resolved vertically correct but horizontally incorrect so I got final pivot force wrong (final answer should have been 278N i think)

4. a) i) show eke to be 1377.8 and mgh=1372 . 1377.8>1372 blah blah blah
ii) got v to be 8.27 m/s
iii) got 1.75 m above ground
b) got P = 30,000W and F=600N.

I've got writings of my working if anybody wants to cross reference :-)

right so that exam went okay for me, hopefully something like 68/72!

my answers are as follows:
1.a) i) I know i got the diagram right
ii) i got 1/12 u towards the left ( same direction as before) (edit: i also got 5/16 for coefficient of restitution)
b) i) v=8 u=9
ii) 87J

2. i) by symmetry x co-ord = 0 & y co-ord = 0.5a. worked out z co-ord to be 0.4a.
ii) (0, 9/16 a, 5/8 a)
iii) 35 degrees
iv) i got all the term correct for this but somehow miscalculated the answer -.- should have been 0.342 and achieved by (4mg-16m)/(4root3 *mg)

3. a) i) y=20 x=0 r=80
ii)ac is thrust, ab tension, bc thrust plus all external forces.
iii) Tbc = 40 root 3
Tba = 20 root 3
Tac = 40
b) i) P = 148
ii) Q = 168
then resolved vertically correct but horizontally incorrect so I got final pivot force wrong (final answer should have been 278N i think)

4. a) i) show eke to be 1377.8 and mgh=1372 . 1377.8>1372 blah blah blah
ii) got v to be 8.27 m/s
iii) got 1.75 m above ground
b) got P = 30,000W and F=600N.

I've got writings of my working if anybody wants to cross reference :-)

Well, I didn't even attempt a quarter of the paper, so I'm doing my bit at lowering the grade boundaries for you. 😂😥

show working for 4B power question since i obtained a tenth of your values.

P = 50F when at constant velocity therefore retarding force is P/50.
when p reduced by 20% acc is -0.08 so F=m.a = 1500*-0.08 = -120N
0.8 * P = 2P/125
P/50 - 2P/125 = 120
2.5P - 2P = 15000
0.5P =15000
P = 30,000W therefore 30,000/50 = 600N (F)

right so that exam went okay for me, hopefully something like 68/72!

my answers are as follows:
1.a) i) I know i got the diagram right
ii) i got 1/12 u towards the left ( same direction as before) (edit: i also got 5/16 for coefficient of restitution)
b) i) v=8 u=9
ii) 87J

2. i) by symmetry x co-ord = 0 & y co-ord = 0.5a. worked out z co-ord to be 0.4a.
ii) (0, 9/16 a, 5/8 a)
iii) 35 degrees
iv) i got all the term correct for this but somehow miscalculated the answer -.- should have been 0.342 and achieved by (4mg-16m)/(4root3 *mg)

3. a) i) y=20 x=0 r=80
ii)ac is thrust, ab tension, bc thrust plus all external forces.
iii) Tbc = 40 root 3
Tba = 20 root 3
Tac = 40
b) i) P = 148
ii) Q = 168
then resolved vertically correct but horizontally incorrect so I got final pivot force wrong (final answer should have been 278N i think)

4. a) i) show eke to be 1377.8 and mgh=1372 . 1377.8>1372 blah blah blah
ii) got v to be 8.27 m/s
iii) got 1.75 m above ground
b) got P = 30,000W and F=600N.

I've got writings of my working if anybody wants to cross reference :-)

Pretty sure this is all correct and the impulse in 1aii was 7.5u I think

that question was 5 marks but i doubt you'll have lost all 5 as you got the z value correct and yes you're right about 2i. doubt they'll drop me any marks for giving more than they asked for though lol.

how many marks do you think it was out of 72 for a standardised 100%?

If the examiner bothers to read all my jumble and understand what I did, maybe I'll get a mark or so but I don't think they'll understand anything, I though what I was doing was wrong, so didn't put much effort into it as I felt like I was running out of time. Came out of the exam and realised although my method was long winded, if I had used the correct value 1-y I would of got 35 anyway.


I did do all MEI past papers from June 2006 but I haven't been checking the grade boundaries, so I don't have much of an idea. The problem with this paper is that most of Q3 and Q4 weren't very hard while Q1 and Q2 shouldn't have been hard but all the algebra would have definitely put people off, like it did to me and cause silly mistakes.

For this reason I think that the standardisation will be weird. e.g. you would need lower marks to get B and C but A and A* will be roughly as should be or slightly lower. I don't understand how MEI works out 100% UMS but I'll be happy with above 90.

yes i got 7.5u too, forgot that bit!

Pretty sure this is all correct and the impulse in 1aii was 7.5u I think

I put it as -7.5u is that still right? Since the momentum of Q decreased?

I think 3/6 marks is reasonable, you should get all the method marks and miss out on the final few.

In 4)a)ii) I think you'd probably get 1, based on past papers I think it would be like 1 mark for attempting to use the correct method (e.g. moments about A or whatever) and then 1 is for all correct terms, then final for correct answer.

I think it would depend on which way you defined as the positive direction of momentum in your diagram and which way round you put the objects, so either positive or negative can be correct, it just has to be consistent with the structure of earlier working