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# MEI Mechanics 2 (M2) - 18 May 2016

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1. Did anyone get 15N for the tension on the very last question
2. (Original post by chemari1)
Was the coefficient of friction 1.57 and was the tension in one of the rods 394?
I got something like 0.86 or something
3. On question 1, not sure what part, when it said why can't P be moving at half the speed in the opposite direction after the collision, how do you show it? It was 3 marks and I don't think I did enough or even did the right thing at all
4. (Original post by 11234)
Did anyone get 15N for the tension on the very last question
Yeah!
5. (Original post by Alex621)
On question 1, not sure what part, when it said why can't P be moving at half the speed in the opposite direction after the collision, how do you show it? It was 3 marks and I don't think I did enough or even did the right thing at all
I think you got e=7/3>1 but 0<e<1
6. Im so glad this is done and over with....
7. (Original post by 11234)
I think you got e=7/3>1 but 0<e<1
did you work out what the new final velocity of Q would be if P's velocity did half and change direction? at first I just recalculated e using the negative version of P's final velocty, but then I crossed it out and worked out the new velocity of Q. I can't remember what I got for e but it was bigger than 1 :/
8. (Original post by Alex621)
did you work out what the new final velocity of Q would be if P's velocity did half and change direction? at first I just recalculated e using the negative version of P's final velocty, but then I crossed it out and worked out the new velocity of Q. I can't remember what I got for e but it was bigger than 1 :/
Yeah I said something like assume its reversed for the sake of contradiction then calculate e based on that which was greater than 1
9. (Original post by 11234)
Yeah I said something like assume its reversed for the sake of contradiction then calculate e based on that which was greater than 1
I did that too but i didnt state that e is greater (although my answer of e was greater than 1) instead i said that new e doesnt equal old e (the one we found the previous part) is that acceptable?

Because e is a constant and it shouldnt change.
10. Answers I got for this paper:
1: PCLM: speed of Q =7/3 ms^-1.e=1/9. If direction of P was reversed as well as halved, there would be a gianin K.E: 4.375J before and 10.375J after (could also show that e>1). Dropping objectoff back of Q does not change the speed as there is no force in the directionof motion of Q so no impulse. Velocity of p after onject is fired off: using PCLM gives 0.5*2=0.05*(V-10)+0.45V. This gives V=3. The wall cannot be smooth as the velocityparallel to the wall is not the same before and after: 10cos60 is not equal to6cos40.
2: K.E converts into work doneagainst resistance: 0.5*0.04*2500=0.2F. F=250N. PCLM gives 0.04*50=4*V.V=0.5ms^-1. Change in K.E of bullet= work done against resistance + gain in KEof block: 0.5*0.04*2500-0.5*4*0.25=250d. d=0.198m. Next was the block on theslope.Find the resistance force and mew (sorry for the lack of symbol).91.5*8-0,5*6*(49-1)-9.8*6*8*sin30=WD (against resistance)=352.8J. Frictionforce =WD/distance =352.8/8=44.1N. F=mewR.44.1/(9.8*6*cos30)=0.866025...=0.866. Finally the power of the tension in thestring when moving at 7 ms^-1. p=fv= 91.5*7=640.5W.
3: centre of mass (0.99,0.42) Rand Q were 180 and 120 (not neccesarily respectively... i am not sure. It was ashow that). To label the diagrams I just added the 180N and 120N and put allthe arrows as tensions. To find X(d) moments about A gives 261N (another showthat). To find the tensions in AB BC and CD, resolve first vertically at b.Gives 468N as tension in AB. Resolve horizontally at B gives 432N (compressionin BC). Resolving horzintally at D (using X(d)=261) gives 435N compression inDC. Resolving vertically at D gives Y(d)=348N. Resolving the exernal forces ofthe framework vertically gives Y(a)=300-348=-48N =48N vertically downwards.
4: The first two parts were show that’s.The second (for 7 marks involved finding tan(alpha) in terms of h and then solvinga quadratic for h=0.2. The last part of the paper was find T. Moments about theedge about which is will tip (was the edge closest to O and further to theright): T*0.5*cosA-T*0.2*sinA-0.1*42=0. (where tanA=3/4 so a 345 triangle offorces) This gives T=15N. Thats the whole paper!
11. (Original post by Rawsonj)
x
Cheers! I think I got all the same as you, except I can't remember what I got for the tensions of the rods for the frameworks question. For Q2, what did the question ask where you have an answer of V=0.5ms^-1, not sure if I missed that part of the question?
12. thats not correct, to keep is momentum constant, its speed must change to balance out the fact that you lost weight.
13. (Original post by Alex621)
Cheers! I think I got all the same as you, except I can't remember what I got for the tensions of the rods for the frameworks question. For Q2, what did the question ask where you have an answer of V=0.5ms^-1, not sure if I missed that part of the question?
It was a two marker (part ii) telling you to use PCLM to find the velocity of the block (withe the bullet embedded) if the block is on a smooth plane.
14. For the very last one, about finding T, I misread it as 'point of toppling' instead of 'point of turning'. When I worked through, I got T=0 which I knew was not right but didn't have enough time to redo it.
Do you think I would get any method marks for this?
15. The last question I didn't multiply the tension components by their distance, but I did for the centre of mass.

NOOOOOOOOOOOOOOOOOOOOooooooooooo ooooooooooOOOOOOOOOOOOOOOOO.
16. (Original post by Rawsonj)
Answers I got for this paper:
1: PCLM: speed of Q =7/3 ms^-1.e=1/9. If direction of P was reversed as well as halved, there would be a gianin K.E: 4.375J before and 10.375J after (could also show that e>1). Dropping objectoff back of Q does not change the speed as there is no force in the directionof motion of Q so no impulse. Velocity of p after onject is fired off: using PCLM gives 0.5*2=0.05*(V-10)+0.45V. This gives V=3. The wall cannot be smooth as the velocityparallel to the wall is not the same before and after: 10cos60 is not equal to6cos40.
2: K.E converts into work doneagainst resistance: 0.5*0.04*2500=0.2F. F=250N. PCLM gives 0.04*50=4*V.V=0.5ms^-1. Change in K.E of bullet= work done against resistance + gain in KEof block: 0.5*0.04*2500-0.5*4*0.25=250d. d=0.198m. Next was the block on theslope.Find the resistance force and mew (sorry for the lack of symbol).91.5*8-0,5*6*(49-1)-9.8*6*8*sin30=WD (against resistance)=352.8J. Frictionforce =WD/distance =352.8/8=44.1N. F=mewR.44.1/(9.8*6*cos30)=0.866025...=0.866. Finally the power of the tension in thestring when moving at 7 ms^-1. p=fv= 91.5*7=640.5W.
3: centre of mass (0.99,0.42) Rand Q were 180 and 120 (not neccesarily respectively... i am not sure. It was ashow that). To label the diagrams I just added the 180N and 120N and put allthe arrows as tensions. To find X(d) moments about A gives 261N (another showthat). To find the tensions in AB BC and CD, resolve first vertically at b.Gives 468N as tension in AB. Resolve horizontally at B gives 432N (compressionin BC). Resolving horzintally at D (using X(d)=261) gives 435N compression inDC. Resolving vertically at D gives Y(d)=348N. Resolving the exernal forces ofthe framework vertically gives Y(a)=300-348=-48N =48N vertically downwards.
4: The first two parts were show that’s.The second (for 7 marks involved finding tan(alpha) in terms of h and then solvinga quadratic for h=0.2. The last part of the paper was find T. Moments about theedge about which is will tip (was the edge closest to O and further to theright): T*0.5*cosA-T*0.2*sinA-0.1*42=0. (where tanA=3/4 so a 345 triangle offorces) This gives T=15N. Thats the whole paper!
I think I got the same answers!!! What do you think boundary wise it will be this year
17. If I put two resasons with the main body being the one I think is right and the secondary one not, will I lose marks
18. I'm so glad that that's over! I can't believe I actually found that paper ok! It seems like I have similar answers to everyone however on the one abot how far the bullet goes...what was the original distance before they included friction

The only question I really stuggled with was q4) i! I'm suprised I got ii and iii!
19. (Original post by 11234)
I think I got the same answers!!! What do you think boundary wise it will be this year
Guessing around 58. But wouldn't be surprised if it was higher for an A.
20. 65/66 what UMS do u think? ( made really stupid mistakes that I could kick myself for!)

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