* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download VU3Motion2009
Brownian motion wikipedia , lookup
Hooke's law wikipedia , lookup
Center of mass wikipedia , lookup
Coriolis force wikipedia , lookup
Specific impulse wikipedia , lookup
Jerk (physics) wikipedia , lookup
Modified Newtonian dynamics wikipedia , lookup
Work (thermodynamics) wikipedia , lookup
Classical mechanics wikipedia , lookup
Newton's theorem of revolving orbits wikipedia , lookup
Fictitious force wikipedia , lookup
Centrifugal force wikipedia , lookup
Hunting oscillation wikipedia , lookup
Rigid body dynamics wikipedia , lookup
Relativistic mechanics wikipedia , lookup
Equations of motion wikipedia , lookup
Seismometer wikipedia , lookup
Classical central-force problem wikipedia , lookup
VCE PHYSICS Unit 3 Topic 1 Motion in 1 & 2 Dimensions Unit Outline To achieve this outcome the student should demonstrate the knowledge and skills to: • apply Newton’s laws of motion to situations involving two or more forces acting along a straight line and in two dimensions; • analyse the uniform circular motion of an object moving in a horizontal plane (FNET = mv2/R) such as a vehicle moving around a circular road; a vehicle moving around a banked track; an object on the end of a string. • Apply Newton’s 2nd Law to circular motion in a vertical plane; consider forces at the highest and lowest positions only; • investigate and analyse the motion of projectiles near the Earth’s surface including a qualitative description of the effects of air resistance; apply laws of energy and momentum conservation in isolated systems; • analyse impulse (momentum transfer) in an isolated system, for collisions between objects moving along a straight line (FΔt = mΔt); • apply the concept of work done by a constant force work done = constant force x distance moved in the direction of the force work done = area under force distance graph • analyse relative velocity of objects along a straight line and in two dimensions; • analyse transformations of energy between: kinetic energy; strain potential energy; gravitational potential energy; and energy dissipated to the environment considered as a combination of heat, sound and deformation of material; kinetic energy i.e. ½ mv2; elastic and inelastic collisions in terms of conservation of kinetic energy strain potential energy i.e. area under force-distance graph including ideal springs obeying Hooke’s Law ½ kx2 gravitational potential energy i.e. mgΔh or from area under force distance graph and area under field distance graph multiplied by mass • apply gravitational field and gravitational force concepts g = GM/r2 and F = GM1M2/r2 • apply the concepts of weight (W = mg), apparent weight (reaction force, N) , weightlessness (W = 0) and apparent weightlessness (N = 0) • model satellite motion (artificial, moon, planet) as uniform circular orbital motion (a = v2/r = 4π2r/T2) • identify and apply safe and responsible practices when working with moving objects and equipment in investigations of motion. Chapter 1 Topics covered: • The S.I. System. • Position. • Scalars &Vectors. • Vector Addition & Components. 1.0 The S. I. System The system of units used in Physics is the “Systeme Internationale d’Units” or more simply the S. I. System. The system has two important characteristics; Different units for the same physical quantity are related by factors of 10.(eg. mm; cm; km) Length: Unit Metre (m) Mass: Unit Kilogram (kg) Velocity: Force: Unit N = ms kg -1ms-2 Time: Unit Second (s) The system is based on 7 Fundamental Units, each of which is strictly defined. S.I. System: 7 Fundamental Units Electric Current: Unit Ampere (A) Velocity Force - Derived Unit All other units, so called DERIVED UNITS, are simply combinations of 2 or more of the Fundamental Units. Temperature: Unit Kelvin (K) Luminous Intensity: Unit Candela (cd) Amount of Substance: Unit Mole (M) 1.1 Position To specify the POSITION of an object, a point of ORIGIN needs to be defined. It is from this point all measurements can be taken. -30 -25 -20 -15 For example on the number line below the point labelled 0 is the origin and all measurements are related to that point. The Number Line 0 -10 -5 5 10 - 15 Units Thus a number called -15 is 15 units to the left of 0 on the number line. 15 20 25 30 + 30 Units A number called +30 is 30 units to the right of 0. 1.2 Scalars & Vectors • Before proceeding, it is important to define two general classes of quantities. • 1. SCALAR QUANTITIES: • These are COMPLETELY specified by: • A MAGNITUDE (ie a NUMBER) • and A UNIT • Examples of Scalar Quantities would be: Temperature (17oC), Age (16 years), Mass (2.5 kg), Distance (150 m). • 2. VECTOR QUANTITIES: • • • • These are COMPLETELY specified by: A MAGNITUDE (ie. A NUMBER) and A UNIT and A DIRECTION • Examples of Vector Quantities would be: Displacement (2.7 km, West), Force (15 N, Downward), Acceleration (1.5 ms-2, S.E.) VECTORS ARE GENERALLY REPRESENTED BY ARROWS: The length of the arrow represents the magnitude of the vector. The orientation of the arrow represents the direction of the vector. N 15 A vector of magnitude 15 units directed East 1.3 Vector Addition VECTOR ADDITION Two Forces act at the Centre of Mass of a body. The first of 4N East and the second of 3N South Which way will the body move ? SINGLE VECTOR DIAGRAM N Centre of Mass 450 4 5 55 3 Magnitude = = A Vector of: Magnitude; 5 units Direction; NE or N45E or 45T 42 + 25 32 3 In a direction, and with a force, that is the sum of the 2 vectors Direction: Sin = 3/5 = Sin-1 3/5 = 36.90 THE RESULTANT FORCE HAS A MAGNITUDE OF 5 N DIRECTED AT E 36.90 S 1.4 Vector Subtraction An object moving East at 8.0 ms-1 changes its velocity to 8.0 ms-1 South The velocity change (v) is given by vf - vi vi (- vI ) is a negative vector. It can be vf 8.0 ms-1 E converted to a positive one by reversing its direction. What is the object’s 8.0 ms-1 S Then, by performing a vector addition, the change in velocity ? velocity change v can be obtained. v Magnitude v = = vf 8.0 ms-1 S 82 + 82 128 = 11.3 ms-1 - vi 8.0 ms-1 W Direction: Tan = 8/8 = 1.0 = Tan-1 1.0 = 450 THE CHANGE IN VELOCITY = 11.3 ms-1 AT S 450W 1.5 Vector Components A “Jump Jet” is launched from a 150 ramp at a velocity of 40 ms-1 What are the vertical and horizontal components of its velocity ? 40 ms-1 Vertical Component: V VERTICAL = 40 Sin 150 = 10.4 ms-1 V VERTICAL 150 VHORIZONTAL V VERTICAL and VHORIZONTAL are the COMPONENTS of the plane’s velocity. Horizontal Component: VHORIZONTAL = 40 Cos 150 = 38.6 ms-1 Motion - Revision Questions Question type: Vectors Adam is testing a trampoline. The diagrams show Adam at successive stages of his downward motion. Figure C shows Adam at a time when he is travelling DOWNWARDS and SLOWING DOWN. A B Q1: What is the direction of Adam’s acceleration at the time shown in Figure C ? Explain your answer. C D A: Acc is UPWARD. In order to meet the requirements set travelling downward BUT slowing down, he must be decelerating ie. Accelerating in a direction opposite to his velocity. Thus acc is upward. Chapter 2 Topics covered: • Distance versus Displacement. • Speed versus Velocity. • Acceleration. • Graphical Representations. 2.0 Distance vs Displacement • • • • • Distance is a Scalar Quantity having a magnitude and a unit. The S.I. unit for Distance is the metre (m) Distance is best thought of as: “How far you have travelled in your journey”. Displacement is a Vector Quantity having a magnitude, a unit and a direction. The S.I. unit for Displacement is the metre (m), plus a direction Displacement is best thought of as: “How far from your starting point you are at the end of your journey”. Distance and Displacement may or may not be numerically equal, depending on the nature of the journey. JOURNEY No 1. Distance = Displacement +ve direction 100 m Start At the end of the run: Distance = 100 m. Displacement = +100 m Finish JOURNEY No 2. Distance Displacement 400 m track Start/Finish At the end of the one lap run: Distance = 400 m. Displacement = 0 m 2.1 Speed vsINSTANTANEOUS Velocity vs AVERAGE • • • • • • • • Speed is defined as the Time Rate • of Change of Distance. Speed is a Scalar Quantity. • Mathematically: Speed = Distance/Time The S.I. unit for Speed is metres/sec (ms-1) • Velocity is defined as the Time Rate of Change of Displacement. • Velocity is a Vector Quantity. • Mathematically: • Velocity = Displacement/Time The S.I. unit for Velocity is metres/sec (ms-1), plus a direction IN ALL CALCULATIONS AND EQUATIONS USED IN THE COURSE, ASSUME INSTANTANEOUS VALUES ARE REQUIRED UNLESS OTHERWISE STATED. VELOCITY The term velocity can be misleading unless a specific label is attached. The label indicates whether the velocity is an Average value calculated over a long period of time OR an Instantaneous value calculated at any instant of time. A simple example illustrates: A journey of 40 km across the suburbs takes 1 hour; VAV = 40/1 = 40 kmh-1 BUT VINST could be anything from 0 kmh-1 (stopped at traffic lights) to VINST = 100 kmh-1 (travelling along the freeway). 2.2 Some Common Speeds Event Speed (ms-1) Speed (kmh-1) 1. Grass Growing 5.0 x 10-8 1.8 x 10-7 1-2 4-8 3. Marathon Runner 5 18 4. 100 m Sprinter 10 36 5. Suburban Speed Limit 16.7 60 6. Freeway Speed Limit 30.6 110 7. Boeing 737 Cruising 246 886 2.99 x 108 1.1 x 109 2. Walking Pace 8. Speed of Light 2.3 Acceleration • • • • • • Acceleration is defined as the Time Rate of Change of Velocity. Acceleration is a Vector Quantity. Mathematically: Acceleration = Velocity/Time The S.I. unit for acceleration is metres/sec/sec (ms-2) Since acceleration is a vector quantity, a body travelling with a constant speed but in a constantly changing direction must be accelerating. So a cyclist travelling around a corner at constant speed is, in fact, accelerating ! (More of this later). ACCELERATING VEHICLE v a The velocity and acceleration are in the same direction DECELERATING VEHICLE v a The velocity and acceleration are in opposite directions. • 2.4 Graphical Representations Much of the information delivered in this Physics course is presented graphically. • Generally, graphs “tell a story” and you need to develop the ability to “read” the story the graph is telling. • There are two basic families of graphs you should be familiar with: (a) Sketch Graphs, paint a broad brush, general picture of the relationship between the quantities graphed. (b) Numerical Graphs from which exact relationships may be deduced and/or exact values may be calculated. Velocity Displacement Distance SKETCH GRAPHS Time The Story: The The Story: AsStory: time passes, the distance of the object The Story: As As time time passes, passes, the the velocity displacement of of thethe of object the is from itspasses, starting point is increasing in aobject As time the distance object increasing isfrom increasing in amore uniform quickly manner (the slope (the slope is This is uniform manner (the slope is constant). This its starting point does not change. constant). increasing This at a aan is constant the graph rate). a constantly isthe thegraph graph object moving at constant is stationary object. accelerating object speed. This is the graph a constantly accelerating object Motion - Revision Questions Question type: Sketch Graphs In a road test, a car was uniformally accelerated from rest over a distance of 400 m in 19 sec. The driver then applied the brakes, stopping in 5.1 sec with constant deceleration. The graphs A to F below should be used to answer the questions below. The horizontal axis represents time and the vertical axis could be velocity or distance. A B C D E F Q2: Which of the graphs, A to F, represents the velocity time graph for the entire journey ? A: Graph B Q3: Which of the graphs, A to F, best represents the distance time graph of the car for the entire journey ? A: Graph E 2.5 Exact Graphical Relationships • • • • • You are required to be familiar with graphs of: Distance or Displacement Versus Time; Speed or Velocity Versus Time and Acceleration Versus Time These graph types and the exact information obtainable from them can be summarised in the table given below. Put this table on your cheat sheet. Graph Type Read directly from Graph Slope of Graph gives: Area under Graph gives: Distance or Displacement versus Time Distance or Displacement Speed or Velocity No Useful Information Speed or Velocity versus Time Speed or Velocity Acceleration Distance or Displacement Acceleration versus Time Acceleration No Useful Information Velocity 2.6 The Equations of Motion • • • • • • • • These are a series of equations linking velocity, acceleration, displacement and time. THE EQUATIONS CAN ONLY BE USED IN SITUATIONS WHERE THE ACCELERATION IS CONSTANT. The 3 most important of these equations are: 1. v = u + at 2. v2 = u2 +2ax 3. x = ut + ½at2 where u = Initial Velocity (ms-1) v = Final Velocity (ms-1) a = Acceleration (ms-2) x = Displacement (m) t = Time (s) u= v= a= x= t= HINTS FOR EQUATIONS’ USE When using these equations always list out the information supplied in the question and what you are required to calculate; then choose the appropriate equation to use. Make a list like this. +ve Sometimes it is necessary to choose a positive direction ie. up or down for vertical motion questions or left or right for horizontal motion questions. Questions are often asked which require a 2 step process to get to the answer, ie. A value for acceleration may be needed before the final velocity can be found. Motion - Revision Questions Question type: Equations of Motion In a road test, a car was uniformally accelerated from rest over a distance of 400 m in 19 sec. The driver then applied the brakes, stopping in 5.1 sec with constant deceleration. Q4: Calculate the acceleration of the car for the first 400 m. A: Firstly, list information: u=0 v=? a=? x = 400 m t = 19 s Choose the appropriate equation: x = ut + ½at2 400 = 0 + ½a(19)2 a = 2.22 ms-2 Motion - Revision Questions Question type: Average Speed In a road test, a car was uniformally accelerated from rest over a distance of 400 m in 19 sec. The driver then applied the brakes, stopping in 5.1 sec with constant deceleration. Q5: Calculate the average speed for the entire journey, covering both the accelerating and braking sections. A: Average Speed = Total Distance Need to know u, the initial speed for the braking section which equals the final Total Time For the accelerated part of journey: speed for the accelerating section. For accelerating section: u=0 v = u + at v=? = 0 + (2.2)(19) -2 a = 2.2 ms = 41.8 ms-1 x = 400 m Now can calc s t = 19 s x = ut + ½at2 Need to calc acc To get braking distance, Braking list = (41.8)(5.1) + ½(-8.2)(5.1)2 v = u + at use Eqns of Motion becomes 0 = 41.8 + a(5.1) = 213.2 - 106.6 u = ? List does not u = 41.8 ms-1 = 106.6 m a = - 8.2 ms-2 v = 0 contain v = 0 Still not a = ? enough info a = ? enough Total Distance = 400 + 106.6 = 506.6 m x = ? the calculate s x = ? info Total Time = 19 + 5.1 = 24.1 s t = 5.1s t = 5.1s So, Average Speed = 506.6/24.1 = 21 ms-1 Distance = 400 m Time = 19 sec For the braking part of journey: Distance = needs to be calculated Time = 5.1 sec Chapter 3 Topics covered: • Newton’s Laws. • Force in Two Dimensions. • Momentum and Impulse. • Conservation of Momentum. 3.0 Newtonian Motion • • • Sir Isaac Newton (1642 - 1727) was unique for a number of reasons, but mostly because he developed a set of laws describing the motion of objects in the universe. Prior to Newton, scientists believed that a set of laws existed which explained motion on Earth and these laws had to be modified to describe motions in all other parts of the universe. Newton was the first scientist to realise that all motion anywhere in the universe could be described by a single set of laws which then had to be modified for use in the friction riddled confines of the Earth. Isaac Newton Aged 26 3.1 Newton’s Laws INERTIA – That property possessed by all bodies with mass whereby they tend to resist changes to their motion. It is associated with an object’s mass – more mass, more inertia. Inertia is NOT a force. Newton developed 3 laws covering motion in the universe, they are: LAW 1. THE LAW OF INERTIA. A body will remain at rest, or in a state of uniform motion, unless acted upon by a net external force. LAW 2. The acceleration of a body is directly proportional to the net force applied and inversely proportional to its mass. (a = F/m) LARGE MASS – LARGE INERTIA LAW 3. For every action there is an equal and opposite reaction small mass – small inertia Motion - Revision Questions Question type: Newton’s Laws A seaplane of mass 2200 kg takes off from a smooth lake as shown. It starts from rest, and is driven by a CONSTANT force generated by the propeller. After travelling a distance of 500 m, the seaplane is travelling at a constant speed, and then it lifts off after travelling a further 100 m. Total Opposing Force (N) 10000 8000 6000 4000 2000 0 The total force opposing the motion of the seaplane is not constant. The graph shows the TOTAL FORCE OPPOSING THE MOTION of the seaplane as a function of the distance travelled. Distance (m) 100 200 300 400 500 600 Q6: What is the magnitude of the net force acting on the seaplane after it has travelled a distance of 500 m from the start ? A: At d = 500 m the plane is travelling at CONSTANT VELOCITY, So ΣF = 0 Motion - Revision Questions Question type: Newton’s 2nd Law Q7: What is the magnitude of the seaplane’s acceleration at the 200 m mark ? A: At d = 500 m the seaplane is subject to 0 net force (see previous question). Thus, Driving Force = Opposing Force = 10,000 N (read from graph). At d = 200 m the total opposing force = 2000 N (read from graph) So ΣF = 10,000 - 2000 = 8000 N Now, we know that ΣF = ma So, a = ΣF/m = 8000/2200 = 3.64 ms-2 Total Opposing Force (N) 10000 8000 6000 4000 2000 0 Distance (m) 100 200 300 400 500 600 Motion - Revision Questions Question type: Work Q8: Estimate the work done by the seaplane against the opposing forces in travelling for a distance of 500 m. A: Work = Force x Distance = Area under F vs d graph. Area needs to be calculated by “counting squares”. Each square has area = 2000 x 100 = 2 x 105 J Total number of squares (up to d = 500 m) = 9 whole squares (x) + 6 part squares (p) = 12 whole squares. So Work done = (12) x (2 x 105) = 2.4 x 106 J Total Opposing Force (N) 10000 p 8000 p x x p x x x x x x 6000 4000 p 2000 p 0 p x 100 200 300 400 500 600 Distance (m) Motion - Revision Questions Question type: Newtons 2nd Law In Figure 2, a car of mass 1000 kg is being towed on a level road by a van of mass 2000 kg. There is a constant retarding force, due to air resistance and friction, of 500 N on the van, and 300 N on the car. The vehicles are travelling at a constant speed. Figure 2 Q9: What is the magnitude of the force driving the van? As the vehicles are travelling at constant speed, a = 0 and thus ΣF = 0 Thus driving force = total retarding force Thus driving Force = 800 N Q10: What is the value of the tension, T, in the towbar? Looking at the towed car alone the forces acting are tension in the towbar and the retarding force. Since the car is travelling at constant velocity, ΣF = 0, so tension = retarding force = 300 N Motion - Revision Questions Question type: Newton’s 3rd Law a The figure shows a cyclist with the bicycle wheels in CONTACT with the road surface. The cyclist is about to start accelerating forward. FTR FRT Q11: Explain, with the aid of a clear force diagram, how the rotation of the wheels result in the cyclist accelerating forwards. A: The wheels rotate in the direction shown. The force labelled FTR is the force the tyre exerts on the road. This force is directed in the opposite direction to the acceleration and thus cannot be the force producing that acceleration. The force labelled FRT is the Newton 3 reaction force arising from the action of FTR. It is this force (directed in the same direction as the acceleration) that actually produces the acceleration of the bike and rider. 3.2 Newton’s Laws Restrictions & Consequences RESTRICTIONS CONSEQUENCES The laws only apply at speeds st law is concerned As far as Newton’s 1 much, much less than the “rest” and “uniform motion” are the same speed of light. The laws apply equally in ALL state. You cannot perform any test which can inertial frames of reference. show whether you are stationary or moving FRAME OF REFERENCE ? at constant velocity. A frame of reference is best The action and reaction law requires there to described as “your point of be TWO bodies interacting, the ACTION force observation.” acting on one body and the REACTION acting An inertial frame of reference on the other. is one that is either Deciding when an Action / Reaction situation stationary or moving with exists can be done by answering the constant velocity. question: A non inertial frame of Does the second (reaction) force disappear reference is accelerating. immediately the first (action) force Humans in non inertial disappears ? frames tend to invent If the answer is yes, you have an action “fictitious forces” to explain reaction pair. their experiences Motion - Revision Questions Question type: Relative Motion A train is travelling at a constant velocity on a level track. Lee is standing in the train, facing the front, and throws a ball vertically up in the air, and observes its motion. Q12: Describe the motion of the ball as seen by Lee. Lee sees the ball move straight up and down. Sam, who is standing at a level crossing, sees Lee throw the ball into the air. Q13: Describe and explain the motion of the ball as seen by Sam. From Sam’s point of view, the ball follows a parabolic path made up of the vertical motion imparted by Lee and the horizontal motion due to the train. 3.3 Force in Two Dimensions AN OBJECT UNDER THE ACTION OF 4 FORCES F1 F4 F3 FRES F2 Perform a vector addition In which direction will of the forces. The the object accelerate ? Resultant Force (FRES ) will give the direction F of the acc. 1 F2 F3 FRES F4 The object will be subject to FRES and accelerate in that direction • A Force is either a Push or a Pull. • A Force is either a CONTACT type force or a FIELD type force. • Force is a Vector quantity having both a magnitude, a unit and a direction. • Force is NOT one of the 7 fundamental units of the S.I. System and thus it is a Derived quantity. • The unit for Force is kgms-2. This was assigned the name the NEWTON (N), in honour of Sir Isaac. • Forces can act in any direction and the TOTAL, NET or RESULTANT force is the vector sum of all forces acting on a body. • The body will then ACCELERATE in the direction of the RESULTANT FORCE, according to Newton 2. Motion - Revision Questions Question type: Force in 1 & 2 Dimensions A cyclist is towing a small trailer along a level bike track (Figure 1). The cyclist and bike have a mass of 90 kg, and the trailer has a mass of 40 kg. There are opposing constant forces of 190 N on the rider and bike, and 70 N on the trailer. These opposing forces do not depend on the speed of the bike. The bike and trailer are initially travelling at a constant speed of 6.0 m s-1. Q14. What driving force is being exerted on the road by the rear tyre of the bicycle? A: Constant speed implies ΣF = 0; So driving force = total retarding force Total retarding force = 190 + 70 = 260 N = driving force 3.4 Momentum & Impulse • Newton called Momentum the “quality of motion” and it is a measure of a body’s translational motion - its tendency to continue moving in a particular direction. • Roughly speaking, a body’s momentum indicates which way the body is heading and just how difficult it was to get the body moving with its current velocity. • Momentum is a Vector Quantity. • Mathematically: Momentum (p) = m.v where, m = mass (kg), v = velocity (ms-1) p = momentum (kgms-1) • Impulse is the “transfer mechanism” for momentum. • In order to change the momentum of a body you need to apply a force for a certain length of time to produce the change. • Impulse is a Vector Quantity. • Mathematically: Impulse (I) = F.t where F = Force (N) t = time (s) I = Impulse (Ns) • From Newton 2 (F = ma) and the definition of acceleration, (a = v/t), we get: F = mv/t Ft = mv • Thus: Impulse = Momentum Motion - Revision Questions Question type: Momentum A small truck of mass 3.0 tonne collides with a stationary car of mass 1.0 tonne. They remain locked together as they move off. The speed immediately after the collision was known to be 7.0 ms-1 from the jammed reading on the car speedometer. Robin, one of the police investigating the crash, uses conservation of momentum to estimate the speed of the truck before the collision. Q15 : What value did Robin obtain? A: PBEFORE = PAFTER PBEFORE = (3000)(x) PAFTER = (3000 + 1000)(7.0) So, 3000x = 28,000 x = 9.3 ms-1 The calculated value is questioned by the other investigator, Chris, who believes that conservation of momentum only applies in elastic collisions. A: Momentum is conserved in ALL Q16: Explain why Chris’s comment is wrong. types of collisions whether they be elastic or inelastic. KE is not conserved in this type (inelastic) collision. Motion - Revision Questions Question type: Momentum A car of mass 1000 kg travelling on a smooth road at 5.0 ms–1 collides with a truck that is stationary at a set of traffic lights. After the collision they are stuck together and move off with a speed of 2.0 ms–1 Q17 : How much momentum did the car transfer to the truck? A: Mom is ALWAYS conserved. Mom of car before = (1000)(5) = 5000 kgms-1 Mom of car after = (1000)(2) = 2000 kgms-1 Since mom is conserved Mom loss by car = Mom gain by truck So, Mom transferred to truck = 3000 kgms-1 Q18 : What is the mass of the truck? A: p = mv so, m = p/v = 3000/2 = 1500 kg Motion - Revision Questions Question type: Impulse A car of mass 1000 kg travelling on a smooth road at 5.0 ms–1 collides with a truck that is stationary at a set of traffic lights. After the collision they are stuck together and move off with a speed of 2.0 ms–1 Q19: If the collision took place over a period of 0.3 s, what was the average force exerted by the car on the truck? For the truck Impulse = Change in momentum Ft = mv F(0.3) = 3000 F = 10,000 N Motion - Revision Questions Question type: Momentum A railway truck (X) of mass 10 tonnes, moving at 6.0 ms-1, collides with a stationary railway truck (Y), of mass 5.0 tonnes. After the collision they are joined together and move off as one. 10 tonnes 5 tonnes -1 6.0 ms Before Collision X Y stationary v ms-1 After Collision X Y Q20: Calculate the final speed of the joined railway trucks after collision. A: In ALL collisions Momentum is conserved. So Mom before collision = Mom after collision Mom before = (10 x 103)(6.0) + (5 x 103)(0) Mom after = (15x 103)(v) So v = 60,000/15,000 = 4.0 ms-1 Motion - Revision Questions Question type: Impulse A railway truck (X) of mass 10 tonnes, moving at 6.0 ms-1, collides with a stationary railway truck (Y), of mass 5.0 tonnes. After the collision they are joined together and move off as one at a speed of 4.0 ms-1. 10 tonnes Before Collision 6.0 5 tonnes ms-1 X Y stationary After Collision X 4.0 ms-1 Y Q21: Calculate the magnitude of the total impulse that truck Y exerts on truck X A: Impulse = Change in Momentum Truck X’s change in momentum = Final momentum – Initial Momentum = (10 x 103)(4.0) – (10 x 103)(6.0) = -2.0 x 104 kgms-1 The mechanism for this change in momentum is the impulse supplied by Truck Y So, I = 2.0 x 104 Ns 3.5 Elastic & Inelastic Collisions All collisions (eg, cars with trees, cyclists with the footpath, neutrons with uranium atoms, bowling balls with pins etc.) fall into one of 2 categories: (a) ELASTIC COLLISIONS, where BOTH Momentum AND Kinetic Energy are conserved. (Very few collisions are of this type). If anywhere, these will most likely occur on the atomic or subatomic level. (b) INELASTIC COLLISIONS, where Momentum is conserved BUT Kinetic Energy is NOT conserved. (Most collisions are of this type). The “lost” Kinetic Energy has been converted to other forms of energy eg, heat, sound, light. U n Neutron INELASTIC COLLISION ELASTIC COLLISION Motion - Revision Questions Question type: Elastic/Inelastic Collisions A small truck of mass 3.0 tonne collides with a stationary car of mass 1.0 tonne. They remain locked together as they move off. The speed immediately after the collision was known to be 7.0 ms-1 from the jammed reading on the car speedometer. Robin, one of the police investigating the crash, uses conservation of momentum to estimate the speed of the truck before the collision at 9.3 ms-1 Q22: Use a calculation to show whether the collision was elastic or inelastic. A: Total KE before collision = ½ mv2 = ½ (3000)(9.3)2 = 129735 J Total KE after collision = ½ mv2 = ½ (4000)(7.0)2 = 98,000 J KE is NOT conserved. So collision is INELASTIC Motion - Revision Questions Question type: Inelastic Collisions A railway truck (X) of mass 10 tonnes, moving at 6.0 ms-1, collides with a stationary railway truck (Y), of mass 5.0 tonnes. After the collision they are joined together and move off as one at 4.0 ms-1. 10 tonnes 5 tonnes -1 6.0 ms Before Collision X Y stationary After Collision X 4.0 ms-1 Y Q23: Explain why this collision is an example of an inelastic collision. Calculate specific numerical values to justify your answer. A: In Inelastic collisions Momentum is conserved BUT Kinetic Energy is not. For this collision pBEFORE = (10 x 103)(6.0) + (5 x 103)(0.0) = 6 x 104 kgms-1 pAFTER = (15 x 103)(4.0) = 6 x 104 kgms-1 KEBEFORE = ½mv2 = ½(10 x 103)(6.0)2 = 1.8 x 105 J KEAFTER = ½mv2 = ½(15 x 103)(4.0)2 = 1.2 x 105 J Thus Mom IS conserved but KE is NOT conserved, therefore this is an inelastic collision 3.5 Conservation of Momentum • The Law of Conservation of Momentum states that in an isolated system, (one subject to no outside influences), the total momentum is conserved. • So in a collision (say a car hitting a tree and coming to a stop), the total momentum before the collision = total momentum after the collision. Mass Mass of of car car plus plus passengers == m m passengers v=0 Velocity = v Momentum Momentum 0 pp==mv THE CONSERVATION LAW DOES NOT ALLOW MOMENTUM TO DISAPPEAR. THE APPARENTLY “LOST” MOMENTUM HAS, IF FACT, BEEN TRANSFERRED THROUGH THE TREE TO THE EARTH. Since the Earth has a huge mass (6.0 x 1024 kg) the change in its velocity is negligible. 3.6 The Physics of Crumple Zones & Air Bags A car crashes into a concrete barrier. The change in momentum suffered by the car (and passengers) is a fixed quantity. So, Impulse, (the product of F and t), is also fixed. However, individual values of F and t can vary as long as their product is always the same. So if t is made longer, consequently F must be smaller. Crumple Zones increase the time (t) of the collision. So, F is reduced and the passengers are less likely to be injured. The same logic can also be applied to Air Bags The air bag increases the time it takes for the person to stop. So the force they must absorb is lessened. So they are less likely to be seriously injured. A further benefit is this lesser force is distributed over a larger area Motion - Revision Questions Question type: Momentum/Impulse In a car the driver’s head is moving horizontally at 8.0 ms-1 and collides with an air bag as shown. The time taken for the driver’s head to come to a complete stop is 1.6 x 10-1 s. This collision may be modelled as a simple horizontal collision between the head of mass 7.0 kg and the air bag. Q24: Calculate the magnitude of the average contact force that the air bag exerts on the driver’s head during this collision. A: Impulse = Change in Momentum FΔt = Δ(mv) So F = Δ(mv)/Δt = (7.0)(8.0)/(1.6 x 10-1) = 350 N Motion - Revision Questions Question type: Air Bags/Crumple Zones In a car the driver’s head is moving horizontally at 8.0 ms-1 and collides with an air bag as shown. The time taken for the driver’s head to come to a complete stop is 1.6 x 10-1 s. This collision may be modelled as a simple horizontal collision between the head of mass 7.0 kg and the air bag. Q25: Explain why the driver is less likely to suffer a head injury in a collision with the air bag than if his head collided with the car dashboard, or other hard surface. A: The change in momentum suffered by the driver’s head is a FIXED quantity no matter how his head is brought to rest. • Therefore the product of F and t (ie Impulse) is also a fixed quantity. • However the individual values of F and t may be varied as long as their product always remains the same. • The air bag increases the time over which the collision occurs, therefore reducing the size of the force the head must absorb so reducing the risk of injury. • The air bag also spreads the force over a larger area, reducing injury risk. • Without the air bag the driver’s head may hit a hard surface decreasing the time to stop his head and necessarily increasing the force experienced and thus the likelihood of injury. • In addition the force will be applied over a much smaller area increasing the likelihood of severe injury Chapter 4 Topics covered: • Centre of Mass. • Weight. • Reaction Force. • Bouncing Balls. • Friction. • Various Force Applications 4.0 Centre of Mass In dealing with large objects it is useful to think of all the object’s mass being concentrated at one point, called the Centre of Mass of the object. The C of M of the object is the point around which it will spin if a torque or turning force is applied to the object. Centre of Mass C of M For regularly shaped objects eg. squares or rectangles, cubes or spheres the Centre of Mass of the object is in the geometric centre of the object For oddly shaped objects eg. a boomerang, the C of M may fall outside the perimeter of the object. 4.1 Centre of Mass - Systems For a system of 2 or more bodies, the position of the C of M may be determined from the formula: XCofM = (m1x1 + m2x2 + m3x3 + …) (m1 + m2 + m3 +…) WHAT IS THE CENTRE OF MASS OF THIS SYSTEM? XC of M is the position of the Centre of Mass of the System as measured from a CHOSEN REFERENCE POINT. m1,m2,m3, etc are the masses of the individual components of the System x1, x2, x3 etc are the distances measured from the CHOSEN REFERENCE POINT to the centres of mass of the system’s components. 5.0 m 1.0 m Centre of Mass of each mass 2.875 m A 2.5 m 3.5 m 1.0 m 30 kg mass CHOSEN REFERENCE POINT IS A X C of M = [(50 x 2.5) + (30 x 3.5)] 50 kg beam Centre of Mass of System (50 + 30) = 2.875 m from A 4.2 Weight Object of Mass (m) Centre of Mass • • • Weight = mg Weight Force acts through the Centre of Mass and is directed toward the Centre of the Earth • • The effect of a Gravitational Field on a Mass is called its WEIGHT. Weight is a FORCE and therefore a Vector quantity. Mathematically: W = mg where W = Weight (N) m = mass (kg) g = Gravitational Field Strength (Nkg-1) Weight acts through the Centre of Mass of the body and is directed along the line joining the centres of the the two bodies between which the Gravitational Field is generated. On Earth, the Gravitational Field of Strength 9.8 Nkg-1 gives any mass under its influence alone an acceleration of 9.8 ms-2 4.3 Reaction Force • • • Any stationary object which is under the influence of the Earth’s Gravitational Field must be subject to a force equal in size, but opposite in direction to, its weight. This equal but opposite force is called the NORMAL REACTION FORCE. The Normal Reaction Force only arises from the action of the weight force and DOES NOT EXIST AS AN ISOLATED FORCE IN ITS OWN RIGHT. R - Normal Reaction Force Object Mass = m Table W - Weight R acts upwards from the boundary between the table and the mass through the Centre of Mass W acts downwards from the Centre of Mass toward the centre of the Earth R and W are NOT an ACTION - REACTION PAIR WHY ? 4.4 Bouncing Balls. When a ball, falling under the action of gravity, arrives at a hard surface, it is the Normal Reaction Force (R), which provides; (a) The Force required to decelerate the ball to a stop and then, (b) The Force needed to accelerate the ball away from the surface. vvv==00 R v v W v v W R W W v v W W v=0 W W W The ball is dropped: a = g Velocity = 0 Acceleration = g a=g a=g a=g aa==gg a = g Ball strikes surface, R aa arises and begins to decelerate ball. R > W (acc now up) 4.5 Friction • Friction is the most unusual of all forces as it cannot start an object moving. • Friction can, however, slow or stop an object once it is moving or prevent it from starting to move. • Slowing a moving object is the result of DYNAMIC FRICTION, while stopping an object from starting to move is the result of STATIC FRICTION. • Generally speaking, for the same pair of surfaces, Static Friction Dynamic Friction. • The size of the frictional force depends upon: (a) The Roughness of the surfaces measured by the Coefficient of Friction () (b) The Separation of the surfaces. • Friction does NOT depend on the AREA in contact. OBJECT HELD STATIONARY BY STATIC FRICTION R Fr F (Static Friction) (Pulling Force) W The size of the Frictional Force Fr is calculated from: Fr = R 4.6 Friction Applications • Friction is NOT always a hindrance to living on Earth, often it is vital for movement over the Earth’s surface. Consider the following: A car is accelerating along a Rough (meaning having Friction) Road Surface. Acceleration • • FTR Force of TYRE on ROAD. ACTION FORCE FRT Direction of rotation Force of Driven Wheel ROAD on TYRE. REACTION FORCE The frictional force between the tyre and the road (FTR) is directed backwards. This force CANNOT provide the forward propulsion. In order for the Car to accelerate in the direction shown a force must exist in that direction. Thus it is the Force of the Road on the Tyre (FRT) which gives rise to the acceleration IT IS THE REACTION FORCE WHICH PRODUCES THE ACCELERATION 4.7 Various Force Situations: Stationary and Falling Bodies MASS - STATIONARY ON A TABLE + ve direction R Normal Reaction Force Acts from the point of contact b/w Mass &Table MASS - FALLING WITH AIR RESISTANCE FR +ve a Centre of Mass a=0 W = mg From Newton 2 F = ma F = R - W = ma but a = 0 So, R - W = 0 and R = W Weight acts directly down from the Centre of Mass W = mg F = ma F = W - FR So, W - FR = ma and FR = ma - mg = m(a - g) 4.8 Various Force Situations: Lifts LIFT - ACCELERATING UPWARDS LIFT - ACCELERATING DOWNWARDS +ve R +ve a W = mg a R is a measure of the Mass (m) on Floor weight” of the of Lift W = mg mass. When accelerating For the Mass (m) upward the mass is F = ma “heavier” than normal, by an F = W - R amount (ma), and when so ,W - R = ma accelerating downward it is and R = W - ma “lighter” than normal, by = mg - ma (ma). Thus R = m(g - a) Mass (m) on Floor “apparent of Lift For the Mass (m) F = ma F = R – W so, R - W = ma and R = W + ma = mg + ma Thus R = m(g + a) R 4.9 Various Force Situations: R Inclined Planes a RR RR mg Cos W = mg mg Sin Body of mass ( m) on an inclined plane Weight and Normal Reaction act triangle on mass Force triangle and plane Weight broken upinclined into 2 components Parallel (to the plane) component = mgSin Parallel Component can be transferred are similar, ’s are equal. IT IS THIS THAT to act through the CCOMPONENT of M. ACCELERATES THE MASS (mg DOWN Perpendicular Component CosTHE ) =PLANE R a a a a W W W W Notice the Component of the Weight parallel to the plane (mg Sin) increases as the plane gets steeper thus increasing the acceleration of the mass down the plane. 4.10 Various Force Situations: Connected Masses Frictionless Pulley Masses connected by Light Inextensible String. (Tension (T) the same everywhere) a +ve Frictionless Table Frictionless Pulley m1 a T m2 T m2 W = m1g +ve For m1: T - m1g = m1a F = ma For m2 : W - T = m2a T m1 T For m1: T = m1a Light inextensible String. (Tension (T) the same everywhere) For m2: m2g - T = m2a W = m2 g W = m2g Solve simultaneous equations to get acceleration Chapter 5 Topics covered: • Work. • Work & Energy. • Power. • Energy Types. • Conservation of Energy. 5.0 Work • • • • • • • The term WORK has a very strict definition in the physics world. WORK DONE BY A VARIABLE FORCE If a FORCE moves an object through a DISTANCE, WORK has Force been done on that object. F Mathematically: W = F.d Area = Work done by the Force where W = Work (Joules) Work = ½F x d F = Force (N) d = Distance (m) This formula can only be used if the d Force remains constant through the distance course of doing the work. Work is a SCALAR quantity. If the Force varies in the course of No work has been done if the force doing the Work (as in stretching a has not caused the object to move. spring), the work done can only be No work has been done if the object calculated from the area under the begins and ends its movement in Force versus distance graph. the same place. ie travels in a circle. 5.1 Work & Energy In the Physics world, Work and Energy are intimately related. Energy is very difficult to define. It is easy to say what energy can do but not so easy to say what it is. Thus energy is defined in terms of work. An object is said to possess energy if it has the ability or capacity to do work. Work is the “Transfer Mechanism” for Energy meaning that if some work has been done on an object the amount of energy it possesses has been changed. WORK DONE = ENERGY TRANSFERRED Motion - Revision Questions Question type: Work and Energy A model rocket of mass 0.20 kg is launched by means of a spring, as shown in Figure 1. The spring is initially compressed by 20 cm, and the rocket leaves the spring as it reaches its natural length. The force-compression characteristic of the spring is shown in Figure 2. Q26 : How much energy is stored in the spring when it is compressed? A: Compressing the spring requires work to be done on it. (this work is stored as elastic potential energy in the spring) Work done = area under graph up to a compression of 0.2 m = ½ (0.2)(1000) = 100 J Motion - Revision Questions Question type: Energy Conversion Q27: What is the speed of the rocket as it leaves the spring? A: All the elastic potential energy stored at max compression (100 J) will be converted to Kinetic Energy at release. So 100 = ½ mv2 v = √1000 = 31.6 ms-1 Motion - Revision Questions Question type: Equations of Motion Q28: What is the maximum height, above the spring, reached by the rocket? You should ignore air resistance on the way up since the rocket is very narrow. +ve u = 31.6 ms-1 v=0 a = -10 ms-2 x=? t=? v2 = u2 +2ax 0 = (31.6)2 + 2(-10)x x = 1000/20 = 50 m Motion - Revision Questions Question type: Newtons 2nd Law Retarding Force (R) (Newtons) time (s) Q29: What is the acceleration of the rocket at a time 5 s after the parachute +ve opens? A: At t = 5.0 s, R = 1.8 N Weight of Rocket W = mg = (0.2)(10) = 2.0 N R W When the rocket reaches its maximum height, the parachute opens and the system begins to fall. In the following questions you should still ignore the effects of air resistance on the rocket, but of course it is critical to the force on the parachute. This retarding force due to the parachute is shown as R in Figure 3, and its variation as a function of time after the parachute opened is shown in Figure 4. ΣF = ma a = ΣF/m = (2.0 – 1.8)/ 0.2 = 1.0ms-2 Motion - Revision Questions Question type: Work and Energy In a storeroom a small box of mass 30.0 kg is loaded onto a slide from the second floor, and slides from rest to the ground floor below, as shown in Figure 4. The slide has a linear length of 6.0 m, and is designed to provide a constant friction force of 50 N on the box. The box reaches the end of the slide with a speed of 8.0 m s–1 Figure 4 Q30: What is the height, h, between the floors? At the second floor the box has Potential Energy = mgh On reaching the ground floor this has been converted to Kinetic Energy (½mv2) plus the work done against friction in moving down the slope thus, mgh = ½ mv2 + Fd (30.0)(10)h = ½ (30)(8.0)2 + 50(6.0) h = 4.2 m 5.2 Energy Types • • KINETIC ENERGY The Energy of Motion. It is the energy possessed by moving objects Mathematically: K.E. = ½mv2 where, K.E. = Kinetic Energy (J ) m = mass (kg) v = velocity (ms-1) ELASTIC POTENTIAL ENERGY •The energy stored in elastic materials (eg. Springs & rubber bands) •Mathematically: Es = ½ kx2 where, Es = Elastic P. E. (J ) k = Spring Constant (N kg-1) x = Extension or Compression (m) GRAVITATIONAL POTENTIAL ENERGY • The Energy of Position. The energy possessed by an object due to its position. • Mathematically: P.E. = mgh where, P.E. = Grav. P. E. (J ) m = mass (kg) g = gravitational field strength (N kg-1) h = height above zero point (m) • The zero point for measuring height is usually, but not always, the surface of the Earth. Energy is not a directional quantity so all these forms of energy are Scalar Quantities. Motion - Revision Questions Question type: Elastic Potential Energy Figure 4 The box then slides along the frictionless floor, and is momentarily stopped by a spring of stiffness 30 000 N m–1 Q31: How far has the spring compressed when the box has come to rest? KE of box is converted to Elastic Potential Energy in the spring Thus ½ mv2 = ½ kx2 ½ (30.0)(8.0)2 = ½ (30,000)(x)2 x = 0.25 m • 5.3 Power Power is defined as the Time rate of doing Work. • Mathematically: P = W/t where, • P = Power (W - Watts) W = Work (J ) t = time (s) • Power is a Scalar Quantity • SINCE WORK DONE = ENERGY TRANSFERRED • Power can also be defined as the Time rate of Energy Transfer. • Mathematically: P = E/t where, P = Power (W - Watts) E = Energy (J ) t = time (s) • It is useful to note that since W = F.d and v = d/t the original power formula P = W/t = (F.d)/t =F.v • This allows the calculation of the power of a body moving at constant velocity 5.4 Conservation of Energy • • The law of conservation of energy states: ENERGY CAN BE NEITHER CREATED NOR DESTROYED BUT ONLY TRANSFORMED FROM ONE FORM TO ANOTHER. Energy conversion processes convert useful, ORDERED energy (energy capable of doing useful work) into useless, DISORDERED energy. This disordered energy is sometimes called Thermal Energy, or Low Grade Unrecoverable Heat. The percentage efficiency of energy transfers is calculated from: %Eff = EOUT/EIN x 100/1 No energy transfer process can be 100% efficient in our friction ridden world. If 100% efficiency could be attained perpetual motion machines would then be possible. 5.5 “Seeing” Energy (1) With a little practice you can “watch” energy flow through a system just as an accountant can “watch” money flow through an economy. The most obvious form of energy is Kinetic Energy, the energy of motion. It is easy to see when kinetic energy is transferred to or from an object. As KE leaves the object it slows down: as in a car slowing down when you lift off the accelerator pedal. A water polo ball speeds up when you do work on it during the action of throwing, you are transferring energy from your body into the ball, where the energy shows up in the motion of the ball. 5.6 Seeing Energy (2) Potential Energy is more difficult to “see”. It can take many different forms, as shown in the table below. In each case nothing is moving; but because the objects still have a great potential to do work, they possess Potential Energy. At certain points all a body's energy may be potential whilst at another all its energy may be kinetic. Recognizing these points will enable the solutions to most problems in this area. Form of Potential Energy Example 1. Gravitational Potential Energy A person standing at the top of a building 2. Elastic Potential Energy A wound clock spring 3. Electrostatic Potential Energy A cloud in a thunderstorm 4. Chemical Potential Energy A firecracker 5. Nuclear Potential Energy Uranium Chapter 6 Topics covered: • The Experience of Acceleration. • Circular Motion. • Centripetal Force. • Vertical Circles. • Projectile Motion. • Projection Angles • Projectile Graphs • Real Life Projectiles 6.0 The Experience of Acceleration Nothing is more central to the understanding of the laws of motion than understanding the relationship between force and acceleration. Up to now we have looked at forces and noticed they can produce accelerations (Newtons 2nd Law). Now we will reverse the process – looking at accelerations and noticing that they require a force. For you to accelerate, something must push or pull on you. Just where and how that force is exerted on you determines how you “feel” when you accelerate. The “backward” force you experience when a car accelerates is caused by your body’s “inertia”, its tendency not to want to accelerate. The car and your seat are accelerating, and since the seat back acts to keep you from falling “through” its surface, it provides a forward support force which causes you to accelerate forward. But the seat can’t exert a force uniformly throughout your body. Instead, it pushes only on your back and your back then pushes on your bones, internal organs and tissues to make them accelerate forward. The experience of the accelerating car seat is very similar to the experience of “gravity” when you stand still on the Earth’s surface. In this situation you feel “heavy” ie. you experience your weight 6.1 Fictitious Forces When the car seat is causing you to accelerate forward, you also feel “heavy”; your body senses all the internal forces needed to accelerate its pieces forward, and you interpret these sensations as “weight”. This time you experience the weight directed toward the back of the car. The gravity-like “force” that you experience as you accelerate is truly indistinguishable from the force of gravity. No laboratory instrument can determine directly whether you are experiencing gravity or simply accelerating. However despite the convincing sensations, the backward, heavy feeling in your gut as you accelerate is not due to a real force. This experience of acceleration is explained by the supposed action of a Fictitious Force. This Fictitious Force always points in the direction opposite to the acceleration that causes it and its strength is proportional to the acceleration. a FFict 6.2 Apparent Weight With the car stationary the only force you are subject to is your Weight As the car accelerates you are “pushed back” in your seat by the fictitious force The vector sum of the Weight and Fictitious force produces the Apparent Weight you experience whilst accelerating. Fictitious Force Apparent Weight a Fictitious Fictitious Force Force aa Apparent Apparent Weight Weight Weight Weight The larger the Acceleration, the larger the Fictitious force, the more “backward” your Apparent Weight becomes. Weight 6.3 Circular Motion • • • • • • Newton’s 1st law says, in part, an object experiencing no net force will travel in a straight line at constant speed. Thus, in order to make an object travel in a circle ie. constantly change the direction of its velocity, a force must be constantly applied to it. This means the object is constantly accelerating. The number of times an object spins around per second is called its Frequency (f). An object travelling around a circle completes one full spin in a time interval called a Period (T). Period and frequency (f) are the inverse of one another. Thus: T = 1/f T =Period (sec) f = Frequency (Hz or sec-1) CIRCLE OF RADIUS = R R Circumference of circle = 2 R Time for one revolution = Period = T Speed = Distance , So v = 2 R = 2 Rf Time T V 6.4 Centripetal Acceleration Since the direction of the velocity at any point is the tangent to the circle at that point, the direction of the velocity is constantly changing, thus an object travelling in a circular path is subject to a constant acceleration. The size of that acceleration (called the CENTRIPETAL (or centre seeking) ACCELERATION) is given by: aC = v2/R where aC =centripetal acc (ms-2) v = linear velocity (ms-1) R = radius of circle (m) Substituting for v = 2R/T gives aC = 4 2 R/T2 CIRCLE VIEWED FROM ABOVE aC = vf - vi aC - vi vi vf THE CENTRIPETAL ACCELERATION, aC ,IS ALWAYS DIRECTED TOWARD THE CENTRE OF THE CIRCLE. 6.5 Centripetal Force The fact that an acceleration exists, directed toward the centre of the circle requires there to be a force acting in the aFCC same direction. This force is called the CENTRIPETAL FORCE (FC) and is defined in terms of Newton’s 2nd law. Mathematically: Girl Riding a FC = maC = mv2/R = m42R/T2 Playground Ride T Centripetal Force is NOT a REAL FORCE in its own right, but is Fictitious supplied by other real, measurable Force FC forces. Centre of Mass In the case opposite, the girl requires a Centripetal Force to Direction of travel her circular path Rotation The Tension (T) of her muscular grip on the pivot pole of the ride provides the Centripetal Force. She, of course, will use the idea of a Fictitious Force, centrifugal force, to Linear Velocity explain the “outward” pull she feels while on the constantly accelerating ride. Motion - Revision Questions Question type: Centripetal Force Mark Webber and his Formula 1 racing car are taking a corner at the Australian Grand Prix. A camera views the racing car head on at point X on the bend where it is travelling at constant speed. At this point the radius of curvature is 36.0 m. The total mass of the car and driver is 800 kg. FC Q32: On the diagram showing the camera’s view of the racing car, draw an arrow to represent the direction of the NET force acting on the racing car at this instant. Camera's head on view of racing car at point X The magnitude of the horizontal force on the car is 6400 N. 36.0 m Q33: Calculate the speed of the car. X A: FC = mv2/R So v =√FCR/m = √(6400)(36.0)/(800) = 17 ms-1 Camera Motion - Revision Questions Question type: Centripetal Force Q34: Referring to the racing car from the previous slide, explain: (a) Why the car needs a horizontal force to turn the corner. (b) Where this force comes from. A: (a) Newton 1 states all objects will continue in a straight line unless acted upon by a net force. In order for the car to travel in a circular path a constant force directed at right angles to the direction of motion (toward the centre of the circle) must exist. (b) The horizontal force arises from the frictional force exerted BY THE ROAD ON THE TYRES and is directed toward the centre of the circle the car is travelling in. Motion - Revision Questions Question type: Centripetal Force Fc The safe speed for a train taking a curve on level ground is determined by the force that the rails can take before they move sideways relative to the ground. From time to time trains derail because they take curves at speeds greater than that recommended for safe travel. Figure 5 shows a train at position P taking a curve on horizontal ground, at a constant speed, in the direction shown by the arrow. Q35: At point P shown on the figure, draw an arrow that shows the direction of the force exerted by the rails on the wheels of the train. Motion - Revision Questions Question type: Centripetal Force The radius of curvature of a track that is safe at 60 km/h is approximately 200 m. A: Centripetal Force Fc = mv2 R Q36: What is the radius of curvature of a track that would be safe at a speed of 120 km/h, This can be solved as a ratio question assuming that the track is constructed to the same strength as for a 60 km/h curve? m(60)2 = m(120)2 200 x x = 800 m Q37: At point Q the driver applies the brakes to slow down the train on the curve. Which of the arrows (A to D) indicates the direction of the net force exerted on the wheels by the rails? A: B 6.6 Centrifugal Force Despite its fictitious nature, centrifugal force creates a compelling sensation of gravity like force. The girl on the ride feels as though gravity is pulling her outward as well as down and must hold the handle tight in order not to fall off. Linear Velocity Fictitious forces, such as Centrifugal Force, do NOT contribute to the net force experienced by an object. Thus if the girl lets go she will fly off the ride in the direction of the linear velocity, NOT in the direction of the fictitious force. Anyone who talks about centrifugal force as being a real, measurable force is talking rubbish and does not understand physics, be wary of ALL they say !!!!! A stationary object on Earth experiences a weight force of 1g. The fictitious centrifugal force experienced by 5 kg of clothes during the spin cycle in a washing machine, travelling in a 0.25 m radius circle at 20 ms-1 is about 163 g’s and they have an apparent weight of 815 kg. 6.7 Banked Corners Race track and road designers often build their tracks or roads with “banking” on the corners. This design feature is used to enhance the safety of the track or road allowing users to round corners at higher speeds (and with a greater margin of safety) than they could if the corner was not banked. Any vehicle travelling around a corner with velocity v needs a centripetal force (FC) acting toward the centre of the corner In the case of the flat, non banked, corner this centripetal force (FC) is supplied by the friction of the road against the tyres (FRT) With the banked corner the centripetal force has an extra component - that being a component of the car’s weight force acting toward the centre of the corner. This gives a larger overall Centripetal Force (FC) larger than in the non R banked case. FC v R F LAT CORNER Component of W acting toward centre of corner BANKED CORNER FC FC FRT FRT W FRT W FRT The larger FC allows the car either a greater margin for safety or a faster speed around a banked corner compared to a flat corner of the same radius 6.8 Vertical Circles Objects travelling in vertical circles are subject to the acceleration due to gravity, thus their speed will vary depending where in their motion observations are made. Analysis of this type of motion is based on energy considerations and the fact that the motion takes place in a uniform gravitational field. In theoretical situations, the TOTAL ENERGY REMAINS CONSTANT, but varies between Kinetic and Potential, depending on where in the circle you choose to look. A soccer ball of mass 0.5 kg moves along a friction free track at 20.0 ms-1 toward a vertical circle 10.0 m in diameter. v = 20 ms-1 PE = 49 J KE = 51 J PE = 24.5 J 10 m KE =100J KE = 75.5 J PE = 0 Its initial K.E. = ½mv2 = ½(0.5)(20)2 = 100 J PE = mgh =(0.5)(9.8)(5.0) = 24.5 J KE = 100 J KE = 100 - 24.5 = 75.5 J PE = 0 J 5.0 m h=0 6.9 Projectile Motion Projectile Motion is the motion which objects, launched at some angle to the Earth’s gravitational field, will undergo. Projectile motion is a combination of TWO INDEPENDENT MOTIONS. (a) HORIZONTAL MOTION which is CONSTANT VELOCITY motion. (b) VERTICAL MOTION which is CONSTANT ACCELERATION motion. vHORIZ vHORIZ vVERT The horizontal motion, being constant velocity motion, is covered by only one equation: v = d/t. The vertical motion, being constantly accelerated motion, is covered by the Equations of Motion The only common factor between the the two motions is THE TIME OF FLIGHT (t). vHORIZ vVERT vHORIZ vHORIZ vVERT vVERT vHORIZ vHORIZ remains constant throughout vVERT increases under the action of gravity vVERT Motion - Revision Questions Question type: Projectile Motion A car takes off from a ramp and the path of its centre of mass through the air is shown below. First model the motion of the car assuming that air resistance is small enough to neglect. Q38: Which of the directions (A - H), best shows the VELOCITY of the car at X ? A: Direction C. Q39: Which of the directions (A - H), best shows the VELOCITY of the car at Y ? A: Direction D A B A B H G C D Direction of Motion H G C X F D E Y F E Motion - Revision Questions Question type: Projectile Motion Q40: Which of the directions (A - H), best shows the ACCELERATION of the car at X ? A: Direction E. Q: Which of the directions (A - H), best shows the ACCELERATION of the car at Y ? A: Direction E. Now suppose that AIR RESISTANCE CANNOT BE NEGLECTED. Q41: Which of the directions (A - H), best shows the ACCELERATION of the car at X ? A: Direction F. A B A B H G C D Direction of Motion H G C X F D E Y F E Motion - Revision Questions Question type: Projectile Motion A bushwalker is stranded while walking. Search and rescue officers drop an emergency package from a helicopter to the bushwalker. They release the package when the helicopter is a height (h) above the ground, and directly above the bushwalker. The helicopter is moving with a velocity of 10 ms–1 at an angle of 30° to the horizontal, as shown in Figure 1. The package lands on the ground 3.0 s after its release. Ignore air resistance in your calculations. Figure 1 Q42: What is the value of h in Figure 1? A: Can approach in a number of ways. One way is: Find how long it takes to rise and fall back to same vertical height (h) Then subtract from total time to give a time for vertical fall under gravity with an initial velocity of 10 ms-1 directed at 30o below the horizontal Motion - Revision Questions Question type: Projectile Motion Figure 1 1. Time to get back to h Since this part of the journey is symmetrical time to reach max height = ½ time for this part of journey. h 2. Value of h again analyse vertical motion Analyse vertical motion u = 10 Sin 30o 2 s = ut + ½ at v = ? o -1 u = - 10 Sin 30 ms +ve a = 10 ms-2 h = (5)(2) + ½ (10)(4) v=0 v = u + at h = 30 m s=h -2 +ve a = 10 ms 0 0 = -10 Sin 30 + 10t t = 2.0 s s=? t = 5/10 = 0.5 sec. t=? So, total time to get back to h = 1.0 s Motion - Revision Questions Question type: Projectile Motion Figure 1 Q43: Assuming that the helicopter continues to fly with its initial velocity, where is it when the package lands? Which one of the statements below is most correct? A. It is directly above the package. B. It is directly above a point that is 15 m beyond the package. C. It is directly above a point that is 26 m beyond the package. D. It is directly above a point that is 30 m from the bushwalker. Both the package and the helicopter have the same (constant) horizontal velocity, so the package will always be directly below the helicopter. So, alternative A is correct Motion - Revision Questions Question type: Sketch Graphs Q44: Which of the graphs below best represents the speed of the package as a function of time? A: The key to the question is SPEED, this is the sum of both vertical and horizontal. Speed never falls to zero (there is always a horizontal component), so C and D are out A shows the object’s acceleration falling as it nears the ground – not so ! So B must be the correct answer Motion - Revision Questions Question type: Projectile Motion Fred is playing tennis on the deck of a moving ship. He serves the ball so that it leaves the racket 3.0 m above the deck and travels perpendicular to the direction of motion of the ship. The ball leaves the racket at an angle of 8° to the horizontal. At its maximum height it has a speed of 30.0 ms-1. You may ignore air resistance in the following questions. Q45: With what speed, relative to the deck, did the ball leave Fred’s racket? Give your answer to three significant figures. A: Assuming projectile behaviour: VHORIZONTAL = 30 ms-1 throughout the flight 30 ms-1 x = initial speed x 8o 30 ms-1 thus, Cos 8 = 30/x x = 30/Cos 8 = 30.3 ms-1 Motion - Revision Questions Question type: Projectile Motion Q46: At its highest point, how far was the ball above the deck? 30 ms-1 x 8o VVERT 30 ms-1 A: Using the equations of motion for the balls flight from racquet to point of max height, where VVERT = 0 +ve u = VVERT = 30.3 Sin 8 v=0 a = -10 ms-2 s=? t=? v2 = u2 + 2as 0 = (30.3 Sin 8)2 + 2(-10)s s = 0.89 m So Height above the DECK = 3.0 + 0.89 = 3.89 m Motion - Revision Questions Question type: Relative Motion The ship is travelling straight ahead at a velocity of 10 ms-1 Q47: When the ball is at its highest point at what speed is it moving relative to the ocean? 30 ms-1 10 ms-1 A: V BALL REL OCEAN = VSHIP + VBALL VSHIP = + VBALL 30 10 θ VBALL REL OCEAN VBALL REL OCEAN = √(30)2 + (10)2 = 31.6 ms-1 Q48: at what angle is the ball travelling relative to the direction of the ship’s travel? Tan θ = 30/10 = 3.0 θ = 71.60 6.10 Projection Angles +ve v vVERT vHORIZ • There are two basic types of vHORIZ = v Cos vVERT = v Sin projectile motion: This does not vary This does vary from (a) Objects initially projected during the flight +v Sin at t = 0 horizontally from some point to zero at t = ½t above the Earth’s surface. As to - v Sin at t = t shown previously . (b) Objects initially projected at Flight can be broken up into some angle above the 2 equal parts each taking exactly horizontal as shown here. ½t to complete v Total flight time = t At this point all velocity is horizontal 6.11 Projectiles – Time, Height & Range v h θ Range (R) An object is fired from ground level with a velocity v at an angle θ to the horizontal It reaches a maximum vertical height, h and finally lands at the same horizontal level as it started, having covered a horizontal distance of R 1. Time to reach Maximum Height (h) Upward is +ve. u = v sin θ use eqns of motion: v=0 v = u + at a = -g 0 = v sin θ – gt s=? t = v sin θ t=? g 2. Total Time of Flight Total time = (time to reach h) x 2 = 2 v sin θ g 3. Maximum Height Upward is +ve u = v sin θ use eqns of motion v=0 v2 = u2 + 2as a = -g 0 = v2 sin2θ – 2gh s=h h = v2 sin2θ t = v sin θ 2g g 4. Range of Projectile Horizontally vH = dH / t vH = v cos θ R = 2v sinθ x v cos θ dH = R g t = 2v sinθ 2 x 2 sin θcos θ = v g g = v2 sin 2θ g 5. Maximum Range occurs when sin 2θ = 1 2θ = 900 or θ = 450 • • 6.12 Real Life Projectiles The mathematical analysis of projectile motion ignores the effects of friction, in particular air resistance. The real world trajectory differs from the theoretical trajectory in three main ways. (1) The actual range (horizontal distance) covered in the real world is less because of the reduction of the horizontal component of velocity due to air resistance. (2) The actual height achieved will be less, due to the effect of air resistance on the upward vertical velocity. (3) The object will fall more steeply than it rises, the path of the projectile is no longer symmetrical around its highest point. Loss of height Theoretical Actual Loss of Range Path no longer symmetrical Chapter 7 Topics covered: • Elastic Materials and Hooke’s Law. • Elastic Potential Energy. 7.0 Elastic Materials & Hooke’s Law Elastic materials such as springs display elastic behaviour. This means that when they are deformed, stretched or compressed, they change their shape or condition. When the force causing the deformation is removed, they return to their original shape or condition. Such materials are able to store energy while deformed and release that energy when allowed to return to their original condition. Mathematically F = - k(∆x) The negative sign indicates that the restoring force F acts in the opposite direction to the extension or compression ∆x The relation between Force and the Extension and/or Compression of an elastic material is called Hooke’s Law: where F = Restoring Force (N), k = Spring Constant (Nm-1) ∆x = Extension and/or Compression (m) • • • 7.1 Potential Energy in Elastic Materials Elastic materials store energy when they are deformed and release that energy when they return to their original condition. The amount of energy stored can be found from the Elastic Potential Energy Formula: ES = ½kx2 where ES = Elastic P. E. (J ) k = Spring Constant (Nm-1) x = extension and or compression (m) For materials which display “irregular” behaviour, the Potential Energy stored can only be found from the area under the Force vs Extension (or Compression) graph. “REGULAR” ELASTIC BEHAVIOUR Slope = Spring Force constant (k) F Extension x Area = ½Fx = ½kx2 = Energy stored up to extension x “IRREGULAR” ELASTIC BEHAVIOUR Area = Energy stored up to x Area is found by “counting squares” Force F Extension x Chapter 8 Topics covered: •Universal Gravitation •Gravitational Attraction •Circular Orbits •Maths of Circular Orbits •Kepler’s Laws •Satellites in Space •Energy Transfers 8.0 Law of Universal Gravitation • • • • • • Gravity is the most well known of all the Natural Forces. We live with the effects of gravity every day and would lead completely different lives if gravity was not present. It is gravity alone which gives us our sense of “up and down”. Gravity is a FIELD force, it acts at a distance. It is ALWAYS an ATTRACTIVE force. Gravity arises from the interaction of 2 masses. The size of the gravitational force is directly proportional to the product of the two masses and inversely proportional to the square of the distance between them. Mathematically: Fg = GM1M2 R2 where Fg = Gravitational Force (N) G = Universal Gravitational Constant (6.67 x 10-11 Nm2kg-2) M1, M2 = masses (kg) R = Separation of the masses (m) EACH of the bodies experiences the SAME FORCE even though they may have vastly different masses. Motion - Revision Questions Question type: Universal Gravitation Newton was the first person to quantify the gravitational force between two masses M and m, with their centres of mass separated by a distance R as F= GMm R2 where G is the universal gravitational constant, and has a value of 6.67 × 10-11 N m2 kg2. For a mass m on the surface of Earth (mass M) this becomes F = gm, where g = GM R2 Q49: Which one of the expressions (A to D) does not describe the term g? A. g is the gravitational field at the surface of Earth. B. g is the force that a mass m feels at the surface of Earth. C. g is the force experienced by a mass of 1 kg at the surface of Earth. D. g is the acceleration of a free body at the surface of Earth. A: B Motion - Revision Questions Question type: Universal Gravitation The radius of the orbit of Earth in its circular motion around the Sun is 1.5 × 1011 m Q50: Indicate on the diagram, with an arrow, the direction of the acceleration of Earth. Q51: Calculate the mass of the Sun. Take the value of the gravitational constant G = 6.67 × 10–11 N m2 kg–2. (Figure 3). Question assumes you know the period of rotation of the earth around the sun ie. 365 days = 365 x 24 x 60 x 60 = 31536000 s (3.15 x 107 s) F C = Fg me4π2 R/T2 = Gmems/R2 ms = 4π2R3 GT2 ms = 4(3.14)2(1.5 x 1011)3 6.67 x 10-11(3.15 x 107)2 ms = 2 x 1030 kg Motion - Revision Questions Question type: Universal Gravitation Nato III is a communication satellite that has a mass of 310 kg and orbits Earth at a constant speed at a radius R = 4.22 x 107 m from the centre of Earth. Q52: What is the speed of Nato III in its orbit ? A: The centripetal force (FC) required by Nato III to complete its circular orbit is supplied by the force of gravitational attraction (Fg) between Earth and Nato III. Thus Fg = FC Thus GMem/R2 = mv2/R So v = √GMe/R = √(6.67 x 10-11)(5.98 x 1024) -------------------------------------(4.22 x 107) = 3074 ms-1 = 3.1 x 103 ms-1 R Earth Nato III Motion - Revision Questions Question type: Circular Motion Q53: Which ONE of the following statements (A - D) about Nato III is correct ? A: The net force acting on Nato III is zero and therefore it does not accelerate. B: The speed is constant and therefore the net force acting on Nato III is zero. C: The is a net force acting on Nato III and therefore it is accelerating. D: There is a net force acting on Nato III, but it has zero acceleration. A: Alt C is correct R Earth Nato III 8.1 Gravitational Attraction What Gravitational Force do the Earth and the Moon experience because of their proximity to one another ? Earth FG R = 384,000 km = 3.84 x 108 m FG Moon Note the distance (R) is measured from the Centre of Mass of each body Mass of Moon MM= 7.34 x 1022 kg FG = GMEMM Mass of Earth ME = 5.98 x 1024 kg R2 = (6.67 x 10-11)(5.98 x 1024)(7.34 x 1022) (3.84 x 108)2 = 1.99 x 1020 N The Earth experiences an attractive force of 1.99 x 1020 N directed toward the Moon while AT THE SAME TIME the Moon experiences the same force directed toward Earth. Motion - Revision Questions Question type: Gravitational Force Q54: What is the magnitude of the force exerted by Earth on a water molecule of mass 3.0 × 10-26 kg at the surface of Earth? A: F = GMm R2 = (6.67 x 10-11)( 5.98 x 1024)(3.0 x 10-26) (6.37 x 106)2 = 2.95 x 10-25 N Motion - Revision Questions Question type: Gravitational Field Strength Nato III is a communication satellite that has a mass of 310 kg and orbits Earth at a constant speed at a radius R = 4.22 x 107 m from the centre of Earth. R Q55: Calculate the magnitude of the Earth’s gravitational field at the orbit radius, R = 4.22 x 107 m, of Nato III. Give your answer to 3 sig figs. You MUST show your working. G = 6.67 x 10-11 Nm2kg-2 Me = 5.98 x 1024 kg. Earth Nato III A: g = GMe/R2 = (6.67 x 10-11)(5.98 x 1024) -----------------------------------------(4.22 x 107)2 = 0.224 Nkg-1 8.2 Circular Orbits under Gravity On the large scale of planets, moons, stars and other bodies in the universe, their motions are determined by the gravitational attractions between them. If, for example, a moon travels in a circular orbit around its host planet, it must be subject to a Centripetal Force. This Centripetal Force must be supplied by some kind of interaction between the planet and its moon. This interaction is the GRAVITATIONAL ATTRACTION between the Planet and its Moon. Thus the Centripetal Force (FC) needed for circular motion is supplied by the Gravitational attraction (Fg) between the planet and its moon. Planet Moon FG FG FC 8.3 Mathematics of Circular Orbits FC FG FG Moon Planet The Centripetal Force (FC) the moon is subject to is given by: FC = MMv2/R Remember the velocity of an object travelling in a circle is: v = 2R/T v2 = 4 2R2/T2 Substituting for v2 we get: FC = MM 42R/T2 This Centripetal Force is supplied by the Gravitational Force between the the Planet and the Moon: FG = GMMMp/R2 GMMMP/R2 = MM42R/T2 Rearranging we get: R3/T2 = GMP/42 The terms G, MP, and are all constants so their ratio is constant. The ratio R3/T2 is also a constant 8.4 Kepler’s Laws Johannes Kepler (1570 - 1630) discovered the laws governing planetary motion which describe the movement of the planets in our solar system. In the previous slide we found the ratio R3/T2 was a constant, (its value is roughly 3.0 x 1018 ). The 3rd law: R3/T2 = GMP/42 can be rewritten as: R= 3 GMPT2 42 Notice that the radius of orbit of any satellite, while it does depend on the mass of the planet around which it circulates, DOES NOT depend its own mass. This is true of any satellites whether man made of naturally occurring. In 1615 Kepler discovered that the ratio of the of the cube of the average sun - planet distance (R3) to the square of its period (T2) was a constant for ALL planets in our solar system, this became known as Kepler’s Third Law. His other 2 laws establish planets’ speeds and the elliptical nature of planet orbits. Motion - Revision Questions Question type: Kepler’s 3rd Law A satellite in a circular orbit of radius 3.8 × 108 m around Earth has a period of 2.36 × 106 s. Q56: Calculate the mass of Earth. You must show your working. A: R3 = GMe T2 4π2 Me = (4π2) (3.8 x 108)3 (6.67 x 10-11) (2.36 x 106)2 = 5.83 x 1024 kg 8.5 Satellites in Space. A satellite moving through space will often use the gravitational field of a planet, like Jupiter or star like our Sun to help propel it through space. A At point A, the satellite comes under the influence of the gravitational field of the planet. The field does Work ON the satellite, accelerating it toward the planet. B By the time it has reached B, the satellite has increased its Kinetic Energy, and hence its Speed, sufficiently to pass around, rather than crash into, the planet. Jupiter The satellite flies past the planet leaving with greater speed, having “stolen” some of the energy stored in the planet’s field. Path of Satellite around Planet This is the normal way of sending satellites to the outer planets and even outside our solar system. 8.6 Energy Transfers in Gravitational Fields The Gravitational Field vs Distance graph for Jupiter showing positions A and B for the satellite is shown. Since Work Done = Energy Transferred, the area under the graph represents the work done by the field on, and the change in energy possessed by, “1 kg of satellite mass” in moving from A to B. In exam questions, the area normally needs to be found by the “counting the squares” method. For the satellite with a mass >1 kg, total energy possessed = area x mass g(Nkg-1) Area gives: 1. Work by the field on 1 kg of satellite moving from A to B. 2. The increase in Kinetic Energy possessed by 1 kg of satellite in moving from A to B. 3. The loss in Potential Energy of 1 kg of satellite moving from A to B RJ B A Distance R (m) Motion - Revision Questions Question type: Weight & Weightlessness The Russian space station MIR (Russian meaning - peace) is in a circular orbit around the Earth at a height where the Gravitational Field Strength is 8.7 Nkg-1 Q57: Calculate the magnitude of the gravitational force exerted by Earth on the astronaut of mass 68 kg on MIR A: W = mg = (68)(8.7) = 592 N When the astronaut wishes to rest he has to lie down and strap himself into bed. Q58: What is the magnitude of the force that the bed exerts on the astronaut before he begins to fasten the strap ? A: 0 N Motion - Revision Questions Question type: Weightlessness Newspaper articles about astronauts in orbit sometimes speak about zero gravity when describing weightlessness. Q59: Explain why the astronaut in the orbiting MIR is not really weightless. A: Weight is the action of a gravitational field acting on a mass. For true weightlessness to exist the gravitational field strength needs to be zero. There is no place in the universe where the gravitational field strength is zero, so nothing is truly “weightless”. The astronaut actually does have weight, as calculated earlier. The astronaut’s “apparent weight” is zero because he is not subject to a reaction force inside the spacecraft since both he and the spacecraft are in a state of constant free fall. Ollie Leitl 2009