You are Here: Home >< Maths

# OCR MEI M2 - 18th May 2016

• View Poll Results: Predicted Grade Boundary for an A?
60<
11
17.74%
59
2
3.23%
58
9
14.52%
57
6
9.68%
56
14
22.58%
55
5
8.06%
54
8
12.90%
53
3
4.84%
<52
4
6.45%

1. (Original post by chemari1)
does anyone remember the numbers for thecoefficient of frition question?
was it
0.5 x 6 x 49 + 91.5 x 8 = 6x9.8x8sin30 + 0.5 x 6x1 + 8 F
ke work done by tensionsion gain in gpe final ke work done against friction

F=80.1, and coefficient= 80.1/ 6x9.8xcos30 = 1.57!!!!!!
I musta been right kiddars
0.5*6*49 - 0.5*6*1 = 91.5*8 - 6*9.8*8sin(30) - 8F

So F=44.1
Coefficient = 44.1/6*9.8cos(30) = 0.866

It looks as if you have you're '49' and '1' in your kinetic energy terms the wrong way around.
2. (Original post by Connorbwfc)
0.5*6*49 - 0.5*6*1 = 91.5*8 - 6*9.8*8sin(30) - 8F

So F=44.1
Coefficient = 44.1/6*9.8cos(30) = 0.866

It looks as if you have you're '49' and '1' in your kinetic energy terms the wrong way around.
I did:
initial energy +work done by tension = final energy + work done against friction

with 91.5*8 in the same side as 0.5*6*49 as they both provide the energy for the movement?
3. (Original post by Connorbwfc)
0.5*6*49 - 0.5*6*1 = 91.5*8 - 6*9.8*8sin(30) - 8F

So F=44.1
Coefficient = 44.1/6*9.8cos(30) = 0.866

It looks as if you have you're '49' and '1' in your kinetic energy terms the wrong way around.
i got the exact same answer.
I thought the first question was a bit messed tbh.

I hope the grade boundary for an A isnt too high , Because i thought this was harder than last year by a bit. Im gonna guess that an A is probably 58 and a B is like 49 or sumthin lol
4. how many marks was question 4 as a whole?
5. (Original post by chemari1)
how many marks was question 4 as a whole?
I think It was 18? A 5, 6 and 7 marker? (Think order was 6,7,5?)
6. Guys you know for the framework lets say i got one of the internal forces, and the rest i did right method and used wrong values, how many marks do you reckon i dropped ? Question was out of 8 marks btw.
7. Can anyone remember their working for the mechanical energy one?
8. (Original post by leaf3)
Can anyone remember their working for the mechanical energy one?
loss of kinetic energy = w.d against friction
9. (Original post by leaf3)
Can anyone remember their working for the mechanical energy one?
0.5 x 6 x 49 + 91.5 x 8 = 6x9.8x8sin30 + 0.5 x 6x1 + 8 F
initial ke + work done by tension =gain in gpe+ final ke +work done against friction F=80.1, and coefficient= 80.1/ (6x9.8xcos30) = 1.57
10. (Original post by Bealzibub)
loss of kinetic energy = w.d against friction
Ah yeah I was just wondering about actual values, probably a bit hopeful though :P
thanks
11. (Original post by chemari1)
0.5 x 6 x 49 + 91.5 x 8 = 6x9.8x8sin30 + 0.5 x 6x1 + 8 F
initial ke + work done by tension =gain in gpe+ final ke +work done against friction F=80.1, and coefficient= 80.1/ (6x9.8xcos30) = 1.57
haha thanks! I actually meant the bullet one though sorry :O
12. (Original post by leaf3)
haha thanks! I actually meant the bullet one though sorry :O
I don't remember that one very well sorry, are you sure about question 4 being worth 18 marks?
13. (Original post by chemari1)
I don't remember that one very well sorry, are you sure about question 4 being worth 18 marks?
Each one of the 4 questions is worth 18 marks.
14. (Original post by StrangeBanana)
Consider this: the small object (call it R) is shot out at 10 ms^-1 relative to P's initial speed, i.e. at 8ms^-1 relative to the ground, with P moving off at 2ms^-1. Why is P's speed going to change? R has already left, it can't influence P's motion any more.

It has to mean 10ms^-1 relative to P's speed after the shot.
By the same logic, if it is fired out backwards at 10ms^-1 relative to P, you are saying that at the instant when it is fired out, P has already increased its speed to its new increased speed (And therefore the speed of separation between P's new speed and the speed of the bullet is 10ms^-1). How would P have increased to this speed at the instant as the bullet was shot out?

Doesn't P only increase to this new speed once it has actually shot out the bullet?

This all comes down to one question; at the exact instant the bullet it shot out, does P have its initial speed or the speed of after the projection?
15. (Original post by Connorbwfc)
Each one of the 4 questions is worth 18 marks.
cheers mate, do you think the boundary for an A is 52 by any chance?
16. (Original post by chemari1)
cheers mate, do you think the boundary for an A is 52 by any chance?
I would say that this is unlikely, but not impossible. Imo the A boundary will probably lie around 55-56.
17. (Original post by Connorbwfc)
I would say that this is unlikely, but not impossible. Imo the A boundary will probably lie around 55-56.
okay, I'm doing further pure 1 on Friday and mechanics 3 next week
18. I loved that paper. And M2 is my weakest module so I'm glad it went well. Took too long on the frameworks question though. Missed out that 2 marker and rushed the one after that. Looking at the answers here, looks like it went well.
19. (Original post by Connorbwfc)
By the same logic, if it is fired out backwards at 10ms^-1 relative to P, you are saying that at the instant when it is fired out, P has already increased its speed to its new increased speed (And therefore the speed of separation between P's new speed and the speed of the bullet is 10ms^-1). How would P have increased to this speed at the instant as the bullet was shot out?

Doesn't P only increase to this new speed once it has actually shot out the bullet?

This all comes down to one question; at the exact instant the bullet it shot out, does P have its initial speed or the speed of after the projection?
No, P's speed has increased at the instant after it has been fired. P's speed isn't really defined at the moment it's fired, because it changes instantaneously from 2 to 3 (obviously wouldn't happen in reality but does in this ideal scenario; it's a vertical line on the velocity-time graph). The 10ms^-1 is being "measured" at the instant after firing, not the instant before.
20. (Original post by StrangeBanana)
No, P's speed has increased at the instant after it has been fired. P's speed isn't really defined at the moment it's fired, because it changes instantaneously from 2 to 3 (obviously wouldn't happen in reality but does in this ideal scenario; it's a vertical line on the velocity-time graph). The 10ms^-1 is being "measured" at the instant after firing, not the instant before.
Thanks for the reply and I think that your answer will end up being correct according to the mark scheme...

However, how do we know when it says 'It is projected with a speed of 10ms^-1" relative to P", that this 10ms^-1 is measured at the instant after firing and not the instant before?

In this situation, what gives the instant after firing priority over the instant before firing?

If this question is out of 4, how many marks do you think I will get for my method? 2?

## 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: August 17, 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

### Lied about being a virgin

Should I come clean?

Poll
Useful resources