MEI Mechanics 2 (M2) - 18 May 2016

Did anyone get 15N for the tension on the very last question

Reply 41

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

Reply 42

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 :colonhash:

Reply 43

Original post by 11234

Did anyone get 15N for the tension on the very last question

Yeah!

Reply 44

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 :colonhash:

I think you got e=7/3>1 but 0<e<1

Reply 45

Im so glad this is done and over with.... :smile:

Reply 46

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 :/

Reply 47

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

Reply 48

mohdstudent

1

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.

Reply 49

Rawsonj

1

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! :biggrin:

(edited 7 years ago)

Reply 50

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? :eek:

Reply 51

reecedad

thats not correct, to keep is momentum constant, its speed must change to balance out the fact that you lost weight.

Reply 52

Rawsonj

1

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? :eek:

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.

Reply 53

tambeshakunt

2

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?

Reply 54

Wunderbarr

16

The last question I didn't multiply the tension components by their distance, but I did for the centre of mass.

NOOOOOOOOOOOOOOOOOOOOoooooooooooooooooooooOOOOOOOOOOOOOOOOO.

Reply 55

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! :biggrin:

I think I got the same answers!!! What do you think boundary wise it will be this year

Reply 56

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

Reply 57

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