exam1-2015-1-8-answer-typo-corrected

2015 F = ma Exam 3 2. A car travels directly north on a straight highway at a constant speed of 80 km/hr for a distance of 25 km. The car then continues directly north at a constant speed of 50 km/hr for a distance of 75 more kilometers. The average speed of the car for the entire journey is closest to (A) 55.2 km/hr ← CORRECT (B) 57.5 km/hr (C) 65 km/hr (D) 69.6 km/hr (E) 72.5 km/hr Solution Average speed is total distance divided by total time. The time required for the first 25 km is 0.31 hours; the time required for the second 75 km is 1.5 hours. 3. The force of friction on an airplane in level flight is given by F f = kv 2 , where k is some constant, and v is the speed of the airplane. When the power output from the engines is P 0 , the plane is able to fly at a speed v 0 . If the power output of the engines is increased by 100% to 2 P 0 , the airplane will be able to fly at a new speed given by (A) 1 . 12 v 0 (B) 1 . 26 v 0 ← CORRECT (C) 1 . 41 v 0 (D) 2 . 82 v 0 (E) 8 v 0 Solution P = Fv , so P = kv 3 . Then v/v 0 = 3 p P/P 0 . Copyright c 2015 American Association of Physics Teachers

2015 F = ma Exam 6 6. Three trolley carts are free to move on a one dimensional frictionless horizontal track. Cart A has a mass of 1.9 kg and an initial speed of 1.7 m/s to the right; Cart B has a mass of 1.1 kg and an initial speed of 2.5 m/s to the left; cart C has a mass of 1.3 kg and is originally at rest. Collisions between carts A and B are perfectly elastic; collisions between carts B and C are perfectly inelastic. Cart A Cart B Cart C 1.9 kg 1.1 kg 1.3 kg 1.7 m/s 2.5 m/s What is the velocity of the center of mass of the system of the three carts after the last collision? (A) 0.11 m/s ← CORRECT (B) 0.16 m/s (C) 1.4 m/s (D) 2.0 m/s (E) 3.23 m/s Solution The center of mass velocity is unchanged by any collisions, so v cm = (1 . 9 kg)(1 . 7 m / s) + (1 . 1 kg)( - 2 . 5 m / s) + (1 . 3 kg)(0 m / s) (1 . 9 kg) + (1 . 1 kg) + (1 . 3 kg) = 0 . 11 m / s Copyright c 2015 American Association of Physics Teachers

2015 F = ma Exam 9 The following information applies to questions 9 and 10 A 0.650 kg ball moving at 5.00 m/s collides with a 0.750 kg ball that is originally at rest. After the collision, the 0.750 kg ball moves off with a speed of 4.00 m/s, and the 0.650 kg ball moves off at a right angle to the final direction of motion of the 0.750 kg ball. 9. What is the final speed of the 0.650 kg ball? (A) 1.92 m/s ← CORRECT (B) 2.32 m/s (C) 3.00 m/s (D) 4.64 m/s (E) 5.77 m/s Solution Yes, we purposefully choose the velocities to satisfy the 3-4-5 triangle, albeit incorrectly. Conserve momentum, sketch a right triangle. Then p 2 f 2 + p 1 f 2 = p 2 1 i , with p 1 i = (0 . 65 kg)(5 . 0 m / s) = 3 . 25 kg · m / s and p 2 f = (0 . 75 kg)(4 . 0 m / s) = 3 . 0 kg · m / s therefore, p 1 f = 1 . 25 kg · m / s and then v 1 f = (1 . 25) / (0 . 65) m / s = 25 / 13 m / s . Copyright c 2015 American Association of Physics Teachers

2015 F = ma Exam 12 13. Consider the pendulum bob when it is at an angle θ = 1 2 θ max on the way up (moving toward θ max ). What is the direction of the acceleration vector? θ max (A) ~a θ max (B) ~a θ max (C) ~a θ max (D) ~a ← CORRECT θ max (E) ~a Solution There must be acceleration directed toward the center of the circle. There must be acceler- ation directed tangent to the circle toward the bottom of the path. Copyright c 2015 American Association of Physics Teachers

2015 F = ma Exam 15 16. Shown below is a graph of potential energy as a function of position for a 0.50 kg object. 1 2 3 4 5 x (cm) 0 10 20 -10 E (J) 6 Which of the following statements is NOT true in the range 0 cm < x < 6 cm? (A) The object could be at equilibrium at either x = 1 cm or x = 3 cm. (B) The minimum possible total energy for this object in the range is -10 J. (C) The magnitude of the force on the object at 4 cm is approximately 1000 N. (D) If the total energy of the particle is 0 J then the object will have a maximum kinetic energy of 10 J. (E) The magnitude of the acceleration of the object at x = 2 cm is approximately 4 cm/s 2 . ← CORRECT Solution Consider each scenario in turn. Copyright c 2015 American Association of Physics Teachers

2015 F = ma Exam 18 The following information applies to questions 19 and 20 A U-tube manometer consists of a uniform diameter cylindrical tube that is bent into a U shape. It is originally filled with water that has a density ρ w . The total length of the column of water is L . Ignore surface tension and viscosity. 19. The water is displaced slightly so that one side moves up a distance x and the other side lowers a distance x . Find the frequency of oscillation. (A) 1 2 π p 2 g/L ← CORRECT (B) 2 π p g/L (C) 1 2 π p 2 L/g (D) 1 2 π p g/ρ w (E) 2 π √ ρ w gL Solution Frequency of oscillation is given by f = 1 2 π p k/m where m is the mass, in this case LAρ , with L the total length of the fluid, and k is an effective spring constant. If the fluid is displaced a distance x , then the net restoring force on the fluid is given by 2 xgAρ , so k = 2 gAρ . Then f = 1 2 π p 2 g/L Copyright c 2015 American Association of Physics Teachers

2015 F = ma Exam 21 22. A solid ball is released from rest down inclines of various inclination angles θ but through a fixed vertical height h . The coefficient of static and kinetic friction are both equal to μ . Which of the following graphs best represents the total kinetic energy of the ball at the bottom of the incline as a function of the angle of the incline? E kinetic θ (A) 0 90 E kinetic θ (B) 0 90 E kinetic θ (C) 0 90 ← CORRECT E kinetic θ (D) 0 90 E kinetic θ (E) 0 90 Solution The ball rolls without slipping so long as θ is less than some critical angle. For θ less than the critical angle no energy is lost to friction. For θ greater than the critical angle the ball will slip, and energy will be lost to friction. If the angle is 90 degrees the ball falls straight down and there is no friction, so no energy lost. There must then be some angle where the most energy is lost to friction. Copyright c 2015 American Association of Physics Teachers

2015 F = ma Exam 22 23. A 2.0 kg object falls from rest a distance of 5.0 meters onto a 6.0 kg object that is supported by a vertical massless spring with spring constant k = 72 N/m. The two objects stick together after the collision, which results in the mass/spring system oscillating. What is the maximum magnitude of the displacement of the 6.0 kg object from its original location before it is struck by the falling object? (A) 0.27 m (B) 1.1 m ← CORRECT (C) 2.5 m (D) 2.8 m (E) 3.1 m Solution Conserve energy to find the speed of the 2.0 kg object at the instant before the collision: m 1 gy = (2 . 0 kg)(10 m / s)(5 . 0 m) = 100 J and this equals the kinetic energy so v = p 2 K 1 /m 1 = 10 . 0 m / s . The collision is inelastic, so v f = v 1 m 1 m 1 + m 2 = 2 . 5 m / s Now conserve energy again K f = 1 2 m t v 2 = 1 2 (8 . 0 kg)(2 . 5 m / s) 2 = 25 J This compresses the spring and changes the gravitational potential energy, so we must solve for x in the equation 1 2 kx 2 + m t gx = K f or 36 x 2 + 20 x - 25 = 0 which has solutions x = - 1 . 15 m and x = 0 . 6 m. It was pointed out by a very astute and determined reader that 1.15 m should have been rounded to 1.2 m as opposed to 1.1 m. Okay, I have received enough complaints about this problem that I will add to the solution. First, there is a typo: the equation 1 2 kx 2 + m t gx = K f Copyright c 2015 American Association of Physics Teachers

2015 F = ma Exam 23 should have read 1 2 kx 2 + m 1 gx = K f but there isn’t a mistake in the physics or the math, as we did indeed use m = 2 . 0 kg in the solution like we should have. So the answer is correct. But a number of persistent individuals have continued to insist that we use m = 8 . 0 kg in this expression. After all, they ponder, since both masses are moving, and we included both masses in the kinetic energy expression, why aren’t we including both masses in the potential energy expression? This is a subtlety attached to the quadratic nature of springs. If we want to consider an energy expression that has the change in potential of the 6.0 kg mass, then we need to measure the spring from the unstretched length before the 6.0 kg mass is suspended from the spring. The energy expression is then ( m 1 + m 2 ) g ( x + h ) + 1 2 k ( x + h ) 2 + U 0 = K f , where both h and x are measured positive if up, so m 2 g = - kh. The quantity U 0 is a constant energy term that makes the two approaches numerically equivalent; we want the total potential energy of the system to be zero before the second mass m 2 hits it, and we want the second mass to also have no potential energy at x = 0. Expand and combine, - kh ( x + h ) + m 1 g ( x + h ) + 1 2 kx 2 + kxh + 1 2 kh 2 + U 0 = K f or 1 2 kx 2 + m 1 gx + m 2 gh - 1 2 kh 2 + U 0 = K f But U 0 is chosen so that the two energy expression approaches are equivalent, so m 2 gh - 1 2 kh 2 + U 0 = 0 and therefore they are: the total potential energy of the system is zero just before the masses collide in either approach. I suspect there are a few readers who still think that slight of hand was performed in this last step. Isn’t physics wonderful? Now I hope this is the last edit I need to make on this solution. Copyright c 2015 American Association of Physics Teachers

2015 F = ma Exam 24 24. The speed of a transverse wave on a long cylindrical steel string is given by v = s T M/L , where T is the tension in the string, M is the mass, and L is the length of the string. Ignore any string stiffness, and assume that it does not stretch when tightened. Consider two steel strings of the same length, the first with radius r 1 and a second thicker string with radius r 2 = 4 r 1 . Each string is tightened to the maximum possible tension without breaking. What is the ratio f 1 /f 2 of the fundamental frequencies of vibration on the two strings? (A) 1 ← CORRECT (B) √ 2 (C) 2 (D) 2 √ 2 (E) 4 Solution The string will break if T > σA , where σ is the tensile strength and A is the cross sectional area, so v = s T M/L = r σAL M = r σ ρ , where ρ is the density. This expression is independent of the length of the string, so the wave speed on the two strings is the same. Since v = fλ and the strings are the same length, they have the same fundamental frequency. Copyright c 2015 American Association of Physics Teachers

2015 F = ma Exam 25 25. Two identical carts A and B each with mass m are connected via a spring with spring constant k . Two additional springs, identical to the first, connect the carts to two fixed points. The carts are free to oscillate under the effect of the springs in one dimensional frictionless motion. Cart A Cart B Under suitable initial conditions, the two carts will oscillate in phase according to x A ( t ) = x 0 sin ω 1 t = x B ( t ) where x A and x B are the locations of carts A and B relative to their respective equilibrium positions. Under other suitable initial conditions, the two carts will oscillate exactly out of phase according to x A ( t ) = x 0 sin ω 2 t = - x B ( t ) Determine the ratio ω 2 /ω 1 (A) √ 3 ← CORRECT (B) 2 (C) 2 √ 2 (D) 3 (E) 5 Solution Though you can solve this problem with matrix methods, it can also be solved by carefully digesting the problem. For the in-phase situation, the separation between the masses is constant, so the force of A on B is constant. In that case, only one spring exerts a varying force, and then the angular frequency must be given by ω 1 = p k/m For the out of phase situation, the separation between the objects is not constant, but there exists a point on the connecting spring that does not move. As such, either object can be though of moving under the influence of two springs in parallel, one with length L , the there with length L/ 2. That is equivalent to a spring with constant k in parallel with a spring of constant 2 k . The net spring constant is then k + 2 k = 3 k . As such, the frequency is given by ω 2 = p 3 k/m = √ 3 ω 1 Copyright c 2015 American Association of Physics Teachers