course 201
3:28pm, 09/11
Class 090909vvvv
Rubber band measurements
You have data that allows you to determine the length of the thin rubber bands and the length of the thick rubber band, for each setup.
Report the span of your hand on the 'ruler' you used to measure the rubber band system, as well as your height in inches. Report as two numbers separated by commas:
&&&& (report your numbers starting on the next line)
33, 68
Report you raw data below (this will be the quantities you actually measured in class; note that you didn't measure the lengths of the rubber bands but the positions of their ends, so your raw data will not include the lengths). :
&&&& Your raw data from class should be reported starting in the next line:
10, 49
A data report needs to be followed by an explanation of how the data was obtained and what it means.
* In the first setup you had a single 'thin' rubber band opposing the single 'thicker' rubber band. You observed this system for three different degrees of stretch.
Beginning in the line below, report the lengths of the 'thin' rubber band and the length of the 'thick' rubber band, as determined from your raw data. Each line should be reported as two numbers, separated by a comma:
&&&& (lengths of 'thin' then 'thick' bands, first trial (least stretch)):
20.5, 16.5
&&&& (lengths of 'thin' then 'thick' bands, second trial (medium stretch)):
24, 19
&&&& (lengths of 'thin' then 'thick' bands, third trial (greatest stretch observed)):
24.7, 19.3
Now sketch a graph of y vs. x, where y = length of 'thin' rubber band and x = length of 'thick' band. Your graph will consist of three points.
These results make perfect sense.
However the only data you reported was in the single line
10, 49.
Your raw data do not support the numbers you reported here.
&&&& What is the slope between the first and second point on your graph (give the slope in the line below, then starting in a new line explain how you calculated the slope)?
1.4
-I got the slope by using the slope formula (y2-y1)/(x2-x1)=m. (24-20.5)/(19-16.5) = 1.4
&&&& What is the slope of the graph between the second and third point on your graph (give the slope in the line below, then starting in a new line explain how you calculated the slope)?
2.33
-Once again, I used the slope formula. (24.7-24)/(19.3-19) = 2.33
good
&&&& Do you believe that the difference in the slopes is due mainly to unavoidable errors in measurement, or to the actual behavior of the rubber bands? Support your answer with the best arguments you can.
-I believe its the nature of the rubber bands. The thicker is harder to stretch since its denser, making the thinner one do most of the stretching, which increases the slope.
* In the second setup you had three 'thin' rubber bands, all stretched between the same two paper clips, opposing the single 'thicker' rubber band. You observed this system for three different degrees of stretch.
Beginning in the line below, report the lengths of the 'thin' rubber bands and the length of the 'thick' rubber band, as determined from your raw data. Each line should be reported as two numbers, separated by a comma:
&&&& (lengths of 'thin' then 'thick' bands, first trial (least stretch)):
14, 16.5
&&&& (lengths of 'thin' then 'thick' bands, second trial (medium stretch)):
15.5, 18.5
&&&& (lengths of 'thin' then 'thick' bands, third trial (greatest stretch observed)):
16, 19.5
Now sketch a graph of y vs. x, where y = common length of 'thin' rubber bands and x = length of 'thick' band. Your graph will consist of three points.
&&&& What is the slope between the first and second point on your graph (give the slope in the line below, then starting in a new line explain how you calculated the slope)?
.75
-I plugged in the numbers for the slope. (15.5-14)/(18.5-16.5) = .75
&&&& What is the slope between the second and third point on your graph (give the slope in the line below, then starting in a new line explain how you calculated the slope)?
.5
-Once again, I plugged in the numbers. (16-15.5)/(19.5-18.5) = .5
&&&& Do you believe that the difference in the slopes is due mainly to unavoidable errors in measurement, or to the actual behavior of the rubber bands? Support your answer with the best arguments you can.
-Again, I believe its due to the behavior of the rubber bands. The thin rubber bands may have been easier to stretch, but with three of them together, they had less elasticity than the thick rubber band. So the thick rubber band would stretch further, making X larger than Y each time.
Question to think about (and answer as best you can): You you think that during the measurement process, the rubber bands were being held apart by force, power, energy or something else, or perhaps by all three? Answer beginning in the line below, and support your answers with your best reasoning. Don't worry about being wrong, but do give it some thought. &&&&
-I believe they were being held apart by force. It wasnt energy that was holding them apart, because the energy in the bands was trying to pull them together, as is their nature.
Acceleration of Toy Cars in Two Opposite Directions
You also measured the motion of a couple of toy cars, moving in two opposite directions. You were asked to obtain information that will tell you the acceleration of each car in each direction.
&&&& Give your raw data in starting in the line below. Be sure to include all information necessary to interpret your data.
Pick 'north' or 'south' for your positive direction. State your choice: &&&&
-North.
For the first trial in which the car moved in the northerly direction, explain how you reason out the acceleration. Show how you reason out your results, starting with the raw data, based on the definitions of rate of change, average velocity and average acceleration. You may assume that the acceleration is uniform, so that the v vs. t graph is in fact trapezoidal. Give you explanation starting in the line below:
&&&& -Time interval = 2 full pendulum cycles(60 cycles per minute), displacement = +77cm
-If the average rate of acceleration is calculated by (Vf-Vo)/deltaT, then itd be (0-7cm)/2 = -38.5cm/s
Repeat for the first trial in the southerly direction:
&&&& - Time interval = 2 full pendulum cycles(60 cycles per minute), displacement = -77cm
-If the average rate of acceleration is calculated by (Vf-Vo)/deltaT, then itd be (0+77cm)/2 = +38.5cm/s
Find the acceleration for the next trial; however you may abbreviate your calculations and don't need to repeat the verbal explanations:
&&&& -deltaT = 2.5 cycles, displacement = -78.5cm
- (0+78.5)/2 = +39.25cm/s
(you may copy the two lines above as many times as necessary to account for all the trials you think it necessary to report your results; for these additional lines your work may be further abbreviated)
Intervals
An interval runs from some initial event to a final event.
Example: f an object's velocity is increasing at it coasts down a ramp, we might speak of the interval between first reaching velocity 2 meters / second and first reaching velocity 5 meters / second.
The initial event would be the event of reaching 2 meters / second for the first time.
The final event would be the event of reaching 5 meters / second for the first time.
In the case of the given information:
2 meters/second would be the initial velocity, which we would designate by v0.
5 meters/second would be the final velocity, which we would designate by vf.
Furthermore we can speak of initial and final positions and clock times, which we might or might not know:
The initial event would occur at some initial clock time, which could be denoted by t0.
The final event would occur at some initial clock time, which could be denoted by tf.
The time interval corresponding to this interval would be the change in clock time, `dt = tf - t0.
The initial event would occur at some position, which could be denoted by s0.
The final event would occur at some final position, which could be denoted by sf.
The change in position on this interval, also called the displacement, would be `ds = sf - s0.
In an experiment, any of the quantities t0, tf, `dt, s0, sf or `ds might be directly measured using a timing device (e.g., a pendulum, a stopwatch, a computerized timer) or a position-measuring device (e.g., a meter stick or a ruler). There would of course be some uncertainty in the measurement. We would not generally measure all these quantities, but we could.
Depending on the object, we might be able to obtain some measure of v0, vf or `dv. More commonly, we would infer these quantities by measuring a series of positions or clock times.
What might be the initial event and the final event in each of the following situations, and what quantities are given for each?
A drag racer completes a 1/4 mile time trial in 12 seconds. &&&&
-Vo = 0m/s, deltaT= 12s, deltaS= .25 miles
A muon created in the upper atmosphere spends 12 milliseconds in the atmosphere before disintegrating. &&&&
-deltaT=12ms, Vf=0m/s
After receiving a handoff at his own 10-yard line, with the game clock reading 3:42, a fullback gains 30 yards before being tackled with the game clock reading 3:37. &&&&
-Vf=0m/s, deltaT=5s, deltaS= 30yd
A mosquito hatched at 9:00 a.m. is eaten by a bat at 3:42 p.m. of the same day, and a point 40 feet north of where it hatched. &&&&
-Vf=0m/s, deltaT=6hrs, 42mins, deltaS=40ft
The quantities you give in all of these questions are correct for the implied intervals.
However the question asked you to identify the initial event and the final event. You've done this implicity (that is, your correct identification of the quantities implies that you know what the interval is), but you also need to explicitly name those events.
Deriving the equations of uniformly accelerated motion from the definitions of rate, velocity and acceleration:
Recall from previous classes that the definitions of rate, velocity and acceleration, along with the assumption of uniform acceleration, lead us to the following equations for uniformly accelerated motion on an interval:
Equation 1: `ds = (vf + v0) / 2 * `dt
Equation 2: a_ave = (vf - v0) / `dt.
Both of these equations are easily derived by finding the slope and area of a v vs. t trapezoid with altitudes v0 and vf, and base `dt.
The five quantities which appear in these equations are:
v0, the velocity at the beginning of the interval
vf, the velocity at the end of the interval
`dt, the change in clock time during the interval
a_ave, the average acceleration on the interval
`ds, the change in position, or displacement, on the interval.
It's worth noting that each of the two equations includes four of these five quantities.
We can derive two additional equations as follows:
Solve the second equation for vf and plug the resulting expression into the first equation, eliminating vf and obtaining the equation
Equation 3: `ds = v0 `dt + .5 a `dt^2
Solve the second equation for `dt and plug the resulting expression into the first equation, eliminating `dt and obtaining the equation
Equation 4: vf^2 = v0^2 + 2 a `ds.
The details of the algebra are given below:
Derivation of Equation 3:
Start with Equation 2, and solve for vf:
a_Ave = (vf - v0) / `dt. Multiply both sides by `dt:
a_Ave * `dt = (vf - v0) / `dt * `dt. The right-hand side rearranges to (vf - v0) * (`dt / `dt), and `dt / `dt = 1 so
a_Ave * `dt = vf - v0. Add v0 to both sides
a_Ave * `dt + v0 = vf - v0 + v0. Since -v0 + v0 = 0 we get
a_Ave * `dt + v0 = vf. Switch right- and left-hand sides of the equation and reverse the order of addition:
vf = v0 + a_Ave * `dt.
Now plug this expression into Equation 1:
`ds = (vf + v0) / 2 * `dt. Substituting v0 + a_Ave * `dt for vf we have
`ds = (v0 + a_Ave * `dt + v0) / 2 * `dt. Simplify the numerator of the right-hand side by adding the two v0 terms:
`ds = ( 2 v0 + a_Ave * `dt) / 2 * `dt. Applying the distributive property of multiplication over addition we have
`ds = (2 v0) / 2 * `dt + a_Ave * `dt / 2 * `dt. Since (2 v0) / 2 = (2 / 2) * v0 = 1 * v0 = v0, `dt * `dt = `dt^2, and 1/2 = .5 we have
`ds = v0 * `dt + .5 a_Ave * `dt^2.
This is Equation 3.
Derivation of Equation 4:
We first solve Equation 2 for `dt.
Starting with
a_Ave = (vf - v0) / `dt, we multiply both sides by `dt to obtain
a_Ave * `dt = vf - v0 (this was also done and fully explained in the derivation of Equation 3 above).
Now divide both sides by a_Ave to get
`dt = (vf - v0) / a_Ave.
Now plug this expression into Equation 1:
`ds = (vf + v0) / 2 * `dt. Substituting (vf - v0) / a_Ave for `dt we get
`ds = (vf + v0) / 2 * (vf - v0) / `a_Ave. This gives us
`ds = (vf + v0) * (vf - v0) / (2 * a_Ave).
Note that by the distributive law (vf + v0) * (vf - v0) = vf * (vf - v0) + v0 * (vf - v0) = vf * vf - vf * v0 + v0 * vf - v0 * v0 = vf^2 - vf * v0 + vf * v0 - v0^2 = vf^2 - v0^2, so our equation becomes
`ds = (vf^2 - v0^2) / (2 * a_Ave) .
We choose to solve this equation for vf^2.
We first multiply both sides by 2 * a_Ave:
`ds * (2 * a_Ave) = (vf^2 - v0^2) . Then add v0^2 to both sides to get
2 * a_Ave * `ds + v0^2 = vf^2 - v0^2 + v0^2. Since -v0^2 + v0^2 = 0, after switching left-and right-hand sides we have
vf^2 = v0^2 + 2 a_Ave `ds.
This is Equation 4.
Using the Equations of Uniformly Accelerated Motion
The four equations of uniformly accelerated motion, as derived above, are:
Equation 1: `ds = (vf + v0) / 2 * `dt
Equation 2: a_Ave = (vf - v0) / `dt
Equation 3: `ds = v0 * `dt + .5 * a_Ave * `dt^2
Equation 4: vf^2 = v0^2 + 2 a_Ave * `dt^2.
Since acceleration is uniform, we don't really need the subscript _Ave in our symbol a_Ave. Instead we can just write a.
Equation 2 is fine the way it appears, but is often more convenient to use rearranged to the form vf = v0 + a_Ave * `dt (starting with Equation 2 as expressed above, multiply both sides by `dt, then add v0 to both sides and switch right- and left-hand sides).
So the equations are more commonly expressed as
Equation 1: `ds = (vf + v0) / 2 * `dt
Equation 2: vf = v0 + a * `dt
Equation 3: `ds = v0 * `dt + .5 * a * `dt^2
Equation 4: vf^2 = v0^2 + 2 a * `dt^2.
We will use the equations in this form. (Note that your text has an equivalent but slightly different form of the equations; we'll explain the difference later).
You should verify the following properties of these equations for yourself:
The equations are expressed in terms of the five variables v0, vf, `dt, a and `ds.
Each equation includes exactly four of these five variables.
Every possible combination of three of the five variables appears in at least one of the four equations (for example, v0, `dt and `ds is one possible combination of three of the five variables, and all appear in Equation 1; in addition all three appear in Equation 3).
It follows that if we know any three of the five quantities v0, vf, `dt, a and `ds, we can find at least one equation in which these quantities all appear; and furthermore since there are only four quantities in any one equation, that equation can be solved to find the value of the fourth quantity.
We will do a number of exercises as we learn to apply this scheme.
* Exercise 1: You have already analyzed your information for the race cars you observed in class. Now do the following:
Write down on your paper the symbols v0, vf, a, `dt, `ds.
From your raw data you can determine `ds and `dt.
The initial and final events can be described as 'car leaves end of finger' and 'car comes to rest'. So you also know the value of which of the five quantities? &&&&
-Vf, deltaT, deltaS
On your paper circle the symbols for the three quantities you know.
Now write down all four equations, and circle the symbols for the three quantities you know (i.e., circle vf, `ds and `dt in all four equations).
Write down the one equation which includes all three symbols, and circle these symbols in the equation. Which equation did you write down, and which symbol was not circled? &&&&
-Equation 1. Vo.
Solve this equation for the non-circled variable and describe the steps necessary to do so. If your algebra is rusty you might find this challenging, but make your best attempt. We will be discussing this much more fully in our next class, but you at least need to get the wheels turning, and your instructor needs to know what you can and cannot do with the algebra. &&&&
- (deltaS = ((Vf+ Vo)/2)deltaT) The first step will be to bring the deltaT over from the right-hand side to the left side by dividing both sides by deltaT. ((deltaS/deltaT) = (Vf+Vo)/2) The next step is get rid of the fraction on the right side by multiplying both sides by 2. (2(deltaS/deltaT) = (Vf + Vo)) The last step is to get Vo by itself by subtracting Vf by both sides. (2(deltaS/deltaT) -Vf = Vo)
Having solved the equation as best you can, substitute the values of the three known quantities vf, `ds and `dt into that equation. Then simplify your expression to get the value of the unknown quantity. Again, this will take some practice and you might have made some errors, but do your best, for the same reasons outlined above. &&&&
-(77cm = ((0cm/s + Vo)/2)2s) Divide by 2s. (36.5cm/s = (0cm/s+Vo)/2) Multiply by 2. (77cm/s = (0cm/s + Vo)) Subtract 0cm/s. (77cm/s = Vo)
* Exercise 2
A ball is released from rest on a ramp of length 4 meters, and is timed from the instant it is released to the instant it reaches the end of the ramp. It requires 2 seconds to reach the end of the ramp.
What are the events that define the beginning and the end of the interval? &&&&
-Letting go of the ball and the ball hitting the end of the ramp.
Write down on your paper the symbols v0, vf, a, `dt, `ds.
From the given information you know the values of three of the five quantities. What are the known quantities? &&&&
-Vo=0m/s, dT = 2s, dS = 4m
On your paper circle the symbols for the three quantities you know.
Now write down all four equations, and circle the symbols for the three quantities you know.
Write down an equation which includes all three symbols, and circle these symbols in the equation. Which equation did you write down, and which symbol was not circled? &&&&
-Equation 1. Vf.
Solve this equation for the non-circled variable and describe the steps necessary to do so. If your algebra is rusty you might find this challenging, but as before make your best attempt. &&&&
-(deltaS = ((Vf+ Vo)/2)deltaT) The first step will be to bring the deltaT over from the right-hand side to the left side by dividing both sides by deltaT. ((deltaS/deltaT) = (Vf+Vo)/2) The next step is get rid of the fraction on the right side by multiplying both sides by 2. (2(deltaS/deltaT) = (Vf + Vo)) The last step is to get Vf by itself by subtracting Vo by both sides. (2(deltaS/deltaT) -Vo = Vf)
Having solved the equation as best you can, substitute the values of the three known quantities into that equation. Then simplify your expression to get the value of the unknown quantity. Again, do your best. &&&&
-(4m = ((Vf + 0m/s)/2)2s) Divide by 2s. (2m/s = (Vf + 0m/s)/2) Multiply by 2. (4m/s = Vf + 0m/s) Subtract 0m/s. (4m/s = Vf)
* Exercise 3
A ball is dropped from rest and falls 2 meters to the floor, accelerating at 10 m/s^2 during its fall.
What are the events that define the beginning and the end of the interval? &&&&
-Letting go of the ball and the ball hitting the ground.
Write down on your paper the symbols v0, vf, a, `dt, `ds.
From the given information you know the values of three of the five quantities. What are the known quantities? &&&&
-A, Vo, dS
On your paper circle the symbols for the three quantities you know.
Now write down all four equations, and circle the symbols for the three quantities you know.
Write down the one equation which includes all three symbols, and circle these symbols in the equation. Which equation did you write down, and which symbol was not circled? &&&&
-Equation 4. Vf.
There are two equations which each contain three of the five symbols. Write down the other equation and circle the three known symbols in the equation. Which equation did you write down, and which symbol was not circled? &&&&
-Equation 3. dT.
One of your equations has `dt as the 'uncircled' variable. You want to avoid that situation (though if you're ambitious you may give it a try). Solve the other equation for its non-circled variable (which should be vf) and describe the steps necessary to do so. If your algebra is rusty you might find this challenging, but as before make your best attempt. &&&&
-You start out with the equation, (Vf^2 = (dS * 2A) + Vo^2). Your first step is to multiply the dS and 2A. (Vf^2 = (2AdS) + Vo^2) The last thing you will do is get the square root of everything. (Vf = (2AdS)^.5 + Vo)
Having solved the equation as best you can, substitute the values of the three known quantities into that equation. Then simplify your expression to get the value of the unknown quantity. Again, do your best. &&&&
-(Vf^2 = (dS * 2A) + Vo^2) Enter data. (Vf^2 = (2m * 2(10m/s^2)) + 0m^2/s^2) Multiply 2 by 10m/s^2. (Vf^2 = (2m * 20m/s^2) + 0m^2/s^2) Multiply 2m by 20m/s^2. (Vf^2 = 40m^2/s^2 + 0m^2/s^2) Square root everything. (Vf = 40m/s + 0m/s) The answer is *Vf = 40m/s.*
Introductory Problem Sets
One very good reason to understand these problems is that they often appear on tests. You don't need to turn them in, but you are expected to have attempted them and studied the solutions. You may of course ask questions about these problems and their solutions.
This is a repeat from last week's class.
Work through Introductory Problem Set 1 (http://vhmthphy.vhcc.edu/ph1introsets/default.htm > Set 1). You should find these problems to be pretty easy, but be sure you understand everything in the given solutions.
You should also preview Introductory Problem Set 2 (http://vhmthphy.vhcc.edu/ph1introsets/default.htm > Set 2). These problems are a bit more challenging, and at this point you might or might not understand everything you see. If you don't understand everything, you should submit at least one question related to something you're not sure you understand.
The difference between formulas and principles:
Question from last week's assignment:
If you have numbers for v0, vf and `dt how would you use them to find the following:
the change in velocity on this interval &$&$
One frequent student response:
Plug in the numbers for the formula vf - vo
Instructor's response:
Your answer is technically correct, but I prefer that answers be expressed in terms of principles and concepts rather than formulas. In this case, I know that you know the principle that the change in a quantity is equal to its final value minus its initial value, and that you could have easily expressed your answer in terms of this principle.
For the benefit of the entire class I'm going to expand on this idea:
One problem physics students often have is that they think they think in terms of memorizing formulas.
However it's much easier to learn the subject by learning a few principles and learning to reason from those principles.
The way I often put this is:
Why learn 1000 formulas when you can learn a dozen principles?
In this case it is better to use the principle that
the change in a quantity is found by subtracting its initial value from its final value
than to quote a formula that applies only to the change in velocity.
At this point in the course you have two alternatives.
The first alternative is to memorize the following separate formulas.
`dx = x_f - x_0
`ds = s_f - s_0
`dt = t_f - t_0
`dv = vf - v0
`da = a_f - a_0
Having memorized these you would then soon have to memorize the following:
`dF = F_f - F_0
`dp = p_f - p_0
`dy = y_f - y_0
`dKE = KE_f - KE-0
`dPE = PE_f - PE_0
`domega = omega_f - omega_0
`dtheta = theta_f - theta_0
`dalpha = alpha_f - alpha_0
as well as about a dozen formulas of this nature.
The second alternative is to learn to apply the principle that
the change in a quantity is found by subtracting its initial value from its final value
and apply this principle to whatever situation arises.
At this point in your course everything you do is done in terms of the concept of an interval, using a few definitions
the definition of average rate of change
the definition of velocity
the definition of acceleration
the definition of a graph trapezoid
the geometry of a graph trapezoid
applied using a few principles
the principle that all quantities with units must always be expressed in terms of units
the principle that the change of a quantity on an interval is found by subtracting its initial value on that interval from its final value on that interval
expressed in terms of some basic mathematics
basic arithmetic (including the arithmetic of fractions)
the basic rules of algebra (including the algebra of fractions)
very basic geometry
"
Very good work.
It's safe to say that you have pretty much got it at this point.
See my notes and try to clear up the last few details.