A ball rolling down a ramp hits a car which moves off uphill. Can you come up with an experiment to predict how far the car will move if the ball is released from any height? What if different masses of balls are used?
Students try to figure out the relationship between the ball's release height and how far the car moves.
For my middle school class, who’ve been dealing with linear relationships all year, they could do this easily if the distance the car moves is directly proportional to height from which the ball was released?
The question ultimately comes down to momentum, but I really didn’t know if the experiment would work out to be a nice linear relationship. If you do the math, you’ll find that release height and the maximum distance the car moves are directly proportional if the momentum transferred to the car by the ball is also directly proportional to the velocity at impact. Given that wooden ball and hard plastic car would probably have a very elastic collision I figured there would be a good chance that this would be the case and the experiment would work.
It worked did well enough. Not perfectly, but well enough.
The bright blue iridescence of the wings of this insect results from the way light refracts through the thin layered membranes of the wings.
When you look at the sunlight reflected off this black insect’s wings at just the right angle, they blaze bright blue. The phenomena is called iridescence, and results from the way different wavelengths of light refract through the wing membrane. Blue light is of just the right wavelength that the light reflected off the top of the membrane and the light that’s refracted through the membrane constructively interfere. The Natural Photonics program at the University of Exeter has an excellent page detailing the physics of iridescence in butterflies (Lepidoptera), and the history of the study of the subject.
Surface tension supports the water strider on the surface of the still water of the creek.
Down at the creek the water striders are out. They can stand, walk and jump on the surface of the water without penetrating the surface because of the force of surface tension that causes water molecules to stick together — it’s the same cohesive force that make water droplets stick to your skin. I got a decent set of photos to illustrate surface tension.
The water striders still create ripples on the surface of the water, even though they never break the surface.
The green canopy that over hangs the creek allows for some nice photographs.
One of the middle-school projects is to build a little solar water heater. By simply pumping water through a black tube that’s sitting in the sun, you can raise the temperature of the water by about 15°C in about 15 minutes.
The solar water heater in action.
Next year I want to try building an actual solar water heater, similar to the passive air heater my students built two years ago, with the tubing in a greenhouse box to see just how efficient we can make it.
The area around the Sun is blue on Mars because the gasses in the thin atmosphere don't scatter much, but the Martian dust does (it scatters the red). Image via NASA.
The dust in Mars’ atmosphere scatters red, while the major gasses in Earth’s atmosphere (Nitrogen and Oxygen) scatter blue light. Longer wavelengths of light, like red, will bounce off (scatter) larger particles like dust, while shorter wavelengths, like blue light, will bounce of smaller particles, like the molecules of gas in the atmosphere. The phenomena is called Rayleigh scattering, and is different from the mechanism where different molecules absorb different wavelengths of light.
Ezra Block and Robert Krulwich go into details on NPR.
Blue sky in the upper right, but the dust scatters the red light.
The discussion started off with projectiles. If you’re floating in space — zero gravity — and throw something, like a basketball, away from you, you’ll be pushed off in the opposite direction. In fact, if you throw something that has the exact same mass as you do away from yourself, you’ll move off in the opposite direction with the exact same speed as the thing you threw.
Then I brought up rockets, and how they’re expelling gas to move them forward. I think it was the phrase, “expelling gas” that did it. The next question, which the student brought up somewhat circumspectly, sidling around the issue and the language, was (more or less), “So if you expel gas in space will you move off in the other direction?”
The simple answer, appropriate to that stage of the discussion, was, of course, “Yes.”
Which lead to to, “What about spitting?”
“Yes.”
“What about, you know, peeing?”
“Yes, except …”
At that point I thought it would be wise to rein it in a little, and make a further point about the whole action-reaction thing.
“You see, if you expel anything, wouldn’t it just be stuck in your spacesuit with you? Then you’re not really expelling it, it’s still attached to you, so you wouldn’t really move. What would be more useful would be to collect it in something like a spray can or a squeeze bottle. Then you can just squirt it out opposite the direction you want to go in to control your movement.”
This produced a moment of thoughtful silence as they figured out the logistics.
Notes
I thought this was a useful conversation to have. The students were interested and animated. And I believe it’ll be memorable too.
An artist's concept depicts the Deep Space 1 probe with its ion engine operating at full thrust. Image via NASA.
P.S.: I’d wanted to talk about ion drives, which operate on the same reaction principle, but are much cooler (after all they’re used in Star Trek). Instead of burning fuel to create the propulsive force ion drives generate an electric field that ejects charged particles; we’d been talking about ions and charged particles earlier in the day. However, I decided on the day that it would just complicate what was a new issue. I’ll probably bring it up this week though as we recurse through Newton’s Laws.
PhD Comics does a wonderful job of explaining of sub-atomic particles: what we know, what we don’t know. What’s particularly great about this video is that it goes into how physicists are using the Large Hadron Collider to try to discover new particles: by making graphs of millions of collisions of particles and looking for the tiniest of differences between different predictions of what might be there.
I also like how clear they make the fact that science is a processes of discovery, and what fascinates scientists is the unknown. Students do experiments all the time and if they don’t find what they expect — if it “doesn’t work” — they’re usually very disappointed. I try my best to let them know that this is really what science is about. When your experiment does not do what you want, and you’re confident you designed it right, then the real excitement, the new discoveries, begin.
I punched a small hole (about 1mm radius) in a one gallon plastic bottle and had my students measure the rate at which water drained. Even though the apparatus and measurement technique was fairly rough, we were able to, with a little calculus, determine the equation for the height of the water in the bottle as a function of time.
Figure 1. A student uses a stopwatch to measure the outflow rate of water from the plastic bottle.
Introduction
Questions about water draining from a tank are a pretty common in calculus textbooks, but there is a significant difference between seeing the problem written down, and having to figure it out from a physical example. The latter is much more challenging because it does not presuppose any relationships for the change in the height of water with time; students must determine the relationship from the data they collect.
The experimental approach mimics the challenges faced by scientists such as Henry Darcy who first determined the formulas for groundwater flow (Darcy, 1733) almost 300 years ago, not long after the development of modern calculus by Newton and Leibniz (O’Conner and Robertson, 1996).
Procedure
Figure 2. The apparatus.
I punched a small hole, about 1mm in radius, in the base of a plastic, one-gallon bottle. I chose this particular type of bottle because the bulk of it was cylindrical in shape.
Students were instructed to figure out how the rate at which water flowed out (outflow rate) changed with time, and how the height of the water (h) in the bottle changed with time. These relationships would allow me to predict the outflow rate at any time, as well as how much water was left in the container at any time.
Data collection:
To measure height versus time, we marked the side of the bottle (within the cylindrical region) in one centimeter increments and recorded the time it took for the water level to drop from one mark to the next. There were a total of 11 marks.
To measure the outflow rate, we intercepted the outflow using a 25 ml beaker (not shown in Figure 2), and measured the time it took to fill to the 25 ml mark.
Results
Time elapsed since last measurement
Height of water in container
Time to fill 25ml beaker
Δt (s)
h (cm)
tf (s)
0.0
11
6.4
45.5
10
7.1
52.3
9
8.5
43.1
8
7.1
56.1
7
7.8
60.6
6
8.3
57.6
5
8.9
65.0
4
9.9
72.8
3
11.2
76.7
2
13.0
91.1
1
14.8
122.0
0
17.3
Table 1: Outflow rate, water height change with time.
To analyze the data, we calculated the total time (the cumulative sum of the elapsed time since the previous measurement) and the outflow rate. The outflow rate is the change in volume with time:
So our data table becomes:
Time
Height of water in container
Outflow rate
t (s)
h (cm)
dV/dt (ml/s)
0.0
11
3.91
45.5
10
3.52
97.8
9
2.94
140.9
8
3.52
197
7
3.21
257
6
3.01
315.1
5
2.81
380.1
4
2.53
452.9
3
2.23
529.6
2
1.92
620.7
1
1.69
742.7
0
1.45
Table 2. Height of water and outflow rate of the bottle.
The graph of the height of the water with time shows a curve, although it is difficult to determine precisely what type of curve. My students started by trying to fit a quadratic equation to it, which should work as we’ll see in a minute.
Figure 3. The decrease in height with time is not a straight line (is non-linear).
The plot of the outflow rate versus time, however, shows a pretty good linear trend. (Note that we do not use the first three datapoints (in Table 2), which we believe are erroneous because we were still sorting out the measuring method.)
Although I’ll note here that the data should ideally be modeled using a square root equation (Torricelli’s Law), that is beyond the present scope of the problem (we’ll try that tomorrow as a follow exercise).
Plotting the data in Excel we could add a linear trendline.
Figure 4. The outflow rate decreases linearly with time.
For the linear trendline, Excel gives the equation:
The y-axis is outflow rate (the change in volume with time), and the x axis is time (t), so the linear equation becomes:
Notice that this is a differential equation.
To determine the rate of at which the height of water in the container is changing, we need to recognize that the container is cylindrical in shape, and the volume (V) of a cylinder depends on its radius (r) and height (h):
which can be substituted into differential in the rate equation:
since π and the radius (r) are constants (since the jug’s shape is a cylinder), they can be pulled out of the differential:
Dividing through by πr2 solves for the rate of change of height:
Isolating the coefficients gives:
This equation should give the rate at which the height changes with time, however, if you look at it carefully you’ll realize that for the time range we’re using (less than 800 seconds) the value of dh/dt will always be positive. We correct this by recognizing that the outflow rate is a loss of water, so a positive outflow should result in a negative change in height, therefore we rewrite the equation as:
which gives:
Now comes the calculus
Now, we can use this rate equation to find the equation for height versus time by integrating with respect to time:
to get:
And all we have to do to the find the constant of integration is substitute in a known point, an initial value. As is often the case, the best point to use is the starting point where t=0 makes the rest of the calculations easier. In our case, when t=0, h=11:
so:
And our final equation becomes:
which is a quadratic equation as my students guessed before we did the calculus.
Does it work?
You will notice that in the math above, we never use the height data in determining equation for height versus time; all the calculations are based on the trendline for the outflow rate versus time.
As a result, we can compare the results of our equation to the actual measurements to see if our calculations are even close. Remarkably, they are.
Figure 5. The results from our equation (modeled) match the measured heights so well, the data points are difficult to distinguish on the graph.
Discussion
Despite all the potential for error, particularly, our relatively crude measurement techniques, and the imperfect cylindrical shape of our plastic bottle, the experiment went remarkably well.
Students found it quite challenging, and required some assistance even though this is a problem they have seen before in their textbook.
The problems in the textbook use Torricelli’s Law, which should much better describe the draining of a tank than the linear equation we find for dV/dt.
Torricelli’s Law:
where a is the area of the outlet hole, and g is the acceleration due to gravity.
In our actual experiment it is difficult to tell that a square root function would work better. Excel does not have an option for matching a square root function, so the calculations would become more involved (although it could be set up using Excel’s iterative solver or Goal Seek function).
Conclusion
Our experiment to use calculus to determine the rate of change of the height of water in a leaking plastic water bottle was a successful exercise even though the roughness of the data collection did not permit identification of the square root law for leakage.
