Link to worked solutions for Jan 2010 M1 paper:

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpNGI1MTc1ZmUtNWQ2OC00NDlkLTkyMTItZDQ4ZTc4MzQ2MjNl&hl=enSupporting notes for worked solutions:

1. The diagram shows the system before and after the collision.

(a) considering just particle A calculate the change in momentum by subtracting the intial momentum from the final momentum - the impulse is the change in momentum

(b) considering the whole system and using the principle of conservation of momentum, the total initial momentum = the total final momentum - and solve for m

2. (a) on a speed-time graph, constant acceleration is a straight line sloping up, constant speed is a horizontal line and constant deceleration is a straight line sloping down

(b) simplest method is to consider the distance as the area under the graph and use area of trapezium to set up an equation with T as the only unknown

- alternatively the journey can be split into three parts and suvat used on the final part of the journey with T as the unknown

- finally solve for T

3. (a) considering horiztonal the tension in B is the only unknown

(b) so then use vertical to find m

4. (a) moments about A eliminates the tension in A to find an expression for tension in C

(b) use up = down and the answer to (a) to find expression for tension in A

(c) answer to (a) equals 8 x answer to (b) and solve for W

5. (a) the given information about the particle's motion enables a suvat equation to be set up to find the accelearation

(b) use F=ma to find the resultant force down the slope

** the remainder of this question is missing and will be updated***

6. (a) use F=ma on only particle A to find the tension

(b) use F=ma on only particle B using tension from (a) to set up an equation from which the value of k can be found

(c) the pulley being smooth means there is no force (eg friction) resisting motion

(d) suvat on particle A falling can find the distance it falls beforce reaching the plane, particle B rises the same amount so ends up double that distance above the plane

7. (a) the displacement of S is the position after 4s minus the initial position (when doing this make sure the initial position is in brackets), then the velocity is the displacement/time and finally the speed is the magnitude of the velocity (use pythagoras on the i and j part of the velocity)

(b) the diagram shows the velocity vector - the angle needed for the bearing is the same as the angle indicated inside the triangle

(c) the position at any given time is the initial position + (time x velocity)

(d) displacement of S from L is (position of S - position of L) so then distance is magnitude of this vector (using pythagoras) subst t=T and distance = 10 and then solve for T