AP Physics 1 2016 Free Response Question 5

Question: 
5. A rope that has uniform thickness is hanging vertically from an oscillator. When the oscillator is switched on and adjusted to a certain frequency, the rope forms a standing wave. 

(a) Let P and Q be two points on the rope as shown above. Explain why the tension at P is greater than the tension at Q.
(b) Explain whether increasing the tension in a rope can increase the speed of waves (or wave speed) along the rope. You may use diagrams or/and equations to explain the standing wave as shown above.

Scoring Guidelines:
(a) Indicate that there is more weight of a rope below one point than the other. [1 mark] Indicate that the tension at a point supports the weight of the rope below the same point. [1 mark]

(b) Indicate that the wavelength is either longer near the top of the rope or shorter near the bottom. [1 mark] 
Indicate that the frequency of the wave is the same throughout the rope. [1 mark] 
Use an equation to conclude that wave speed is greater near the top of the rope or lesser near the bottom, based on a difference in wavelength. [1 mark] 
Indicate that the tension is greater near the top of the rope or lesser near the bottom. [1 mark] 
Provide a well-structured answer [1 mark]. 
(https://secure-media.collegeboard.org/digitalServices/pdf/ap/apcentral/ap16_physics_1_q5.pdf)

Possible answers: 
(a) The small section of rope at the location P supports the length of rope immediately below P. The rope at P supports more weight as compared to the rope at Q and thus, the tension at P is higher. 

(b) The wavelength near the top of the rope is observed to be longer as compared to the wavelength near the bottom. 
By using the equation v = fλ, the wave speed (v) is greater at a higher location because the frequency (f) is the same throughout the rope and the wavelength (λ) is longer near the top. Thus, we can deduce that higher tension near the top of the rope results in higher wave speed. 
Alternatively, we can explain the wave speed by using the equation, v = √(F_T/μ), in which F_T is the tension of the string and μ is the linear mass density (or mass per unit length) of the string.

Feynman’s insights or goofs?: 
Feynman provides some insights on the wave equation, ∂²χ/∂x² = (1/c_s²) ∂²χ/∂t², which describes the behavior of sound in matter. In his own words, “If we take a thin slab of air of length Δx and of unit area perpendicular to x, then the mass of air in this slab is ρ₀Δx and it has the acceleration ∂²χ/∂t², so the mass times the acceleration for this slab of matter is ρ₀Δx(∂²χ/∂t²). (It makes no difference for small Δx whether the acceleration ∂²χ/∂t² is evaluated at an edge of the slab or at some intermediate position.) If now we find the force on this matter for a unit area perpendicular to x, it will then be equal to ρ₀Δx(∂²χ/∂t²). We have the force in the +x-direction, at x, of amount P(x, t) per unit area, and we have the force in the opposite direction, at x+Δx, of amount P(x+Δx, t) per unit area: P(x, t) − P(x+Δx, t) = −(∂P/∂x)Δx... (Feynman et al., 1963, section 47–3 The wave equation).” That is, the wave equation can be understood and derived by using Newton’s second law of motion.

In short, Feynman mentions that “sound is a branch of mechanics, and so it is to be understood in terms of Newton’s laws (Feynman et al., 1963, section 47–1 Waves).” Moreover, he suggests that the physics of sound waves in a gas involves three steps: First, the movement of the gas molecules changes the density of the gas. Second, the change in density of the gas corresponds to a change in the gas’s pressure. Third, the differences in pressure of the gas result in movement of the gas molecules. In other words, the movement of the gas molecules is closely related to the differences in pressure of the gas.

Similarly, we can derive the wave equation for a vibrating string in 3 steps as shown below. 
Step 1: By using Newton’s second law of motion (F = ma) on a small element of the string moving in the vertical direction, F_T = m d²y/dt² = μΔx d²y/dt² in which μ is mass per unit length of the string. 

Step 2: The tension of the string acting on the small element of the string moving in the vertical direction, F_T sin θ₂ – F_T sin θ₁ ≈ F_T tan θ₂ – F_T tan θ₁   
(Note: sin θ ≈ tan θ for a small angle, θ.) 
In other words, the net vertical force acting on the small element of the string = F_T(Δy₂/Δx – Δy₁/Δx) = F_T (slope at x + Δx) – F_T (slope at x) 

Step 3: By equating the two expression: F_T(Δy₂/Δx – Δy₁/Δx) = μΔx d²y/dt² 
Thus, F_T(Δy₂/Δx – Δy₁/Δx)/Δx = μ d²y/dt² 
As Δx approaches zero, we have F_T∂²y/∂x²= μ ∂²y/∂t². 

The wave equation of the string can be expressed as ∂²y/∂x² = (1/[F_T/μ]) ∂²y/∂t² and the wave speed of the string can be represented by the equation, v = √(F_T/μ).

     In summary, if we represent the wave equation as F_T∂²y/∂x²= μ ∂²y/∂t², the right-hand side of the equation μ ∂²y/∂t² is about the vertical acceleration of an element of a string times its mass per unit length, whereas the left-hand side of the equation (F_T∂²y/∂x²) is related to the rate of change of the slope of the string with respect to distance times the tension of the string.

     Importantly, we can explain that an increase in the tension (F_T) of a rope can increase the wave speed of the rope by using the wave equation or v = √(F_T/μ). Alternatively, we can apply Newton’s second law of motion on a small element of the vertical rope, F_T = m d²x/dt² = μΔy d²x/dt² in which μ is the mass per unit length of the rope. In other words, the increase in the tension of the rope can increase the speed of waves along the rope.

Reference: 
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.

AP Physics 1 2016 Free Response Question 4

Question: 4. A circuit contains a battery and four identical resistors arranged as shown in the diagram below.

(a) Rank the electrical potential difference across each resistor in descending order. If any resistors have potential differences with the same magnitude, state that explicitly. Explain your answer.

(b) When resistor B is removed in the electrical circuit, what happens to the electric current through resistor A? Explain your answer.

(c) When resistor B is removed in the electrical circuit, what happens to the electric current through resistor C? Explain your answer. 

Scoring Guidelines: 

(a) Ranking: R_A = R_D > R_B = R_C. 

Indicate that the electrical potential difference is the same in R_A and R_D because the electric current is the same through R_A and R_D. [1 mark] 

Indicate that the electrical potential difference is the same in R_B and R_C because R_B and R_C are in parallel. [1 mark] 

Indicate that the electrical potential difference is less in R_B (and/or R_C) than in R_A (and/or R_D) because the electric current splits through R_B and R_C. [1 mark] 

(b) Indicate that the effective resistance of the circuit increases. [1 mark] 

Explain correctly why the electric current through R_A decreases because of the change in electric current or electrical potential difference in the circuit. [1 mark] 

(c) Indicate that the electric current through the battery is the same as electric current through R_C and there is no splitting of electric current. [1 mark] 
(https://secure-media.collegeboard.org/digitalServices/pdf/ap/apcentral/ap16_physics_1_q4.pdf)

Possible answers:

The electrical circuit can be redrawn as follows;

(a) Let the electromotive force of the battery be E and the resistance of each resistor be r. Total resistance = R_A + R_B,C + R_D = r + r/2 + r = 5r/2. 

By using Ohm’s law, I = V/R = E/(5r/2) = (2/5)(E/r) = 0.40 (E/r). 

Thus, ΔV_A = ΔV_D = Ir = 0.40 E and ΔV_B = ΔV_C = 0.20 E. 

The ranking of electrical potential difference can be presented as follows: 

ΔV_A = ΔV_D > ΔV_B = ΔV_C. 

(b) After removing the resistor B, the electrical circuit can be redrawn as follows:

Total resistance = R_A + R_C + R_D = r + r + r = 3r. 

Electric current, I_a = V/R = E/3r = 0.33 (E/r). 

The electric current is decreased from 0.40 (E/r) to 0.33 (E/r). 

Thus, the answer is “decrease.” 

(c) The electric current through resistor C also equals to 0.33 (E/r) because the three resistors A, B, and C are in series and have the same resistance r. 

However, the electric current through resistor C in the original circuit is I_A/2 = 0.20 (E/r) because the electric current through resistor A is equally split in resistor C and resistor B. 

The electric current is increased from 0.20 (E/r) to 0.33 (E/r). 

Thus, the answer is “increase.” 

Feynman’s insights or goofs?: 

In Feynman’s own words, “[t]he second kind of circuit element is called a resistor; it offers resistance to the flow of electrical current. It turns out that metallic wires and many other substances resist the flow of electricity in this manner: if there is a voltage difference across a piece of some substance, there exists an electric current I = dq/dt that is proportional to the electric voltage difference: V = RI = Rdq/dt. The proportionality coefficient is called the resistance R. This relationship may already be familiar to you; it is Ohm’s law (Feynman et al., 1963, section 23–3 Electrical resonance).” Essentially, Feynman explains that Ohm’s law is about the mathematical relationship, V = RI = Rdq/dt. However, some physics educators may not like the phrase “flow of electric current” as used by Feynman. This is because electric current is a flow of charge carriers and thus, the flow of electric current is “the flow of flow of charge carriers.”

In his Nobel lecture, Wilczek (2004) humorously says that “Ohm’s first law is V = IR. Ohm’s second law is I = V/R. I’ll leave it to you to reconstruct Ohm’s third law (p. 413).” Historically speaking, there are at least two different versions of Ohm’s Law: ‘the law for a part of a circuit’ and ‘the law for a whole circuit’ (Ashford and Kempson 1908; Kipnis 2009). Ohm’s law for a part of a circuit means that an electric current through an electrical conductor is directly proportional to the potential difference across the conductor, and the resistance of the conductor is assumed to be constant. This version of Ohm’s law can be represented by the equation, I = ΔV/R. Alternatively, Ohm’s law for a whole circuit means that an electric current through a conductor is directly proportional to the potential difference across the conductor and it is inversely proportional to its resistance. This version of Ohm’s laws can be represented by the following equation, I = E/(R + r), in which E is the electromotive force of a power supply and r is the internal resistance of the power supply. 

To be more precise, Ohm’s law can be defined by stating the operating conditions such as the electric current is not too high. When the electrical current is sufficiently high, this can lead to a significant heating effect. First, a heating of the electrical conductor may increase the temperature of the conductor. Second, the increase in temperature can result in an increase in the length of the conductor. Third, when the temperature of the conductor is being increased, the electrical resistance is also increased, and this can lead to a reduction in the electric current through the conductor. Thus, Ohm’s law does not always hold because electrical properties of the conductor are dependent on the operating conditions. 

References: 

1. Ashford, C. E., & Kempson, E. W. E. (1908). The elementary theory of direct current dynamo electric machinery. Cambridge: The University Press. 

2. 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. 

3. Kipnis, N. (2009). A Law of Physics in the Classroom: The Case of Ohm’s Law. Science & Education, 18(3-4), 349-382. 

4. Wilczek, F. (2004). Asymptotic freedom: From Paradox to Paradigm. In F., Wilczek, & B., Devine, (Eds.). (2006). Fantastic Realities: 49 Mind Journeys and a Trip to Stockholm. Singapore: World Scientific.

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 (v_avg) 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 = v_avg 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: v_avg = 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 v_avg 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 (v) as 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 (v_avg = 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.