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