Mechanics1 Worked solutions May 2010 paper

Link to worked solutions for May 2010 M1 paper:

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpZTE2YTNjODMtOTI0Zi00MDJhLWFlYTEtYzk1NzcyYzllN2U5&hl=en


Further supporting notes for the worked solutions:

1. position at time t=6s subtract 4s x velocity to find position at t=2s then magnitude of the position vector to find the distance (using pythagoras)

2. diagams shows speeds and directions before and after collision
(a) use principle of conservation of momentum - considering whole system - initial momentum (mass x velocity) = final momentum - when equation for initial=final is set up, m and u can be eliminated so k can be found
(b) considering only particle P find the change in momentum (final momemtum - initial momentum) and this is equal to the impulse

3. considering vertical forces allows expression for normal reaction to be found which then allows expression for the frictional force to be found by considering max friction = coefficient of friction x reaction force
- then considering horizontal forces allows an equation to be set up which can be solved for m

4. the unknown distance is set up as x being the distance from Tom to the pivot (to make calculations relatively simple) which so as the pivot is 1m away from the end of the beam Sophie is x+1 from Tom (double distance from end of beam)
- using the given info about the reactions and considering up=down allows the reaction forces to be calculated
then moments about D has x as the only unknown
- not forgetting that the distance from Tom to the end of the beam is x+1 when stating the final answer

5. (a) both P and Q are horizontal (for the period of constant speed) and then a straight line diagonally down to meet at the same point (for the period of deceleration) - the lines must cross so that there is a period of time where Q is faster than P so that it catches up
(b) area under a speed-time graph represents distance - both have travelled 800m - use area of trapezium on Q to find X and then same for P to find T

6. (a) suvat on journey from starting point to max height to find distance - then add the 49m to find height above ground
(b)&(c) suvat on whole journey until hits floor to find speed at that point and time taken

7. because P is horizontal it is most efficient to work vertically first to eliminate P to find R and then horizontally to find P

8. direction of motion indicated is because A is heavier than B
(a)&(b) F=ma on each particle finds two equations in a and T which can then be solved simultaneously
(c) suvat on motion for 0.5 seconds before string breaks to find distance travelled in that time - add this distance to 1m for height above ground when string breaks
then suvat for motion after string breaks (travelling freely under gravity)
solve quadratic for t using quadratic formula - the negative solution for t does not apply to the situation