Welling Maths Year 13

Monday, 10 January 2011

Mechanics1 Numerical solutions - all papers

Friday, 7 January 2011

"EVERYTHING YOU NEED TO KNOW FOR MECHANICS 1"

http://dl.dropbox.com/u/15976318/Everything_You_Need_to_Know_for_Mechanics_1.doc

Something I made for a past A-Level revision seesion I ran at another school.

Make sure you use it as a guide, not a cheat sheet. That is, you should be attempting exams papers independently only refering to it when you are stuck. As you do more exam papers, you should not have to refer to the sheet as much.

Thursday, 6 January 2011

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

Wednesday, 5 January 2011

Mechanics1 Worked solutions Jan 2010 paper

Link to worked solutions for Jan 2010 M1 paper:

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpNGI1MTc1ZmUtNWQ2OC00NDlkLTkyMTItZDQ4ZTc4MzQ2MjNl&hl=en

Supporting 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

Mechanics1 Worked solutions May 2009 paper

Link to worked solutions for May 2009 M1 paper:

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpNWI3NGU2ZjQtZTk4OC00ZDVlLWIyZWUtZDkyYzFhM2Y4Y2Yy&hl=en

Supporting notes for worked solutions:

1. For journeys P to Q and P to R, distance and time are known so for each journey an equation in terms and u & a can be set up - then solve the equations simultaneously

2. (a) the vector i + 2j is parallel to F2 so can be used to find the angle - it is the angle with the vertical that is needed
(b) To find the resultant of F1 & F2 just add them - then for R to be parallel to i the j part is zero - then solve for p

3. Because the impulse is given work with each particle separately - define right as positive - particle A's initial momentum is positive and the impulse acts left so negative - particle B's initial momentum is negative and the impulse acts right so positive - then as momentum = mass x velocity, velocity = momentum/mass

4. Aside about the angle - given info can be used with a 3,4,5 triangle to find the sin and cos (could work out the angle with inverse tan but this will be a long decimal)
- perpendicular to the plane to find normal reaction which is then used with coefficient of friction to find the friction force - then work out the resultant force down the slope and use F=ma to find the acceleration

5. both vertical and horizontal can set up an expression for R in terms of P which are then equated to find P

6. (a) consider the whole system (eliminating the tension forces) using F=ma to find the acceleration
(b) consider just the car (could equally have considered just the trailer) using F=ma to find the tension in the towbar
(c) thrust forces in the towbar act towards the car and trailer (opposite to the tension in (a) and (b)) - considering the trailer only using F=ma enable the acceleration (actually deceleration) to be found which the braking force to be calculated using F=ma on the car only

7. (a) moments about Q enables the unknown tension at Q to eliminated to the the tension in P can be found in terms of x
(b) considering up=down enables the tension in Q to be found in terms of x
(c) the question states x is between 0 and 1.4 so the minimum tension in Q & maximum tension in P will be x=0 (ie at P) and the maximum tension in Q & minimum tension in P will be x=1.4 (ie at Q)
(d) use the information given to state that Q=3P and substitue the answers to (a) and (b) and solve for x

8. (a) velocity of H is given so speed is jsut pythagoras on the i and j parts of the velocity (ie finding the magnitude of the velocity)
(b) the position vector for H is the inital position (100j) plus time x velocity
(c) the position vector for K is the inital position plus time x velocity then the displacement of K from H is the position vector for K subtract the position vector for H
(d) for H & K to meet both i and j parts of the displacement found in (c) must be zero at the same value of t - then substitute the found value of t into one of the position vectors

Thursday, 30 December 2010

Mechanics1 Worked solutions June 2006 paper

Link to worked solutions for June 2006 M1 paper:

https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0B2J1T-insOmpZTQyZGU3NDUtN2QzNy00MDE1LTg1YzAtODM1ZWRmYmVjNzQx&hl=en

Notes to support worked solutions:

1. (a) the gradient of a speed-time graph is acceleration so a straight line (so constant gradient) will represent constant acceleration
(b) so a horizontal line has gradient = 0 so acceleration = 0 so constant speed
(c) area under a speed time graph represents distance travelled

2. (a) considering the whole system the inital momentum will be equal to the final momentum - remember momentum = mass x velocity - make sure directions of velocities are used correctly - in this case right is defined as positive - the only unknown v is found - the fact that it is found to b positive shows that the direction is to the right so the direction of the particle is unchanged
(b) considering just a single particle the change in momentum (final momentum - initial momentum) is equal to the impulse acting on it

3.(a) for the journey A to B, s&u&t are known so a can be found
(b) for the whole journey A to C, s&u are known and a was found in (a) so v can be found
(c) again using whole journey A to C the only remaining unknown t can be found - don't forget to subtract 2 seconds to give the required answer of the time from B to C

4. aside about the angle of elevation - tan of the angle is given as 3/4 which could be used to find the angle although this would be a long decimal - it is simpler for the question to find that sin of the angle = 3/5 and cos of the angle = 4/5 gby using a 3,4,5 triangle
(a) perpendicular to the plane finds the normal reaction
parallel to the plane finds the frictional force
these are combined to find the coefficient of friction
(b) frictional force is now up the slope althought its size will not have changed from (a) because the normal reaction won't have changed because no forces perpendicular to the slope have changed
- find the resultant force down the slope (ie parallel to the slope)
- use F=ma to find the acceleration down the slope

5. (a) the information given about the tensions and just using up=down lets us find the size of the tensions
(b) the diagram was set up with x being half the length AB - by taking moments about this length can be found
(c) taking moments about B eliminates the unknown W so the tension can be found then up = down finds W

6. (a) considering whole system (tensions cancel each other out) and using resultant force = mass x acceleration the acceleration is found
(b) by considering just the car the tension can be found (could have equally considered just the trailer
(c) as diagram shows the only forces acting on car are the engine and the resistance - again use F = ma to find the acceleration - followed by suvat to find the distance
(d)

7. (a) speed is the magnitude of the velocity vector so just use pythagoras on the i and j values
(b) make sure the correct angle is identified for the bearing (clockwise from north)
(c) the position after 3 hours is the start position plus 3 lots of the velocity
(d) position after 2 hours is the start position plus 2 lots of the velocity
- the new velocity just adds 5 to the j every hour
(e) to be due east of a point means having the same vertical distance (ie the j component as that point) - be careful translating 1.2 hours into 1 hour and 12 minutes (12 mins is 2/10 of an hour)
(f) 2 hours after the new velocity just adds 10 the the j
the displacement vector from R is found by subtracting R
then the distance is the magnitude of the displacement vector so just use pythagoras on the i and j components

Tuesday, 21 December 2010

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