Mechanics1 Worked solutions Jan 2006 paper

The link is to the worked solutions for the January 2006 Mechanics 1 paper and then there are some notes below explaining the solutions.

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpZTBiYmFiMzEtYjBjNS00ZDdkLWI3NDUtODg3ZTdiOTFkNjA5&hl=en

1. The particle is going to travel up to a maximum height and then falll to the ground. Known suvat variables for the whole journey are u,a,t so h and v can be found to answer the the two parts of the question.

2. (a) Remember momentum = mass x velocity and remember initial momemtum = final momentum if whole system considered. Only unknown is the veolicty to be found.
(b) (i) Again only unknown considering momentum of whole system is the mass of Q to be found.
(ii) Remember impulse = change in momentum. Interested in impulse on Q so consider particle Q only and find change in momentum. Remember the velocities are in opposite directions so must have opposite signs.

3. (a) Considering beam as uniform so weight of beam through pivot. By taking moments about the pivot the weight of the beam and the reaction at the beam are not involved.
(b) Uniform means weight acts at midpoint
(c) Now weight (15g) of beam moves away from centre. It doesn't matter which side it is put as a negative solution will mean you put it thw wrong side. Again moments about pivot eliminates reaction at pivot.

4. Resultant of P and Q is R. So diagrams shows P followed by Q being the same as R. If needed the cosine rule is given in formula book under Core2. Make sure solutions are sensible.

5. Best to work perpendicular and parallel to the plane so that one of the unknowns can be eliminated (F when working perpendicular). So P and the weight need to be split into two components.
(a) perpendicular to the plane P is only unknown
(b) parallel to the plane - now P is known F is only unknown - and as R is known, the coefficient of friction is known (remember max friction = coefficient of friction x normal reaction)
(c) best to redraw diagram with P removed. Remember friction will now be up the slope as it is on point of moving down slope. Normal reaction will have changed so recalculated. Coefficient of friction is a function of the particle and slope so doesn't change (use value calculated in (b)). The maximum friction force is greater than the force down the slope so the particle doesn't move.

6. (a) speed is magnitude of velocity
(b) remember to find angle and then form into a bearing
(c) to find position after time t is initial position + t x velocity
for the A and B to collide the i and j parts of their position must both match at the same time (they do at t=7)
then substitute t=7 into either position (should give same solution)

7. Remember tension will be the same on both sides. Larger particle will move down the slope and smaller particle up.
(a) considering just particle A (smooth slope)
only force down is component of weight
only force up is tension
using Newtons second law
resultant force down slope = mass x acceleration
(b) now particle B (rough slope)
perpendicular to find expression for normal reaction
particle is moving up
only force up is tension (known from (a))
forces down are component of weight and friction (reaction known so coefficient of friction is only unknown)
resultant force up slope = mass x acceleration
(c) forces on pulley are vertical components of tension