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# OCR MEI M2 - 18th May 2016

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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
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53
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<52
4
6.45%

1. (Original post by gemdarkstone)
For question 2bi) it asked for the power of the tension in the string whilst going 7 ms-1.the power would have to over come the the tensions as well so it would've been p=fv
p=(91.5+44.1+6gsin(30)) x 7 = 1165
but i'm not sure
No, power of a force is just force x velocity unless the 91.5 was resultant force
2. (Original post by gemdarkstone)
For question 2bi) it asked for the power of the tension in the string whilst going 7 ms-1.the power would have to over come the the tensions as well so it would've been p=fv
p=(91.5+44.1+6gsin(30)) x 7 = 1165
but i'm not sure
It just said power of string though, so surely you dont need to take into account the resistive forces? If it said power of friction, I dont think you would consider the other forces acting the other way. Not sure though, you might be right.
3. (Original post by StrangeBanana)
This isn't quite right. It's moving at 10ms^-1 after it has left P, so you have to let P's speed be u (say), then the small object's speed will be 10 - u, relative to the ground. Then you solve using conservation of momentum as usual; it gives you u = 3ms^-1.
The object was shot backwards with a speed of 10ms^-1 relative to P wasn't it? So relative to the ground it was shot back with a speed of 8ms^-1.
The original velocity of P was 2ms^-1 wasn't it?

So I did 0.5*2 = 0.05*-8 + 0.45*V.
and solved to get V = 28/9 = 3.11ms^-1 (3sf)
4. The one with mass dropping off speed stays same
5. was the tension shown to be drawn on at an angle and labelled with the 91.5 being the angled force? I can't remember
6. (Original post by Connorbwfc)
The object was shot backwards with a speed of 10ms^-1 relative to P wasn't it? So relative to the ground it was shot back with a speed of 8ms^-1.
The original velocity of P was 2ms^-1 wasn't it?

So I did 0.5*2 = 0.05*-8 + 0.45*V.
and solved to get V = 28/9 = 3.11ms^-1 (3sf)
No I didn't consider it like that, I did final velocities to be 10-v to the left and then v to the right. PCLM solve, resulted in 3 ms-1. I am definite this is right.
7. (Original post by Connorbwfc)
The object was shot backwards with a speed of 10ms^-1 relative to P wasn't it? So relative to the ground it was shot back with a speed of 8ms^-1.
The original velocity of P was 2ms^-1 wasn't it?

So I did 0.5*2 = 0.05*-8 + 0.45*V.
and solved to get V = 28/9 = 3.11ms^-1 (3sf)
Anyone else in the taking 10m/s as the object's velocity rather than its relative separation club?
8. (Original post by Bealzibub)
The one with mass dropping off speed stays same
are you sure?
because conservation of momentum:
0.75x7/3ms-1 = 0.1x0 + 0.65xv

as you can see, the Q object has lost mass, so the velocity would be different... that my working out though, i could be wrong. what did you do?
9. (Original post by StrangeBanana)
Connor you're missing part (iii) of question 1; a small mass falls off of Q after the collision, at negligible speed relative to Q, and it asked you to find Q's speed after that. It was 7/3 ms^-1 - no change from before.
It does change, the mass of Q changes because the small object falls off but the momentum still remains the same.

E.G. The mass before = 1kg and velocity = 5ms^-1 making the momentum 5Ns, Part of the mass (0.2kg) falls off and stops, but the momentum of the system still remains at 5Ns so the remaining 0.8kg must have a velocity of 6.25ms^-1 (5 / 0.8).

Sorry I can't remember the actual numbers
No I didn't consider it like that, I did final velocities to be 10-v to the left and then v to the right. PCLM solve, resulted in 3 ms-1. I am definite this is right.
It didn't say that they seperate at 10ms^-1.

It said that the object is shot of with a speed relative to P didn't it? At the point when it is shot off P is travelling at 2ms^-1 to the right, so the block is shot off at 8ms^-1 to the left surely?
11. (Original post by gemdarkstone)
are you sure?
because conservation of momentum:
0.75x7/3ms-1 = 0.1x0 + 0.65xv

as you can see, the Q object has lost mass, so the velocity would be different... that my working out though, i could be wrong. what did you do?
When it drops it has same velocity Q not 0
12. (Original post by gemdarkstone)
are you sure?
because conservation of momentum:
0.75x7/3ms-1 = 0.1x0 + 0.65xv

as you can see, the Q object has lost mass, so the velocity would be different... that my working out though, i could be wrong. what did you do?
When the thing is dropped off it still carries on travelling with the horizontal velocity that Q had as no force is applied.

Therefore Q carries on at the same speed.
13. (Original post by Connorbwfc)
When the thing is dropped off it still carries on travelling with the horizontal velocity that Q had as no force is applied.

Therefore Q carries on at the same speed.
Yeah, got Q carries at same speed as well
14. (Original post by Connorbwfc)
When the thing is dropped off it still carries on travelling with the horizontal velocity that Q had as no force is applied.

Therefore Q carries on at the same speed.
This is right, http://www.mei.org.uk/files/papers/m207ja_hneuc6.pdf check question 1. Same wording.
15. For the question which said that the object of mass 0.05kg is shot of with a speed of 10ms^-1 relative to Q. Does it mean this is relative to the speed of Q before it was shot or or after it was shot off. This is what I am having problems understanding.
16. (Original post by Connorbwfc)
It didn't say that they seperate at 10ms^-1.

It said that the object is shot of with a speed relative to P didn't it? At the point when it is shot off P is travelling at 2ms^-1 to the right, so the block is shot off at 8ms^-1 to the left surely?
Wasn't Q travelling at 4ms-1? The speed of object fired had velocity relative to q so object had velocity 10-4=6ms-1

I thought q or p was travelling at 4ms-1, whichever one it was hence speed of object being 6.
17. TBH as soon as you see the wording, explain your reasoning you can just tell its gonna be something dirty like the speed remains the same.
18. (Original post by decombatwombat)
This is right, http://www.mei.org.uk/files/papers/m207ja_hneuc6.pdf check question 1. Same wording.
"Very few candidates could obtain any credit for this part. Many did notappreciate the vector nature of the problem and merely stated (incorrectly)that the sledge would speed up because the mass had decreased andmomentum had to be conserved. A small number of candidates appreciatedthat there would be no change in the velocity of the sledge but could notgive a valid reason for this. Few mentioned that there was no force on thesledge in the direction of motion."

I feel like even 9 years later, it'll still be a minority that gets it. I mean hopefully...
19. What are the boundaries gonna be like ? High like 60 + or around 57 ?
What are the boundaries gonna be like ? High like 60 + or around 57 ?
Wouldn't be surprised if we had the exact same grade boundaries as last year: 62, 56, 51 for A*, A, B

Edit: my mistake, I was looking at a different subject on my spreadsheet of grade boundaries, the actual boundaries last year were 61 54 47

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