Sunday, June 26, 2016

AP Physics 1 2015 Free Response Question 1

Question
This question is based on a simple Atwood’s machine: two blocks are connected by a massless string passing over massless pulleys which are frictionless (See figure 1 below). Block 2’s mass (m2) is greater than block 1’s mass (m1). Students have to derive the magnitude of the acceleration of block 2 when the two blocks are released from rest. The answer should be expressed in terms of m1, m2, and g.


Fig. 1

It could be of interest to mention that a similar question can be found in Feynman’s Tips on Physics. The question involves the calculation of frictional force: “Two masses, m1 = 4 kg and m3 = 2 kg, are connected with cords of negligible weight over essentially frictionless pulleys to a third mass, m2 = 2 kg. The mass m2 moves on a long table with a coefficient of friction…(Feynman et al., 2006, p. 144).”

Scoring Guidelines:
Using Newton’s second law for block 1
m1a = Tm1g
1 point

Using Newton’s second law for block 2
m2a = m2gT
1 point
Eliminate T and derive the acceleration:
T = m1a + m1g
m2a = m2gm1am1g
(m2 + m1)a = (m2m1)g
a = (m2m1)g / (m1 + m2)
1 point
(Source: http://apcentral.collegeboard.com/home)

Comments:

The main purpose of the question is to assess students’ ability to apply Newton’s second law of motion to a modified Atwood’s machine. Students are expected to have a basic knowledge of free body diagrams and an understanding of factors that determine the acceleration of the system. However, they may have difficulty with the use of conventions for force and acceleration because one block is moving upward and the other block is moving downward.

We may expect this question to be simply answered by using Newton’s Second law of motion as shown below:
To find the acceleration of block 2, we can visualize an “equivalent question” in which the forces acting on block 1 and block 2 are shown in figure 2 below:

Fig. 2

The larger force, m2g, acting on block 2 is the driving force and the smaller force, m1g, acting on block 1 is the opposing force.
For block 2: the driving force (m2g) is in the direction to the right, whereas the tension (T) is in the direction to the left.
By using Newton’s Second Law of motion, ΣF = ma
we have m2gT = m2a ------ (1)
(Convention: The forces that are in the direction to the right are assigned positive.)

For Block 1: The tension (T) is in the direction to the right and the opposing force (m1g) is in the direction to the left.
By using Newton’s Second Law again, we have T m1g = m1a ------(2)
By solving the two equations (1) and (2), a = (m2 m1)g / (m1 + m2)

However, we can visualize block 1 and block 2 as a system.
In this case, the total mass is m1 + m2 and the external forces on the system are m2gm1g. Thus, a = F/m = (m2 m1)g / (m1 + m2)
Would students be penalized by using only one equation?


Feynman's insights?:
Alternatively, Feynman explains that “the laws of Newton could be stated not in the form F = ma but in the form: the average kinetic energy less the average potential energy is as little as possible for the path of an object going from one point to another (Feynman et al., 1964, section 19–1 A special lecture—almost verbatim).” That is, it is possible to use Euler-Lagrange’s equation to answer this question. Interestingly, Mr. Bader introduced the principle of least action to his student, Feynman. In addition, Dias, Araújo, Silva, Santos, Barros, & Carvalho-Santos (2012), for example, propose to introduce Euler-Lagrange’s equation in introductory physics. However, it is unclear whether students who use a more advanced method could be awarded extra credit or penalized instead. (I happen to know of a high school student who was able to apply Euler-Lagrange’s equation to solve mechanics problems that are even more difficult.)

It is not difficult to use Euler-Lagrange’s equation to solve this question on Atwood’s machine. It can be accomplished by using only a few steps as shown below: 
The kinetic energy of the system (T) is given by T = ½ m12 + ½ m2
The potential energy of the system (U) is given by U– m2gx – m1g(l – x
Thus, the Lagrangian can be written as 
TU = ½m12 + ½m22 + m2gx + m1g(lx)
By using Euler-Lagranges equation of motion, d/dt(∂L/∂) – ∂L/∂x = 0
We get (m1 + m2) – (m2m1)g = 0
Therefore, = (m2m1)g/(m1 + m2)

(Would some physics teachers penalize students for using Euler-Lagrange’s equation to get the correct answer?)

Note:
1. In Feynman’s words, “when I was in high school, my physics teacher—whose name was Mr. Bader—called me down one day after physics class and said, ‘You look bored; I want to tell you something interesting.’ Then he told me something which I found absolutely fascinating, and have, since then, always found fascinating. Every time the subject comes up, I work on it. In fact, when I began to prepare this lecture I found myself making more analyses on the thing. Instead of worrying about the lecture, I got involved in a new problem. The subject is this—the principle of least action (Feynman et al., 1964, section 19–1 A special lecture—almost verbatim).”

2. For another discussion of this question, you can visit the following websites:
https://www.youtube.com/watch?v=NrNG6OPijeY
https://www.youtube.com/watch?v=ff6SNrgUFt0

References:
1. Dias, C. F., Araújo, M. A., Silva, G. M., Santos, C. A., Barros Jr, P., & Carvalho-Santos, V. L. (2012). Adaptation of the Euler-Lagrange equation for studying one-dimensional motions in a constant force. arXiv preprint arXiv:1209.2197.
2. Feynman, R. P., Gottlieb, M. A., Leighton, R. (2006). Feynman's tips on physics: reflections, advice, insights, practice: a problem-solving supplement to the Feynman lectures on physics. San Francisco: Pearson Addison-Wesley.
3. Feynman, R. P., Leighton, R. B., & Sands, M. L. (1964). The Feynman Lectures on Physics, Vol IIMainly electromagnetism and matter. Reading, MA: Addison-Wesley.

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