"How to Study Physics" by David R. Hubin and Charles Riddell, was published by the
You, like many students, may view college level physics as difficult. You, again like many students, may seem overwhelmed by new terms and equations. You may not have had extensive experience with problem-solving and may get lost when trying to apply information from your textbook and classes to an actual physics problem. It is hoped that this pamphlet will help!
It's designed to help you stay out of the difficulties that come when you think small and get too involved in memorizing formulas or other specific details without understanding the underlying principles. It will guide you in understanding how to apply specific knowledge to the problems, how to start, how to seek help, how to check your answer. In short, it will help you develop the study skills that are important not just in physics but in all of your courses.
It's important to recognize that physics is a problem-solving discipline. Your physics teacher will stress major themes and principles, and one major goal is that you, the student, will be able to apply these principles to understand and solve problems. You should focus on this fact, that in a physics course, you are expected to solve problems.
An overview of your course can help you organize your efforts and increase your efficiency. To understand and retain data or formulas, you should see the underlying principles and connecting themes. It is almost inevitable that you will sometimes forget a formula, and an understanding of the underlying principle can help you generate the formula for yourself.
Take these steps to getting an overview early in the term so that all subsequent material can be integrated into your overview:
It's important that you be well prepared for class in order to use its potential fully for integrating the course material. To prepare for the class, you should do the following:
Come to the class on time and stay till the very end. Often teachers give helpful hints in the first and last minutes of the lecture. Unfortunately, these times are when a lot of people are not listening.
Reading the text and solving homework problems is a cycle: Questions lead to answers that lead back to more questions. An entire chapter will often be devoted to the consequences of a single basic principle. You should look for these basic principles. These Laws of Nature give order to the physicists' view of the universe. Moreover, nearly all of the problems that you will be faced with in a physics course can be analyzed by means of one or more of these laws.
When looking for relationships among topics, you may note that in many instances a specific problem is first analyzed in great detail. Then the setting of the problem is generalized into more abstract results. When such generalizations are made, you should refer back to the case that was previously cited and make sure that you understand how the general theory applies to the specific problem. Then see if you can think of other problems to which that general principle applies. Some suggestions for your physics reading:
You may now be like many students a novice problem solver. The goal of this section is to help you become an expert problem solver. Effective, expert problem solving involves answering five questions:
You, the expert, will decide, "this is an energy
problem," or, "this is a
In a physics course it's important to remember a couple of things about physicists and physics professors:
1. Read the problem. Look up the meanings of any terms that you do not know. Answer for yourself the question, "What's this about?" Make sure you understand what is being asked, what the question is. It is very helpful if you reexpress the problem in your own words or if you tell a friend what the problem is about.
2. Make a drawing of the problem. Even a poor drawing can be helpful, but for a truly good drawing include the following:
A. Give a title that identifies the quantity you are seeking in the problem or that describes the problem.
B. Label the drawing, including the parameters or variables on which the solution depends and that are given in the problem. Write down the given values of these parameters on the drawing.
C. Label any unknown parameters that must be calculated along the way or obtained from the text in order to find the desired solution.
D. Always give the units of measure for all quantities in the problem. If the drawing is a graph, be sure to give both the units and the scale of the axes.
E. Include on the drawing information that is assumed and not given in the problem (such as g, the value of the acceleration due to gravity), and whether air resistance and friction are neglected.
3. Establish which general principle relates the given parameters to the quantity that you are seeking. Usually your picture will suggest the correct techniques and formulas. At times it may be necessary to obtain further information from your textbook or notes before the proper formulas can be chosen. It often happens that further information is needed when the problem has a solution that must be calculated indirectly from the given information. If further information is needed or if intermediate quantities must be computed, it is here that they are often identified.
4. Draw a second picture that identifies the coordinate system and origin that will be used in relating the data to the equations. In some situations this second picture may be a graph, free body diagram, or vector diagram rather than a picture of a physical situation.
5. Even an expert will often use the concrete method of working a problem. In this method you do the calculation using the given values from the start, so that the algebra gives numerical values at each intermediate step on the way to the final solution. The disadvantage of this method is that because of the large number of numerical calculations involved, mistakes are likely, and so you should take special care with significant figures. However this method has the advantage that you can see, at every step of the way, how the problem is progressing. It also is more direct and often makes it easier to locate a mistake if you do make one.
6. As an expert, you will more and more use the formal method of working a problem. In this method, you calculate the solution by doing as much as possible without using specific numbers. In other words, do as much of the algebra as you can before substituting the specific given values of the data. In long and complicated problems terms may cancel or expressions simplify. Our advice: gain experience in problem solving by substituting the numbers when you start physics, but gradually adopt the formal approach as you become more confident; many people adopt a compromise approach where they substitute some values but retain others as symbols (for example, "g" for the acceleration due to gravity).
7. Criticize your solution: Ask yourself, "Does it make sense?" Compare your solution to any available examples or to previous problems you have done. Often you can check yourself by doing an approximate calculation. Many times a calculation error will result in an answer that is obviously wrong. Be sure to check the units of your solution to see that they are appropriate. This examination will develop your physical intuition about the correctness of solutions, and this intuition will be very valuable for later problems and on exams.
An important thing to remember in working physics problems is that by showing all of your work you can much more easily locate and correct mistakes. You will also find it easier to read the problems when you prepare for exams if you show all your work.
8. In an examination, you may have to do problems under a strict time limitation. Therefore, when you are finished with a homework problem, practice doing it again faster, in order to build up your speed and your confidence.
When you have completed a problem, you should be able, at some later time, to read the solution and to understand it without referring to the text. You should therefore write up the problem so as to include a description of what is wanted, the principle you have applied, and the steps you have taken. If, when you read your own answer to the problem, you come to a step that you do not understand, then you have either omitted a step that is necessary to the logical development of the solution, or you need to put down more extensive notes in your write-up to remind you of the reasons for each step.
It takes more time to write careful and complete solutions to homework problems. Writing down what you are doing and thinking slows you down, but more important it makes you behave more like an expert. You will be well paid back by the assurance that you are not overlooking essential information. These careful write-ups will provide excellent review material for exam preparation.
This problem is stated and the solution written down as you would work it out for homework.
In 1947 Bob Feller, former
1. What does the problem ask for, and what is given? Answer: The speed of the baseball is given, and what is wanted is the height that the ball would reach if it were thrown straight up with the given initial speed. You should double check that whoever wrote the problem correctly calculated that 98.6 miles/hr is equal to 44.1 m/s. You should state explicitly, in words, that you will use the 44.1 m/s figure and that you will assume the baseball is thrown from an initial height of zero (ground level). You should also state explicitly what value of g you will use, for example, g = 9.81 m/s2. You should also state that you assume that air resistance can be neglected. Since you don't know the mass of the baseball, say that you don't (you won't need it, anyway).
2. Make a drawing:
3. The general principles to be applied here are those of uniformly accelerated motion. In this case, the initial velocity vo decreases linearly in time because of the gravitational acceleration. The maximum height ym occurs at the time tm when the velocity reaches zero. The average velocity during from t = 0 to t = tm is the average of the initial velocity v = vo and the final velocity v = 0, or half the initial velocity.
4. Make a second drawing. In this case, try a graph of velocity as a function of time:
Notice that the graph is fairly accurate: You can approximate the value of g as 10 m/s2, so that the velocity decreases to zero in about 4.5 s. Therefore, even before you use your calculator, you have a good idea of about the value of tm.
5. The concrete method can now be applied: An initial velocity of 44.1 m/s will decrease at the rate of 9.81 m/s2 to zero in a time tm given by
tm = 44.1 / 9.81 = 4.4954 s .
During that time, the average velocity is vav = 44.1 / 2 = 22.05 m/s. Therefore the height is given by
ym = vav tm = 99.12 = 99.1 m .
Notice that for all "internal" calculations, more than the correct number of significant figures were kept; only when the final answer was obtained was it put into the correct number of significant figures, in this case three.
6. To do this problem in a formal method, use the formula for distance y as a function of t if the acceleration a is constant. Do not substitute numbers, but work only with symbols until the very end:
y = yo + vo t + a t2 / 2 ,
where yo = 0 is the initial position, vo = 44.1 m/s is the initial velocity, and a = - g = - 9.81 m/s2 is the constant acceleration. However, do not use the numerical figures at this point in the calculation. The maximum value of y is when its derivative is zero; the time tm of zero derivative is given by:
dy/dt = vo + a tm = 0 --> tm = - vo / a .
The maximum height ym is given by putting this value of tm into the equation for y:
ym = yo + vo ( - vo / a ) + a ( - vo / a )2 / 2 = yo - vo2 / 2a .
Now substitute: yo = 0, vo = 44.1, a = - 9.81. The result is
ym = 0 + 0.5 (44.1)2 / 9.81 = 99.1 m .
7. Look over this problem and ask yourself if the answer makes sense. After all, throwing a ball almost 100 m in the air is basically impossible in practice, but Bob Feller did have a very fast fast ball pitch!
There is another matter: If this same problem had been given in a chapter dealing with conservation of energy, you should not solve it as outlined above. Instead, you should calculate what the initial and final kinetic energy KE and potential energy PE are in order to find the total energy. Here, the initial PE is zero, and the initial KE is m vo2 / 2. The final PE is m g ym and the final KE is zero. Equate the initial KE to the final PE to see that the unknown mass m cancels from both sides of the equation. You can then solve for ym, and of course you will get the same answer as before but in a more sophisticated manner.
8. To prepare for an exam, look over this problem and ask yourself how you can solve it as quickly as possible. You may be more comfortable with the concrete approach or with the formal approach; practice will tell. On an actual exam, you might not have time for a complete drawing or a complete listing of principles. By working this problem a couple of times, even after you've gotten the answer once, you will become very familiar with it. Even better, explain the problem to a friend of yours, and that way you really will be an expert!
Again, this problem is stated and the solution written down as you would work it out for homework. As in Sample Problem #1, we go through the eight steps of the general outline.
A one kilogram block rests on a plane inclined at 27o to the horizontal. The coefficient of friction between the block and the plane is 0.19. Find the acceleration of the block down the plane.
1. The problem asks for the acceleration, not the position of the block nor how long it takes to go down the plane nor anything else. No mention is made of the difference between static or kinetic coefficients of friction, so assume they are the same. The mass is given, but you will eventually find that it doesn't matter what the mass is. (If the mass had not been given, that would be an indication that it doesn't matter, but even in that case you may find it easier to assume a value for the mass in order to guide your thoughts as you do the problem.)
2. Here is the first picture. Note that the angle is labeled (θ) , and the coefficient of friction is labeled (m) . In addition, the use of m for the mass and a|| for the acceleration down the plane are defined in the picture.
3. There are two general principles that apply here. The first is
F = m a ,
where F is the net
force, a vector, and a is acceleration, another vector; the two vectors
are in the same direction. The mass m will eventually be found not to make any
difference, and in that case, you might be tempted to write this law as a
= F / m, since a is what you want to
find. However, the easiest way to remember
The second principle is that the frictional force is proportional to the normal force (the component of the force on the block due to the plane that is perpendicular to the plane). The frictional force is along the plane and always opposes the motion. Since the block is initially at rest but will accelerate down the plane, the frictional force will be up along the plane. The coefficient of friction, which is used in this proportionality relation, is .
4. It is now time to draw the second picture. It helps to redraw the first picture and add information to it. In this case a vector diagram is drawn and various forces are defined.
Note that in the vector diagram, the block has been replaced by a dot at the center of the vectors. The relevant forces are drawn in (all except the net force). Even the value assumed for the gravitational acceleration has been included. Some effort has been made to draw them to scale: The normal force is drawn equal in magnitude and opposite in direction to the component of the gravity force that is perpendicular to the plane. Also, the friction force has been drawn in parallel to the plane and opposing the motion; it has been drawn in smaller than the normal force. The angles of the normal and parallel forces have been carefully drawn in relation to the inclined plane. This sub-drawing has a title and labels, as all drawings should.
5. We will do this problem using the formal approach, leaving the concrete method for a check (see below).
6. Now for calculation using the formal approach, where you work with algebra and symbols rather than with numbers. First state in words what you are doing, and then write down the equation:
FG|| = m g sin(θ) and FGN = m g cos (θ) .
FN = m g cos (θ) .
Ff = m m g cos (θ) .
F = m g sin (θ) - m m g cos (θ) .
a|| = g sin (θ) - m g cos (θ) = g ( sin (θ) - m cos (θ) ) .
a = (9.8 m/s2) (sin 27o - 0.19 cos 27o) = (9.8) (0.454 - 0.19 x 0.891) = 2.79 = 2.8 m/s2 .
7. When you look over this answer to see if it makes sense, try doing the problem by substituting numbers in at each step (the concrete approach). The weight of a kilogram, for example is 9.8 N. The normal (perpendicular to the plane) component of the gravitational force is 9.8 times cos 27o or 8.73 N. This makes sense, for if the angle were very small, the normal component of the gravitational force would be almost equal to 9.8 itself. Notice that although the final answer should be given to two significant figures, you should keep three in these intermediate calculations.
The parallel component of the gravitational force is 9.8 sin 27o = 4.45 N. The normal force due to the plane is equal in magnitude to the gravitational normal force (but opposite in direction), and so the frictional force is 0.19 times 8.73 or 1.66 N. The net force is down the plane and equal to the difference 4.45 - 1.66 = 2.79 N. Divide this value by 1 kg to get the acceleration 2.79 m/s2 (which is rounded off to 2.8 m/s2).
Again examine your solution. It says that the block does accelerate down the plane because the final answer is positive. The acceleration is less than g, again a reasonable result. Notice that if the angle were more than 27o, then its sine would be larger and its cosine smaller, so the acceleration would be greater. If the angle were less than 27o then the opposite would be true, and the acceleration, as calculated above, could become negative. But a negative value for acceleration would be wrong, because that would say that the block would accelerate up the plane because the frictional force dominates, and that is impossible. Instead, if the calculation had produced a negative value for a, you would have had to change the solution to a = 0, meaning that the frictional force was enough to prevent sliding.
8. Now anticipate how you'd do this problem on an exam. Is the concrete approach faster and easier for you? Or would you be more comfortable using the formal approach on an exam? It is a good idea to practice doing this problem when you study for an exam, if you think a similar problem will be asked.
If you have followed an active approach to study similar to the one suggested in this handout, your preparation for exams will not be overly difficult. If you haven't been very active in studying, your preparation will be somewhat harder, but the same principles still apply. Always remember: Physics courses, and therefore physics exams, involve problem solving. Hence, your approach to studying for exams should stress problem solving.
Here are some principles:
For example: If velocity and acceleration principles have been emphasized in the course, look over all of your homework problems to see if they illustrate these principles, even partially. Then if you also can anticipate an emphasis on friction and inertia, once again review all of your homework problems to see if they illustrate those principles.
These tips are based on a list
"17 Tips that UT Seniors Wish They'd Known as Freshmen"
by Dr. John Trimble, a professor in the English Department. He is a member of
The University of Texas's