MEI Mechanics 2 (M2) - 18 May 2016 Watch
- 24-05-2016 21:57
(Original post by Rawsonj)
- 17-06-2016 00:13
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!