You are Here: Home >< Maths

# MEI Mechanics 2 (M2) - 18 May 2016 Watch

1. (Original post by SigCig)
Yep Mech 3 tomorrow. Good luck to everyone doing it

Posted from TSR Mobile
Have a feeling its gonna be a hard paper...
2. (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!
for question 2, in that of the block on the slope, how come you are taking the initial KE from the work done by the rope pulling it up the slope? I would say they both provide the energy for the posterior translation of this energy as work done against friction, gain in GPE and a final KE

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: June 17, 2016
Today on TSR

### What is the latest you've left an assignment

And actually passed?

### Simply having a wonderful Christmas time...

Discussions on TSR

• Latest
• ## See more of what you like on The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

• Poll
Useful resources

Can you help? Study help unanswered threadsStudy Help rules and posting guidelinesLaTex guide for writing equations on TSR

## Groups associated with this forum:

View associated groups
Discussions on TSR

• Latest
• ## See more of what you like on The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

• The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.