Thursday, December 1, 2016

AP Physics 1 2016 Free Response Question 3

Question
3. A track is inclined at an angle θ to the horizontal. It has over 100 small speed bumps that are evenly spaced at a distance (d) apart. Assume that a cart of mass (M) is initially moving from rest at the top of the track and its average speed is slowly increased to a maximum value (vavg) after reaching the 40th bump. In other words, the time interval needed to move from one bump to the next bump has become constant.

(a) Consider the motion of the cart between the 41st bump and 44th bump.
(i) Sketch a graph of the cart’s velocity (v) with respect to time from the moment it reaches 41st bump to 44th bump.
(ii) Draw a dashed horizontal line at v = vavg within the same period of time.

(b) Suppose the distance between the two neighboring bumps is increased but the inclined angle of the track remains the same. Is the maximum speed of the cart higher or lower than the speed when the bumps are closer together? Explain your answer.

(c) With the bumps having the original spacing, the track is tilted to a greater angle. Is the maximum speed of the cart higher or lower than the speed when the ramp angle was smaller? Explain your answer.

(d) Assume a student guesses the following equation for the cart's maximum average speed: vavg = C Mgsin θ/d, where C is a positive constant. 
(i) To verify the equation, the cart’s speed is measured by varying the mass (M) of the cart. A graph of the maximum average speed of the cart is plotted against the cart’s mass by using a motion detector (See below). Explain whether the data are consistent with the equation. 


(ii) Is it possible that the equation makes physical sense? Explain your answer.

Scoring guidelines: 
(a) (i) For drawing a constant upward slope in each segment between bumps. [1] 
For drawing the graph that shape is like a sawtooth function. [1] 
For drawing the same maximum velocity in each cycle that occurs whenever the cart reaches a bump. [1] 
(ii) For drawing the vavg line that is horizontal and it coincides with the average velocity. [1] 

(b) Greater than. [1] 
For explaining the cart has more time to accelerate between the bumps. [1] 

(c) Greater than. For explaining that the acceleration of the cart is increased. [1] 
For explaining that the component of the gravitational force is increased. [1] 

(d) (i) No. For explaining that the maximum average velocity is not proportional to M. [1] 
For relating the equation to the graphical data. [1] 
(ii) No. For explaining that the maximum average velocity is not inversely proportional to the spacing, d. [1]
(https://secure-media.collegeboard.org/digitalServices/pdf/ap/apcentral/ap16_physics_1_q3.pdf)

Possible answers
(a) A graph of the cart’s velocity (vas a function of time is sketched below. 

(b) The maximum speed of the cart is relatively higher because it can accelerate for a longer duration of time or a longer distance. We can use the equation v = u + at to explain that the maximum speed is greater because the acceleration (a) is constant. (Note that we have assumed the force acting on the cart is constant, and thus, the cart’s acceleration is constant.) 

(c) The maximum speed of the cart is relatively higher because the component of gravitational force acting on the cart along the track is increased. There is a greater change of gravitational potential energy into kinetic energy, and thus, an increase in the final velocity of the cart. (We can also use the equation v = u + at to explain that the maximum speed is higher because the acceleration is increased.) 

(d) (i) No. 
From the graph, if the mass of the cart is close to zero, its average speed is approximately equal to 1 m/s. This is different from the equation (vavg = C Mgsin θ/d) which indicates its average speed should be zero. Furthermore, the average speed of the cart is about 2 m/s for M = 2.0 kg and 1.5 m/s for M = 1.0 kg (from the graph). That is, the cart’s average speed is not directly proportional to its mass. 

(ii) The equation does not make physical sense because the average speed should be increased with a longer distance d between bumps. However, based on this equation, the average speed would decrease if the distance between bumps is increased. 

Feynman’s insights or goofs?
Feynman has some insights on constraint forces: “[a]nother interesting feature of forces and work is this: suppose that we have a sloping or a curved track, and a particle that must move along the track, but without friction. Or we may have a pendulum with a string and a weight; the string constrains the weight to move in a circle about the pivot point. The pivot point may be changed by having the string hit a peg, so that the path of the weight is along two circles of different radii. These are examples of what we call fixed, frictionless constraints. In motion with a fixed frictionless constraint, no work is done by the constraint because the forces of constraint are always at right angles to the motion. By the “forces of constraint” we mean those forces which are applied to the object directly by the constraint itself — the contact force with the track, or the tension in the string (Feynman et al., 1963, section 14–2 Constrained motion).”

For “forces of constraint,” Feynman’s explanations can be analyzed from the perspectives of “system,” “condition,” “characteristic,” and “effect.” First, the “system” may be specified as a particle on a track or a pendulum with a string. Second, the “conditions” could include a track that is fixed and frictionless. Third, a “characteristic” of constraint forces may refer to the contact force (or reaction force) that is always at right angles to the motion. Fourth, the “effects” of constraint forces are the particle is in contact with the track (or a specified surface) and there is no work done. Thus, for a cart moving down the track, we can understand that there is no work done by the normal force. 
On the contrary, some academics explain that constraint forces such as a frictional force can do work on the system in virtual displacements (Kalaba & Udwadia, 2001). However, this does not mean that Feynman has erred in excluding the frictional force. It is just a matter of idealization in which there is no work done by the contact force on a cart that is moving down a smooth track.

Furthermore, Feynman elaborates that “[t]he forces involved in the motion of a particle on a slope moving under the influence of gravity are quite complicated, since there is a constraint force, a gravitational force, and so on. However, if we base our calculation of the motion on conservation of energy and the gravitational force alone, we get the right result. This seems rather strange, because it is not strictly the right way to do it — we should use the resultant force. Nevertheless, the work done by the gravitational force alone will turn out to be the change in the kinetic energy, because the work done by the constraint part of the force is zero (Feynman et al., 1963, section 14–2 Constrained motion).” Therefore, some students may interpret that the work done by the gravitational force only result in the change in the kinetic energy.


In some examination questions, it is not always clear to students whether a force acting on an object will have no work done or only result in a change in the kinetic energy. Importantly, for a cart moving down the track, there is also a change in the gravitational potential energy. In general, the work done by a gravitational force may lead to a change of energy in different forms depending on how we idealize the physical context. Furthermore, students may have difficulties answering a question because the work-energy theorem is sometimes expressed as a gain in kinetic energy as a result of work done. Similarly, Feynman’s statement only specifies the change in the kinetic energy. However, the work done by a gravitational force can result in a change in the kinetic energy and gravitational potential energy.


Importantly, the term work is potentially misleading to students because it has a daily meaning which is related to physiological work. Thus, some physics educators propose to abandon the term work and replace it with “force-times-distance” or “force-displacement product” (Hilborn, 2000). Nevertheless, the phrase “force-displacement product” is awkward and it can still be confusing to students. This is because work is defined as the scalar product of a force and the displacement of an object in a specified direction. It is also incomplete to merely use the phrase “work done” as it is sometimes unclear whether it is due to an external force, gravitational force, or electric force. You may find similar situations in some textbooks or examination questions.


Note: Goldstein, Poole, & Safko (2001) state that “[w]e now restrict ourselves to systems for which the net virtual work of forces of constraints is zero. We have seen that this condition holds true for rigid bodies and it is valid for a large number of other constraints. Thus, if a particle is constrained to move on a surface, the force of the constraint is perpendicular to the surface, while the virtual displacement must be tangent to it, and hence the virtual work vanishes. This is no longer true if sliding friction forces are present, and we must exclude such systems from our formulation. The restriction is not usually happening since friction is usually a macroscopic phenomenon (p. 17).”

References: 
1. Feynman, R. P., Leighton, R. B., & Sands, M. (1963). The Feynman Lectures on Physics, Vol I: Mainly mechanics, radiation, and heat. Reading, MA: Addison-Wesley. 
2. Goldstein, H., Poole, C. P., & Safko, J. L. (2001). Classical Mechanics, 3rd edition. Reading, MA: Addison-Wesley.
3. Hilborn, R. C. (2000). Let’s ban work from physics!. The Physics Teacher, 38(7), 447. 
4. Kalaba, R., & Udwadia, F. (2001). Analytical dynamics with constraint forces that do work in virtual displacements. Applied Mathematics and Computation, 121(2), 211-217.

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