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# OCR MEI M2 - 18th May 2016

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1. (Original post by decombatwombat)
Yeah I got 0.866 and 44.1 also. For the first question did you get e=1/9 and v = 7/3?
Yup, got that. You know for the question what is the speed of P after its top mass is dropped when its moving with 10ms relative velocity question. The 4 marker. I got it to be 3 ms-1. You ?
2. (Original post by decombatwombat)
Yeah I got 0.866 and 44.1 also. For the first question did you get e=1/9 and v = 7/3?
Ah hopefully I got that then, I was only positive on remembering the 8 part, friction sounds familiar to me too and I got both of those other values also
3. (Original post by decombatwombat)
Yeah I got 0.866 and 44.1 also. For the first question did you get e=1/9 and v = 7/3?
Yup, got that. You know for the question what is the speed of P after its top mass is dropped with relative velocity of 10 ms-1. Did u get 3 ms-1 ?
Yup, got that. You know for the question what is the speed of P after its top mass is dropped when its moving with 10ms relative velocity question. The 4 marker. I got it to be 3 ms-1. You ?
OMG Thank god, yeah I got that, took me a while to realise it meant separation speed was 10, not just speed of the object.

I think the only other strange ones were the last part of question 1 and part ii I thinkg. I just compared the horizontal components before and after, and because they weren't equal, the plane wasn't smooth. For part 2, I found the impulse on Q if the direction was reversed, which gave Q
5. I'll add to the questions from earlier

Q1

i) 7/3 i think e=1/9
ii) Show the plane cannot be smooth
ii) If P was -2ms-1, then e would be 7/3 which is greater than 1, therefore impossible.
iv) The one where it fires of think I got 28/9, because it was fired off with a speed of 10ms^-1 relative to P, which was moving at 2ms^-1 so it dropped off at 8ms^-1 relative to the ground.
v) The one where it dropped off Q kept the same velocity, because it would drop off with the same velocity as Q, so momentum has to be conserved and no force applied in the direction of motion.

Q2

i) F = 250N
ii) 0.5ms-1 for the block and bullet
iii) 0.198m
bi) Power one. 640.5 Watts

Q3

i) x = 0.99 and y = 0.42
ii) other Y was 180 I think and show that = 120
iii) Diagram
iv) Show Xd = 261N
v) DC in 435N of compression. AC in 285N of compression. CB in 432N of compression. AB in 468N of tension. YD = 348N. XD = 48N
(You only needed to find the forces in three of the rods for the question, but not sure which three)

Q4

i) Show that COM. You need to times the area of the circle at the top by 4 as it had four times the density.
ii) "Show that" Quadratic and then solving for h
iii) 15N for T

Anything missing?
Yup, got that. You know for the question what is the speed of P after its top mass is dropped when its moving with 10ms relative velocity question. The 4 marker. I got it to be 3 ms-1. You ?
OMG Thank god, yeah I got that, took me a while to realise it meant separation speed was 10, not just speed of the object.

I think the only other strange ones were the last part of question 1 and part ii I thinkg. I just compared the horizontal components before and after, and because they weren't equal, the plane wasn't smooth. For part 2, I found the impulse on Q if the direction was reversed, which gave Q v=5, and so an overall gain in kinetic energy in the system, which isn't possible without an external force. Not so sure about that part though.
7. (Original post by leaf3)
Got the same for most of these
got very confused in the ME question so didn't get that correct
I think I got μ=0.833... or something?
Also for the one where it's dropped off, no change in velocity as no horizontal component applied? (thought it was like the snowball past exam one of you can remember it?)
Was getting answers in the 400's and 300's regions for the tensions frameworks question (I think), definitely got the 468N one, not sure about the 111N though
Yeah, you're right for the dropped off one, I messed that one up.
8. When it said explain your reasoning, and it said like negligible relative velocity. I got that to have the same speed as before, is this just me? the two marker? So wait, definitely 75 N compression in the frameworks isn't familiar to anyone ?
When it said explain your reasoning, and it said like negligible relative velocity. I got that to have the same speed as before, is this just me? the two marker? So wait, definitely 75 N compression in the frameworks isn't familiar to anyone ?
Yeah, I got the same speed as before. I dont remember 75N compression, but I remember 2 of the rods having compression, and one having tension. That 75 might be for the fourth rod though, but you only had to work out three, so you might still be fine.
10. Oops repost.
11. The frameworks one sucked because they decided to print the diagram on the other side of the page meaning you had to keep flipping
12. For 2i I got F = 2500N and 0.01988m for next part, anyone else?
13. 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.
14. (Original post by Bealzibub)
For 2i I got F = 2500N and 0.01988m for next part, anyone else?
No, sorry, got 250 N as resistance and then 0.198 m. I feel like you might have made an arithmetic error.
15. (Original post by Connorbwfc)
I'll add to the questions from earlier

Q1

i) 7/3 i think e=1/9
ii) Show the plane cannot be smooth
ii) If P was -2ms-1, then e would be 7/3 which is greater than 1, therefore impossible.
iv) The one where it drops of think I got 28/9, because it dropped off with a speed of 10ms^-1 relative to P, which was moving at 2ms^-1 so it dropped off at 8ms^-1 relative to the ground.

Q2

i) F = 250N
ii) 0.5ms-1 for the block and bullet
iii) 0.198m
bi) Power one. 640.5 Watts

Q3

i) x = 0.99 and y = 0.42
ii) other Y was 180 I think and show that = 120
iii) Diagram
iv) Show Xd = 261N
v) DC in 435N of compression. AC in 285N of compression. CB in 432N of compression. AB in 468N of tension. YD = 348N. XD = 48N
(You only needed to find the forces in three of the rods for the question, but not sure which three)

Q4

i) Show that COM. You need to times the area of the circle at the top by 4 as it had four times the density.
ii) "Show that" Quadratic and then solving for h
iii) 15N for T

Anything missing?
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
16. (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
Definitely was thinking this, but for 2 marks- i just did 91.5 times by 7. :/
17. (Original post by Connorbwfc)
iv) The one where it drops of think I got 28/9, because it dropped off with a speed of 10ms^-1 relative to P, which was moving at 2ms^-1 so it dropped off at 8ms^-1 relative to the ground.
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.
18. (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.
19. (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
Power = Driving Force x Velocity, so you don't need to take into account the other forces :<
20. (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 asked for the power of the tension I think.

So I just did the force of the tension (91.5m) multiplied by 7 (the velocity)

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