STEP Support Programme Mechanics STEP Questions This is a selection of mainly STEP I questions with a couple of STEP II questions at the end. STEP I and STEP II papers follow the same specification, the only difference is that STEP II questions are intended to be more difficult. 2012 S1 Q11 1 Preparation If tan θ = 5 12 and 0 θ < π 2, find cos θ and sin θ. A block of mass 20kg is lying on a slope which makes an angle of 30 to the horizontal. The coefficient of friction between the block and the slope is 0.7. Take g = 10. Find the minimum value needed to: (a) pull the block up the slope; (b) push the block down the slope. Leave your answers in surd form. Draw a good big diagram, putting in all the forces! Remember that friction counteracts the tendency to move. (iii) Two weights, with masses m 1 and m 2 where m 1 > m 2, are connected by a light inextensible string over a smooth pulley. Find the acceleration of the weights in terms of m 1, m 2 and g. Mechanics STEP Questions 1
2 The Mechanics question The diagram shows two particles, A of mass 5m and B of mass 3m, connected by a light inextensible string which passes over two smooth, light, fixed pulleys, Q and R, and under a smooth pulley P which has mass M and is free to move vertically. Particles A and B lie on fixed rough planes inclined to the horizontal at angles of arctan 7 24 and arctan 4 3 respectively. The segments AQ and RB of the string are parallel to their respective planes, and segments QP and P R are vertical. The coefficient of friction between each particle and its plane is µ. Given that the system is in equilibrium, with both A and B on the point of moving up their planes, determine the value of µ and show that M = 6m. In the case when M = 9m, determine the initial accelerations of A, B and P in terms of g. Discussion Note that the second part of the question the three particles can have different accelerations (unlike question 1(iii)) but they are related because the length of the string is fixed. Mechanics STEP Questions 2
2008 S1 Q9 3 Preparation We need some basic ideas to tackle this question: If ( you ) express the position of a particle (of mass m) at time t as the position vector (ẋ ) x, then the velocity (a vector, of course) is obtained by differentiation: v =, y ẏ where the dot denotes differentiation with respect to t. If you know the velocity in the above form, then the speed of the particle is just the length of the velocity vector, which is ẋ 2 + ẏ 2. The kinetic energy is then 1 2 m(ẋ2 +ẏ 2 ). It looks as if you are adding the kinetic energy in the x direction to the kinetic energy in the y direction, but you shouldn t think of it like this: energy is not a vector, so doesn t have components in different directions. The potential energy of a particle of mass m in a uniform gravitational field (the gravitational field of the Earth can be considered to be uniform over relatively short distances) can be written as mgy where y is the height of the particle above some chosen level. Total energy for a system of particles is the sum of the kinetic and potential energy of each particle. If there are no dissipative forces (friction, for example), then the total energy of a system is conserved which means that it is constant. If you differentiate this energy with respect to time, you get zero. A particle moves on the circle x 2 + (y a) 2 = a 2. It starts at the origin. Draw a diagram and write down x and y coordinates of the particle, in terms of a and θ, when the radius from the centre to the particle makes an angle θ with the downward vertical (assume x 0). Obtain the velocity vector of the particle given that θ = ω. Show that the speed of the particle is aω (for ω > 0). A thin circular hoop of radius a rolls without slipping in a straight line on a horizontal plane, with the plane of the hoop vertical. You are given that θ = ω, which is constant. When the hoop has rolled through θ radians, how far has its centre moved? Find the position vector and the velocity vector of the centre. 4 The Mechanics question Two identical particles P and Q, each of mass m, are attached to the ends of a diameter of a light thin circular hoop of radius a. The hoop rolls without slipping along a straight line on a horizontal table with the plane of the hoop vertical. Initially, P is in contact with the table. At time t, the hoop has rotated through an angle θ. Write down the position at time t of P, relative to its starting point, in cartesian coordinates, and determine its speed in terms of a, θ and θ. Show that the total kinetic energy of the two particles is 2ma 2 θ2. Given that the only external forces on the system are gravity and the vertical reaction of the table on the hoop, show that the hoop rolls with constant speed. Mechanics STEP Questions 3
2000 S1 Q10 5 Preparation Generally, and rather surprisingly, momentum is normally conserved in a collision (even in an explosion). But considering the conservation of momentum is not usually enough to determine the motion after the collision. For example, in one dimension following a collision of two particles, there are two unknowns, namely the velocities of the two particles. Conservation of momentum gives just one equation, so we need one more equation. This is supplied by what is known as Newton s experimental law. Newton s Experimental law is: e = v B v A u A u B where u A and u B are the initial velocities of the two particles and v A and v B are the final velocities. 1 Think of this as Speed of Separation e = Speed of Approach. Note that these velocities can both be positive or both negative (if the particles are moving in the same direction) or they can have different signs if they are moving towards each other. 2 Here e is a constant which depends on the particles, called the coefficient of restitution. It describes how the speed of separation of two particles after a collision is related to their speed of approach. Normally, kinetic energy is lost is a collision, being converted to heat or sound. A perfectly elastic collision is one where e = 1, and none of the kinetic energy is dissipated. A perfectly inelastic collision has e = 0 and the particles stick together. All collisions have 0 e 1. Two particles have a head on collision (which means they are travelling in opposite directions). If the first particle has mass 5kg and initial speed 3ms 1 and the second particle has mass 2kg and initial speed 5ms 1 and the coefficient of restitution is e = 0.5 find an equation connecting v A and v B. Be careful: the particles are initially travelling in opposite directions. You need to decide which direction is going to be positive. A good diagram of before and after will be helpful. Given that linear momentum is conserved, write down another equation connecting v A and v B. Hence find the final speeds. (iii) Show that the change in kinetic energy resulting from the collision is 1680 49 (You may find the difference of squares (twice) useful. Or you may not.) Joules. 1 We call them velocities, even though we are only considering motion in one dimension (along a line), because they can be going in either direction (to the left or to the right). It is normal to use the term speed as the magnitude of velocity, so it is never negative. 2 The signs can get confusing; it is best to rely on common sense rather than formulae. Mechanics STEP Questions 4
6 The Mechanics question Three particles P 1, P 2 and P 3 of masses m 1, m 2 and m 3 respectively lie at rest in a straight line on a smooth horizontal table. P 1 is projected with speed v towards P 2 and brought to rest by the collision. After P 2 collides with P 3, the latter moves forward with speed v. The coefficients of restitution in the first and second collisions are e and e, respectively. Show that e = m 2 + m 3 m 1 m 1. Show that 2m 1 m 2 + m 3 m 1 for such collisions to be possible. If m 1, m 3 and v are fixed, find, in terms of m 1, m 3 and v, the largest and smallest possible values for the final energy of the system. It is important to think of a good way of naming the velocities of the particles before and after each collision; this of course must be carefully set out in your answer. Don t forget, when you are trying to derive the inequalities, that 0 e 1. Questions 2009 S1 Q11 and 2011 S2 Q9 are similar. Mechanics STEP Questions 5
2010 S1 Q10 7 Preparation This mechanics question only requires fairly simple manipulations (differentiation, scalar product) of vectors, so there is not really any preparation to be done; just get on with it! Of course, to find the derivative of a vector, you just differentiate its components. 8 The Mechanics question A particle P moves so that, at time t, its displacement r from a fixed origin is given by r = ( e t cos t ) i + ( e t sin t ) j. Show that the velocity of the particle always makes an angle of π 4 with the particle s displacement, and that the acceleration of the particle is always perpendicular to its displacement. Sketch the path of the particle for 0 t π. A second particle Q moves on the same path, passing through each point on the path a fixed time T after P does. Show that the distance between P and Q is proportional to e t. Discussion The statement made in the preparation section: Of course, to find the derivative of a vector, you just differentiate its components. is not actually an of course statement. You need to think about, for example. r(t + h) r(t) lim h 0 h which is the definition of the derivative. Then you write it out in terms of components to get the result stated above. You are not expected to do that here. The result may be obvious, but does need justification; in fact, it would not be correct if the vector were written in polar coordinates and axes. Mechanics STEP Questions 6
1993 S1 Q11 9 Preparation Two equal thin rods AB and BC, each of length 2a, are joined at B so that ABC = 90. Let G be the centre of gravity of the combined rods. Find the coordinates of G with respect to an origin at B, where A has coordinates (0, 2a) and C has coordinates (2a, 0). Draw a diagram showing the rods. Let X be the point on AB such that GX is perpendicular to AB. Find the values of tan BGX and tan AGX. (iii) On your diagram, draw the line through B that would be vertical if the rods were hung up by the point B. On the same diagram, draw the line through A that would be vertical if the rods were hung up by the point A Let α be the angle between these two lines. Show that tan α = 2. 10 The Mechanics question A piece of uniform wire is bent into three sides of a square ABCD so that the side AD is missing. Show that if it is first hung up by the point A and then by the point B then the angle between the two directions of BC is tan 1 18. Discussion Note how quickly a mechanics question can become a geometry question! Mechanics STEP Questions 7
2006 S2 Q11 11 Preparation A particle of mass m moves on a smooth horizontal surface. It experiences a constant force of magnitude F in the positive x-direction. Initially, the particle is at the origin and is moving with speed u in a direction that makes an angle of α with the positive x-axis (α = 1 2 π corresponds to the positive y-axis). Write down expressions for ẍ and ÿ and, by integration and using the initial conditions, obtain the position of the particle at time t. (a) If α = π, find the time taken for the particle to return to the origin. (b) If α = 3 4π, find the time taken for the particle to reach the y-axis. (c) If α = 1 2π, find the time taken for the particle to reach the line y = x. Write g sin 2θ + f cos 2θ in the form R sin(2θ + β), and hence find its maximum value. 12 The Mechanics question A projectile of unit mass is fired in a northerly direction from a point on a horizontal plain at speed u and an angle θ above the horizontal. It lands at a point A on the plain. In flight, the projectile experiences two forces: gravity, of magnitude g; and a horizontal force of constant magnitude f due to a wind blowing from North to South. Derive an expression, in terms of u, g, f and θ for the distance OA. Determine the angle α such that, for all θ > α, the wind starts to blow the projectile back towards O before it lands at A. Blow the projectile back towards O means that the projectile s horizontal velocity is towards the origin. An identical projectile, which experiences the same forces, is fired from O in a northerly direction at speed u and angle 45 above the horizontal and lands at a point B on the plain. Given that θ is chosen to maximise OA, show that OB OA = g f g 2 + f 2 f. Describe carefully the motion of the second projectile when f = g. Mechanics STEP Questions 8
2008 S2 Q11 13 Preparation If tan θ = 3 4 and θ is acute, find cos θ and sin θ. Here are some thoughts about solving this sort of mechanics problem. The first thing to do is to draw a BIG diagram. Draw the wedge with θ much less than 45 so that you can easily see which angles in the diagram are equal to θ and which are equal to 90 θ. Put all the forces in your diagram. You may want to draw separate diagrams for the particle and the wedge, so you are not confused about which force acts on what. Remember Newton s third law. For example, if the particle experiences a frictional force, then the wedge experiences an equal and opposite frictional force. Remember that frictional forces oppose the motion. If you are not sure what direction a frictional force acts in, try to work out which direction the particle would move in if there were no friction. For static friction, the equation F = µr holds only in limiting equilibrium, when the system is just about to move; but for kinematic friction, when the system is already in motion, F = µr is assumed to hold. Write down the equations of motion (Newton s second law) for the various objects in the system. But before doing so, have a good look at the result you are aiming for, because it might help you to decide which directions to resolve the forces. For example, there is no g in the first displayed equation in the Mechanics question below: if you weren t sure whether to resolve forces horizontally and vertically, or parallel and perpendicular to the wedge, the absence of g in the equation is a pretty big clue. Mechanics STEP Questions 9
14 The Mechanics question A wedge of mass km has the shape (in cross-section) of a right-angled triangle. It stands on a smooth horizontal surface with one face vertical. The inclined face makes an angle θ with the horizontal surface. A particle P, of mass m, is placed on the inclined face and released from rest. The horizontal face of the wedge is smooth, but the inclined face is rough and the coefficient of friction between P and this face is µ. When P is released, it slides down the inclined plane at an acceleration a relative to the wedge. Show that the acceleration of the wedge is a cos θ k + 1. To a stationary observer, P appears to descend along a straight line inclined at an angle 45 to the horizontal. Show that tan θ = k k + 1. In the case k = 3, find an expression for a in terms of g and µ. What happens when P is released if tan θ µ? Discussion If you obtained the first displayed equation by resolving forces horizontally for the particle and the wedge separately, you were probably pleased that the reactions and the frictional forces cancelled out. There is (of course) a reason for this. Even though there is friction in the system, so energy (KE+PE) is not conserved, the total momentum of the system is zero, because there are no external horizontal forces acting. Writing down the total horizontal momentum gives the displayed equation. The vertical momentum is not conserved because gravity and the reaction of the horizontal surface on the wedge are external forces. Mechanics STEP Questions 10