M2
hints - Jan 2008 |
||

If you spot any mistakes, please let me know, so that I can correct this page. |
||

Question 1 First, read the page on Energy Methods. Part (b) may be tackled by equating the kinetic energy with which the ball sets off to the gravitational potential energy it has at its maximum height. Part (c) is approached by equating the initial kinetic energy with the sum of the reduced kinetic energy and the acquired gravitational potential energy at the 3m position. |
||

Question 2 Read the page on Calculus Methods. The acceleration is the derivative of velocity, so differentiate the given expression with respect to t. The velocity is the derivative of the position, so the position is the integral of the velocity. Integrate the given expression, not forgetting the arbitrary constant. Then plug in position = 0 and t=0, which will give you an equation from which you can calculate the value of the arbitrary constant. |
||

Question 3 Start by drawing a fresh diagram showing just the ladder. Now you have to add som force arrows. Start with the weight of the ladder at its centre, and a second down arrow at C (which is at some random point to the left of the centre of the ladder), labelling it with a suitable letter to represent the weight of the man. Also, since we know that th wall is smooth, we know that the only force acting on the top of the ladder is a force a right angles to the wall (even though the wall doesn't appear on this diagram!). So mark a horizontal force arrow at the top of the ladder, and christen it suitably. The ladder is in equilibrium so there must also be an upward force and a force acting to the left, in order to balance the forces already drawn. The only possible place for them to act is at A. So draw them in and choose appropriate letters - probably F (standing for Friction) for the horizontal one? Resolve vertically to find the vertical contact force at A. The maximum frictional force that there can be is given by F = μ × the vertical contact force, where μ is the coefficient of friction (given in the question). We know that we are dealing with the maximum available frictional force becasue we are told that the ladder is on the point of slipping. As there are only two horizontal forces, youo automatically know what the force on the |
||

Question 4 Read the page on Calculus Methods. Differentiating all terms with respect to t (which results, of course, in some terms disappearing), you will obtain a velocity vector with new i and j components. For part (b), you need first to differentiate the velocity vector that you obtained in (a), in order to find the acceleration. Then you multiply this by the mass to find the force F (as a function of t). For part (c) you must read the question to discover which of i and j represent north, and then require that the coefficient of the other one be zero. |
||

Question 5 Begin, as usual, by drawing a force diagram for the particle. There are three force: the weight down, a slanting force up to the right along PA and a horizontal force to the right. Because the particle is moving in a circle, it must be accelerating towards B with an acceleration of v ^{2}/r, so that can be calculated quite easily.For part (b), reflect on the fact that there is no vertical acceleration, meaning that you can equate the weight with the vertical component of the tension in AP. (You'll probably want to do some trig to establish the angles in ABP first) For part (c) you use the fact that the combined effect of the horizontal component of the tension in PA together with the tension in PB are what provides the centripetal mass-acceleration mv ^{2}/r. As mv^{2}/r and the horizontal component of the tension in PA are easy to calculate, the result should follow without too much trouble. |
||

Question 6 First, read the page on Energy Methods. Start by extending the string down the slope with a dotted line until it reaches the bottom, and pretend that this is 5.5m from A. The principle to use here is that the total energy stays constant. At the start you have only elastic energy. This means that you know the constant value of the energy. At the point dealt with in (b), you have kinetic energy (unknown), no elastic energy (because it's gone slack) and GPE. To calculate the GPE, you need the height through which the particle has risen. You get at this by knowing that the elastic is at that moment 4m long (its unstretched length), so that the particle must have slid 1.5m up the slope. Simple trig then gives you the height risen. For (c), you have to consider whether the particle has enough KE at (b) to balance the GPE it will need in order to slide the remaining 4m up the slope (another height calculation needed). |
||

Question 7 First, read the page on Energy Methods. Part (a) involves noticing that the KE at B plus the GPE acquired in falling a distance 2a to A is equal to the KE at A. To tackle (b), draw a force diagram for the particle at A. There will be two forces: a weight (known) down and a tension (unknown, so call it T) up. Write down the resultant of these forces, and require it to be equal to the mv ^{2}/r needed to maintain circular motion (using the speed calculated in (a)) |
||

Question 8 OK: this is a bit tougher. Sorry about that. First read the page on Calculus Methods. The first thing is to write an expression for the driving force F from P = Fv, where P is the power of the engine and v the speed of the car. Now draw a force diagram, using a blob to represent the car. There will be two horizontal forces: the driving force F in one direction and the resistive force (given in the question) in the other. So you can now write an expression for the resultant force. Set this equal to the mass-acceleration, but instead of writing this as ma, as you normally would, write the a as the derivative of v. You should then find that the result drops out. For part (b), you need to realise that when the engine is turned off, the power drops to zero, so the term in the equation that had to do with the driving force disappears, giving you the result required. Now cross multiply to bring the v terms to the left and the dt to the right, stick am integral sign in front of each side, and you will end up with an integration that you can do using ordinary polynomial methods. |
||