Problem Solving. Undergraduate Physics

Transcription

1 in Undergraduate Physics In this brief chapter, we will learn (or possibly review) several techniques for successful problem solving in undergraduate physics courses Slide 1

2 Procedure for The following flowchart will be quite useful, particularly for more advanced problems STEP 1: Read the problem carefully. Convince yourself that you understand the physical situation that is present, and what you are being asked to figure out. STEP 2: Draw a diagram, if appropriate. This will help you to visualize the problem. It s also a good opportunity to establish a coordinate system. STEP 3: Write down the given data, as well as the information that you are expected to find. We usually refer to these as the knowns and the unknowns. Some unit conversions might be necessary at this point. STEP 4: Determine which physical principle(s) is (are) applicable, and write down the relevant equation(s). Then, rearrange your equation(s) in order to express the requested unknown quantities as a function of the known quantities. STEP 5: Perform your calculations STEP 6: Consider whether or not your results are reasonable. If not, return to step 1. Slide 2

3 Procedure for Step 4 is where most of the physics occurs. It is generally the most time-consuming step in problem solving. For exam problems, most part-marks will come from this step. Slide 3

4 Procedure for Example (from PHYS 211) A 1-kg ball is thrown upward with an initial speed of 20 m/s. What is the ball s maximum height above the release point? Step 1 When a ball is thrown upward, the downward force of gravity acts to reduce its upward velocity. Eventually, this velocity will become zero, and then negative. The point at which velocity is zero is, by definition, the maximum height of the ball s trajectory. The question asks us to find this point, expressed as a distance above the release point. Step 2 m= 1 kg Final Position: y f is unknown this is what we re trying to solve Final Speed: v f = 0 m/s (from step 1, this is the condition that defines the maximum height of the ball) Initial Position: y i = 0 m (this establishes the coordinate system) Initial Speed: v i = 20 m/s Slide 4

5 Procedure for Step 3 Knowns The problem explicitly provides two knowns: v i = 20 m/s m = 1 kg Furthermore, in steps 1 and 2, we effectively produced two additional knowns: v f = 0 m/s y i = 0 m But wait, there s more! The speed of the ball is affected by gravity. The acceleration due to gravity is another known: a y = g = 9.8 m/s 2, directed downward Unknowns y f, which happens to be the information that we are seeking Slide 5

6 Procedure for Step 4 There are a couple of different (but definitely connected) ways to approach this problem. I ll proceed using concepts of energy. To this end, the relevant physical principle is that gravity is a conservative force, and thus the total mechanical energy of the ball-earth system is conserved. The sum of kinetic energy (K) and potential energy (U) is the same at the release point as it is at the point of maximum height. That is, K i + U i = K f + U f. Next, we recall that any mass m moving with a speed v has a kinetic energy equal to K = 1 2 mv2. For the present problem, we can write K i = 1 2 mv i 2 and K f = 1 2 mv f 2 Slide 6

7 Procedure for Step 4 cont As for the potential energy, recall that we get to choose the reference level at which we define U to equal zero (only changes in potential energy have any physical meaning). We will choose the reference level to occur at the release point. Thus, U i = 0. Furthermore, we know that the gravitational potential energy of a mass m at a height y f above a reference level y i is U f = mg(y f y i ) Then, conservation of mechanical energy tells us K i + U i = K f + U f or, 1 2 mv i = 1 2 mv f 2 + mg(y f y i ) Slide 7

8 Procedure for Step 4 cont Now we must rearrange this equation to express the desired unknown y f as a function of the knowns: 1 2 mv i = 1 2 mv f 2 + mg(y f y i ) 1 2 m v i 2 v 2 f = mg y f y i v 2 2 i v f = y 2g f y i y f = y i + v i 2 2 v f This is where your algebra skills will come in handy (or, where your lack of algebra skills will spell your doom). 2g Slide 8

9 Procedure for Step 5 We are finally ready to plug in our known numerical values for v i, v f, y i, and g: y f = y i + v i 2 2 v f = 0 m + 20 m/s 2 0 m/s 2 2g 2(9.8 m = 20.4 m s 2) Step 6 An athletic person can throw a ball with an initial speed of 20 m/s, and it s not unreasonable for such a throw to reach a height of 20.4 m. Slide 9

10 Further Hints for 1. Solve problems algebraically, only inserting numbers (and using your calculator) at the very last moment. This is useful in two regards: When you perform numerical calculations at each intermediate step, you are accumulating round-off errors. After multiple steps, your final answer might be significantly off from the correct answer. The algebraic result is much more useful than the numerical result. Recalling the motorcycle example from earlier, the more profound result is that any object moving for a time t at an average speed v will travel a distance d = vt. The numerical result of d = 6 km is really only valid for this particular motorcycle in this one instance. Furthermore, in the case of the ball, our algebraic approach indicated that the maximum height is independent of the ball s mass it simply cancelled out! Slide 10

11 Further Hints for 2. When answering multi-part questions, use the appropriate number of significant figures in your answer, but carry all figures to the next part of the question. This also helps to prevent the accumulation of round-off errors. 3. Check the dimensions of your intermediate steps and of your answer. This won t necessarily confirm that you have the right answer, but it can immediately tell you if your answer is incorrect. Furthermore, keep in mind that (i) quantities can only be added to or subtracted from each other if they have the same dimensions, and (ii) the argument of trigonometric, exponential, and logarithmic functions must be dimensionless. This is a good time to remind ourselves of the difference between dimensions and units. Dimensions tell us if we are referring to a length, a mass, a time, etc. Units are various conventions for expressing a given dimension. Each dimension can be expressed in different units, but a unit can not refer to different dimensions. For example, the dimension of length can have units of meters, kilometers, feet, cubits, furlongs, etc. Slide 11

12 Further Hints for 4. If you can think of two different methods to solve a problem, try them both. If they lead to the same answer, you can be even more confident that you are correct. Also, the mere fact that the two methods are equivalent is itself quite an important result! As an example, the previous problem with the ball can be solved entirely using constant-acceleration kinematic equations instead of energy conservation. This must be the case, since the latter is governed by Newton s 2 nd law, which in this case leads precisely to a constant-acceleration motion. Slide 12

13 Further Hints for 5. Check the order of magnitude of your answer. This is otherwise known as the sanity check. The orbital radius of a planet around the sun isn t a few millimeters, and nobody can throw a baseball 1,000 km. If your dimensions are correct, and your numerical answer is non-sensical, you have probably misplaced some powers of 10 (for instance, converting km to meters). 6. Check limiting and/or special cases. Many physical problems have simple or intuitive solutions for certain values of the parameters or the dependent variable(s). When you think that you have solved a problem, check to see that it satisfies these solutions. The ability to perform these checks may be a bit difficult at first, but it will become easier with practice. Slide 13

14 Further Hints for Example: Two masses colliding in one dimension A mass m with velocity v approaches a stationary mass M. The masses collide elastically in one dimension. You calculate that the velocities after the collision are v m = m M m+m v and v M = 2m m+m v First of all, it should be easy to check that both equations have the proper dimensions of velocity. Then, check these limiting cases: 1. If m = M, our answer suggests that m stops and M attains a velocity v. This is quite believable, especially if you ve ever played pool. 2. If m is much less massive than M, our answer suggests that M basically remains stationary while m bounces backwards with nearly the initial speed v. This also makes sense, since from m s point of view, M is essentially a brick wall. Slide 14