MATHEMATICAL TECHNIQUES FOR A- LEVEL PHYSICS

Size: px

Start display at page:

Download "MATHEMATICAL TECHNIQUES FOR A- LEVEL PHYSICS"

Erin Parsons
6 years ago
Views:

1 Name:. MTHEMTICL TECHNIQUES FOR - LEVEL PHYSICS (TO CCOMPNY THE DVNCING PHYSICS COURSE) Mr Green, Mrs Wynne, Mr Clark, Mr Harris Kesgrave High School (Based on a version by Maria Pavlidou, Cleeve school)

2 Index Introduction...3 cknowledgements...4 Chapter : Rearranging equations...5 Chapter : Ratios...8 Chapter 3: How to use and convert prefixes...3 Chapter 4: How to solve a lengthy problem... Chapter 5: Graphs...4 Chapter 6: Trigonometry 0 References...4 To assess your learning you will need to: ) Create an account by signing up on IsaacPhysics at ) Join Mr Green s Summer Task Questions group by clicking on: 3) Do all 8 sections from through to 8 by st week in September. 4) Mr Green at dgreen@kesgrave.suffolk.sch.uk if you have any problems or issues over the summer. Version 4 Maths for Physics May, 06

3 Introduction The idea for this booklet is to make the dvancing Physics course more accessible to students, especially the ones who do not have the high mathematical skills required for the course. Many students understand the physics but struggle with maths and as a result do not achieve high grades in Physics. One reason Physics is seen as a difficult subject is not only because it often deals with difficult physical concepts but because it also requires very good skills in manipulating equations, in understanding and decoding relationships and patterns between variables and in processing lengthy calculations. Our hope is that this book will become a first step towards lifting the fear of Physics as an - level course and help students achieve the grades they truly deserve. You must try to answer all the questions (although miss the logs chapter if you are really struggling) and there will be a test at the beginning of term to see how much you know. You MUST get a minimum % score or you will have to come along to Physics Club to practise and do a retake. To get a decent grade in Physics your maths must be reasonable especially with rearranging formulae, using standard form and doing trigonometry. Mr Green, Mrs Wynne, Mr Clark, Mr Harris May 06 Version 4 Maths for Physics May, 06

4 cknowledgements This booklet is based on text developed by Maria Pavlidou of Cleeve school and was made available on Teachbox.net. Version 4 Maths for Physics May, 06

5 Chapter : Rearranging equations The first step in learning to manipulate an equation is your ability to see how it is done once and then repeat the process again and again until it becomes second nature to you. In order to show the process once I will be using letters rather than physical concepts. You can rearrange an equation a bc with b as the subject (divide both sides by c) b a c or c as the subject (divide both sides by b) c a b ny of these three symbols a b, c, can be itself a summation, a subtraction, a multiplication, a division, or a combination of all. So, when you see a more complicated equation, try to identify its three individual parts a, b, c before you start rearranging it. Worked examples v v Equation First Rearrangement Second Rearrangement f T v f T f f u f v u f v f v f T u f THINK! s you can see from the third worked example, not all rearrangements are useful. In fact, for the lens equation only the second rearrangement can be useful in problems. So, in order to improve your critical thinking and know which rearrangement is the most useful in every situation, you must practise with as many equations as you can. Version 4 Maths for Physics May, 06

6 NOW TRY THIS! From now on the multiplication sign will not be shown, so written as a bc a bc Fill in the blank spaces with your own examples. will be simply Equation First Rearrangement Second Rearrangement (Power of lens) P f f v (Magnification of lens) m u v u c (refractive index) n v c v Q (current) I (electric potential) (power) t E P t E V Q (power) P VI (conductance) (resistance) (resistance) I G V V R I R G (power) (power) (stress) P I R V P R F F (strain) x l x l (Young s modulus) E (conductance) G L (resistance) R L (resistivity) Version 4 Maths for Physics May, 06

7 (phase angle) (displacement) ft y asin f t a sin (Young s interference) (quantum energy) E hf (electron wavelength) h mv (suvat) v = u + at L x d f h a = t = (suvat) v = u + as s = u = (suvat) a = If u = 0, t = s = ut + ½ at Version 4 Maths for Physics May, 06

8 Version 4 Maths for Physics May, 06 Chapter : Ratios. Manipulating ratios These are the most useful ways of manipulating ratios. They will help you when you rearrange equations. First: C B D D C B Second: C D B D C B Third: D D C B B D C B Fourth: D D C B B D C B and finally, this is how you simplify a complex fraction (turn the bottom fraction upside down and multiply): c b d a d c b a THINK! It is always useful to convert a complex fraction into a simple one, using the form above, before you try anything else. If b or d are missing, substitute them with number. NOW TRY THIS! Here are some examples to do: [6] 3 4

9 Molecules of DN have been stretched using optical tweezers. Using the following data find the cross sectional area () of a DN strand, if the Young s Modulus (E) of DN is 0 8 Pa and a load (F) of 4 x 0-0 N results in 0% strain (ε = 0.). [] Here is how you start: manipulate the equation until you have on the left-hand side and only at this point substitute all the symbols for their equivalent numbers. E E F Version 4 Maths for Physics May, 06

10 Chapter 3: How to use and convert prefixes Mathematical Prefixes (LERN THESE!!) [] Prefix Symbol Name Multiplier femto f quadrillionth 0-5 pico p trillionth 0 - nano n billionth 0-9 micro µ millionth 0-6 milli m thousandth 0-3 centi c hundredth 0 - deci d tenth 0 - deka da ten 0 hecto h hundred 0 kilo k thousand 0 3 mega M million 0 6 giga G billion 0 9 tera T trillion 0 peta P quadrillion 0 5 When you are given a variable with a prefix you must convert it into its numerical equivalent in standard form before you use it in an equation. Remember x 0-3 is 0.00, x 0 3 is 000. FOLLOW THIS! lways start by replacing the prefix symbol with its equivalent multiplier. For example: 0.6 μ = 0.6 x 0-6 ( =.6 x 0-7 = ) 3 km = 3 x 0 3 m ( = 3000m ) 0 ns = 0 x 0-9 s ( = x 0-8 s ) DO NOT get tempted to follow this further (for example: 0.6 x 0-6 =.6 x 0-7 and also 0 x 0-9 s = 0-8 s) unless you are absolutely confident that you will do it correctly. It is always safer to stop at the first step (0 x 0-9 s) and type it like this into your calculator. NOW TRY THIS!.4 kw = 0 μc = 4 cm = 340 MW = 46 pf = 0.03 m = 5 Gbytes = 43 kω = 0.03 MN = Version 4 Maths for Physics May, 06

11 Chapter 4: How to solve a lengthy problem (This is VERY important to get maximum marks and method marks in questions) Problems in physics can appear to be difficult at first sight. However, once you analyse the problem in well-defined steps you should be able to solve it without any difficulty. The steps you need to follow are:. Identify the variables you are given and the ones you are asked to find i.e. write down WHT YOU KNOW and WHT YOU NEED TO FIND OUT!. Convert all units given to SI units 3. Give a different symbol to each variable; try to stick to the well known symbols. To simplify, write the values for each symbol (m=600) but don t worry about writing the units at this stage. 4. Recognise and then WRITE DOWN which equation/s to use. You do this by looking at what variables are available to you and what variables you are asked to find. This is a critical stage; experience is the most important factor here. This is why you need to practise again and again... This is also why you need to KNOW LL YOUR EQUTIONS VERY WELL! 5. PLUG THE NUMBERS INTO THE EQUTION gain, experience is very important here! You may need to rearrange the equation! 6. Write down the final answer and add the correct units. EXMPLE car of mass 600 kg is travelling at 0 ms -. When the brakes are applied, it comes to rest in 0.0 km. What is the average force exerted by the brakes? [3] (Note: there are many ways to solve this problem and in a shorter method will be introduced.) STEP. mass= 600 kg initial velocity = 0 m s - final velocity = 0 m s - stopping distance = 0.0 km force =? STEP. re they all in SI units? No. So... stopping distance = 0.0 km = 0.0 x 0 3 m = 0 m Version 4 Maths for Physics May, 06

12 STEP 3. m 600 u v s F? STEP 4. E W (Change in energy = work done) W Fs (Work done = force x distance in the direction of the force) EK mv (Kinetic energy = 0.5 x mass x velocity x velocity) STEP 5. Find the initial kinetic energy: E K mu Find the final kinetic energy: E K mv Find the change in kinetic energy: E E K E K Equalise the change in kinetic energy with the work done and rearrange to find the force: E W E Fs F E s F F 3000 STEP 6. The final answer is F 3000 N The negative sign shows that the force is in opposite direction to the velocity, i.e. it is the frictional force that stops the car. Version 4 Maths for Physics May, 06

13 NOW TRY THIS! girl diving from a 5 m platform wishes to know how fast she enters the water. She is in the air for.75 s and dives from rest (with an initial speed of zero). What can you tell her about her entry speed? (acceleration due to gravity = 9.8 m s - ) [7]. Use the equation v = u + at where: v (final speed) = u (initial speed) + a (acceleration due to gravity = - 9.8) x t (time in seconds) REMEMBER! Follow all steps! Do not try to rush through! STEP. STEP. STEP 3. STEP 4. STEP 5. STEP 6. Version 4 Maths for Physics May, 06

14 Chapter 5: Graphs 5. Types of graphs Graphs are VERY important in physics because they show patterns between variables. straight line graph that starts from the (0,0) point is the best proof that two variables are directly proportional. Straight line graph You should know from your maths that the general equation for a straight line is y ax b where a is the gradient of the graph and b is the point that the line, cuts the y-axis (you may know this as y = mx + c). y b a x It is also useful to know from your maths the equation for a hyperbola, a parabola, an ellipse etc. Make your own table including all the graph shapes you know and their functions as you come across them in the course. Make sure you include the logarithmic and exponential functions. Function Graph Example in physics y ax b (formula for a straight line which you may alternatively know as y = mx + c where m is the gradient and c the intercept) Version 4 Maths for Physics May, 06

15 5. How to choose the right graph for plotting lthough the above list is important, when it comes to finding a relationship between two variables the only graph that can show this very clearly is the straight line graph. EXMPLE Let s say that you want to prove the relationship between the kinetic energy of an object and its velocity. You plot velocity on the x-axis (the independent variable) and kinetic energy on the y-axis (the dependent variable). You will get a curve which as you now is a parabola (since the kinetic energy is directly proportional to the square of the velocity). Now let s say you do another experiment that, unknown to you, also follows the same pattern. You will also get a curve when you plot the graph. Will you be able to recognise that this is a parabola? What if it is a curve that is very close to a parabola but not quite? What can you do to be sure that you have cracked the relationship? Think again about the example above. If instead of plotting kinetic energy against velocity you plot kinetic energy against velocity squared what will you get? You will get a straight line through zero! Moreover, you will be certain that the relationship is that: the kinetic energy is directly proportional to the velocity squared. K.E. = ½ mv so plotting K.E. on the y axis (the dependent variable what you measure) and v on the x axis (the independent variable what you change) you will get a straight line with the gradient = ½ m. So what have we learned so far? LWYS IM T PLOTTING TWO VRIBLES THT WILL GIVE YOU STRIGHT LINE! Here are some examples: To prove that resistance R is inversely proportional to cross sectional area, plot R against. This should give you a straight line. To prove that the square of the period T of a pendulum is directly proportional to its length l plot either T against l or T against l Version 4 Maths for Physics May, 06

16 NOW TRY THIS!. The pendulum equation is: T l g a) What variables should you plot against each other in order to prove that the period of the pendulum does not depend on its mass? What will the shape of this graph be? b) What variables should you plot against each other to prove that the period depends on the gravitational field strength as shown by the equation?. The universal gravitational law is given by the equation: mm F G r a) What variables should you plot against each other in order to prove that the attractive force is directly proportional to both masses (i.e. m x M) of the objects? b) What variables should you plot against each other in order to prove that the attractive force is inversely proportional to the distance squared between the objects? Version 4 Maths for Physics May, 06

17 5.3The significance of the gradient (VERY important!) During your coursework you will be asked to decide which graphs to plot in order to show a relationship or to calculate a physical constant. We have already noted how important it is to aim at plotting a graph that will end up being a straight line. This gives you a definite answer about the relationship between the two variables. But there is more to it. The gradient of this line will give you information about a constant in your experiment. EXMPLE Let s say that you want to measure the gravitational field strength of Earth with a pendulum. T against l. The You vary the length and measure the period. You then decide to plot graph will be a straight line. What will its gradient be? To find this, compare the pendulum equation with the straight line equation as shown below: T 4 l g y ax b (or y = mx + c if you are more familiar with this as the equation of a straight line) I hope you can see that y (the dependent variable) corresponds to T, x (the independent variable) corresponds to l, b (the intercept) corresponds to zero, and a (the gradient) corresponds to 4 g from your graph you will know the value of g from this as:. This tells you that once you measure the gradient 4 g and you will then be able to calculate gradient 4 4 g g gradient Version 4 Maths for Physics May, 06

18 NOW TRY THIS! Try to find the gradient in all the situations listed below. The first three have been done for you. Equation Plot y against x gradient Constant V x axis : current gradient R R R I y axis : voltage (V = R x I Compare to (for a fixed resistor) y = ax + b or y = mx + c if you prefer and you can see V is the y and I is the x with R the gradient and the intercept is zero) V x axis : voltage R R gradient (for a fixed resistor) I y axis : current R x axis : force l gradient y axis : extension E l E E gradient R L R L F x l mv Fs (stopping distancevelocity relationship) xd L (double slit interference) xd L (double slit interference) x axis : L y axis : R x axis : y axis : R x axis : v y axis : s x axis : d y axis : x x axis : L y axis : x gradient (Young s modulus) (resistivity) gradient gradient F gradient gradient (resistivity) (fricition) (wavelength) (wavelength) part from its use as explained above, the gradient in all lines (curved or straight) corresponds to the derivative of the function you plot (but don t worry, you don t need to differentiate for the S course). This is why if you plot time on the x-axis and displacement Version 4 Maths for Physics May, 06

19 on the y-axis the gradient corresponds to the velocity of the object. If the line is curved the gradient does not stay the same, which means that it is equal to the instantaneous velocity of the object. For the same reason if you plot time on the x-axis and velocity on the y-axis the gradient corresponds to the acceleration of the object. If the line is curved the gradient does not stay the same, which means that it is equal to the instantaneous acceleration of the object. 5.4The significance of the area under a graph The area between a graph of y = f(x) and the x-axis is equal to the definite integral of the function. This formula gives a positive result for a graph above the x-axis, and a negative result for a graph below the x-axis. [9] This is why the area under a velocity-time graph is equal to the distance covered by the object. In simpler language, if you count the squares below a speed/time graph you will get the distance travelled. If the graph is a straight line then the area under can be calculated very precisely as the area of a triangle or trapezium etc. If the line is a curve, the area is often estimated to a good precision before it can give you some useful information. You will use this idea more at level so for the moment we will leave it at this stage. Version 4 Maths for Physics May, 06

Chapter 6: Trigonometry 6. Basic Trig You will need to know how to perform basic trigonometry using SOH CH TO.

Opposite side divided by the Hypotenuse. Using CH, the Cosine of the angle is equal to the djacent side divided by the Hypotenuse.

20 Chapter 6: Trigonometry 6. Basic Trig You will need to know how to perform basic trigonometry using SOH CH TO. This will be needed to work out how vectors like Velocity, Force and Displacement combine together in a right angled triangle to give a Resultant vector. First of all, remember that in a right angled triangle the sides are labelled as follows: Using SOH, the Sine of the angle is equal to the Opposite side divided by the Hypotenuse. Using CH, the Cosine of the angle is equal to the djacent side divided by the Hypotenuse. Using TO, the Tangent of the angle is equal to the Opposite side divided by the djacent side. The angle is measured in Degrees or something called RDINS. Radians are often used as they are easily manipulated as a number (they are the ratio of the arc of a circle divided by the radius) see the diagram below: Version 4 Maths for Physics May, 06

For a complete circle, the angle is 360 0. The rc distance = circumference = r where r = Radius. The angle in Radians is therefore = r/r = Radians.

(about 0.08 Radians or less) tan is approximately equal to sin is approximately equal to in Radians. Using trigonometry, you can work out angles and sides in right angle triangles.

21 For a complete circle, the angle is The rc distance = circumference = r where r = Radius. The angle in Radians is therefore = r/r = Radians. You need to understand this so that you can use the following in some Physics calculations: = Radians 80 0 = Radians 90 0 = / Radians Using your calculator, you can see that for a small angle (about 0.08 Radians or less) tan is approximately equal to sin is approximately equal to in Radians. Using trigonometry, you can work out angles and sides in right angle triangles. You will find right angle triangles in lots of Physics problems and they are used to work out velocities, forces and displacements. Here is an example of forces. One (3N) is pushing horizontally on an object and the other (4N) is pushing vertically on the object. dding these vectors nose to tail as shown can show the Resultant of these forces. This is how you can work out the resultant force and the direction the object will move as a result of this resultant force (the angle will give us the direction the object will be pushed in to the horizontal). The Resultant can be calculated by using Pythagoras (Resultant = ) or using Trig once we know the angle. We need to decide whether to use SOH, CH or TO. We know the Opposite and djacent sides so we need to use TO. Tan = ¾ = Using your calculator you type in 0.75 tan - and get the answer = 36.9 degrees. Using trig again, we can then calculate the resultant by using SOH or CH we know and Opposite or djacent and we need to find the Hypotenuse. Let s use CH. Cos 36.9 = djacent/hypotenuse therefore: Hypotenuse = djacent/cos 36.9 = 4/0.8 = 5N lternatively use Pythagoras! Version 4 Maths for Physics May, 06

the following triangles:. n object has forces acting on it one horizontally and one vertically.

From the problem, redraw the force vectors as a right angle triangle (draw them nose to tail or don t take

4N Calculate the horizontal force on the object (hint: use Pythagoras) and then work out the angle.

22 NOW TRY THIS! Practise calculating angles, Opposite sides, djacent sides or Hypotenuse using Sine, Cosine and Tangent for the following triangles:. n object has forces acting on it one horizontally and one vertically. The vertical force is 4N and the resultant force is 7N. From the problem, redraw the force vectors as a right angle triangle (draw them nose to tail or don t take your pen off the page when adding the vectors to get the Resultant. 4N Calculate the horizontal force on the object (hint: use Pythagoras) and then work out the angle.. n aircraft is flying due North (= a bearing of 0 degrees) at 00m/s and there is a cross wind blowing due West (bearing of 90 degrees) of 0m/s. Calculate the resultant velocity and the new heading (these are given as Bearings shown as degrees from North for example a heading of 45 degrees is N.E.) Version 4 Maths for Physics May, 06

23 3. Calculate the resultant velocities and the angles for the following: Version 4 Maths for Physics May, 06

24 References [] dvancing Physics S, Institute of Physics 008, page 5. [] [3] Physics for you, Keith Johnson, Nelson Thornes Ltd., 004, Extension sheet, page 7. [4] Maths for scientists, D. Galloway et.al, Mills &Boon Ltd.,980, pages 4-7 [5] [6] [7] dvancing Physics S, CD, IoP 008, Chapter 8 Q0S Distance, time and speed calculations Version 4 Maths for Physics May, 06

Summer Work package for IB and A level prospective physics students.

Summer Work package for IB and A level prospective physics students. Task 1. Read the first 3 sections of Mathematical techniques for A level physics (rearranging equations, ratios and prefixes). (This