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## VCE PHYSICS

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**VCE PHYSICS**Unit 3 Topic 1 Motion in 1 & 2 Dimensions**Unit Outline**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.**S.I. System:**7 Fundamental Units Length: Unit Metre (m) Mass: Unit Kilogram (kg) Time: Unit Second (s) Electric Current: Unit Ampere (A) Temperature: Unit Kelvin (K) Luminous Intensity: Unit Candela (cd) Amount of Substance: Unit Mole (M) Velocity: Unit ms-1 Force: Unit N = kg ms-2 Velocity - Derived Unit Force - Derived Unit 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) The system is based on 7 Fundamental Units, each of which is strictly defined. All other units, so called DERIVED UNITS, are simply combinations of 2 or more of the Fundamental Units.**- 15 Units**+ 30 Units 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. For example on the number line below the point labelled 0 is the origin and all measurements are related to that point. Thus a number called -15 is 15 units to the left of 0 on the number line. A number called +30 is 30 units to the right of 0.**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.) N 15 A vector of magnitude 15 units directed East 1.2 Scalars & Vectors 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.**4**4 Centre of Mass 450 N 5 3 3 Magnitude = 42 + 32 5 = 25 5 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 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 A Vector of: Magnitude; 5 units Direction; NE or N45E or 45T THE RESULTANT FORCE HAS A MAGNITUDE OF 5 N DIRECTED AT E 36.90 S**8.0 ms-1 E**8.0 ms-1 S v vf 8.0 ms-1 S Magnitude v = 82 + 82 vi = 128 - vi vf = 11.3 ms-1 8.0 ms-1 W 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 ) is a negative vector. It can be converted to a positive one by reversing its direction. Then, by performing a vector addition, the velocity change v can be obtained. What is the object’s change in velocity ? Direction: Tan = 8/8 = 1.0 = Tan-1 1.0 = 450 THE CHANGE IN VELOCITY = 11.3 ms-1 AT S 450W**40 ms-1**V VERTICAL 150 VHORIZONTAL 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 ? Vertical Component: V VERTICAL = 40 Sin 150 = 10.4 ms-1 Horizontal Component: VHORIZONTAL = 40 Cos 150 = 38.6 ms-1 V VERTICAL and VHORIZONTALare the COMPONENTS of the plane’s velocity.**A**B C D Vectors Motion - Revision QuestionsQuestion type: 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: 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. Q1: What is the direction of Adam’s acceleration at the time shown in Figure C ? Explain your answer.**Chapter 2**Topics covered: • Distance versus Displacement. • Speed versus Velocity. • Acceleration. • Graphical Representations.**+ve direction**400 m track 100 m Start Finish Start/Finish 2.0 Distance vs Displacement JOURNEY No 1. Distance = 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. At the end of the run: Distance = 100 m. Displacement = +100 m JOURNEY No 2. Distance Displacement At the end of the one lap run: Distance = 400 m. Displacement = 0 m**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 INSTANTANEOUS vs AVERAGE 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 anInstantaneous 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). 2.1 Speed vs Velocity IN ALL CALCULATIONS AND EQUATIONS USED IN THE COURSE, ASSUME INSTANTANEOUS VALUES ARE REQUIRED UNLESS OTHERWISE STATED.**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 2. Walking Pace 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 8. Speed of Light 2.99 x 108 1.1 x 109**v**a v a 2.3 Acceleration ACCELERATING VEHICLE • 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). The velocity and acceleration are in the same direction DECELERATING VEHICLE The velocity and acceleration are in opposite directions.**Displacement**Time Distance Distance Velocity Time Time Time 2.4 Graphical Representations SKETCH GRAPHS • 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. The Story: As time passes, the distance of the object from its starting point is increasing in a uniform manner (the slope is constant). This is the graph an object moving at constant speed. The Story: As time passes, the velocity of the object is increasing in a uniform manner (the slope is constant). This is the graph a constantly accelerating object The Story: As time passes, the displacement of the object is increasing more quickly (the slope is increasing at a constant rate). The Story: As time passes, the distance of the object from its starting point does not change. This is the graph a stationary object. This is the graph a constantly accelerating object**A**F C D E B Sketch Graphs Motion - Revision QuestionsQuestion type: 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. 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**Read directly**from Graph Area under Graph gives: Graph Type Slope of Graph gives: 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. Distance or Displacement versus Time Distance or Displacement Speed or Velocity No Useful Information Speed or Velocity Distance or Displacement Speed or Velocity versus Time Acceleration Acceleration versus Time No Useful Information Velocity Acceleration**Make a list like this.**+ve 2.6 The Equations of Motion • HINTS FOR EQUATIONS’ USE u = v = a = x = t = 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. • 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) 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.**Equations of Motion**Motion - Revision QuestionsQuestion type: 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**Average Speed**Motion - Revision QuestionsQuestion type: 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. Need to know u, the initial speed for the braking section which equals the final speed for the accelerating section. A: Average Speed = Total Distance Total Time For the accelerated part of journey: Distance = 400 m Time = 19 sec For the braking part of journey: Distance = needs to be calculated Time = 5.1 sec For accelerating section: u = 0 v = ? a = 2.2 ms-2 x = 400 m t = 19 s v = u + at = 0 + (2.2)(19) = 41.8 ms-1 Now can calc s x = ut + ½at2 = (41.8)(5.1) + ½(-8.2)(5.1)2 = 213.2 - 106.6 = 106.6 m Need to calc acc v = u + at 0 = 41.8 + a(5.1) a = - 8.2 ms-2 To get braking distance, use Eqns of Motion u = ? v = 0 a = ? x = ? t = 5.1s Braking list becomes u = 41.8 ms-1 v = 0 a = ? x = ? t = 5.1s List does not contain enough info the calculate s Still not enough info Total Distance = 400 + 106.6 = 506.6 m Total Time = 19 + 5.1 = 24.1 s So, Average Speed = 506.6/24.1 = 21 ms-1**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**LARGE MASS – LARGE INERTIA**small mass – small inertia 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 3. For every action there is an equal and opposite reaction 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)**Newton’s Laws**Motion - Revision QuestionsQuestion type: TotalOpposing 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. Force (N) 10000 8000 6000 4000 2000 Distance (m) 0 100 200 300 400 500 600 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. 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**Newton’s 2nd Law**Motion - Revision QuestionsQuestion type: 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 Distance (m) 0 100 200 300 400 500 600**x x**x x x x x x p p p p p x p Work Motion - Revision QuestionsQuestion type: 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 Q8: Estimate the work done by the seaplane against the opposing forces in travelling for a distance of 500 m. 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**Figure 2**Newtons 2nd Law Motion - Revision QuestionsQuestion type: 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. 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**FTR**FRT a Newton’s 3rd Law Motion - Revision QuestionsQuestion type: The figure shows a cyclist with the bicycle wheels in CONTACT with the road surface. The cyclist is about to start accelerating forward. 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 The laws only apply at speeds much, much less than the speed of light. The laws apply equally in ALL inertial frames of reference. CONSEQUENCES As far as Newton’s 1st law is concerned “rest” and “uniform motion” are the same state. You cannot perform any test which can show whether you are stationary or moving at constant velocity. FRAME OF REFERENCE ? A frame of reference is best described as “your point of observation.” An inertial frame of reference is one that is either stationary or moving with constant velocity. A non inertial frame of reference is accelerating. Humans in non inertial frames tend to invent “fictitious forces” to explain their experiences The action and reaction law requires there to be TWO bodies interacting, the ACTION force acting on one body and the REACTION acting on the other. Deciding when an Action / Reaction situation exists can be done by answering the question: Does the second (reaction) force disappear immediately the first (action) force disappears ? If the answer is yes, you have an action reaction pair.**Relative Motion**Motion - Revision QuestionsQuestion type: • 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.**F1**F1 F4 F4 F3 F3 F2 F2 FRES FRES 3.3 Force in Two Dimensions AN OBJECT UNDER THE ACTION OF 4 FORCES • 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. Perform a vector addition of the forces. The Resultant Force (FRES ) will give the direction of the acc. In which direction will the object accelerate ? The object will be subject to FRES and accelerate in that direction**Motion - Revision QuestionsQuestion 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**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 3.4 Momentum & Impulse**Momentum**Motion - Revision QuestionsQuestion type: 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. Q16: Explain why Chris’s comment is wrong. A: Momentum is conserved in ALL types of collisions whether they be elastic or inelastic. KE is not conserved in this type (inelastic) collision.**Momentum**Motion - Revision QuestionsQuestion type: • 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**Impulse**Motion - Revision QuestionsQuestion type: 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**10 tonnes**5 tonnes 6.0 ms-1 Before Collision X Y stationary v ms-1 After Collision X Y Momentum Motion - Revision QuestionsQuestion type: 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. 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**10 tonnes**5 tonnes 6.0 ms-1 Before Collision X Y stationary 4.0 ms-1 After Collision X Y Impulse Motion - Revision QuestionsQuestion type: 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. 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**U**n Neutron ELASTIC COLLISION INELASTIC COLLISION 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: (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. (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.**Elastic/Inelastic Collisions**Motion - Revision QuestionsQuestion type: 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**10 tonnes**5 tonnes 6.0 ms-1 Before Collision X Y stationary 4.0 ms-1 After Collision X Y Inelastic Collisions Motion - Revision QuestionsQuestion type: 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. 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**Mass of car plus**passengers = m Velocity = v Momentum p = 0 v = 0 3.5 Conservation of Momentum Mass of car plus passengers = m • 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. Momentum p = 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**Momentum/Impulse**Motion - Revision QuestionsQuestion type: 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**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 Air Bags/Crumple Zones Motion - Revision QuestionsQuestion type: 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.**Chapter 4**Topics covered: • Centre of Mass. • Weight. • Reaction Force. • Bouncing Balls. • Friction. • Various Force Applications**C of M**4.0 Centre of Mass 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. For oddly shaped objects eg. a boomerang, the C of M may fall outside the perimeter of the object. 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**XCofM = (m1x1 + m2x2 + m3x3 + …)**(m1 + m2 + m3 +…) 2.875 m 5.0 m 1.0 m 1.0 m Centre of Mass of each mass X C of M = [(50 x 2.5) + (30 x 3.5)] (50 + 30) A 2.5 m Centre of Mass of System 3.5 m 4.1 Centre of Mass - Systems 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. WHAT IS THE CENTRE OF MASS OF THIS SYSTEM? For a system of 2 or more bodies, the position of the C of M may be determined from the formula: CHOSEN REFERENCE POINT IS A 30 kg mass 50 kg beam = 2.875 m from A**Centre of Mass**Weight = mg 4.2 Weight Object of Mass (m) • 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 Weight Force acts through the Centre of Mass and is directed toward the Centre of the Earth