2008 Free Response

Question 1

Mech. 1. A skier of mass M is skiing down a frictionless hill that makes an angle e with the horizontal, as shown in the diagram. The skier starts from rest at time t = O and is subject to a velocity-dependent drag force due to air resistance of the form F = —bv, where V is the velocity of the skier and b is a positive constant. Express all algebraic answers in terms of M, b, e, and fundamental constants.

(d) Solve the differential equation in part (b) to determine the velocity of the skier as a function of time, showing all your steps.

(d) 3 points For taking the differential equation from part (b) and correctly separating the variables in preparation for integration (definite or indefinite integral) dv = Mgsin9 — bv M dt du dt M sin9-bv M For correct integration of both sides of equation For example, using a method involving an indefinite integral Letting u = Mgsin9 — bv, so du = —b dv 1 du dt b Inu — -bt/M u = Ce gsin9 — bv = Ce-bt/M Using t) = O at t = 0 Mgsin9 = C -bt/M Mgsin9 — bv = Mgsin9 e bt/M —bv = Mgsin9 e- — Mg sin 9 For a correct final expression for v (t) bt/M)

Question 3

Mech. 3. In an experiment to determine the spring constant of an elastic cord of length 0.60 m, a student hangs the cord from a rod as represented above and then attaches a variety of weights to the cord. For each weight, the student allows the weight to hang in equilibrium and then measures the entire length of the cord. The data are recorded in the table below:

The student now attaches an object of unknown mass m to the cord and holds the object adjacent to the point at which the top of the cord is tied to the rod, as represented above. When the object is released from rest, it falls 1.5 m before stopping and turning around. Assume that air resistance is negligible.

iii. Calculate the maximum speed of the object.

(iii) 2 points For a correct ener ex ssion 1 = —kx2 + —mv 2 mgyumax 2 2 max = 3.8 m/s 1 2 2 max 1 —kx2 = mgyvmax 2 Dmax2 = 2gyumax m For correct substitution of values previously obtained (especially those from part (d)(i)) (25 N/m) = 2(9.8 m) - (0.27 (0.69 kg) um ax 2 = 14.4 (m/s)2

2009 Free Response

Question 1

Mech. 1. A 3.0 kg object is moving along the x-axis in a region where its potential energy as a function of x is given as U (x) = 4.0x2, where U is in joules and x is in meters. When the object passes the point x = —0.50 m, its velocity is +2.0 m/s. All forces acting on the object are conservative.

x For a minimum of one complete cycle of a sine curve starting at the origin on the x versus t graph For a minimum of one complete cycle of a cosine squared curve starting at the maximum value on the K versus t graph For maxima and minima of the x graph matching the zeroes of the K graph I point I point 1 point

Question 2

x Mech. 2. You are given a long, thin, rectangular bar of known mass M and length t with a pivot attached to one end. The bar has a nonuniform mass density, and the center of mass is located a known distance x from the end with the pivot. You are to determine the rotational inertia 1b of the bar about the pivot by suspending the bar from the pivot, as shown above, and allowing it to swing. Express all algebraic answers in terms of 1b , the given quantities, and fundamental constants. (a) i. By applying the appropriate equation of motion to the bar, write the differential equation for the angle 9 the bar makes with the vertical. ii. By applying the small-angle approximation to your differential equation, calculate the period of the bar's motion. (b) Describe the experimental procedure you would use to make the additional measurements needed to determine 1b . Include how you would use your measurements to obtain 1b and how you would minimize experimental error. (c) Now suppose that you were not given the location of the center of mass of the bar. Describe an experimental procedure that you could use to determine it, including the equipment that you would need.

(a) (i) 4 points For the rotational form of Newton's second law = la For a correct expression of the magnitude of torque For correctly labeling the torque as negative —Mgxsine = Iba For expressing a as the second time derivative of 9 d29 —Mgxsine = 1b dt2 (ii) 4 points For the appropriate small angle approximation For small an les sing = 9 d2e —Mgx9 = 1b dt2 d29 Mgx dt2 2 For recognizing that the coefficient of 9 is 0) Mgx 2 For the relationship between T and (this point was awarded for the equation alone or with relevant work, but NOT as part of multiple random equations) 2m For the final expression for T (this point was awarded for the final correct answer with no supporting work) T = 27t Mgx

(b) (c) 5 points For an experimental procedure that includes: A valid approach How the variables will be measured or calculated, including equipment to be used How these variables will be used to determine 1B How to minimize error Example 1 : Displace the bar by a small angle and release it to To reduce errors, time 10 complete oscillations with a stopwatch. Calculate the average value of the time for 10 oscillations and then divide by 10 to determine the period T . Calculate IR from T 21tJ77ME , using known values of M and x. Example 2: Locate a photogate at the bottom of the bar's swing; set it to measure the amount of time the photogate is blocked. While the bar is hanging from its pivot point, displace the bar to a horizontal position and measure the height of the center of mass above the position of the photogate with a meter stick. Allow the bar to swing through the photogate and obtain the time the gate is blocked. To reduce errors, repeat this procedure five times and obtain an average time. Measure the width of the bar and use this and the time to determine the speed of the bar at the bottom of the swing, D = width/time . Calculate the angular speed of the bar from o = v/t . Apply conservation of energy to the bar: Mgh = IB02 2. Calculate 1B from 1B 2Mgh m , using known values of M, measured value of h, and calculated value of m . 2 points For a valid procedure to locate the center of mass For specifying the equipment to be used Example 1 : Place the bar on top of the top of a prism. Adjust the position of the bar until it is balanced horizontally. The point at which this occurs is the center of mass. Example 2: Place the bar near the edge of a desk or table. Slowly push the bar so it hangs off the table until it is just ready to tip. The point at which this occurs is the center of mass.

2010 Free Response

Question 3

Mech. 3. A skier of mass m will be pulled up a hill by a rope, as shown above. The magnitude of the acceleration of the skier as a function of time t can be modeled by the equations It t a sin where amax and T are constants. The hill is inclined at an angle 9 above the horizontal, and friction between the skis and the snow is negligible. Express your answers in terms of given quantities and fundamental constants. (a) Derive an expression for the velocity of the skier as a function of time during the acceleration. Assume the skier starts from rest. (b) Derive an expression for the work done by the net force on the skier from rest until terminal speed is reached. (c) Determine the magnitude of the force exerted by the rope on the skier at terminal speed. (d) Derive an expression for the total impulse imparted to the skier during the acceleration. (e) Suppose that the magnitude of the acceleration is instead modeled as a = am axe-rt/2T for all t > O , where amax and T are the same as in the original model. On the axes below, sketch the graphs of the force exerted by the rope on the skier for the two models, from t = O to a time t > T . Label the original model Fl and the new model F2 .

(a) 4 points For a correct relationship between velocity and acceleration du OR D = OR dt For a correct substitution of the expression for acceleration into the integral relationship sing-I dt OR sin@ dt (0 < t < T) a For a correct evaluation of the integral, with an integration constant or correct limits amaxT Itt cos— +C OR D l) amaxT Itt cos — For a correct determination ofthe integration constant or evaluation between the limits 0(0) = amaxT 1 — cos amaxT OR - cos;t -1

(b) 2 points For indicating that the work done by the net force is equal to the change in kinetic ener For a correct substitution of velocity from (a) into the work-energy expression 2amaxT = gma€(l — cosz) vo = — cos0) = 0 2amaxT 2ma2ÄT

(d) 2 points J =JFdt For a correct substitution of force into the impulse-time relationship sin dt J = ma Itt — cos For evaluation at the limits of integration mamaxT [— cos 0] 2mamaxT

(e) 6 points Itt Fl = mg sine + ma sin max T —Rt/2T F2 = mg sing + mamax e mg sin 9 For a graph labeled Fl : for starting at mg sine for half a sine wave with a maximum at — T/2 for returning to original starting point at t = T for a horizontal line at the original starting point for t > T For a graph labeled F2 : for starting on the vertical axis at a point above the starting point of Fl (if there is no Fl graph, this point was awarded if the F2 graph starts above mg sine ) for an exponential decay graph 1 point 1 point 1 point 1 point 1 point 1 point

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