2015 USA Physics Olympiad Exam Part A 6 Out of symmetry,the current through each vertical resistance of 2R must be the same,as well as the currents through each vertical resistance of 4R.This will give us a few equations: Is 12+I4 I2=I6+I4 144R 122R+166R Solve, Is 16+214 414=2(I6+I4)+616 or I4=416 and Is=916 or 6= c.It turns out that it is possible to replace the above circuit with a new circuit as follows: B From the point of view of any passive resistance that is connected between A and B the circuits are identical.You don't need to prove this statement,but you do need to find It and Rt in terms of any or all of R and Is. Solution Use the previous results.For a short,all of the current will flow through AB,so It=I6=Is For an open circuit,the potential across AB will be IsR,so Rt=9R. Copyright C2015 American Association of Physics Teachers
2015 USA Physics Olympiad Exam Part A 6 Out of symmetry, the current through each vertical resistance of 2R must be the same, as well as the currents through each vertical resistance of 4R. This will give us a few equations: Is = I2 + I4 I2 = I6 + I4 I44R = I22R + I66R Solve, Is = I6 + 2I4 4I4 = 2 (I6 + I4) + 6I6 or I4 = 4I6 and Is = 9I6 or I6 = 1 9 Is c. It turns out that it is possible to replace the above circuit with a new circuit as follows: It Rt A B From the point of view of any passive resistance that is connected between A and B the circuits are identical. You don’t need to prove this statement, but you do need to find It and Rt in terms of any or all of R and Is. Solution Use the previous results. For a short, all of the current will flow through AB, so It = I6 = 1 9 Is. For an open circuit, the potential across AB will be IsR, so Rt = 9R. Copyright c 2015 American Association of Physics Teachers
2015 USA Physics Olympiad Exam Part A > Question A3 A large block of mass mo is located on a horizontal frictionless surface.A second block of mass mt is located on top of the first block;the coefficient of friction (both static and kinetic)between the two blocks is given by u.All surfaces are horizontal;all motion is effectively one dimensional. A spring with spring constant k is connected to the top block only;the spring obeys Hooke's Law equally in both extension and compression.Assume that the top block never falls off of the bottom block;you may assume that the bottom block is very,very long.The top block is moved a distance A away from the equilibrium position and then released from rest. WWWWMG mb a.Depending on the value of A,the motion can be divided into two types:motion that expe- riences no frictional energy losses and motion that does.Find the value Ac that divides the two motion types.Write your answer in terms of any or all of u,the acceleration of gravity g,the masses mt and mb,and the spring constant k. Solution The maximum possible acceleration of the top block without slipping is moamax=umtg.If the top block is not slipping then the angular frequency is given by k w2=1 mi+mb amax≥Aw22 or Ae= mt b.Consider now the scenario A>Ac.In this scenario the amplitude of the oscillation of the top block as measured against the original equilibrium position will change with time.Determine the magnitude of the change in amplitude,AA,after one complete oscillation,as a function of any or all of A,u,g,and the angular frequency of oscillation of the top block wt. Solution There are several ways to do this. The energy of an oscillation is approximately equal to B=kA2. Copyright C2015 American Association of Physics Teachers
2015 USA Physics Olympiad Exam Part A 7 Question A3 A large block of mass mb is located on a horizontal frictionless surface. A second block of mass mt is located on top of the first block; the coefficient of friction (both static and kinetic) between the two blocks is given by µ. All surfaces are horizontal; all motion is effectively one dimensional. A spring with spring constant k is connected to the top block only; the spring obeys Hooke’s Law equally in both extension and compression. Assume that the top block never falls off of the bottom block; you may assume that the bottom block is very, very long. The top block is moved a distance A away from the equilibrium position and then released from rest. mb mt a. Depending on the value of A, the motion can be divided into two types: motion that experiences no frictional energy losses and motion that does. Find the value Ac that divides the two motion types. Write your answer in terms of any or all of µ, the acceleration of gravity g, the masses mt and mb, and the spring constant k. Solution The maximum possible acceleration of the top block without slipping is mbamax = µmtg. If the top block is not slipping then the angular frequency is given by ω2 = s k mt + mb , so amax ≥ Aω2 2 or Ac = µg mt k 1 + mt mb b. Consider now the scenario A Ac. In this scenario the amplitude of the oscillation of the top block as measured against the original equilibrium position will change with time. Determine the magnitude of the change in amplitude, ∆A, after one complete oscillation, as a function of any or all of A, µ, g, and the angular frequency of oscillation of the top block ωt . Solution There are several ways to do this. The energy of an oscillation is approximately equal to E = 1 2 kA2 . Copyright c 2015 American Association of Physics Teachers
2015 USA Physics Olympiad Exam Part A 8 Take the derivative,and △E=kA△A That energy is lost to friction.if AAc then the top block has almost completed a complete half cycle before the bottom block catches up with it (speed),so the energy lost in half a cycle is approximately 1 △E=2Af=2Aμmtg 2 where f is the frictional force. Combine, 4μmtg=k△A, or △A=4m9=49 42 c.Assume still that A>Ac.What is the maximum speed of the bottom block during the first complete oscillation cycle of the upper block? Solution The bottom block accelerates according to mt a=μ mb The upper block oscillates as if it is free,since the bottom block exerts a constant force on it,so wt=Vk/mt gives a half period of t=TVmt/k. The maximum speed is then mt b=μTVmt/R mh Note that the maximum speed of the top block is Ut Awt The ratio is =μ9mbkA mtmt t Remember that Ae=μ9K t mt 1+ b %=Am A mb+mt ←1 Copyright C2015 American Association of Physics Teachers
2015 USA Physics Olympiad Exam Part A 8 Take the derivative, and ∆E = kA∆A. That energy is lost to friction. if A Ac then the top block has almost completed a complete half cycle before the bottom block catches up with it (speed), so the energy lost in half a cycle is approximately 1 2 ∆E = 2Af = 2Aµmtg where f is the frictional force. Combine, 4µmtg = k∆A, or ∆A = 4µmtg k = 4 µg ωt 2 c. Assume still that A Ac. What is the maximum speed of the bottom block during the first complete oscillation cycle of the upper block? Solution The bottom block accelerates according to a = µg mt mb The upper block oscillates as if it is free, since the bottom block exerts a constant force on it, so ωt = p k/mt gives a half period of t = π p mt/k. The maximum speed is then vb = µg mt mb π p mt/k Note that the maximum speed of the top block is vt = Aωt The ratio is vb vt = µg mt mb mt kA Remember that Ac = µg mt k 1 + mt mb , so vb vt = Ac A mt mb + mt 1 Copyright c 2015 American Association of Physics Teachers