JC A Level H2 Physics Dynamics Notes
- 1. Physics SSP 2012 - DYNAMICS 1 A Parachutist is moving downwards at constant velocity. Derive an expression of drag force in terms of g. 1. Show proper free-body diagram. 2. Key Points: constant velocity hence acceleration is zero, which means net force on parachutist is zero. 3. F = ma mg – Fdrag = m(0) Fdrag = mg B A car is moving at constant speed. Draw the free-body diagram of the car and include all the possible forces acting on the car. 1. Free body diagrams to include: - weight of car - force of ground on car (acting on tyres) - Driving force acting on front wheels (or both wheels) - Frictional force acting on wheels - Drag Force 1.2. Newton’s Second Law 1.2.1. Common problems involving Fnet = ma (15 mins) A (acceleration of object along a slope) A box of mass m travels up a slope with acceleration a. A constant external force F acting on the box pushes it up the slope. A constant frictional force f acts on the box’s bottom surface. The slope is angled at o . Derive an expression of the external force F in terms of m, f , a and . F – (mgsin + f) = ma F = (mgsin + f) + ma Additional Point: Can discuss power provided by engine. i.e. Power = Fv = (mgsin + f)v + mav
- 2. Physics SSP 2012 - DYNAMICS 2 B (two accelerating objects in contact) Two boxes A and B of equal mass m resting on the floor are in contact. External force F acts on A and as a result A and B undergoes acceleration a. Derive an expression for the magnitude of force F of B on A. Body A and B: F = (2m)a a = F/2m Body B only: F A on B = (m)a = m(F/2m) = F/2 By Newton’s 3rd law, Magnitude of F B on A = F/2 C (atwood pulley) A rope goes over a fixed pulley and two masses A and B are secured to the two free ends of the rope. Mass A has an acceleration a upwards. Mass A is m and Mass B is M. Derive an expression for the tension in the rope. Working: 1.Mass A: T – mg = ma Mass B: Mg – T = Ma 2.Solve for T D (numerical example) A car of mass 1000 kg travelling along a straight horizontal road has an acceleration of 1.8 m s–2 when a driving force of 2.6 kN acts in the forward direction. What is the resistive force acting backwards on the car? A 0.8 kN B 1.8 kN C 2.6 kN D 4.4 kN F = ma 2600 - Fr = (1000)(1.8) Fr = 0.8 kN E The diagram shows a parachutist of mass 82 kg falling towards the Earth. In each case, determine the net force and the acceleration of the parachutist. a b c
- 3. Physics SSP 2012 - DYNAMICS 3 For (a) Net force = 500 N downwards F = ma 500 = 82a a = 6.1 m s-2 (downwards) parachutist is speeding up For (b) Net force = 0 a = 0 m s-2 Constant velocity (unless parachutist changes his body shape) For (c) Net force = 700 N upwards 700 = 82a a = 8.5 m s-2 (upwards) parachutist is slowing down F The diagram shows an 80 kg person in a lift. The normal contact force acting on the person from the base of the lift is R. Determine the magnitude of R when the lift: (a) is travelling upwards at a constant velocity of 2.0 m s−1 (b) is accelerated upwards at 2.3 m s−2 . (a) R – mg = m(0) R = mg (b) R - mg = m(2.3) R = m(2.3) + mg = 968.8 N Man feels heavier 1.2.2. Common problems involving Fave = mv / t (10 mins) A Tennis ball of mass m travelling horizontally at constant speed v hits a rigid wall and rebounds horizontally with the same speed. The duration of impact is t. Derive an expression for the average force Fave of wall on the ball. Fave = -2mv/t Negative means that the direction of Fave is away from the wall B An object with mass m drops on the floor and rebounds with the same speed. Derive an expression for the average force Fave of floor on the object in terms of the object’s weight and rate of change of momentum of object. Fave - W = 2mv/t Fave = 2mv/t + W
- 4. Physics SSP 2012 - DYNAMICS 4 1.3. Newton’s Third Law (5 mins) A Describe three examples/situations of Newton’s third law. Draw out the free-body diagrams for each example/situation. Emphasise that force is of same nature Example: Gravitational Force of Earth on Man and Man on Earth Example: Force of rocket on fuel/air mixture, force of fuel/air mixture (thrust) on rocket Example: Force of water on object (upthrust) and force of object on water 1.4. Impulse (10 mins) A A moving box initially travelling at constant velocity hits a wall and comes to a stop. Draw the possible force-time graph for each of the scenarios Hard Wall : F-t diagram is hill-shaped with large maximum force over small duration of time Wall has a piece of thick sponge attached: F-t diagram is hill-shaped with smaller maximum force over larger duration of time Discuss the features of the two graphs. Similar feature: change of momentum of box is same hence area under F-t graphs are the same Derive an expression for the initial velocity of the box in terms of average force of wall on box, Fave and duration of impact, t. Fave = (0-mv)/t v = -(Favet) / m negative sign indicates that Fave is opposite direction to v
- 5. Physics SSP 2012 - DYNAMICS 5 2. Conservation of Momentum 2.1. Fill in the blanks (True or False) and any relevant comments. For the last column, state the equations that can be used for problem-solving. (15 mins) Total momentum of two objects before and after collision are equal Total momentum of two objects before, during and after collision are equal Total KE of two objects before and after collision are equal Total KE of two objects before, during and after collision are equal Equations to be used Elastic Collision Yes. Always. Yes. Always. Yes. Always. No. During collision, Total KE before collision is converted to other forms of energy (eg. EPE) Total Initial Momentum = Total Final Momentum Total Initial KE = Total Final KE Rel speed of approach = Rel speed of separation Inelastic collision Yes. Always. Yes. Always. No. Part of Total KE before collision is lost during the collision No Total Initial Momentum = Total Final Momentum Perfectly inelastic collision Yes. Always. Yes. Always. No. More or all of Total KE before collision is lost during the collision No Total Initial Momentum = Total Final Momentum And common velocities after collision
- 6. Physics SSP 2012 - DYNAMICS 6 2.2. Graphical Problem (10 mins) Scenario: Two masses m and 2m approaches each other with velocities 1 m s-1 and 2 m s-1 respectively. They collide elastically and the impact of duration is 1 second. Sketch the velocity-time graphs for the two masses. After drawing graph, check that at every time interval, total momentum is constant. 2.3. Practice Questions (10 mins) 1 A body moving with speed v collides elastically with another body travelling in the opposite direction with speed 2 v . Which row in the table below correctly gives the relative velocity of approach and the relative velocity of separation of the two bodies? Relative velocity of approach Relative velocity of separation A 2 v 2 v B 2 v 2 3v C 2 3v 2 v D 2 3v 2 3v Ans: D
- 7. Physics SSP 2012 - DYNAMICS 7 2 A body of mass m travels with a velocity 3v and collides with another particle of mass 2m which is initially stationary. After the collision, the two particles move with the same velocity. Which row in the table gives the final velocity of the two particles and the loss in kinetic energy during the collision? Final velocity Loss in kinetic energy A v mv2 B v 3mv2 C 1.5 v mv2 D 1.5 v 3mv2 Ans: B 3 In an inelastic collision, which quantities are conserved? A total kinetic energy and total momentum but not total energy B total kinetic energy and total energy but not total momentum C total momentum and total energy but not total kinetic energy D total kinetic energy, total momentum and total energy Ans: C 4 What is a correct statement of the principle of conservation of momentum? A in an inelastic collision the total kinetic energy and momentum are constant B in any collision the total momentum of an isolated system is constant C in any isolated system the force on a body equals the rate of change of momentum D momentum is constant when mass and velocity are constant Ans: B