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SOLUTIONS Aug 2016 exam TFY4102 1) In a perfectly ELASTIC collision between two perfectly rigid objects A) the momentum of each object is conserved. B) the kinetic energy of each object is conserved. C) the momentum of the system is conserved but the kinetic energy of the system is not conserved. D) both the momentum and the kinetic energy of the system are conserved. E) the kinetic energy of the system is conserved, but the momentum of the system is not conserved. Answer: D 2) In a double-slit experiment, if the slit separation is increased, which of the following happens to the interference pattern shown on the screen? A) The minima get closer together. B) The maxima stay at the same position. C) The minima and maxima stay at the same position. D) The minima stay at the same position. E) The maxima get further apart. Answer: A 3) Fluid fills the container shown in the figure. At which of the indicated points is the pressure greatest? A) A B) B C) C D) D E) The pressure is the same at each of the labeled points. Answer: E 4) Which of the following statements about Gauss's law are correct? (There may be more than one correct choice.) A) Gauss's law is valid only for symmetric charge distributions, such as spheres and cylinders. B) If there is no charge inside of a Gaussian surface, the electric field must be zero at points of that surface. C) Only charge enclosed within a Gaussian surface can produce an electric field at points on that surface. D) If a Gaussian surface is completely inside an electrostatic conductor, the electric field must always be zero at all points on that surface. E) The electric flux passing through a Gaussian surface depends only on the amount of charge inside that surface, not on its size or shape. Answer: D, E (A is incorrect; it is just easier to calculate the field in this case. B is violated if the surface is placed inside a uniform field, C is false because there can be a field from outside charges, but there will be no net flux through the surface from these) 5) An object is executing simple harmonic motion. What is true about the acceleration of this object? (There may be more than one correct choice.) A) The acceleration is a maximum when the displacement of the object is a maximum. B) The acceleration is a maximum when the speed of the object is a maximum. C) The acceleration is a maximum when the displacement of the object is zero. D) The acceleration is zero when the speed of the object is a maximum. E) The acceleration is a maximum when the object is instantaneously at rest. Answer: A, D, E use the fact that the acceleration is the second time derivative and the speed is the magnitude of the first derivative of displacement; x(t) = sint or cost MECHANICS An inclined plane that makes an angle of 30 degrees with the horizontal has a spring with spring constant 4500N/m at the bottom. A 2.2 kg block released near the top moves down the plane and compresses the spring to a maximum of 0.024 m from its relaxed length (point A). a) ignoring friction, calculate the distance the lower edge of the block moves from the instant it is released until it comes to a momentary stop against the compressed spring. Conservation of energy: initial potential energy = final elastic energy mgy=1/2 kx2 , where x is the compression amount and y is the height change. b) what is the momentum of the block when it reaches point A (the point where the block first met the spring) after the spring pushes the block back up the plane? There is no friction, so the momentum is the same when it reaches point A going up as it was going down, and you can use either conservation of energy (PE=KE) or the acceleration due to gravity along the slope to solve this one. Momentum is a vector, so you have to specify components to get full credit. | | ⃗ ̂ ̂ ̂ ̂ c) Now include the effects of friction, with µk= 0.10 If the block again compresses the spring by 0.024m, how far does the front(lower) edge of the block travel from release until the moment it stops against the compressed spring? Conservation of energy again, but this time there is a loss(work) term due to gravity. B THERMODYNAMICS A sample of monatomic ideal gas is in a thermally insulated cylinder fitted with a piston that can change the cylinder volume. Initially, the cylinder volume is V1= 2.00 x 10-3 m3, and the temperature is 290 K. The piston then moves until the cylinder volume is 0.200 V1. If the compression is quasi-static and the entropy change is zero, How much work is done on the gas? Q=0; PV= NkT and the change in thermal energy = Work = NCv∆T Being monatomic, each gas particle has three degrees of freedom, so CV=2/3 kB= 2.07 x 10-23J/K C. ELECTROSTATICS (b) 2 cos( ) Ex 2 sin( ) and E y 1 2 2 0 0 where 30. Hence at (3 m, 1 m, 0) (90.0 109 C/m 2 ) Ex sin(30) 2.54 103 N/C and 12 2 2 2(8.85 10 C /N m ) 9 2 9 (130.0 10 C/m ) (90.0 10 C/m 2 )cos(30 ) 3 Ey 2.94 10 N/C, 12 2 2 2(8.85 10 C /N m ) D. WAVES You are blindfolded and given one end of a rope that has two sections joined end to end; the two sections has different linear mass densities, but identical radii and lengths. One end is attached to a pole, but you do not know if it is fixed or sliding. You are able to detect the time and direction of rope displacements that result from inputs that you make with your hand. a) How can you determine whether you are holding the less or more dense rope? There will be reflections from both the interface between the ropes, and from the end. The first pulse to come back will be from the interface; if the reflected pulse is in the same direction as the input, you are holding the heavier rope (no phase change upon reflection). b) How can you determine if the far end of the rope is fixed or sliding? You have to wait for a second pulse to come back – one that has gone all the way to the far end and back. The pulse that is transmitted through the interface and reaches the far end will always be in the direction of the input. If the end of the second rope is tied in place, it will be inverted. If the far end of the rope is free, the second pulse that returns will be in the same direction as the input. c) Find an expression for the ratio of the linear mass densities of the ropes We can estimate the relative densities by measuring the times for the different pulses to return. If the input pulse is at t=0, and the first pulse that returns (from the interface) is arrives at t1 and the second pulse (from the far end) arrives at t2, we can write these times as √ √ √ √ √ ⇒ ( ) E) CIRCUITS The battery has been connected to the circuit in the figure for several minutes. R1=R2=5.0 Ω R3= 4.0 Ω R4= 6.0 Ω R7 = 1.0 Ω R6 =R8= 0.50 Ω There was a misprint and the value of R5 was not given; either use symbol or assume value – no credit was lost by either approach. a) Determine the current through each resistor Let I n be the current through resistor Rn . Since the circuit has been closed a long time, the capacitors will be fully charged and no additional current will flow to or from them. In particular, this means no current will flow across R6 or the 30.0 F capacitor. Current can only flow through that parallel arrangement involving the capacitors by flowing through R7 and R8 , which puts them in series with R1 and R2 . This drastically simplifies the equivalent resistance of the circuit to 1 1 1 Req R1 R2 R7 R8 R3 R4 R5 This allows us to determine the current supplied by the battery: I batt Vbatt Req Note that this must also be the current through resistors R1 , R2 , R7 , and R8 . b) Determine the charge on either plate of each capacitor. The potential difference across the 30.0 F capacitor must be the same as the potential difference across R7 , because they are in parallel. Thus, we can write q30.0 F C30.0 FV30.0 F C30.0 F I R7 R7 . Similarly, the potential difference across the 20.0 F capacitor must be the same as the potential difference across the series combination of R7 and R8 , because those paths are in parallel. Thus q20.0 F C20.0 FV20.0 F C20.0 F I R7 (R7 R8 ) Collecting these results, and assuming R5 = 10 ohms: I battery I1 I 2 I 7 I8 I3 I 4 I5 3.03 A, and I 6 0. The charge on the 30 F capacitor is 1.82 104 C. The charge on the 20 F capacitor is also 1.82 104 C. (you did not need to provide a numerical answer) c) Which capacitor charges more quickly?The 30 µf capacitor will have a lower voltage across it during charging due to the parallel resistor, and the capacitance is larger, so it will take longer to charge up.