IB Physics DP - Mechanics Notes PDF

Head to savemyexams.co.uk for more awesome resources YOUR NOTES IB Physics DP  2. Mechanics CO...

Head to savemyexams.co.uk for more awesome resources YOUR NOTES IB Physics DP  2. Mechanics CONTENTS 2.1 Motion 2.1.1 Distance & Displacement 2.1.2 Speed & Velocity 2.1.3 Acceleration 2.1.4 Graphs Describing Motion 2.1.5 Uniform Acceleration 2.1.6 Acceleration of Free Fall Experiment 2.1.7 Projectile Motion 2.1.8 Terminal Speed 2.2 Forces 2.2.1 Free-Body Diagrams 2.2.2 Newton’s First Law 2.2.3 Newton’s Second Law 2.2.4 Newton’s Third Law 2.2.5 Applying Newton’s Laws of Motion 2.2.6 Friction 2.3 Work, Energy & Power 2.3.1 Kinetic Energy 2.3.2 Gravitational Potential Energy 2.3.3 Elastic Potential Energy 2.3.4 Work Done 2.3.5 Power 2.3.6 Principle of Conservation of Energy 2.3.7 Efficiency 2.4 Momentum & Impulse 2.4.1 Force & Momentum 2.4.2 Impulse 2.4.3 Conservation of Linear Momentum 2.4.4 Collisions & Explosions Page 1 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources 2.1 Motion YOUR NOTES  2.1.1 Distance & Displacement Distance & Displacement Distance is a measure of how far an object travels It is a scalar quantity - in other words, the direction is not important The athletes run a total distance of 300 m Consider a 300 m race From start to finish, the distance travelled by the athletes is 300 m Displacement Displacement is a measure of how far something is from its starting position, along with its direction It is a vector quantity - it describes both magnitude and direction The athletes run a total distance of 300 m, but end up 100 m from where they started Page 2 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources Consider the same 300 m race again YOUR NOTES The athletes have still run a total distance of 300 m (this is indicated by the arrow in  red) However, their displacement at the end of the race is 100 m to the right (this is indicated by the arrow in green) If they had run the full 400 m, their final displacement would be zero Page 3 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources 2.1.2 Speed & Velocity YOUR NOTES  Speed & Velocity Speed The speed of an object is the distance it travels every second Speed is a scalar quantity This is because it only contains a magnitude (without a direction) The average speed of an object is given by the equation: total distance average speed = time taken The SI units for speed are meters per second, m/s, but speed can often be measured in alternative units when it is more appropriate for the situation  Worked Example Florence Griffith Joyner set the women’s 100 m world record in 1988, with a time of 10.49 s. Calculate her average speed during the race. Step 1: List the known quantities Distance, s = 100 m Time, t = 10.49 s Step 2: Write the relevant equation Sprinters typically speed up out of the blocks up to some maximum speed Because Florence’s speed changes over the race, we can calculate her average speed using the equation: average speed = total distance ÷ time taken Step 3: Check any unit conversions Check that all quantities given in the question are in standard units In this example, they are all in standard units Step 4: Substitute the values for total distance and time Average speed = 100 ÷ 10.49 = 9.53288... = 9.53 m/s Velocity The velocity of a moving object is similar to its speed, except it also describes the direction of the motion The speed of an object only contains a magnitude - it’s a scalar quantity The velocity of an object contains both magnitude and direction, e.g. ‘15 m/s south’ or ‘250 km/h on a bearing of 030°’ Page 4 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources Velocity is, therefore, a vector quantity because it describes both magnitude and direction YOUR NOTES  The cars in the diagram above have the same speed (a scalar quantity) but different velocities (a vector quantity). Fear not, they are in different lanes! Instantaneous Speed / Velocity The instantaneous speed (or velocity) is the speed (or velocity) of an object at any given point in time This could be for an object moving at a constant velocity or accelerating An object accelerating is shown by a curved line on a displacement – time graph An accelerating object will have a changing velocity To find the instantaneous velocity on a displacement-time graph: Draw a tangent at the required time Calculate the gradient of that tangent The instantaneous velocity is found by drawing a tangent on the displacement time graph Average Speed / Velocity The average speed (or velocity) is the total distance (or displacement) divided by the total time Page 5 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources To find the average velocity on a displacement-time graph, divide the total displacement YOUR NOTES (on the y-axis) by the total time (on the x-axis)  This method can be used for both a curved or a straight line on a displacement-time graph  Exam Tip When you draw a tangent to a curve, make sure it just touches the point at which you wish to calculate the gradient. The angle between the curve and the tangent line should be roughly equal on both sides of the point. Page 6 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources 2.1.3 Acceleration YOUR NOTES  Acceleration Acceleration is defined as the rate of change in velocity In other words, it describes how much an object's velocity changes every second The equation below is used to calculate the average acceleration of an object: change in velocity acceleration = time taken ∆v a= t Where: a = acceleration in metres per second squared (m s–2) Δv = change in velocity in metres per second (m s–1) t = time taken in seconds (s) The change in velocity is the difference between the initial and final velocity, as written below: change in velocity = final velocity − initial velocity Page 7 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources Equations linking displacement, velocity, and acceleration YOUR NOTES  Speeding Up and Slowing Down An object that speeds up is accelerating An object that slows down is decelerating The acceleration of an object can be positive or negative, depending on whether the object is speeding up or slowing down If an object is speeding up, its acceleration is positive If an object is slowing down, its acceleration is negative (deceleration) Page 8 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources YOUR NOTES  A rocket speeding up (accelerating) and a car slowing down (decelerating)  Worked Example A Japanese bullet train decelerates at a constant rate in a straight line. The velocity of the train decreases from 50 m s–1 to 42 m s–1 in 30 seconds. (a) Calculate the change in velocity of the train. (b) Calculate the deceleration of the train, and explain how your answer shows the train is slowing down. Part (a) Step 1: List the known quantities Initial velocity = 50 m s–1 Final velocity = 42 m s–1 Step 2: Write the relevant equation change in velocity = final velocity − initial velocity Step 3: Substitute values for final and initial velocity change in velocity = 42 − 50 = −8 m s–1 Part (b) Page 9 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources Step 1: List the known quantities YOUR NOTES Change in velocity, Δv = −8 m s–1  Time taken, t = 30 s Step 2: Write the relevant equation ∆v a= t Step 3: Substitute the values for change in velocity and time −8 a= = − 0. 27 m s −1 30 Step 4: Interpret the value for deceleration The answer is negative, which indicates the train is slowing down  Exam Tip Remember the units for acceleration are metres per second squared, m s–2. In other words, acceleration measures how much the velocity (in m s–1) changes every second, m s–1 s–1. Page 10 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources 2.1.4 Graphs Describing Motion YOUR NOTES  Motion Graphs Three types of graphs that can represent motion are Displacement-time graphs Velocity-time graphs Acceleration-time graphs Displacement-Time Graph On a displacement-time graph… Slope equals velocity The y-intercept equals the initial displacement A straight (diagonal) line represents a constant velocity A curved line represents an acceleration A positive slope represents motion in the positive direction A negative slope represents motion in the negative direction A zero slope (horizontal line) represents a state of rest The area under the curve is meaningless Displacement-time graphs displacing difference velocities Velocity-Time Graph On a velocity-time graph… Slope equals acceleration The y-intercept equals the initial velocity A straight (diagonal) line represents uniform acceleration A curved line represents non-uniform acceleration A positive slope represents acceleration in the positive direction A negative slope represents acceleration in the negative direction Page 11 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources A zero slope (horizontal line) represents motion with constant velocity YOUR NOTES The area under the curve equals the change in displacement  Velocity-time graphs displacing different acceleration Acceleration-Time Graph On an acceleration-time graph… Slope is meaningless The y-intercept equals the initial acceleration A zero slope (horizontal line) represents an object undergoing constant acceleration The area under the curve equals the change in velocity How displacement, velocity and acceleration graphs relate to each other Page 12 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources YOUR NOTES  Worked Example  Tora is training for a cycling tournament. The velocity-time graph below shows her motion as she cycles along a flat, straight road. (a) In which section (A, B, C, D, or E) of the velocity-time graph is Tora’s acceleration the largest? (b) Calculate Tora’s acceleration between 5 and 10 seconds. Part (a) Step 1: Recall that the slope of a velocity-time graph represents the magnitude of acceleration The slope of a velocity-time graph indicates the magnitude of acceleration Therefore, the only sections of the graph where Tora is accelerating is section B and section D Sections A, C, and E are flat – in other words, Tora is moving at a constant velocity (i.e. not accelerating) Step 2: Identify the section with the steepest slope Section D of the graph has the steepest slope Hence, the largest acceleration is shown in section D Page 13 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources Part (b) YOUR NOTES Step 1: Recall that the gradient of a velocity-time graph gives the acceleration  Calculating the gradient of a slope on a velocity-time graph gives the acceleration for that time period Step 2: Draw a large gradient triangle at the appropriate section of the graph A gradient triangle is drawn for the time period between 5 and 10 seconds below: Step 3: Calculate the size of the gradient and state this as the acceleration The acceleration is given by the gradient, which can be calculated using: 5 acceleration = gradient = = 1 m s −2 5 Therefore, Tora accelerated at 1 m s−2 between 5 and 10 seconds Motion of a Bouncing Ball For a bouncing ball, the acceleration due to gravity is always in the same direction (in a uniform gravitational field such as the Earth's surface) This is assuming there are no other forces on the ball, such as air resistance Since the ball changes its direction when it reaches its highest and lowest point, the direction of the velocity will change at these points The vector nature of velocity means the ball will sometimes have a: Positive velocity if it is traveling in the positive direction Negative velocity if it is traveling in the negative direction Page 14 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources An example could be a ball bouncing from the ground back upwards and back down again YOUR NOTES The positive direction is taken as upwards  This will be either stated in the question or can be chosen, as long as the direction is consistent throughout Ignoring the effect of air resistance, the ball will reach the same height every time before bouncing from the ground again When the ball is traveling upwards, it has a positive velocity which slowly decreases (decelerates) until it reaches its highest point Page 15 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources YOUR NOTES  At point A (the highest point): The ball is at its maximum displacement The ball momentarily has zero velocity The velocity changes from positive to negative as the ball changes direction The acceleration, g, is still constant and directed vertically downwards Page 16 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources At point B (the lowest point): YOUR NOTES The ball is at its minimum displacement (on the ground)  Its velocity changes instantaneously from negative to positive, but its speed (magnitude) remains the same The change in direction causes a momentary acceleration (since acceleration = change in velocity / time)  Worked Example The velocity-time graph of a vehicle travelling with uniform acceleration is shown in the diagram below. Calculate the displacement of the vehicle at 40 s. Page 17 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources YOUR NOTES  Page 18 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources 2.1.5 Uniform Acceleration YOUR NOTES  Equations of Motion for Uniform Acceleration Deriving Kinematic Equations of Motion The kinematic equations of motion are a set of four equations that can describe any object moving with constant acceleration They relate the five variables: s = displacement u = initial velocity v = final velocity a = acceleration t = time interval Knowing where these equations come from and how they are derived helps to understand them: Page 19 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources YOUR NOTES  Derivation of v = u + at Page 20 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources YOUR NOTES  The average velocity is halfway between u and v Page 21 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources YOUR NOTES  The two terms ut and ½at2 make up the area under the graph Page 22 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources YOUR NOTES  This final equation can be derived from two of the others Summary of the four equations of uniformly accelerated motion Key Takeaways These are all given in the IB DP Physics data booklet The key terms to look out for are: 'Starts from rest', This means u = 0 and t = 0 This can also be assumed if the initial velocity is not mentioned 'Falling due to gravity' This means a = g = 9.81 m s–2 It doesn't matter which way is positive or negative for the scenario, as long as it is consistent for all the vector quantities Page 23 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources For example, if downwards is negative then for a ball travelling upwards, s must be positive YOUR NOTES and a must be negative  'Constant acceleration in a straight line' This is a key indication SUVAT equations are intended to be used For example, an object falling in a uniform gravitational field without air resistance How to use the SUVAT equations Step 1: Write out the variables that are given in the question, both known and unknown, and use the context of the question to deduce any quantities that aren’t explicitly given e.g. for vertical motion a = ± 9.81 m s–2, an object which starts or finishes at rest will have u = 0 or v = 0 Step 2: Choose the equation which contains the quantities you have listed e.g. the equation that links s, u, a and t is s = ut + ½at2 Step 3: Convert any units to SI units and then insert the quantities into the equation and rearrange algebraically to determine the answer  Worked Example The diagram shows an arrangement to stop trains that are travelling too fast. At marker 1, the driver must apply the brakes so that the train decelerates uniformly in order to pass marker 2 at no more than 10 m s–1. The train carries a detector that notes the times when the train passes each marker and will apply an emergency brake if the time between passing marker 1 and marker 2 is less than 20 s. Trains coming from the left travel at a speed of 50 m s–1. Determine how far marker 1 should be placed from marker 2. Page 24 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources YOUR NOTES   Worked Example A cyclist is travelling directly east through a village, which is completely flat, at a velocity of 6 m s–1 east. They then start to constantly accelerate at 2 m s–2 for 4 seconds. a) Calculate the distance that the cyclist covers in the 4 second acceleration period. b) Calculate the cyclist's final velocity after the 4-second interval of acceleration. Later on in their journey, this cyclist (cyclist A) is now cycling through a different village, still heading east at a constant velocity of 18 m s–1. Cyclist A passes a friend (cyclist B) who begins accelerating from rest at a constant acceleration of 1.5 m s–2 in the same direction as cyclist A at the moment they pass. c) Calculate how long it takes for cyclist B to catch up to cyclist A. Part (a) Step 1: List the known quantities Initial velocity, u = 6 m s–1 East Acceleration, a = 2 m s–2 East Time, t = 4 s Displacement, s = ? (this needs to be calculated) Step 2: Identify and write down the equation to use Page 25 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources Since the question states constant acceleration, SUVAT equations can be used YOUR NOTES In this problem, the equation that links s, u, a, and t is  s = (u × t) + (½ × a × t2) Step 3: Substitute known quantities into the equation and simplify where possible s = (6 × 4) + (0.5 × 2 × 42) This can be simplified to: s = 24 + 16 = 40 m Part (b) Step 1: List the known quantities Initial velocity, u = 6 m s–1 East Acceleration, a = 2 m s–2 East Time, t = 4 s final velocity, v = ? (this needs to be calculated) Step 2: Identify and write down the equation to use Since the question states constant acceleration - SUVAT equation(s) - can be used In this problem, the equation that links v, u, a, and t is v = u + (a × t) Step 3: Substitute known quantities into the equation and simplify where possible v = 6 + (2 × 4) This can be simplified to: v = 14 m s–1 Part (c) Step 1: List the known quantities for cyclist A Initial velocity, u = 18 m s–1 East Acceleration, a = 0 m s–2 East Final velocity, v = 18 m s–1 East Time, t = ? Displacement, s = ? Step 2: List the known quantities for cyclist B Initial velocity, u = 0 m s–1 East Acceleration, a = 1.5 m s–2 East Final velocity, v = ? Time, t = ? Displacement, s = ? Step 3: Express the situation for cyclist A and B in terms of displacement, s Page 26 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources Cyclist A can have their situation expressed by: YOUR NOTES sA = (u × t) + (½ × a × t2)  sA = (18 × t) + (½ × 0 × t2) = (18 × t) Cyclist B can have their situation expressed by: sB = (u × t) + (½ × a × t2) sB = (0 × t) + (½ × 1.5 × t2) = (0.75 × t2) Step 4: Equate the two equations and solve for t The two equations describe the displacement of each cyclist respectively When equating them, this will find the time when the cyclists are at the same location sA= sB = (18 × t) = (0.75 × t2) 0 = (0.75 × t2) – (18 × t) 0 = (0.75 × t – 18) × t Therefore, solving for t, it can be two possible answers: t = 0 s or 18 / 0.75 = 24 s Step 4: State the final answer Since the question is seeking the time when the two cyclists meet after first passing each other, the final answer is 24 s  Exam Tip This is one of the most important sections of this topic - usually, there will be one, or more, questions in the exam about solving problems with SUVAT equations. The best way to master this section is to practice as many questions as possible! Page 27 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources 2.1.6 Acceleration of Free Fall Experiment YOUR NOTES  Acceleration of Free Fall Experiment The acceleration of free fall, g, is defined as: The acceleration of any object in response to the gravitational attraction between the Earth and the object Any object released on the Earth will accelerate downwards to the centre of the Earth as long as there are no external forces acting on it On Earth, the acceleration of free fall is equal to g = 9.81 m s–2 Determining g in the Laboratory Aims of the Experiment The overall aim of the experiment is to calculate the value of the acceleration due to gravity, g This is done by measuring the time it takes for a ball-bearing to fall a certain distance The acceleration can then be calculated using an equation of motion Variables Independent variable = height, h Dependent variable = time, t Control variables: Same steel ball–bearing Same electromagnet Distance between ball-bearing and top of the glass tube Equipment List Page 28 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources YOUR NOTES  Resolution of measuring equipment: Metre ruler = 1 mm Timer = 0.01 s Method Page 29 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources YOUR NOTES  Apparatus setup to measure the distance and time for the ball bearing to drop This method is an example of the procedure for varying the height the ball-bearing falls and determining the time taken – this is just one possible relationship that can be tested 1. Set up the apparatus by attaching the electromagnet to the top of a tall clamp stand. Do not switch on the current till everything is set up 2. Place the glass tube directly underneath the electromagnet, leaving space for the ball- bearing. Make sure it faces directly downwards and not at an angle 3. Attach both light gates around the glass tube at a starting distance of around 10 cm 4. Measure this distance between the two light gates as the height, h with a metre ruler 5. Place the cushion directly underneath the end of the glass tube to catch the ball-bearing when it falls through 6. Switch on the current to the electromagnet and place the ball-bearing directly underneath so it is attracted to it 7. Turn the current to the electromagnet off. The ball should drop 8. When the ball drops through the first light gate, the timer starts 9. When the ball drops through the second light gate, the timer stops 10. Read the time on the timer and record this as time, t 11. Increase h (eg. by 5 cm) and repeat the experiment. At least 5 – 10 values for h should be used 12. Repeat this method at least 3 times for each value of h and calculate an average t for each An example of a table with some possible heights would look like this: Example Table of Results Page 30 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources YOUR NOTES  Analysis of Results The acceleration is found by using one of the SUVAT equations The known quantities are Displacement s = h Time taken = t Initial velocity u = u Acceleration a = g The following SUVAT equation can be rearranged: Substituting in the values and rearranging to fit the straight line equation gives: Comparing this to the equation of a straight line: y = mx + c y = 2h/t (m s–1) x=t Gradient, m = a = g (m s–2) y-intercept = 2u 1. Plot a graph of the 2h/t against t 2. Draw a line of best fit 3. Calculate the gradient - this is the acceleration due to gravity g Page 31 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources 4. Assess the uncertainties in the measurements of h and t. Carry out any calculations needed YOUR NOTES to determine the uncertainty in g due to these  The graph of 2h/t against t produces a straight-line graph where the acceleration is the gradient Evaluating the Experiment Systematic Errors: Residue magnetism after the electromagnet is switched off may cause t to be recorded as longer than it should be Random Errors: Large uncertainty in h from using a metre rule with a precision of 1 mm Parallax error from reading h The ball may not fall accurately down the centre of each light gate Random errors are reduced through repeating the experiment for each value of h at least 3- 5 times and finding an average time, t Safety Considerations The electromagnet requires current Care must be taken to not have any liquids near electrical equipment To reduce the risk of electrocution, only switch on the current to the electromagnet once everything is set up A cushion or a soft surface must be used to catch the ball-bearing so it doesn’t roll off / damage the surface The tall clamp stand needs to be attached to a surface with a G clamp so it stays rigid Page 32 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources YOUR NOTES  Worked Example  A student investigates the relationship between the height that a ball-bearing is dropped between two light gates and the time taken for it to drop. Calculate the value of g from the table. Step 1: Complete the table Calculate the average time for each height Add an extra column 2h / t Step 2: Draw graph of 2h/t against time t Page 33 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources YOUR NOTES  Make sure the axes are properly labeled and the line of best fit is drawn with a ruler Step 3: Calculate the gradient of the graph Page 34 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources The gradient is calculated by: YOUR NOTES  Page 35 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources 2.1.7 Projectile Motion YOUR NOTES  Projectile Motion The trajectory of an object undergoing projectile motion consists of a vertical component and a horizontal component These need to be evaluated separately Some key terms to know, and how to calculate them, are: Time of flight: how long the projectile is in the air Maximum height attained: the height at which the projectile is momentarily at rest Range: the horizontal distance travelled by the projectile Page 36 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources YOUR NOTES  How to find the time of flight, maximum height and range Problems involving projectile motion There are two main considerations for solving problems involving two-dimensional motion of a projectile Constant velocity in the horizontal direction Constant acceleration in a perpendicular direction The only force acting on the projectile, after it has been released, is the gravitational pull of the Earth on the object, or weight There are three possible scenarios for projectile motion: Vertical projection Horizontal projection Projection at an angle Page 37 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources YOUR NOTES  Worked Example  To calculate vertical projection (free fall) A science museum designed an experiment to show the fall of a feather in a vertical glass vacuum tube. The time of fall from rest is 0.5 s. What is the length of the tube, L? Page 38 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources YOUR NOTES  Worked Example  To calculate horizontal projection A motorcycle stunt-rider moving horizontally takes off from a point 1.25 m above the ground, landing 10 m away as shown. What was the speed at take-off? (ignoring air resistance) Page 39 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources YOUR NOTES  Worked Example  To calculate projection at an angle A ball is thrown from a point P with an initial velocity u of 12 m s-1 at 50° to the horizontal. What is the value of the maximum height at Q? (ignoring air resistance)  Exam Tip Make sure you don’t make these common mistakes: Forgetting that deceleration is negative as the object rises Confusing the direction of sin θ and cos θ Not converting units (mm, cm, km etc.) to metres Further, it is worth noting that projectile motion is typically symmetrical when air resistance is ignored allowing for use of the peak to find the time of total flight or total horizontal distance by doubling the amount to get from the start point to the peak. Page 40 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources 2.1.8 Terminal Speed YOUR NOTES  Fluid Resistance & Terminal Speed Fluid Resistance When an object moves through a fluid (a gas or a liquid) then there are resistive forces for that movement. These forces are known as drag forces Examples of drag forces are friction and air resistance Drag forces: Are always in the opposite direction to the motion of the object Never speed an object up or start them moving Slow down an object or keep them moving at a constant speed Transfer energy from the kinetic store of the object to the thermal stores of the objects and the surroundings Lift is an upwards force on an object moving through a fluid. It is perpendicular to the fluid flow For example, as an aeroplane moves through the air, the aeroplane pushes down on the air to change its direction This causes an equal and opposite reaction as the air pushes upwards on the wings of the aeroplane (lift) due to Newton's third law Drag forces are always in the opposite direction to the thrust (direction of motion). Lift is always in the opposite direction to the weight A key component of drag forces is that they increase with the speed of the object This is shown in the diagram below: Page 41 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources YOUR NOTES  Frictional forces on a car increase with speed Air Resistance & Projectile Motion Air resistance decreases the horizontal component of the velocity of a projectile This means both its range and maximum height is decreased compared to no air resistance Page 42 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources YOUR NOTES  A projectile with air resistance travels a smaller distance and has a lower maximum height than one without air resistance The angle and speed of release of a projectile is varied to produce either a longer flight path or cover a larger distance, depending on the situation For sports such as the long jump or javelin, an optimum angle against air resistance is used to produce the greatest distance For gymnastics or a ski jumper, the initial vertical velocity is made as large as possible to reach a greater height and longer flight path Terminal Velocity For a body in free fall, the only force acting is weight, and its acceleration g is only due to gravity. The drag force increases as the body accelerates This increase in velocity means the drag force also increases Due to Newton’s Second Law, this means the resultant force and therefore acceleration decreases (recall F = ma) When the drag force is equal to the gravitational pull on the body, the body will no longer accelerate and will fall at a constant velocity This velocity is called the terminal velocity Terminal velocity can occur for objects falling through a gas or a liquid Page 43 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources YOUR NOTES  Page 44 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources A skydiver in freefall reaching terminal velocity YOUR NOTES The graph shows how the velocity of the skydiver varies with time  Since the acceleration is equal to the gradient of a velocity-time graph, the acceleration decreases and eventually becomes zero when terminal velocity is reached After the skydiver deploys their parachute, they decelerate to a lower terminal velocity to reduce the impact on landing This is demonstrated by the graph below: A graph showing the changes in speed of the skydiver throughout their entire journey in freefall Page 45 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources YOUR NOTES  Worked Example  Skydivers jump out of a plane at intervals of a few seconds. Skydivers A and B want to join up as they fall. If A is heavier than B, who should jump first? Skydiver B should jump first since he will take longer to reach terminal velocity This is because skydiver A has a higher mass, and hence, weight More weight means a greater speed, therefore, A will reach terminal velocity faster than B  Exam Tip A common misconception is that skydivers move upwards when their parachutes are deployed - however, this is not the case, they are in fact decelerating to a lower terminal velocity. If a question considers air resistance to be ‘negligible’ this means in that question, air resistance is taken to be so small it will not make a difference to the motion of the body. You can take this to mean there are no drag forces acting on the body. Page 46 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources 2.2 Forces YOUR NOTES  2.2.1 Free-Body Diagrams Vectors In physics, during force interactions, it is common to represent situations as simply as possible without losing information When considering force interactions objects can be represented as point particles These point particles should be placed at the center of mass of the object Force vectors that act upon that objects should be drawn with their tail on that point particle The length of the force vector corresponds to its strength The longer the vector, the greater the force magnitude The below example shows the forces acting on an object when pushed to the right over a rough surface Point particle representation of the forces acting on a moving object The below example shows an object sitting on a slope in equilibrium Page 47 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources YOUR NOTES  Three forces on an object in equilibrium form a closed vector triangle The below example shows the forces acting on an object suspended from a stationary rope Free-body diagram of an object suspended from a stationary rope Page 48 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources Exam Tip YOUR NOTES  When labeling force vectors, it is important to use conventional and appropriate  naming or symbols such as: w or Weight force or mg N or R for normal reaction force (depending on your local context either of these could be acceptable) Using unexpected notation can lead to losing marks so try to be consistent with expected conventions. Page 49 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources Free-Body Diagrams YOUR NOTES Free body diagrams are useful for modeling the forces that are acting on an object  Each force is represented as a vector arrow, where each arrow: Is scaled to the magnitude of the force it represents Points in the direction that the force acts Is labelled with the name of the force it represents or an appropriate symbol Free body diagrams can be used: To identify which forces act in which plane To resolve the net force in a particular direction The rules for drawing a free-body diagram are the following: Rule 1: Draw a point in the centre of mass of the body Rule 2: Draw the body free from contact with any other object Rule 3: Draw the forces acting on that body using vectors with correct direction and proportional length Free body diagrams can be used to show the various forces acting on objects You must be able to apply the following forces with their symbols to free-body diagrams: Weight (W) Tension (T) Normal Reaction Force (N) Upthrust (U) Frictional Forces (Fr) Page 50 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources YOUR NOTES  Worked Example  Draw free-body diagrams for the following scenarios: a) A picture frame hanging from a nail b) A box sliding down a slope Part (a) A picture frame hanging from a nail: The size of the arrows should be such that the 3 forces would make a closed triangle as they are balanced Part (b) A box sliding down a slope: Page 51 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources There are three forces acting on the box: YOUR NOTES The normal contact force, R, acts perpendicular to the slope  Friction, F, acts parallel to the slope and in the opposite direction to the direction of motion Weight, W, acts down towards the Earth  Worked Example Draw a free-body diagram of a toy sailboat with weight 30 N floating in water that is being pulled to the right by an applied force of 35 N with a total resistive force of 5 N. Step 1: Draw the object in a simplified diagram Step 2: Identify all of the forces acting upon the object in the question, including any forces that may be implied Weight: 30 N down Upthrust from the water (since the object is floating): 30 N up Applied force: 35 N to the right Resistive force: 5 N to the left Step 3: Draw in all of the force vectors (arrows), making sure the arrows start at the object and are directed away Page 52 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources An approximation can be made as to the final resultant force due to all of the forces YOUR NOTES Decide whether the resultant force is approximately up or down  Decide whether the resultant force is approximately left or right For example, the resultant force is directed up and to the right Page 53 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources 2.2.2 Newton’s First Law YOUR NOTES  Newton’s First Law Newton's First Law states: A body will remain at rest or move with constant velocity unless acted on by a resultant force If the net force acting on an object is zero it is said to be in translational equilibrium If the forces on a body are balanced (the resultant force is 0), the body must be either: At rest Moving at a constant velocity Since force is a vector, it is easier to split the forces into horizontal and vertical forces If the forces are balanced: The forces to the left = the forces to the right The forces up = the forces down The resultant force is the single force obtained by combining all the forces on the body  Worked Example If there are no external forces acting on the car, other than friction, and it is moving at a constant velocity, what is the value of the frictional force F? Page 54 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources 2.2.3 Newton’s Second Law YOUR NOTES  Newton's Second Law Newton's Second Law states: The resultant force is equal to the rate of change in momentum. The change in momentum is in the same direction as the resultant force This can also be written as: This relationship means that objects will accelerate if there is a resultant force acting upon them This is derived from the definition of momentum as follows: An unbalanced force on a body means it experiences a resultant force If the resultant force is along the direction of motion, it will speed up (accelerate) or slow down (decelerate) the body If the resultant force is at an angle, it will change the direction of the body Page 55 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources YOUR NOTES  Worked Example  A girl is riding her skateboard down the road and increases her speed from 1 m s–1 to 4 m s–1 in 2.5 s.If the force driving her forward is 72 N, calculate the combined mass of the girl and the skateboard. Resultant Force Since force is a vector, every force on a body has a magnitude and direction The resultant force is, therefore, the vector sum of all the forces acting on the body The direction is given by either the positive or negative direction as shown in the examples below Page 56 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources YOUR NOTES  Resultant forces on a body can be positive or negative depending on their direction The resultant force could also be at an angle, in which case, the magnitude and direction of the resultant force can be determined using either: Calculation (usually simple geometry, such as Pythagoras Theorem or the sine and cosine rules) Scale drawing Acceleration Since acceleration is a vector, it can be either positive or negative depending on the direction of the resultant force If the resultant force is in the same direction as the motion of an object, the acceleration is positive If the resultant force is in the opposite direction to the motion of an object, the acceleration is negative An object may continue in the same direction however with a resultant force in the opposite direction to its motion, it will slow down and eventually come to a stop If drag forces are ignored, or severely reduced, the acceleration is independent of the mass of an object This has been shown in experiments by astronauts who have dropped a feather and a hammer on the Moon from the same height Both the hammer and feather drop to the Moon's surface at the same time Page 57 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources YOUR NOTES  Worked Example  Three forces, 4 N, 8 N, and 24 N act on an object with a mass of 5 kg. Which acceleration is not possible with any combination of these three forces? A. 1 m s–2 B. 4 m s–2 C. 7 m s–2 2 D. 10 m s– Step 1: List the values given Three possible forces at any angle of choice: 4 N, 8 N, and 24 N Mass of object = 5 kg Step 2: Consider the relevant equation Newton's second law: F=m×a Step 3: Rearrange to make acceleration the focus a=F÷m Step 4: Investigate the minimum possible acceleration It is best to consider the edges of this problem before dealing with more difficult combinations that is why it is prudent to check the minimum and maximum acceleration The minimum would occur when the forces were acting against each other The minimum possible would be when only the 4 N force was acting on the body Now check the acceleration: 4 N ÷ 5 kg = 0.8 m s–2 Step 4: Investigate the maximum possible acceleration The maximum would occur when all three forces are acting in the same direction Therefore adding to: 4 + 8 + 24 = 36 N Now check the acceleration: 36 N ÷ 5 kg = 7.2 m s–2 Step 5: Consider this range and the options Since option D is higher than 7.2 m s–2; it is not possible that these three forces can produce 10 m s–2 acceleration for this mass Page 58 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources Option D is the correct answer YOUR NOTES   Worked Example A rocket produces an upward thrust of 15 MN and has a weight of 8 MN. A. When in flight, the force due to air resistance is 500 kN. Determine what is the resultant force on the rocket. B. The mass of the rocket is 0.8 × 105 kg. Calculate the magnitude and direction of the acceleration of the rocket. Page 59 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources YOUR NOTES   Exam Tip The direction you consider positive is your choice, as long as the signs of the numbers (positive or negative) are consistent throughout the question.It is a general rule to consider the direction the object is initially travelling in as positive. Therefore all vectors in the direction of motion will be positive and opposing vectors, such as drag forces, will be negative. Page 60 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources 2.2.4 Newton’s Third Law YOUR NOTES  Newton’s Third Law Newton’s third law of motion states: Whenever two bodies interact, the forces they exert on each other are equal and opposite Newton’s third law explains the following important principles about forces: All forces arise in pairs – if object A exerts a force on object B, then object B exerts an equal and opposite force on object A Force pairs are of the same type – for example, if object A exerts a gravitational force on object B, then object B exerts an equal and opposite gravitational force on object A Newton’s third law explains the forces that enable someone to walk The image below shows an example of a pair of equal and opposite forces acting on two objects (the ground and a foot): Newton's Third Law: The foot pushes the ground backwards, and the ground pushes the foot forwards One force is from the foot that pushes the ground backwards The other is an equal and opposite force from the ground that pushes the foot forwards Page 61 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources YOUR NOTES  Worked Example  A physics textbook is at rest on a dining room table.Eugene draws a free body force diagram for the book and labels the forces acting on it. Eugene says the diagram is an example of Newton's third law of motion. William disagrees with Eugene and says the diagram is an example of Newton's first law of motion.By referring to the free-body force diagram, state and explain who is correct. Step 1: State Newton's first law of motion Objects will remain at rest, or move with a constant velocity unless acted on by a resultant force Step 2: State Newton's third law of motion Whenever two bodies interact, the forces they exert on each other are equal and opposite Step 3: Check if the diagram satisfies the two conditions for identifying Newton's third law In each case, Newton's third law identifies pairs of equal and opposite forces, of the same type, acting on two different objects The diagram only involves one object Furthermore, the forces acting on the object are different types of force - one is a contact force (from the table) and the other is a gravitational force on the book (from the Earth) - its weight The image below shows how to apply Newton's third law correctly in this case, considering the pairs of forces acting: Page 62 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources YOUR NOTES  Step 4: Conclude which person is correct In this case, William is correct The free-body force diagram in the question is an example of Newton's first law The book is at rest because the two forces acting on it are balanced - i.e. there is no resultant force Page 63 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources Exam Tip YOUR NOTES  Remember that pairs of equal and opposite forces in Newton's third law act on two  different objects. It's a really common mistake to confuse Newton's third law with Newton's first law, so applying this check will help you distinguish between them. Newton's first law involves forces acting on a single object.These differences are shown in Scenario 1 (Newton's first law) vs. Scenario 2 (Newton's third law) Page 64 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources 2.2.5 Applying Newton’s Laws of Motion YOUR NOTES  Applying Newton’s Laws of Motion Newton's laws of motion are strong tools to understand the motion of objects with forces acting upon them Below are two worked examples demonstrating different situations involving the application of Newton's laws  Worked Example Two blocks of mass 1 kg and 4 kg respectively are attached by a tight massless rope between them. The 1 kg block sits on the left and the 4 kg block sits on the right. The 1 kg mass has a 100 Newton force applied to it pulling it to the left. What is the acceleration of both blocks and the tension in the rope as they move across a frictionless surface? A The acceleration is 15 m s–2 to the left and the tension is 20 N B The acceleration is 20 m s–2 to the left and the tension is 40 N C The acceleration is 15 m s–2 to the left and the tension is 60 N D The acceleration is 20 m s–2 to the left and the tension is 80 N Step 1: Consider the whole of the system Together the 1 kg and 4 kg blocks are both being pulled along by the 100 N force (since the rope is tight) This is a frictionless flat surface, therefore the only forces in the system are the pulling force and the tension force(s) in the rope Therefore the acceleration can be found using Newton's second law Step 2: Find the acceleration Using Newton's second law: F=m×a Rearrange for a a=F÷m a = 100 ÷ 5 = 20 m s–2 to the left Page 65 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources Step 3: Examine the 4 kg mass only YOUR NOTES Since the system is moving with an acceleration of 20 m s–2 to the left, the force on  only the 4 kg mass can be found F=m×a F = 4 × 20 = 80 N to the left The only force acting on the 4 kg mass is the rope and therefore tension force This means there is 20 N pulling force on the 1 kg block only Step 4: State your answer Since acceleration, a = 20 m s–2 to the left and the tension force = 80 N to the left Therefore, the answer is option D  Worked Example A stationary object is subject to a 300 N force towards the left and at 55 degrees leftwards with respect to the vertical and a 450 N force to the right and 35° downwards with respect to the horizontal.Calculate what is the magnitude and direction of the third force that would make this object remain stationary (to the nearest N). Step 1: Recall Newton's first law Newton's First Law states: A body will remain at rest or move with constant velocity unless acted on by a resultant force Therefore, for this object to remain stationary, the resultant force must have a magnitude of 0 Newtons Step 2: Resolve the 300 N force into its horizontal and vertical components Page 66 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources The horizontal component can be resolved from: YOUR NOTES sin(55°) × 300 = 246 N  This is directed to the left The vertical component can be resolved from: cos(55°) × 300 = 172 N This is in an upwards direction Step 3: Resolve the 450 N force into its horizontal and vertical components The horizontal component can be resolved from: cos(35°) × 450 = 369 N This is directed to the right The vertical component can be resolved from: sin(35°) × 450 = 258 N This is in a downwards direction Step 4: Combine the horizontal components The two forces provide 369 N to the right and 246 N to the left Page 67 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources Therefore, since these are opposing directions: YOUR NOTES 369 – 246 = 123 N to the right  Step 5: Combine the vertical components The two forces provide 258 N downwards and 172 N upwards Therefore, since these are opposing directions: 258 – 172 = 86 N downwards Step 6: Make a right angle triangle using these two force vectors There should be a longer size 123 N magnitude vector arrow to the right and then a 86 N magnitude vector arrow downwards These can be connected from start to finish by a third vector which is the hypotenuse of this right-angled triangle Step 7: Use the two vectors magnitudes to find the angle from the horizon Since the vectors are to the right and downwards in this right-angle triangle, neither is the hypotenuse Therefore the angle from the horizontal downwards can be found by using tan: tan(θ°) = Opp. ÷ Adj. = 86 ÷ 123 θ° = tan–1(86 ÷ 123) = tan–1(0.699) θ° ≈ 35° Step 8: Use Pythagoras theorem or trigonometry to find the magnitude of the resultant force Method 1: Using Pythagoras theorem c2 = b2 + a2 where b and a are the vector magnitudes for the horizontal and vertical components found so far and c is the hypotenuse magnitude c2 = 862 + 1232 = 7396 + 15129 = 22525 c = √22 525 ≈ 150 N Method 2: Using trigonometry Using the horizontal component and the angle found in step 7, cos can be used cos(35°) = Adj. ÷ Hyp. = 123 ÷ Hyp. Page 68 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources Therefore: YOUR NOTES 123 ÷ cos(35°) = Hyp. ≈ 150 N  Step 9: State the final answer The third force which would cause this object to remain stationary is 150 N right and 35° downwards from the horizontal  Exam Tip You are expected to know Newton's three laws of motion from memory and how they apply to physical situations. So be sure to practice and use them without having to review them before carrying out a problem. Page 69 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources 2.2.6 Friction YOUR NOTES  Friction Friction is a force that works in opposition to the motion of an object It occurs between two solid bodies that are in contact with one another The opposition of friction slows down the motion of the object When friction is present, energy is transferred in the form of heat This raises the temperature (thermal energy) of the object and its surroundings The work done against the frictional forces causes this rise in the temperature Imperfections at the interface between the object and the surface bump into and rub up against each other Not only does this slow the object down but also causes an increase in thermal energy The interface between the ground and the sled is bumpy which is the source of the frictional force Static & Dynamic Friction There are two kinds of friction to consider for IB DP Physics Static friction occurs when two solid objects are in contact and no movement is occurring between the two objects Dynamic friction occurs once one of the objects is moving past the other, such as in the sled example above Both of these forms of friction depend on the normal reaction force of the object sitting upon the other Page 70 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources Static friction will match any pushing force that acts against until it can no longer hold the YOUR NOTES two objects stationary  Static friction increases in magnitude until movement begins and dynamic friction occurs For any given situation, static friction should reach a maximum value that is larger than that of dynamic friction For a constant pushing force, dynamic friction will be a constant This is because there are more forces at work keeping an object stationary than there are forces working to resist an object once it is in motion The relationship between frictional forces and motion The equation for static friction is given by: F ≤ μS × R Where: F = static frictional force (N) μS = coefficient of static friction R = normal reaction force (N) The coefficient of static friction is a number between 0 and 1 but does not include those numbers It is a ratio of the force of static friction and the normal force The larger the coefficient of static friction, the harder it is to move those two objects past one another The equation for dynamic friction is given by: F = μD × R Where: F = dynamic frictional force (N) μD = coefficient of dynamic friction Page 71 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources R = normal reaction force (N) YOUR NOTES The coefficient of dynamic friction has similar properties to that of static friction  Yet, dynamic friction has a definite force value for a given situation Whereas the force of static friction has an increasing force value  Worked Example An 8.0 kg block sits on an incline of 20 degrees from the horizontal. It is stationary and does have a frictional force acting upon it. Determine the minimum possible value of the coefficient of static friction. Step 1: List the known quantities Mass of the block, m = 8.0 kg Angle between the slope and the horizontal, θ = 20° Step 2: Determine the weight of the block The weight will act directly downwards and comes from the interaction of mass and acceleration due to gravity W=m×g 8.0 × 9.8 = 78.4 N downwards Step 3: Break the weight down into components based on the slope angle The component that is parallel to the slope and provides a force moving the block down the slope can be found from: Page 72 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources sin(20°) × W = F YOUR NOTES F = sin(20°) × 78.4 = 26.8 N  The component that is perpendicular to the slope and the same magnitude as the normal reaction force can be found from: cos(20°) × W = F F = cos(20°) × 78.4 = 73.7 N Step 4: Use the equation of static friction to find the minimum value of the coefficient of static friction The equation for static friction is: F ≤ μS × R In this case, the F is the 26.8 N pushing the block down the slope The R is the normal reaction force which has the same magnitude as the perpendicular component of the weight force which is 73.7 N Therefore the value can be added and μS solved for: 26.8 ≤ μS × 73.7 Rearrange for μS 26.8 ÷ 73.7 ≤ μS 0.36 ≤ μS Step 5: State the final answer The coefficient for static friction must be at least 0.36 or greater for this situation Page 73 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources 2.3 Work, Energy & Power YOUR NOTES  2.3.1 Kinetic Energy Kinetic Energy Kinetic energy (Ek) is the energy an object has due to its motion (or velocity) The faster an object is moving, the greater its kinetic energy When an object is falling, it is gaining kinetic energy since it is gaining speed This energy transferred from the gravitational potential energy it is losing An object will maintain this kinetic energy unless its speed changes Kinetic energy can be calculated using the following equation: Kinetic energy (KE): The energy an object has when it is moving Derivation of Kinetic Energy Equation A force can make an object accelerate; work is done by the force and energy is transferred to the object Using this concept of work done and an equation of motion, the extra work done due to an object's speed can be derived The derivation for this equation is shown below: Page 74 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources YOUR NOTES  Page 75 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources YOUR NOTES  Worked Example  A body travelling with a speed of 12 m s–1 has kinetic energy 1650 J.If the speed of the body is increased to 45 m s–1, estimate what is its new kinetic energy.  Exam Tip When using the kinetic energy equation, note that only the speed is squared, not the mass or the ½.If a question asks about the ‘loss of kinetic energy’, remember not to include a negative sign since energy is a scalar quantity. Page 76 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources 2.3.2 Gravitational Potential Energy YOUR NOTES  Gravitational Potential Energy Gravitational potential energy (Ep) is energy stored in a mass due to its position in a gravitational field If a mass is lifted up, it will gain Ep (converted from other forms of energy) If a mass falls, it will lose Ep (and be converted to other forms of energy) The equation for gravitational potential energy for energy changes in a uniform gravitational field is: Gravitational potential energy (GPE): The energy an object has when lifted up The potential energy on the Earth’s surface at ground level is taken to be equal to 0 This equation is only relevant for energy changes in a uniform gravitational field (such as near the Earth’s surface) Derivation of GPE Equation When a heavy object is lifted, work is done since the object is provided with an upward force against the downward force of gravity Therefore energy is transferred to the object This equation can therefore be derived from the work done Page 77 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources YOUR NOTES   Worked Example To get to his apartment a man has to climb five flights of stairs. The height of each flight is 3.7 m and the man has a mass of 74 kg. What is the approximate gain in the man's gravitational potential energy during the climb? A. 13 000 J B. 2700 J C. 1500 J D. 12 500 J Page 78 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources YOUR NOTES  GPE vs Height The two graphs below show how GPE changes with height for a ball being thrown up in the air and when falling down (ignoring air resistance) Page 79 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources YOUR NOTES  Graphs showing the linear relationship between GPE and height Since the graphs are straight lines, GPE and height are said to have a linear relationship These graphs would be identical for GPE against time instead of height Relationship between GPE & KE There are many scenarios that involve the transfer of kinetic energy into gravitational potential, or vice versa Some examples are: A swinging pendulum Objects in freefall Sports that involve falling, such as skiing and skydiving Using the principle of conservation of energy, and taking any drag forces as negligible: Loss in potential energy = Gain in kinetic energy Page 80 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources YOUR NOTES  Worked Example  The diagram below shows a skier on a slope descending 750 m at an angle of 25° to the horizontal. Calculate the final speed of the skier, assuming that he starts from rest and 15% of his initial gravitational potential energy is not transferred to kinetic energy. Step 1: Write down the known quantities Vertical height, h = 750 sin 25° Ek = 0.85 Ep Step 2: Equate the equations for Ek and Ep Page 81 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources Ek = 0.85 Ep YOUR NOTES ½ mv2 = 0.85 × mgh  Step 3: Rearrange for final speed, v Step 4: Calculate the final speed, v Page 82 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources 2.3.3 Elastic Potential Energy YOUR NOTES  Elastic Potential Energy Elastic potential energy is defined as The energy stored within a material (e.g. in a spring) when it is stretched or compressed It can be found from the area under the force-extension graph for a material deformed within its limit of proportionality A material within its limit of proportionality obeys Hooke’s law Therefore, for a material obeying Hooke’s Law, elastic potential energy can be calculated using: Where: k = force constant of the spring (N m–1) x = extension (m) In your data booklet the extension x is written as Δx This just means the change in x It is very dangerous if a wire under large stress suddenly breaks This is because the elastic potential energy of the strained wire is converted into kinetic energy EPE = KE ½ kx2 = ½ mv2 v∝x This equation shows that the greater the extension of a wire, x, the greater the speed, v, it will have on breaking Page 83 of 139 © 2015-2023 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to savemyexams.co.uk for more awesome resources