TOPIC B: MOMENTUM SPRING 2021 1. Newton’s laws of motion 2. Forms of the equation of motion 2.1 Force, impulse and energy 2.2 Derivation from Newton’s second law 2.3 Examples 3. Conservation of momentum 3.1 Conservation of momentum in collisions 3.2 Restitution 3.3 Oblique collisions 4. Friction 4.1 Types of friction 4.2 Coefficient of friction 4.3 Object on a sloping plane 5. Systems of particles and finite objects 5.1 Dynamics of a system 5.2 The centre of mass 5.3 Important properties of the centre of mass 5.4 Finding the centre of mass 5.5 Some useful centres of mass Mechanics Topic B (Momentum) - 1 David Apsley 1. NEWTON’S LAWS OF MOTION For a particle, classical dynamics is governed by Newton’s Laws of Motion: Law 1. Every particle continues in a state of rest or uniform motion in a straight line unless acted on by a resultant external force. Law 2. When acted upon by an unbalanced force F a particle of mass 푚 accelerates at a rate a such that F = 푚a 1. Law 3. The forces of action and reaction between interacting particles are equal in magnitude, opposite in direction and collinear. Notes (i) The laws strictly apply to particles. They may be extended by summation to rigid bodies if there is no rotation (so that each component moves with the same velocity). (ii) Law 1 requires the concept of inertial reference frames (those that aren’t accelerating relative to u each other). For example, an object perceived as travelling in a straight line by a stationary observer will appear to be travelling in a curved path by somebody in a rotating reference In inertial frame In rotating frame frame. (iii) Law 3 is sometimes referred to as “equal and opposite reactions”. The form above is appropriate for a purely mechanical system, where the objects concerned are in direct contact, but without care it can at least appear to be violated in electrodynamics. (For example, the magnetic forces between two non-parallel current-carrying wires, or between moving charged particles, are not opposite in direction.) However, its corollary for particles in contact – which in fact holds in general – is that total momentum is conserved. (In the case cited this is because electromagnetic fields can be regarded as having momentum, and the wires or moving charged particles are not in direct contact but interact via the intermediate electromagnetic field.) I I I I F F F (=BIL) F 1 Newton’s version was strictly F = d(mv)/dt or “force = rate of change of momentum”, since what he described as the “quantity of motion” was “mass velocity”, what we would today call momentum. Mechanics Topic B (Momentum) - 2 David Apsley 2. FORMS OF THE EQUATION OF MOTION 2.1 Force, Impulse and Energy The Second Law is usually what is referred to as the equation of motion. Initially formulated for particles, by summation it can be applied to whole systems: for example, in continuum mechanics (e.g. fluid mechanics) and in cases of non-constant mass (e.g. rockets). The Second Law can be written in a number of forms, notably: Force: force = mass acceleration, or force = rate of change of momentum Impulse (force time): impulse = change of momentum Energy (force distance): work done = change of kinetic energy Other formulations include Lagrangian and Hamiltonian mechanics. These can be generalised to other physical laws but are too mathematically complex to consider here. 2.2 Derivation From Newton’s Second Law Newton’s Second Law, as we usually write it today, is, for individual particles: F = 푚a force = mass acceleration (1) This gives the instantaneous relationship between the net force acting on a body and the resulting acceleration. If integrated it gives velocity and displacement as a function of time. If 푚 is constant and dv/d푡 = a then we have d(푚v) F = force = rate of change of momentum (2) d푡 This is the form we often use in continuum mechanics; e.g. fluid mechanics. Impulse Integrating (2) w.r.t. time: end ∫ F d푡 =[푚v]start Mechanics Topic B (Momentum) - 3 David Apsley ∫ F d푡 = (푚v)end − (푚v)start impulse = change of momentum (3) This gives the total change in velocity over a period of time. Energy Integrate (2) w.r.t. displacement: dv ∫ F • d퐱 = ∫ 푚 • dx d푡 Since dx = vd푡 this can be written dv ∫ F • dx = ∫ 푚 • vd푡 d푡 1 d = ∫ 푚 (v • v)d푡 2 d푡 d 1 = ∫ ( 푚푣2) d푡 d푡 2 1 end ∫ F • dx = [ 푚푣2] 2 start 1 1 ∫ F • dx = ( 푚푣2) − ( 푚푣2) work done = change of kinetic energy (4) 2 end 2 start This gives the total change in kinetic energy (and hence velocity) in a displacement. Advantages of the energy form (to be considered in detail in Topic C) are: ● it is a scalar – not vector – equation; ● it relates start and end states without having to consider intermediate values; ● it gives velocities directly, without having to integrate acceleration; ● many actual forces (e.g. elasticity) depend on position, not time; ● the work done by important conservative forces (e.g. gravity, elasticity) can be conveniently written in terms of changes in potential energy. All forms of the equation of motion are equivalent and can be derived from each other. They can also, by summation, be extended to whole systems, rather than individual particles (see Section 5 below). Mechanics Topic B (Momentum) - 4 David Apsley 2.3 Examples The following examples apply “F = 푚a” to multi-body systems. We shall adopt several of the following strategies. • Where a system has distinct parts we can explode it: i.e. draw elements a small distance apart, even if they are actually adjacent or connected. • Mechanical systems do not have eyes, so as far as any one component is concerned it only knows about the other parts because of the forces they exert on it. Thus we can formally replace any physical connections by the reaction forces they transmit. • If the system is not rigid then different parts may move at different speeds. For example, in a pulley system one part may move at twice (or any other multiple) the speed of another. The relative speed of parts must be determined from the geometry. • For each part with distinct acceleration one should write “F = 푚a” separately. This allows us to find both internal reactions and accelerations. Example 1. Two boxes with masses 25 kg and 35 kg are placed on a smooth frictionless surface. The boxes are in contact and a force of 30 N pushes horizontally on the smaller box. What is the force that the smaller box exerts on the larger box? 30 N 35 kg 25 kg Example 2. Masses are connected by light, inelastic cables passing round light, smooth pulleys in different configurations, as shown below. Find the direction and magnitude of the acceleration of each mass, and the tension in each cable, for each configuration. (Give your answers in kg-m-s units, as multiples of .) 30 20 30 20 kg kg kg kg (a) (b) Mechanics Topic B (Momentum) - 5 David Apsley 3. CONSERVATION OF MOMENTUM 3.1 Conservation of Momentum in Collisions2 When two bodies A and B collide: the force exerted by A on B is equal and opposite to the force exerted by B on A; the change in momentum of B is equal and opposite to the change in momentum of A; the total momentum is conserved. vB B uB vA uA A Before After For the system shown, with velocities u퐴 and u퐵 before, and velocities v퐴 and v퐵 after, collision conservation of momentum gives: ⏟푚 u + 푚 u = ⏟푚 v + 푚 v 퐴 퐴 퐵 퐵 퐴 퐴 퐵 퐵 (5) total momentum before total momentum after Example 3. A vehicle A of mass 500 kg travelling at 30 m s–1 collides with a vehicle B of mass 900 kg travelling at 21 m s–1 along the same line. After the collision the vehicles remain stuck together for a short distance. What is the speed and direction of the combined assembly: (a) if the vehicles are originally travelling in the same direction? (b) if the vehicles are originally travelling in opposite directions? Momentum is a vector quantity. If there are different directions involved then conservation of momentum must be applied in each direction. If only the velocities before collision are known, this is insufficient to determine the post- collision state. Further conditions are needed: restitution and, in more than one dimension, unchanged individual momentum components in directions perpendicular to the line of approach. 2 Strictly, “interactions” rather than “collisions” – there is no need for the objects actually to touch. An example is the interaction between charged particles. Mechanics Topic B (Momentum) - 6 David Apsley 3.2 Restitution When objects collide they are slightly compressed. Kinetic energy is A B transformed to elastic potential energy, rather like a (very stiff) spring. In the decompression phase this elastic energy is transformed back to kinetic energy as the objects move apart. Because of internal friction some mechanical energy may be dissipated and the amount of kinetic energy after the collision is less than that before. This is usually modelled using a coefficient of restitution e such that: relative velocity of separation 푒 = (6) relative velocity of approach In all circumstances where no external forces act: 0 ≤ 푒 ≤ 1. For a perfectly elastic collision (no energy loss): 푒 = 1 For a perfectly inelastic collision (i.e. objects stick together): 푒 = 0 uA uB Before A B After vA vB For the interacting particles shown above, which are confined to travel along a straight line, momentum conservation gives 푚퐴푢퐴 + 푚퐵푢퐵 = 푚퐴푣퐴 + 푚퐵푣퐵 (total momentum before = total momentum after) whilst restitution gives 푣퐵 − 푣퐴 = 푒(푢퐴 − 푢퐵) (velocity of separation = e velocity of approach) If two of the velocities are known, then these give two equations for the remaining two unknowns.