# Why can't you do it like that?

1. Mar 28, 2016

### Joe90

Hi. I have a question about the best way to explain why you can't solve a physics problem the way one of my students tried to. Here's the problem:

Lauren is running at 3.2 m s-1 when the road starts to go downhill. When she reaches the bottom of the hill she has dropped a vertical distance of 15 m. If all the gravitational potential energy she loses on the way down had been changed to kinetic energy, what would Lauren’s speed have been at the bottom of the hill?

Instead of calculating the total kinetic energy at the bottom = Ek(top) + Ep(lost) and finding v from that (= 17.6 ms-1), my student calculated the "added speed" from the gravitational potential energy lost (17.3 ms-1) and then added the initial speed (3.2 ms-1) to get 20.5 ms-1.

What's the best way of explaining why this doesn't work out?

Thanks for any help.

2. Mar 28, 2016

### PeroK

This is quite a common mistake. If you think about the force of gravity, then the faster you are running to begin with, the less time it takes you to run down the hill, so the less time gravity has to accelerate you. In other words:

Added velocity = Force x Time

Time will be less for someone already running.

A better scenario is simply to look at objects falling a set distance. 5m, say.

If the object starts from rest, it takes 1 sec to fall 5m, so it gains 10m/s.

If the object starts at 100m/s, then it takes about 1/20 th of a second to fall 5m, so gains very little speed in this time. It ends up not much more than the original 100m/s.

3. Mar 28, 2016

### vela

Staff Emeritus
It bothers me when students ask why they can't solve a problem using wrong methods. There are a ton of ways to do a problem incorrectly and only a few ways to do it right. It shouldn't be your job to explain why they can't use any method they like; they need to justify why their method of choice is correct.

For this particular problem, you could explain that $v^2 = v_0^2 - 2g(y-y_0)$ isn't the same as $v = v_0 + \sqrt{-2g(y-y_0)}$, but is that really going to give the student any insight? If the student simply doesn't know $\sqrt{a^2+b^2} \ne a+b$, that's one thing, but it sounds like in this case, your student is asking why a "common sense" solution doesn't work—that is, why the second equation is wrong. How do you explain why it's wrong other than to say it's not right because it doesn't accurately describe how objects move in the universe?

Imagine if the student does this hundreds of times, throwing increasingly random equations at you. You might finally throw up your hands and ask, "Why do you think these equations are right?" You might want to cut straight to the chase and ask that question first. The answer needs to be better than, "Well, it makes sense to me." Common sense can help you check your answers, and intuition can help guide your thinking. But neither is a replacement for a valid analysis.

4. Apr 1, 2016

### mpresic

It is interesting to note that if the student's solution was correct, the total mechanical energy (potential + kinetic) energy would be greater at the bottom of the hill, than it would be at the top of the hill. The student would have a perpetual motion machine.

5. Apr 1, 2016

### mpresic

I disagree with Vela. PeroK does a OK job in explaining why the student's solution is wrong. The idea that it is not your job to address incorrect solutions is glib. Ideally a good teacher presented with an incorrect solution guides the student into establishing the logic and methods to come up with the correct one. More often, the teacher shows why the student is incorrect, and this is not as good. A far distant third best would be to tell the student the correct method, and absolve himself totally of entertaining any logical difficulties from the student, or to tell them, "the universe doesn't work that way".

Vela suggests, "For this particular problem, you could explain that v2=v02&#x2212;2g(y&#x2212;y0)'>v2=v20−2g(y−y0)v2=v02−2g(y−y0) v^2 = v_0^2 - 2g(y-y_0) isn't the same as v=v0+&#x2212;2g(y&#x2212;y0)'>v=v0+√−2g(y−y0)v=v0+−2g(y−y0) v = v_0 + \sqrt{-2g(y-y_0)}" . (Did not copy correctly)

The student should reply, I never thought it would. This equation is not at the root of the misconception, PeroK has got it right (see below).

I think it is great when a student comes to you to address a misconception. The students who do badly are the ones that do not address the misconception until they meet the problem during the exam. Then it is too late. It is not entirely obvious why the student is wrong. PeroK is correct in his analysis, that the time over which gravity acts is shorter in the case when Lauren has a velocity with a downward component, than when she runs horizontally. Thus the "delta v" is smaller. When I taught, I encouraged my students to think about the answer and method after they calculated it, and to test whether the answer made sense.

One (better teacher than I) told me he once let a "good" student leave his office with the incorrect answer. Upon leaving the teacher warned the student to come back if he had any trouble. When the student came back after thinking critically about the answer, the teacher guided him to the correct solution. The student complained, "Why didn't you show me this in the first place?" The teacher said, "You would not have learned as much unless I let you make a mistake".

This is somewhat dangerous pedagogy. Suppose the student never came back to the professor, and was graded down on an exam. The professor (teacher) has to take responsibility for that too.

6. Apr 2, 2016

### vela

Staff Emeritus
You missed my point (and I'll admit I'm probably reading more into Joe90's original post than he intended). I'm just saying that students need to be able to justify their methods. The attitude shouldn't be that their common sense methods are right and that the burden is on others to show why their methods are wrong.

7. Apr 2, 2016

Staff Emeritus
"Show me where the wrong idea gives me the wrong answer" is a natural response to the wrong starting point, but I don't think it's pedagogically useful. The better question is why is the student starting from the wrong place. In the above example, the student would say "the addition of speed rule", and I would press the student to find such a rule in the book. Eventually they make the connection between a plan for problem solving and what is being taught.

8. Apr 2, 2016

### mpresic

But shouldn't a teacher teaching this material to students for perhaps the first time demonstrate to the students why their common sense methods fail when you apply physical principles to them? Am I missing the point again? I do think we are close to agreement, but I am not sure I am interpreting you correctly,

9. Apr 2, 2016

### vela

Staff Emeritus
It depends. The fact that the student got an incorrect answer already tells them that their common-sense method failed. The teacher may need to model the behavior he or she expects of the students, but it needs to be made clear that it's really the students' job to find the mistakes in their reasoning, otherwise they will probably not deal with their misconceptions. Good students tend to know to do this already, which is why they're good at learning. Poor students, though, may feel like they get it because they understood the teacher's explanation, but it never occurs to them to think about why they made the mistake in the first place so they can avoid repeating the mistake.

10. Apr 3, 2016

### mpresic

The OP asked how they could reply to a student who supplied the wrong analysis to a physics problem. PeroK supplied a (correct) observation suggesting why the student's analysis was faulty. It is easy to jump to the conclusion, since the student got the wrong answer, that his common sense led him to the mistake.

In point of fact, if the student conducted a careful analysis, he or she could calculate the downward (vertical) velocity in running down a hill of constant incline. (S)He could add this to the horizontal velocity (vectorially), and after a trigonometric identity: Voila: You get the correct equation that you note in your post. The fact that the angle of the incline does not show up in the final answer suggests this is a general result, and it even applies to a hill with varying inclination.

The common sense ( along with the physical principles) of the student is sufficient if properly applied.

If you try this analysis in a class, you would have to devote 15 - 30 minutes to it. When I started teaching in 1979, I taught engineering majors and got through probably 6 or 7 out of 10 problems assigned per recitation. Later in 1988, I taught pre-meds at a strong state university and I felt it was probably better to get through fewer problems in greater depth. (I got worse reviews). later in 2003, I taught pre-meds again, at less selective state university. (I got worse reviews still).

I have to leave it to you educators (and your students) to what extent you want to treat many problems more superficially or whether to put a few problems, really under the microscope. If this problem were developed in great detail comparing the energy method with this method, it would demonstrate the power of the energy method. The energy method is certainly more concise.

I hope that students in the age of SIRI do not literally hunt for equations to apply haphazardly and inundate their instructors with red herrings. I am also aware that you cannot be expected to examine all (or even most) student solutions in sufficient depth to determine why they are wrong. If the students are seeing this material for the first time, and the mistake is an instructive one (as I believe this is). The 15-30 minutes may be well spent.