AP2 Exam 1

1. The dots in the figure above represent two identical spheres, X and Y, that are fixed in place with

their centers in the plane of the page. Both spheres are charged, and the charge on sphere X is positive. The lines are isolines of electric potential, also in the plane of the page, with a potential difference of 5 V between each set of adjacent lines. The absolute value of the electric potential of the outermost line is 10 V

(A) Indicate the values of the potentials VA, VB, and VC, including the signs, at the labeled

points A, B, and C. Explain your reasoning. VA = +15 V, VB = +25 V, VC = +35 V There is a 5-V potential difference between isolines, and electric potential increases as a positive point charge is approached from afar. I counted lines, starting from 10 V at the outermost line.

(B) Compare the magnitude and signs of the charges of sphere X and sphere Y. Justify your answer. Both sphere X and sphere Y are positive, and sphere X has a greater charge than sphere Y. If sphere Y were negative, there would be a zero-V isoline passing between the two spheres. Also, a positive isoline encloses both charges. Negative charges create negative potential. Each isoline is closer to sphere Y than sphere X, indicating a lower potential at the same

distance. Since V = kq/r, a smaller value for r indicates a smaller q.

(C) The spheres at points X and Y have masses in the same ratio as the magnitudes of their charges. The isolines of gravitational potential for the spheres have shapes similar to those of the isolines shown. Explain why the two sets of isolines have similar shapes. Both gravitational and electrical forces have a direct relationship to the property creating them, mass for gravitational forces and charge for electrical forces. Both have an inverse-square relationship to the distance from a point source. Isolines determine the amount of potential energy an object would have in proximity to the point sources, and potential energy is derived from force. Thus, the isoline profiles are the same. The download has D missing. In the review. E in the download is D in the review, and so on.

A proton with charge plus +q and mass m is released from rest at point B.

(D) Describe how the magnitude of the acceleration of the proton changes after being released. Explain your reasoning with reference to electric potential. Positive charges accelerate in the direction of decreasing potential. The charge would pass across successive isolines away from sphere X. The spacing between these lines is increasing, indicating a progressively weaker electric field and thus a weaker force on the particle. Its acceleration would decrease.

(E) At some time after being released from rest at point B, the proton has moved through a potential difference of magnitude 20 V. Two students are discussing how and why the kinetic energy of the proton would change after it is released.

● Student 1 says that if the system is defined as the proton and the spheres, the increase in the

proton’s kinetic energy is due to a change in the system’s potential energy as the proton moves through the 20 V potential difference.

● Student 2 says that if the system is defined as only the proton, the kinetic energy of the proton increases because positive work is done on the proton by the electric field as the proton moves through the 20 V potential difference.

Discuss each student’s claims, explaining why each is correct or incorrect.

Student 1 is correct. The proton-sphere system contains electric potential energy as a result of the repulsive forces between the charges, which can be converted to kinetic energy of objects within the system. Student 2 is also correct. Viewed as an object in an electric field created by the spheres, there would be an external force doing work on the proton, and increasing its kinetic energy in accordance with the work-energy theorem.

The spheres are separated by . Let be the location of sphere X and be the location of sphere Y. A student sketches the graph above of the electric field as a function of x along the line that joins the two spheres. For this graph, positive values of E represent electric field directed to the right.

(F) State one aspect of the graph that correctly represents the electric field and explain your reasoning.

The electric field is a vector; thus, its sign indicates direction. The electric field points away from positive charges. It would point right in the vicinity of sphere X (+), and left in the vicinity of sphere Y (-)

(G) State one aspect of the graph that incorrectly represents the electric field and explain your reasoning. The electric field will be stronger at equal distances near sphere X vs. sphere Y, because sphere X has a greater charge. The graph shows a weaker field at equivalent distances. Specifically, an E of zero would be closer to sphere Y than sphere X.

2. Two students observe water flowing from left to right through the section of pipe shown above, which decreases in diameter and increases in elevation. The pipe ends on the right, where the water exits vertically into the bottom of a large, deep pool.

(A) A bubble of air passes point A and later point B without changing temperature. In a

clear, coherent, paragraph-length response that may reference equations, state whether the volume of the bubble at point B is greater than, less than, or equal to the volume of the bubble at point A. According to Bernoulli’s equation, pressure decreases when the speed of a fluid increases. Pressure also decreases when depth decreases. According to the equation of continuity, the speed of a fluid increases when the cross-sectional area of the pipe carrying it decreases. Between A and B, the cross-sectional area of the pipe decreases, so the fluid will be moving faster. In addition, the depth decreases. Both of these factors will result in a lower pressure at point B. The ideal gas law indicates that pressure and volume are inversely related when temperature is held constant. Thus, the volume of the bubble will increase.

(B) Suppose the pipe remained a constant diameter, but the elevation of the pipe changed the same amount as in part (A). Would the difference in pressure between points A and B be greater than, less than, or equal to the difference in pressure between points A and B when the pipe narrowed as in part (A)? Justify your answer.

The difference in pressure would be less than before. The decrease in depth would result in a decrease in pressure. But the other factor decreasing pressure, the increase in speed of the water, would be absent.

A triangular prism is fully submerged in a tank of water while suspended from a rope without touching the bottom of the tank as shown in Figure 1. At time t = 0 the rope pulls the prism upwards at a constant speed. At time t1 the top of the prism reaches the surface of the water, and at time t2 the prism has been completely removed from the water, shown in Figure 2. The prism always moves with constant speed. The graph below shows the tension in the rope as a function of time.

(C) In terms of relevant physics principles, explain the behavior of the tension vs. time graph from

t = 0 until a time after t2. The three forces acting on the triangle are gravity down, tension up, and the buoyant force up. The forces are balanced, because the triangle is moving at a constant velocity. According to Archimedes’ Principle, the buoyant force is equal to the weight of the displaced fluid. Thus, there will be no change in the balance of forces until the triangle begins to emerge from the water (< t1). As the triangle is emerging (t1 to t2), less buoyant force would be present, so tension would need to increase to maintain constant speed. Once removed from the water (> t2), the buoyant force would be gone entirely, and tension would equal the constant force of gravity.

AP2 Exam 2

Note : Figure not drawn to scale (A) The figure above shows two cylindrical containers with samples of ideal monatomic gas, each

fitted with a moveable piston of negligible mass. Each sample of ideal gas has the same volume and same number of moles. The cylinder on the left has diameter D, and the cylinder on the right has diameter 2D. On top of each piston is a mass. Both samples of gas have been in the same room for a long time.

(A) In which container, if either, is the pressure greater, the left container, or the right container?

Justify your answer. According to the ideal gas law, PV = nRT, under identical temperature and volume and number of moles, two monatomic gas samples would have the same pressure. We know temperature is the same, because both samples have been in the same place for a long time. The pressure is equal in both containers.

(B) Which mass, if either, is greater, Mass A or Mass B? Justify your answer. Mass B is larger, because it must maintain equilibrium with a larger upward force due to the pressure of the gas (F = PA). The area of the piston on the right is four times the area of the piston on the left. My answer originally said, “Mass 2.” Don’t know if they’d give it to me, probably not. Also, I originally said area was twice as much, not four times as much. Don’t be like me.

(C) A student trying to model the force the gas exerts against the bottom of the piston derives the following equation

𝐹 = 𝐶𝑛𝑇𝐷!

𝑉

in which F is the force exerted by the gas on the piston, n is the number of moles, T is the temperature of the gas, D is the diameter of the cylinder, V is the volume of gas in the cylinder, and C is a constant with appropriate units. Briefly explain whether or not this equation agrees with your reasoning in part (B). The equation agrees with my reasoning. The only change is in the diameter of the piston, which must be squared to yield cross-sectional area. In both F = PA and the above equation, force and area are directly related.

(D) The pressure-volume diagram above shows two processes, A and B, represented by dotted or solid lines as indicated. Both processes bring a sample of gas from initial state I to final state F. Two students are arguing about which process does more work and in which process the most heat energy is added.

Student 1 : The gas would do more work in process A, which means a greater

change in the internal energy. Doing more work requires more heat energy, so the gas would also absorb more heat energy in total during process 2A.

Student 2 : The change in internal energy is the same for both processes, so the

amount of energy transferred to each gas as heat must be the same.

With which aspects of student 1’s reasoning, if any, do you agree? Explain why. More work is done by the gas in process A, because the area under the P-V curve is greater. With which aspect of student 1’s reasoning, if any, do you disagree? Explain why. Doing more work only requires more heat energy if the internal energy of the system is not being allowed to decrease, in accordance with ∆𝑈 = 𝑊 +𝑄.

With which aspect of student 2’s reasoning, if any, do you agree? Explain why. The change in the internal energy in both processes is the same, because their endpoints are the same. Internal energy is proportional to temperature, and the ideal gas law indicates that a sample will have the same temperature if pressure and volume are the same. With which aspect of student 2’s reasoning, if any, do you disagree? Explain why. The student is ignoring the fact that work done by the gas can change internal energy as well as heat.

Two 30 cm long bars with the same cross-sectional area are joined together, as shown in the figure above. The bar on the left is made of Metal A, and the bar on the right is made of Metal B. The left and right ends of the two-bar combination are maintained at 100˚C and 20˚C respectively, and the bars reach equilibrium. The student then measures the temperature T along the bars and produces the data shown in the graph, where x = 0 is the left end and 60 cm is the right end.

(E) Which material, metal A or metal B, has greater thermal conductivity? Justify your answer.

Area, length, and power are the same for both metals. Thermal conductivity is inversely related to change in temperature (power = Q/∆t = kA∆T/L). Thus, metal A has the greater conductivity, because the graph indicates that its ∆T is less.

(F) Independent of the thermal conductivity of the materials, Metal A is found to have a greater electrical resistivity. When electrical current passes through the rods as shown in the figure above, in which location, 15 cm or 45 cm, will the electrical current be greater? Justify your answer.

The current will be the same in both locations, because the two metals are in series, and every electron passing through metal A must also pass through metal B.

2. The circuit shown above contains two bulbs, A and B, an ideal battery, a capacitor, C, and a switch

that is initially open. The switch is eventually closed.

(A) Which of the graphs above best shows the brightness of bulb B before and after the switch is

closed? Briefly explain your answer. Graph A accurately depicts the change in brightness, which corresponds to change in current. There will be no current in the branch with the capacitor so long as the switch is open, so the current would be determined by the emf of the battery and the two bulbs in series. The instant the switch is closed, the ∆V across the capacitor is zero, because it does not contain charge. Kirchoff’s loop rule dictates that the ∆V across the bulb is also zero at this point. All of the current is charging the capacitor, and none is passing through the bulb. As the capacitor charges, its ∆V and that of the bulb gradually increase to a value equal to the ∆V in the bulb before the switch was closed. Thus, the current and brightness gradually approach the original values.

(B) The circuit is used to fully charge two identical capacitors, C1 and C2, one at a time. After

being charged, each capacitor is removed from the circuit. The plates of C2 are then further separated without being discharged until its plates are twice as far apart as the plates of C1, as shown above. In a clear, coherent, paragraph-length response, explain which capacitor, C1 or C2, when used in the circuit below, would cause the bulb to be brighter immediately after the switch is closed.

Both capacitors start out with the same charge in accordance with Q = C∆V, and they retain that charge after being separated from the circuit. Pulling the plates of C2 apart does work on the charged plates and increases both the electric potential and the electric potential energy in the capacitor. When first connected, the greater electric potential of capacitor C2 will cause the bulb to have greater current and to glow more brightly. I missed the fact that the plates were being separated after being disconnected at first and almost got this wrong. Don’t be like me!