Physics Unit 3 2010 Newton‟s laws of motion 1 of 17 Kinematics Study Design Apply Newton‟s three laws of motion in situations where two or more coplanar forces act along a straight line and in two dimensions; Introduction ………………………………. You will be familiar with many of the concepts of motion from your Physics Unit 2 studies. You must know the definitions of the key terms, such as distance, displacement, speed, velocity, acceleration, force, momentum and energy. In Unit 3 you will be expected to use many of the same techniques to solve problems as in Year 11. This repeated content is illustrated in the diagram below. …………………….. Momentum Kinematics Conservation of x,u,v,a,t formulae momentum graphs Impulse Graphical Solutions Motion Energy Types (kinetic, Forces gravitational Newtons Laws potential, elastic potential, Work) Vector in one and two dimensions Calculations using Conservation of energy When solving Year 12 problems, you will have to choose which concept of physics (Kinematics, Momentum, Energy or Forces) will be most useful in solving the problems. Some problems you will be only able to solve using one concept, others can be solved in more than one way, but may be very simple using a particular technique. Hence you must be able to solve problems using all of the concepts, and you must develop an instinct for choosing the most efficient path. You will also need to decide whether to use a graphical or numerical technique to solve problems. In addition there are a few new topics, specifically motion in more than on dimension (projectile motion), motion in different inertial frames, circular motion and Newton‟s law of Universal Gravitation. ………………………………. Physics Unit 3 2010 Newton‟s laws of motion 2 of 17 Definitions From your prior studies of motion you should be familiar with the following kinematic definitions: Distance Travelled - How far an object has moved in total during its motion. (m). Displacement – How far an object is at from a reference position. (m) Speed – How fast an object is moving. (m/s) Velocity – How fast an object and what direction an object is moving in. (m/s) Acceleration – The rate at which the velocity of an object is changing i.e. how many (m/s) the velocity of an object is changing by every second. Acceleration also has a direction These physical quantities can be divided into two categories, scalars and vectors. Vectors: Vectors are quantities that have a magnitude and a direction. E.g. displacement, velocity, and acceleration. Scalars: Scalars are quantities that only have a magnitude, E.g. speed and distance travelled. Average Quantities You will occasionally be asked to determine average quantities. For example, you may be asked to determine the velocity, on average, at which a car was moving between two times. Average velocity and acceleration are determined using the following formulae. total displacement x 2 - x1 average velocity = vav = time taken Δt change in velocity v -u average acceleration = aav = time taken Δt where x2 is the final position, x1 is the initial position, t is the time period, v is the final velocity and u is the initial velocity. Graphs You need to be able to use a wide range of graphs. When given a graph in the exam, look for the following on the graph before even reading the question: type of graph (F – d, F – v, Energy – distance, F – t etc.). the units on the axis. the limit reading on each axis. Look at the scale on both axes, be aware for anything non-standard Think about what is given by a direct reading from the graph, the gradient of the graph and the area under the graph In Year 11 Physics it is typical to restrict the types of graphs that you experience to those with „time‟ usually on the horizontal axis. Expect to find „distance‟ „speed‟ and many others on the horizontal axis in Year 12. Physics Unit 3 2010 Newton‟s laws of motion 3 of 17 Graphical Techniques In kinematics you can be asked to interpret several graphs. Graphs can be used to determine instantaneous quantities i.e. the value of a quantity at a specific time. For example, a velocity time graph (v-t) can be used to determine how fast an object was moving at a specific time. It could also be used to determine how far the object has moved up to that time (by finding the area under the curve) or its acceleration (by determining the gradient at a specific point). The type of information that can be determined from different graphs is summarised in the following table. Graph type x-t v-t a–t Found from Direct reading 'x' at any 't' 'v' at any 't' 'a' at any 't' 't' at any 'x' 't' at any 'v' 't' at any 'a' Gradient Instantaneous Instantaneous 'a' Meaningless velocity at any Average 'a' point. Vav between any two points Area under graph Meaningless x v The gradient at a particular time is determined by drawing a tangent line to the curve at that point, and then determining the gradient of the tangent line. Constant Acceleration Consider the following series of graphs. These illustrate the relationships mentioned in the table above. Notice that the velocity – time graph is the gradient of the displacement – time graph, and the acceleration – time graph is the gradient of the velocity – time graph. Displacement Velocity Acceleration v g u time time time t Physics Unit 3 2010 Newton‟s laws of motion 4 of 17 Examples The figure below appeared in a newspaper featuring skydiving from an aircraft. In this particular example the total mass of the skydiver and equipment is 100 kg. The skydiver jumps from a height of 3000 m above the ground and reaches a constant terminal velocity of 190 km h-1 in a time of 15 s. She then falls at this constant speed of 190 km h-1 for a further 35 s before opening the parachute. Example 1 Question 1 (1998) Convert 190 km h-1 into m s-1. Example 2 Question 2 (1998) On the set of axes, sketch a graph of the motion of the skydiver for the first 50 s of falling. (Air resistance cannot be neglected.) Example 3 Question 3 (1998) Explain why the speed remains constant between 15 s and 50 s of the motion. Physics Unit 3 2010 Newton‟s laws of motion 5 of 17 An object moves along a straight path. Below is a graph of velocity verses time of the object‟s motion. velocity -1 ms 2 1 2 0 1 2 3 4 5 6 7 8 9 time (s) -1 Example 4 Question 1 (1983) What is the average speed of the object‟s motion during the first 3.0 second? Example 5 Question 2 (1983) What is the object‟s distance from the starting point at 9.0 second? Example 6 Question 3 (1983) What is the acceleration of the object at 8.5 second (magnitude and sign)? Physics Unit 3 2010 Newton‟s laws of motion 6 of 17 Example 7 Which one or more of the following graphs (A, B, C, D, E and F) represents the motion under constant non-zero acceleration? (one or more answers) velocity displacement acceleration A B C time time time velocity displacement acceleration D E F Constant time time time Acceleration Formulae The constant acceleration formulae only apply when the acceleration of the object does not change during its entire motion. The most common v u at example is motion under gravity. The constant acceleration formulae are in v 2 u 2 2ax the box. 1 x ut at 2 x is the displacement u is the initial velocity 2 v is the final velocity a is the acceleration 1 t is the time period in question x vt at 2 2 Note that t is a time interval, not a specific time. (u v)t x When using these formulae to solve problems it is best to write down 2 everything that you know from the question, and then write down the thing that you wish to find and then find a formula that relates what you have to what you need. If you cannot find such a formula directly, determine anything you can, and re-read the question to ensure that you have not missed any vital information. Some other facts to consider are: t = 0 is the beginning of the time interval being considered, i.e. the instant at which 'u' occurs. a negative answer for 't' indicates a time previous to 't' = 0. x is not necessarily the same measure as the total distance travelled a body that is travelling in one direction and accelerating in the opposite direction is slowing down. when given the distance travelled in a certain time interval, this distance is the instantaneous velocity halfway through the time interval. E.g. If a body travels 14 m in the seventh second ('t' = 6 to 't' = 7 sec) then the actual velocity at 6.5 seconds is 14 m/s. for motion along the horizontal it is usual to take 'to the right positive' for vector sense for vertical motion (bodies projected vertically or dropped from rest) the direction of the initial displacement is usually taken as positive for vertical motion, the acceleration (symbolised by 'g') is 10 m/s2 vertically downwards at all times, even if the body is momentarily at the top of its vertical flight. Physics Unit 3 2010 Newton‟s laws of motion 7 of 17 The „standing 400 m‟ time for a car is the time that it takes to travel 400 m on a level road, accelerating from rest. The standing 400 m time of a car was 16.0 s. Example 8 Question 1 (2000) Calculate the acceleration of the car, assuming constant acceleration for the entire journey. A man at the top of a building 20m high releases a stone from rest; 0.60 second later he throws a marble vertically downwards with an initial velocity of 8.0 ms-1. Example 9 Question 1 (1980) How long does it take the stone to reach the ground? Example 10 Question 2 (1980) Which of the following best represents the velocity-time graphs for the stone (S) and the marble (M)? velocity velocity velocity A B C M M M S S S time time time velocity velocity D E M M S S time time ***Example 11 (difficult) Question 3 (1980) How long after the stone was dropped does the marble pass the stone? Physics Unit 3 2010 Newton‟s laws of motion 8 of 17 Graphs of velocity versus time are shown below for a car and a motorcycle travelling along the same road. The car passes the stationary motorcycle at t = 0. Velocity -1 km hr Motor cycle 100 Car 80 60 40 20 Time (sec) 0 10 20 30 40 Example 12 Question 1 (1978) What is the average acceleration (in km hr-1 s-1) of the motorcycle during the first 15 seconds? Example 13 Question 2 (1978) At the instant t = 10 sec, the motorcycle‟s A acceleration increases, velocity decreases B acceleration decreases, velocity increases C acceleration and velocity both increase D acceleration and velocity both decrease ***Example 14 Question 3 (1978) At what time does the motorcycle overtake the car? Forces The relationship between a force and the acceleration it causes was first understood by Isaac Newton(1672 – 1727). Newton summarised all motion by three laws: Types of forces Forces can be divided into two major categories, field forces and contact forces . Forces that act at a distance are called Forces created by travelling bodies are FIELD FORCES, (gravitational, called CONTACT FORCES. electrical or magnetic) Physics Unit 3 2010 Newton‟s laws of motion 9 of 17 Newtons First Law Newtons 1st law of motion An important consequence of this law was the If an object has zero net force acting realisation that an object can be in motion without a on it, it will remain at rest, or continue force being constantly applied to it. When you throw a moving with an unchanged velocity. ball, you exert a force to accelerate the ball, but once it is moving, no force is necessary to keep it moving. Prior to this realisation it was believed that a constant force was necessary, and that this force was supplied by that the air pinching in behind the ball. This model, first conceived by Aristotle, proved tenacious, and students still fall into the trap of using it. Newton‟s first law is commonly tested on the exam. This is achieved by the inclusion of statements such as “An object is moving with a constant velocity” within questions. Whenever you see the key words constant velocity in a question, you should highlight them. The realisation that the object is travelling at a constant velocity, and hence that the net force on the object is zero, will be essential for solving the problem. Newton’s Second Law In words, Newton‟s Second Law states that a force on Newtons 2nd law of motion an object causes the object to accelerate (change its velocity). The amount of acceleration that occurs This law relates to the sum total of the depends on the size of the force and the mass of the forces on the body ( ΣF ) the body's mass object. Large forces cause large accelerations. (m) and the acceleration produced (a) Objects with large mass accelerate less when they experience the same force as a small mass. The ΣF = ma. acceleration of the object is in the same direction as F the net force on the object. a= m Note ΣF must have the same direction as 'a'. Newton’s Third Law Newtons 3rd law of motion This law is the most commonly misunderstood. For every action force acting on one object, You need to appreciate that these action/reaction there is an equal but opposite reaction force forces act on DIFFERENT OBJECTS and so you acting on the other object. do not add them to find a resultant force. For example, consider a book resting on a table top as shown in the diagram below. There are two forces acting on the book: Gravity is pulling the book downward and the tabletop is pushing the book upwards. These forces are the same size, and are in opposite directions but THEY ARE NOT a Newton‟s thirds law pair, because they both act on the same object. N The best way of avoiding making a mistake using Newton‟s third law is to use the following statement. W FA on B = - FB on A In the example of the book on the table the Force Table on Book is a Newton third law pair with the Force Book on Table. Notice the first force is on the book and the second force is on the table. They do Physics Unit 3 2010 Newton‟s laws of motion 10 of 17 not act on the same object. Similarly the weight force, which is the gravitational attraction of the earth on the book, is a Newton third law pair with the gravitational force of the book on the earth. The gravitational effect of the book on the earth is not apparent because the earth is so massive that no acceleration is noticeable. Drawing Force Diagrams You will often be asked to draw diagrams illustrating forces. There are several considerations when drawing force diagrams: The arrows that represent the forces should point in the direction of applied force. The length of the arrow represents the strength of the force, so some effort should be made to draw the arrows to scale. An arrow representing a field force should begin at the centre of the object. An arrow representing a contact force should begin at the point on contact where the force is applied. All forces should be labelled. Some sample force diagrams of common situations are drawn below. Mass on a string Mass in free flight T m mg mg Velocity v = 0, so T = mg F = mg = ma Velocity v = constant upwards, so T = mg Velocity v = constant downwards, so T = mg Accelerating Upwards, T - mg = ma. Acceleration Downwards, mg - T = ma. Mass pulled along a plane Smooth (No Friction) Rough (Friction) N a N a T Fr m T m m m mg mg T = ma, N + mg = 0 T - F = ma, N + mg = 0 Bodies with parallel forces acting a a a Physics Unit 3 2010 Newton‟s laws of motion 11 of 17 F1 F1 m m F2 m F2 m F2 m F1 m F1 + F2 = ma F2 – F1 = ma F1 + F2 = ma Bodies with non-parallel forces acting a a a F1 F2 m m F1 m F1 m F2 m m F2 F1 + F2 = ma F1 + F2 = ma F1 + F2 = ma The vectors need to be resolved in order to solve for the acceleration. Inclined planes Another example of forces acting at angles to each other is an object on an incline plane. There are only three different types of examples of a body on an incline plane without a driving force. A body accelerating The component of the weight force acting down the plane is larger then the frictional forces. (This is also true if there are no frictional forces). For these situations you would take down the plane to be positive, the reason for this is that the acceleration is down the plane. Forces perpendicular to the plane Fnet = mgcos - N = 0 N mg Forces perpendicular to the plane Fnet = mgsin - F = ma Thus the acceleration is down the plane. If there is not friction then F the acceleration is gsin A body travelling at constant speed This can be the when an object is not changing its speed whilst travelling down an incline or when the object is at rest on the incline plane. Forces perpendicular to the plane Fnet = mgcos - N = 0 N Forces parallel to the plane mg Fnet = mgsin - F = 0 Thus the acceleration is zero F Physics Unit 3 2010 Newton‟s laws of motion 12 of 17 A body decelerating For these situations you would choose up the plane to be positive, this is because this is the direction of acceleration. Forces perpendicular to the plane Fnet = mgcos - N = 0 N Forces parallel to the plane mg Fnet = F - mgsin = ma Thus the acceleration is up F the plane. A recent Transport Accident Commission television advertisement explains the significant difference between car stopping distances when travelling at 30 kmh-1 and 60 kmh-1. Example 15 2000 Question 10 The stopping distance, from when the brakes are applied, for a car travelling at 30 kmh-1 is 10 m. Which one (A – D) is the best estimate of the stopping distance for the same car, under the same braking, but travelling at 60 kmh-1? A. 20 m B. 30 m C. 40 m D. 90 m Reproduced by permission of the Victorian Curriculum and Assessment Authority, Victoria, Australia. A car of mass 1300 kg has a caravan of mass 900 kg attached to it. The car and caravan move off from rest. They have an initial acceleration of 1.25 m s-2. Example 16 2000 Question 11 What is the net force acting on the total system of car and caravan as it moves off from rest? Example 17 2000 Question 12 What is the tension in the coupling between the car and the caravan as they start to accelerate? Physics Unit 3 2010 Newton‟s laws of motion 13 of 17 After some time the car reaches a speed of 100 kmh-1, and the driver adjusts the engine power to maintain this constant speed. At this speed, the total retarding force on the car is 1300 N, and on the caravan 1100 N. Example 18 2000 Question 13 What driving force is being exerted by the car at this speed? Reproduced by permission of the Victorian Curriculum and Assessment Authority, Victoria, Australia. Anna is jumping on a trampoline. The figure below shows Anna at successive stages of her downward motion. a b c d Figure c shows Anna at a time when she is travelling downwards and slowing down. Example 19 1999 Question 6 What is the direction of Anna‟s acceleration at the time shown in Figure 4c? Explain your answer. Example 20 1999 Question 7 On Figure c draw arrows that show the two individual forces acting on Anna at this instant. Label each arrow with the name of the force and indicate the relative magnitudes of the forces by the lengths of the arrows you draw. Reproduced by permission of the Victorian Curriculum and Assessment Authority, Victoria, Australia. Physics Unit 3 2010 Newton‟s laws of motion 14 of 17 Solutions Example 1 53 m/s To convert from km/hr to m/s you need to divide by 3.6. (This should be on your cheat sheet) 190 3.6 = 52.8 m/s Example 2 vel (km/hr) 200 100 0 10 20 30 40 50 60 to get full marks your graph had to show: that the terminal velocity was reached after 15 sec the velocity increased from 0 to 190 km/hr in the first 15 secs that there was a smooth transition from acceleration to a terminal velocity where the acceleration was zero. Examiner’s comment Example 2 The 3 available marks were allocated as follows: • 1 mark for the terminal velocity section after 15 s. • 1 mark for a velocity increase from zero to 190 km h-1 between 0–15 s. • 1 mark for a graph that shows a smooth transition from increasing velocity to terminal velocity. The average mark for this question was 2.1/3, with the main error being a failure to recognise the smooth transition, resulting in a graph as shown below. Such an answer scored only 2 marks. Air resistance Example 3 Constant speed (velocity) implies a net force of zero. At terminal velocity, the air resistance (upwards) is equal in magnitude but opposite in direction to the weight force (downwards). It is always a good idea to include a diagram in this type of answer. The weight force must come from the centre of mass. The length of the vectors must be the same. Weight = mg Physics Unit 3 2010 Newton‟s laws of motion 15 of 17 Examiner’s comment Example 3 The required explanation needed to cover the points: • Constant speed (velocity) implies a net force of zero. • At terminal velocity, the air resistance force (upwards) is equal in magnitude, but opposite in direction, to the weight force (downwards). Students who addressed this by using force diagrams were generally successful. However, a number of written explanations simply gave the meaning of terminal velocity, rather than addressing the physics of why the velocity remains constant. 32% of students were able to score the full 3 marks, which is relatively low, considering the fairly straightforward nature of this question. Example 4 distance travelled The average speed is always given by . time taken 1 0.5 2 2 1.5 1 1 2 area under the graph 4.5 In this case it is = 2 2 = = 1.5 ms-1. time taken 3 3 Example 5 The distance from the starting point is called the displacement. Therefore this answer requires you to consider vectors, so the direction of travel is important. In the first 3 seconds the object travels 4.5 m in one direction. From 3.5 (sec) to 9.0 (sec) it travels 2 1.0 1 1 3.5 2 1 1 = 4.5 m in the 1 1 opposite direction. after 9 (sec) the net displacement is zero. Example 6 The acceleration is given by the gradient of the velocity – time graph. In this case the gradient is 1 positive, and its value is = 1. +1 ms-2 1 Example 7 Constant non-zero acceleration means that the velocity – time graph is an oblique straight line. A and D Example 8 The distance travelled was 400m, and it took 16.0 seconds. The initial speed was 0 m/s. You need to find the acceleration. This is given by substituting into the equation x = ut + ½at2. 400 = 0 16 + ½ a 162 400 400 = 128 a a= a = 3.125 ms-2 128 This is probably best written as 3.13 ms-2 (correct to 3 sig figs) Example 9 The time it takes for the stone to reach the ground is given by x = ut + ½at2 In this case u = 0, a = 10 and x = 20. 20 = ½ 10 t2 20 = 5 t2 4 = t2 t = 2. The stone take 2 seconds to reach the ground. Physics Unit 3 2010 Newton‟s laws of motion 16 of 17 Example 10 Since both the stone and the marble will both have the same acceleration, the gradient of the graphs must be the same. Either A, C, D, E The marble starts after the stone, so it must be either C or E. The marble starts at v = 8.0 ms-1, it must be graph C ***Example 11 For the marble to pass the stone, it must catch up with the stone. You need to equate two equations of motion and solve for time (t). If you let the marble start at time t = 0, then the time for the stone needs to be t – 0.6. This is because it will have travelled for less time than the marble. Stone x = ut + ½at2 becomes x = 0 t + ½ 10 t2 x = 5t2 Marble x = ut + ½at becomes x = 8 (t – 0.6) + ½ 10 (t – 0.6)2 2 x = 8(t – 0.6) + 5(t – 0.6)2 For the marble to pass the stone both equations must be equal. 5t2 = 8(t – 0.6) + 5(t – 0.6)2 5t2 = 8t – 4.8 + 5(t2 – 1.2t + 0.36) 5t2 = 8t – 4.8 + 5t2 – 6t + 1.8 0 = 2t – 3 2t = 3 t = 1.5 secs. Example 12 You must make sure that you read this question very carefully, because it is actually very easy. The units for the answer are very unusual. change in velocity 100 The average acceleration is given by = = 6.7 kmhr-1s-1 time taken 15 Example 13 B At t = 10s, the gradient (acceleration) decreases but the velocity continues to increase. ***Example 14 The distance travelled by the two vehicles needs to be the same. The distance travelled is the area under the graphs. For the motorcycle the distance travelled in the first 15 seconds = 2 10 80 2 (80 100) 5 = 1 1 850m. During the first 15 secs, the car travels 80 15 = 1200m. So at time t = 15, the car is 350m in front of the motorcycle. After t = 15, the motorcycle catches up at a rate of 100m every 5 seconds. A further 17.5 seconds to make up the 350m distance. The motorcycle will catch the car at 15 + 17.5 = 32.5 seconds. Example 15 C This question was testing you understanding of how speed and stopping distance is related. Let‟s assume the mass of the car, and the force stopping it, are constant. That is, if the car has a kinetic energy of 2 mv 2 and it requires a force by a distance to stop it then: 1 1 2 mv 2 = F d if the speed of the car is doubled then the stopping distance in 4 times larger. This is known as a square relationship. For this question the speed of the car is 30kmh-1 and the stopping distance is 10m. The speed of the car is then doubled to 60kmh-1 then because the speed has doubled the stopping distance is now 4 times the 10m, which is 40m so C is the correct answer. Physics Unit 3 2010 Newton‟s laws of motion 17 of 17 Example 16 Just Fnet = ma Using the mass of the system. Fnet = (1300 + 900) 1.25 = 2.75 103N Example 17 For this question think of the caravan as an object that is accelerating that is pulling it. So F = ma = 900 1.25 = 1.13 103N Example 18 If the car is travelling at a constant velocity then there is no net force acting on the car. Therefore the magnitude of the driving force from the motor equals the retarding force of the car and the caravan. Fd = 1300 + 1100 = 2400N Example 19 Up At this time Anna is travelling downwards and slowing down. This means that her acceleration is up, because it is opposing the motion (she is slowing down). Example 20 Reaction force from the trampoline. Weight = mg The reaction force > weight, because she is slowing down, ie. a net upward force. Examiner’s comment Example 20 Students were expected to draw two force arrows on Figure c. One arrow was the weight force, acting through Anna‟s centre of mass and the other arrow being the normal contact force, acting at her feet. The arrow for the normal contact force should have been longer than the weight force arrow. The average mark for this question was 1.5/3, with the most common errors being in either choosing the incorrect point of application for each force or in not indicating the relative magnitudes as specifically asked in the question.