2016 USA Physics Olympiad Exam Part A 6 b.Consider the following circuit.All four resistors have identical resistance R.Determine Vout in terms of any or all of Vi,V2,and R. R R Vout R Solution For this part,we will assume Va =V.Again Va=0,and if current I flows through the bottom resistor(below the b and g terminals)then V2=2V,since the voltage drop across the bottom two resistors must be equal.Similarly,the voltage drop across the top two resistors is equal, so Vi+Vout =2Va.Then Vout 2Va-Vi=V2-Vi. This circuit is a subtractor. c.Consider the following circuit.The circuit has a capacitor C and a resistor R with time constant RC =r.The source on the left provides variable,but bounded voltage.Assume Vin is a function of time.Determine Vout as a function of Vin,and any or all of time t and T. R Vout Solution We again set Va=V=0.Then the capacitor charge and current satisfy Q=CVin, 9--6u R where the second result follows from Ohm's law.Then → Vin R dt Vout=-T- dt This circuit is a differentiator. Copyright C2016 American Association of Physics Teachers
2016 USA Physics Olympiad Exam Part A 6 b. Consider the following circuit. All four resistors have identical resistance R. Determine Vout in terms of any or all of V1, V2, and R. a b o g R R Vout V1 V2 R R Solution For this part, we will assume Va = Vb. Again Vg = 0, and if current I flows through the bottom resistor (below the b and g terminals) then V2 = 2Vb, since the voltage drop across the bottom two resistors must be equal. Similarly, the voltage drop across the top two resistors is equal, so V1 + Vout = 2Va. Then Vout = 2Va − V1 = V2 − V1. This circuit is a subtractor. c. Consider the following circuit. The circuit has a capacitor C and a resistor R with time constant RC = τ . The source on the left provides variable, but bounded voltage. Assume Vin is a function of time. Determine Vout as a function of Vin, and any or all of time t and τ . a b o g R Vout Vin C Solution We again set Va = Vb = 0. Then the capacitor charge and current satisfy Q = CVin, Q˙ = − Vout R where the second result follows from Ohm’s law. Then Vout R = −C dVin dt ⇒ Vout = −τ dVin dt . This circuit is a differentiator. Copyright c 2016 American Association of Physics Teachers
2016 USA Physics Olympiad Exam Part A Question A3 Throughout this problem the inertial rest frame of the rod will be referred to as the rod's frame, while the inertial frame of the cylinder will be referred to as the cylinder's frame. A rod is traveling at a constant speed of v=c to the right relative to a hollow cylinder.The rod passes through the cylinder,and then out the other side.The left end of the rod aligns with the left end of the cylinder at time t=0 and x =0 in the cylinder's frame and time t'=0 and x'=0 in the rod's frame. The left end of the rod aligns with the left end of the cylinder at the same time as the right end of the rod aligns with the right end of the cylinder in the cylinder's frame;in this reference frame the length of the cylinder is 15 m. For the following,sketch accurate,scale diagrams of the motions of the ends of the rod and the cylinder on the graphs provided.The horizontal axis corresponds to t,the vertical axis corresponds to ct,where c is the speed of light.Both the vertical and horizontal gridlines have 5.0 meter spacing. a.Sketch the world lines of the left end of the rod(RL),left end of the cylinder (CL),right end of the rod(RR),and right end of the cylinder(CR)in the cylinder's frame. b.Do the same in the rod's frame. c.On both diagrams clearly indicate the following four events by the letters A,B,C,and D. A:The left end of the rod is at the same point as the left end of the cylinder B:The right end of the rod is at the same point as the right end of the cylinder C:The left end of the rod is at the same point as the right end of the cylinder D:The right end of the rod is at the same point as the left end of the cylinder d.At event B a small particle P is emitted that travels to the left at a constant speed vp=e in the cylinder's frame. i.Sketch the world line of P in the cylinder's frame. ii.Sketch the world line of P in the rod's frame. e.Now consider the following in the cylinder's frame.The right end of the rod stops instanta- neously at event B and emits a flash of light,and the left end of the rod stops instantaneously when the light reaches it.Determine the final length of the rod after it has stopped.You can assume the rod compresses uniformly with no other deformation. Any computation that you do must be shown on a separate sheet of paper,and not on the graphs.Graphical work that does not have supporting computation might not receive full credit. Solution The graphs are shown below,where the yellow line is the particle P and the green line is the flash of light.Solutions that used Galilean relativity received partial credit,as long as they were self-consistent.The final length of the rod is simply the distance between the line CR and the intersection of RL and the green line,i.e.25/3 m.There's no need to apply length contraction,as we're already in the rest frame of the rod at this point.Nonetheless,the way we have chosen to stop the rod has squeezed it shorter. Copyright C2016 American Association of Physics Teachers
2016 USA Physics Olympiad Exam Part A 7 Question A3 Throughout this problem the inertial rest frame of the rod will be referred to as the rod’s frame, while the inertial frame of the cylinder will be referred to as the cylinder’s frame. A rod is traveling at a constant speed of v = 4 5 c to the right relative to a hollow cylinder. The rod passes through the cylinder, and then out the other side. The left end of the rod aligns with the left end of the cylinder at time t = 0 and x = 0 in the cylinder’s frame and time t 0 = 0 and x 0 = 0 in the rod’s frame. The left end of the rod aligns with the left end of the cylinder at the same time as the right end of the rod aligns with the right end of the cylinder in the cylinder’s frame; in this reference frame the length of the cylinder is 15 m. For the following, sketch accurate, scale diagrams of the motions of the ends of the rod and the cylinder on the graphs provided. The horizontal axis corresponds to x, the vertical axis corresponds to ct, where c is the speed of light. Both the vertical and horizontal gridlines have 5.0 meter spacing. a. Sketch the world lines of the left end of the rod (RL), left end of the cylinder (CL), right end of the rod (RR), and right end of the cylinder (CR) in the cylinder’s frame. b. Do the same in the rod’s frame. c. On both diagrams clearly indicate the following four events by the letters A, B, C, and D. A: The left end of the rod is at the same point as the left end of the cylinder B: The right end of the rod is at the same point as the right end of the cylinder C: The left end of the rod is at the same point as the right end of the cylinder D: The right end of the rod is at the same point as the left end of the cylinder d. At event B a small particle P is emitted that travels to the left at a constant speed vP = 4 5 c in the cylinder’s frame. i. Sketch the world line of P in the cylinder’s frame. ii. Sketch the world line of P in the rod’s frame. e. Now consider the following in the cylinder’s frame. The right end of the rod stops instantaneously at event B and emits a flash of light, and the left end of the rod stops instantaneously when the light reaches it. Determine the final length of the rod after it has stopped. You can assume the rod compresses uniformly with no other deformation. Any computation that you do must be shown on a separate sheet of paper, and not on the graphs. Graphical work that does not have supporting computation might not receive full credit. Solution The graphs are shown below, where the yellow line is the particle P and the green line is the flash of light. Solutions that used Galilean relativity received partial credit, as long as they were self-consistent. The final length of the rod is simply the distance between the line CR and the intersection of RL and the green line, i.e. 25/3 m. There’s no need to apply length contraction, as we’re already in the rest frame of the rod at this point. Nonetheless, the way we have chosen to stop the rod has squeezed it shorter. Copyright c 2016 American Association of Physics Teachers
2016 USA Physics Olympiad Exam Part A 8 The Cylinder's Frame ct. C RL RR Copyright C2016 American Association of Physics Teachers
2016 USA Physics Olympiad Exam Part A 8 The Cylinder’s Frame x ctCL CR RL RR Copyright c 2016 American Association of Physics Teachers