2011 Free Response

Question 1

Launching Projectile Device Mech. 1. A projectile is fired horizontally from a launching device, exiting with a speed . While the projectile is in the launching device, the impulse imparted to it is J , and the average force on it is Favg . Assume the force becomes zero just as the projectile reaches the end of the launching device. Express your answers to parts (a) and (b) in terms of v J F and fundamental constants, as appropriate. x p' avg

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For the correct relationship between impulse and the change in momentum J = Ap = mAv Jp = m(vx — O) = mvx For the correct answer m = Jp/vx

The projectile is fired horizontally into a block of wood that is clamped to a tabletop so that it cannot move. The projectile travels a distance d into the block before it stops. Express all algebraic answers to the following in terms of d and the given quantities previously indicated, as appropriate. (c) Derive an expression for the work done in stopping the projectile. (d) Derive an expression for the average force Fb exerted on the projectile as it comes to rest in the block.

(c) 3 points For using the work-energy theorem O — — mvx2 For substituting the expression for m from part (b) For an indication that the work done is negative Alternate Solution Using kinematics and Newton 's second law to determine the average net force 2 — = 2aavgd = 2aavgd —v avg Fug — ma F For substituting this expression for the force into the equation for work W = JF.dr = Favgd - 2d For substituting the expression for m from part (b) J 02 For an indication that the work done is negative

Now a new projectile and block are used, identical to the first ones, but the block is not clamped to the table. The projectile is again fired into the block of wood and travels a new distance dn into the block while the block slides across the table a short distance D. Assume the following: the projectile enters the block with speed , the average force Fb between the projectile and the block has the same value as determined in part (d), the average force of friction between the table and the block is fT , and the collision is instantaneous so the frictional force is negligible during the collision.

(f) 2 points For a correct application of conservation of momentum to the block-projectile collision mvx (M + The of the block/projectile system immediately after the collision is equal to the work done by friction in stopping it. —(M + m) V2 fTD 2 For substituting for V 2 fTD 1 = fTD m = fTD From part (c) the kinetic energy factor in the equation above is equal to the total work done. From part (d) that work is equal to Fbd. m Fbd fTD Using the expression Fbdn = Fbd — fTD from part (e) to substitute for fTD m m Fbd

2012 Free Response

Question 2

You are to perform an experiment investigating the conservation of mechanical energy involving a transformation from initial gravitational potential energy to translational kinetic energy. (a) You are given the equipment listed below, all the supports required to hold the equipment, and a lab table. On the list below, indicate each piece of equipment you would use by checking the line next to each item. Track Cart String Meterstick Electronic balance Stopwatch Set of objects of different masses Lightweight low-friction pulley (b) Outline a procedure for performing the experiment. Include a diagram of your experimental setup. Label the equipment in your diagram. Also include a description of the measurements you would make and a symbol for each measurement. (c) Give a detailed account of the calculations of gravitational potential energy and translational kinetic energy both before and after the transformation, in terms of the quantities measured in part (b). (d) After your first trial, your calculations show that the energy increased during the experiment. Assuming you made no mathematical errors, give a reasonable explanation for this result. (e) On all other trials, your calculations show that the energy decreased during the experiment. Assuming you made no mathematical errors, give a reasonable physical explanation for the fact that the average energy you determined decreased. Include references to conservative and nonconservative forces, as appropriate.

(a) (b) 1 point For choosing the meterstick and stopwatch, regardless of what else is checked 4 points For a procedure that indicates the height needed to calculate energv For a procedure that indicates distance and time measurements to calculate For a diagram and a clear indication of the height measurement For a diagram and a clear indication of the distance measurement le #1 m mb h Use the electronic balance to determine the mass mc of the cart and the mass mb of one object. Attach the object to the cart using the string. Place the cart on the track and hang the object so that the st-ing passes through the pulley. Allow the object to fall a distance h from its initial position to the floor, using the meterstick to measure the distance fallen. Use the stopwatch to measure the time t it takes the object to fall the distance h. Measure the height H of the table. 1 point 1 point 1 point 1 point 1 point

(b) continued • • d h Use the electronic balance to determine the mass m of the cart. Set the track at an incline, and measure the height h of the incline. Place the cart at the top of the incline, and release from rest. Using the stopwatch, measure the time t it takes for the cart to move down the incline. Measure the distance d that the cart moves down the incline.

(c) 6 points For a clear indication of the For a clear indication of the For a clear indication of the tl potentla ener of the system I potential ener of the system •tial kinetic energy f the system 1 point 1 point 1 point 1 point 2 points Fora clear indication of the mal kinetic ener o the system For a correct calculation of the system Example #1 Initial gravitational potential energy: U go = mcgH + ntbgh Final gravitational potential energy: UM = mcgH Initial kinetic energy: Ko = O 2 Final kinetic energy: K f = -j(mc + mb)Df Acceleration is constant, so d = —(00 + , where d is the distance along the track.

(c) (d) (e) continued Example #2 Initial gravitational potential energy: U go = mgh Final gravitational potential energy Ugf = 0 Initial kinetic energy Ko = O Final kinetic energy K f = 2 Acceleration is constant, so d = 2 points For identifying a reasonable cause for the increase in energy For a reasonable explanation related to the cause identified Example An unintentional push was applied to the cart, thus increasing the initial energy. 2 points For a reasonable cause for the decrease in energy related to the nonconservative forces acting on the system For a reasonable explanation related to the cause identified Example Friction acting on the object decreases the speed, thereby decreasing the energy.

Question 3

Sliding Only Sliding and Rotating Rotating with No Sliding Friction —9— Friction With Coefficient g Mech. 3. A ring of mass M, radius R, and rotational inertia MR is initially sliding on a frictionless surface at constant velocity vo to the right, as shown above. At time t = O it encounters a surface with coefficient of friction g and begins sliding and rotating. After traveling a distance L, the ring begins rolling without sliding. Express all answers to the following in terms of M, R, vo , g, and fundamental constants, as appropriate.

(c) Derive an expression for the time it takes the ring to travel the distance L. (d) Derive an expression for the magnitude of the velocity of the ring immediately after it has traveled the distance L. (e) Derive an expression for the distance L.

(c) (d) 2 points For indicating that the linear speed is equal to Ro when the slipping stops ggt vo — ggt = R For the correct answer — 2gg 1 point For substituting the time found in part (c) into a correct kinematics equation -t) = Do — gg 2 gg

2013 Free Response

Question 3

Note: Figure not drawn to scale. Mech 3. A disk of mass M = 2.0 kg and radius R = 0.10 m is supported by a rope of negligible mass, as shown above. The rope is attached to the ceiling at one end and passes under the disk. The other end of the rope is pulled upward with a force FA . The rotational inertia of the disk around its center is MR2 2 . (a) Calculate the magnitude of the force FA necessary to hold the disk at rest. At time t = O, the force FA is increased to 12 N, causing the disk to accelerate upward. The rope does not slip on the disk as the disk rotates. (b) Calculate the linear acceleration of the disk. (c) Calculate the angular speed of the disk at t = 3.0 s. (d) Calculate the increase in total mechanical energy of the disk from t = O to t = 3.0 s. (e) The disk is replaced by a hoop of the same mass and radius. Indicate whether the linear acceleration of the hoop is greater than, less than, or the same as the linear acceleration of the disk. Greater than Less than The same as Justify your answer.

(a) (b) 2 points For a correct expresslon indicating that F -Mg=o FA = (2.0 m/s2)/2 For a correct answer FA = 9.8 N (or 10 N using g = 10 m/s2 ) 5 points For a correct expression of Newton's second law for translational motion FA + T Mg = Ma (equation 1) For using a correct expression of torque in Newton's second law for rotational motion r = la or FAR-TR=1a For substitutlng for the angular acceleration in terms of the linear acceleratlon (a = a/R) FAR-TR= MR2/2 (a/R) F — T = Ma 2 (equation 2 For combinlng equations 1 and 2 to solve for the linear acceleration Add the two equations 2FÅ - Mg = (3/2)Ma a = N)/2.o kg) - 9.8 m/s2) For a correct answer, with units a = 1.47 m/s2 (1.33m s uslng g = 10 m/s2) 1 point 1 point Ipoint 1 point 1 point 1 point 1 point

2 points For using the relationship between linear and angular acceleration in the equation for angular speed 0) = 00 + at and a = a/ R = 0 , so m = at/R m = (1.47 s) /(0.10 m)

For an answer with units, consistent with previous work o = 44 rad/s (40 rad/s using g = 10 m/s2)

(d) 4 points Express the change in mechanical energy as the sum of the change in potential energy and the change in kinetic energy AE = AUg + AK For a correct expression for the change in kinetic energy including both translational and rotational kinetic energy AK = —M v2 — vo + —1m — For a correct expression for the change in potential energy, includlng a correct t h In terms of the time AUg = Mgh = Mg —at2 vo and 00 are zero, so AE = Mg Lat2 + —mu + —1m 2 For simplifying the expression uslng the relationship between linear velocity and angular velocity AE = Mg Lat2 + + AE = —Mgat2 + —MR0)2 For correctly substituting given values and answers from previous parts into a correct expresslon AE = 52.0 s) 2 + rad/s)2 AE = 159 J (144 J uslngg = 10 m/s2 ) 2 points For selecting "Less than" For a correct justification Example The rotational inertia of a hoop is greater than that of a solid dlSk of the same mass and radius, therefore the acceleration of the hoop would be less. 1 point 1 t 1 point 1 point 1 point 1 polnt

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