PHYSICS 221, FALL 2011 EXAM #2 SOLUTIONS WEDNESDAY, NOVEMBER 2, 2011


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1 PHYSICS 1, FALL 011 EXAM SOLUTIONS WEDNESDAY, NOVEMBER, 011 Note: The unit vectors in the +x, +y, and +z directions of a righthanded Cartesian coordinate system are î, ĵ, and ˆk, respectively. In this exam, assume that the magnitude of the acceleration due to earth s gravity at the surface of the earth is g = 9.80 m/s. Problems 31 through 45 are worth points each 31. Which one of the following five statements is false? A. The gravitational force is a conservative force. B. The mechanical energy of a particle is conserved over a time interval if a kinetic friction force acts on the particle during that time interval. C. A force on a particle is a conservative force if the net work done on the particle by the force over a time interval does not depend on the path between the initial and final positions of the particle. D. The force of a Hooke s law spring on a particle is a conservative force. E. If there are no nonconservative forces acting on a particle during a time interval, then the particle s mechanical energy is conserved over that time interval. Answer B is false because mechanical energy is not conserved when nonconservative (kinetic friction) forces are present. 3. A particle moves along the xaxis under the influence of a conservative force for which the potential energy function of position x is U(x) = A Bx + Cx, where A, B and C are nonzero positive constants. The force on the particle versus x is F x =. A. 0 B. Bx Cx C. B Cx D. Cx E. (A / x) + B Cx F x = du / dx = B Cx. 33. The figure shows the potential energy U of a particle moving along the xaxis as a function of the position x of the particle. The position of unstable equilibrium of the particle is at x = m. A. 0 B. 1 C. D. 1 E. 1 U(J) U(x) = (3 J/m)x (1 J/m 3 )x x (m)
2 34. Which one of the following five statements is false? A. If a truck has mass 5000 kg and speed 10 m/s and a car has mass 000 kg and speed 30 m/s, then the truck has a smaller magnitude of momentum than the car. B. The momentum of each of the colliding particles in a collision is not in general conserved in the collision. C. In an elastic collision between two particles, in general the kinetic energy of each particle is the same before and after the collision. D. If the net external force on a system of two colliding particles is zero, then the total momentum of the two colliding particles is conserved in the collision. E. If a putty ball with an initial momentum p 1 collides with a stationary rigid wall and sticks to it, the wall exerts an impulse of p 1 on the putty ball during the collision. Answer C is false because in an elastic collision only the total kinetic energy is conserved. 35. Kinetic friction forces are present within and between two balls when they collide. The total kinetic energy of the two balls after the collision is the total kinetic energy of the two balls before the collision. A. greater than B. equal to C. less than The WorkEnergy Theorem states that the net work done on a system during a time interval is equal to the change in the kinetic energy of the system during that time interval: W net = K. Therefore, since kinetic friction does negative work on the balls, their total kinetic energy decreases as a result of the collision. 36. A thin circular ring of mass M and radius R is rotating about an axis parallel to its continuous symmetry axis as shown in the figure. This axis of rotation (the vertical solid line on the left) passes through the ring and is perpendicular to the plane of the ring. The moment of inertia of the ring about this axis of rotation is. A. MR B. MR C. ( / 5)MR D. MR / E. 4MR The moment of inertia of the ring about its continuous rotational symmetry axis is I cm = MR. The ParallelAxis Theorem then gives I = I cm + Md = MR + MR = MR.
3 37. A rigid circular disk is rotating at angular speed ω about its continuous rotational symmetry axis. Two small particles A and B of equal mass are rigidly attached to the disk at radii r A and r B = r A, respectively. If the kinetic energy of particle A is K A, then the kinetic energy of particle B is K B =. A. K A / 4 B. K A / C. K A D. K A E. 4K A v B = r B = r A = v A K B = 1 mv B = 1 m(v " A ) = 4 1 mv % A & ' = 4K A 38. A particle of mass m is going around a circle of radius R at constant speed v in a clockwise direction. The position and velocity of the particle at some instant of time are shown in the figure. The magnitude and direction of the angular momentum of the particle with respect to the center of the circle at this instant of time are and, respectively. A. mvr, into the page B. mvr, out of the page C. π mvr, toward the left on the page D. mv, downwards on the page E. mv, toward the right on the page The position vector r of the particle always points from the center of the circle to the particle and hence is always perpendicular to the velocity vector. Thus the magnitude of the angular momentum L is L = pr sin = mvrsin(90 ) = mvr. From the righthand rule, the direction of L = m r v is into the page. 39. A solid cylindrical drum with mass M and radius R has a massless cable wound around it. The drum rotates on an axle directed perpendicular to the page. Gravity acts downward on the drum. The cable is being pulled horizontally toward the right by a force F as shown in the figure. The magnitude and direction of the net torque on the drum about its axis of rotation are and, respectively. A. MgR + RF, upwards on the page B. RF/, out of the page C. RF, into the page D. MgR, downwards on the page E. RF, out of the page Gravity and the normal force of the axle on the drum exert no torque on it about the axle because their moment arms are zero. The magnitude of the torque = r " F of the force F is = rf sin(90 ) = RF. The direction is given by the righthand rule as into the page. 3
4 40. One end of a string is tied to the ceiling and the other end is tied to a horizontal massless rod at a distance d 1 from one end and d from the other. Two masses m 1 and m are hanging at rest from the ends of the rod by strings, as shown in the figure. The mass m is given by. A. m d 1 / d B. m C. m d / d 1 D. m d / d 1 E. m d 1 / d Setting the net torque on the rod due to the hanging weights, about the attachment point of the upper vertical string to the rod, equal to zero gives the lever rule m 1 d 1 = m d, which in turn gives m d 1 / d. The torque of the tension force of the upper vertical string on the rod, about its attachment point to the rod, is zero because the moment arm of that tension force is zero. 41. The centers of two uniform solid spheres with equal masses m are a distance d apart. The magnitude of the gravitational force of each sphere on the other is F. If m is doubled for each sphere and d is also doubled, the new magnitude of the gravitational force of each sphere on the other is F =. A. F B. F C. F/ D. 8F E. F/8 F = G m d. F = G (m) (d) = G m d = F. 4. Which one of the following five statements is false? A. The orbit of a planet around the sun is an ellipse with the sun at one focus of the ellipse. B. The gravitational force of the sun on the earth exerts a torque on the earth about the sun. C. For a small planet orbiting a large star in a circular orbit, the period of the orbit increases as the radius of the orbit increases. D. As the planet mars orbits the sun, the straight line between the sun and mars sweeps out equal areas in equal times. E. The angular momentum with respect to the sun of a planet orbiting the sun is conserved. Answer B is false. A central force (the gravitational force in this case) between two particles has no moment arm about either of the particles. Hence the torque is zero. 4
5 43. The period of oscillation of a simple pendulum is measured near the surface of the earth as a function of the angular amplitude of oscillation of the pendulum. The period of oscillation for an angular amplitude of 80 degrees is the period for an angular oscillation amplitude of 5 degrees. A. the same as B. smaller than C. larger than We found from the Pendulum Lab written homework problem that Answer C is true. 44. The period of oscillation of a mass connected to a Hooke s law spring for an oscillation amplitude of 0.01 m is the period for an oscillation amplitude of 0.1 m. A. the same as B. smaller than C. larger than The period of oscillation is T = oscillation. m / k, which does not depend on the amplitude of 45. The angular frequency of oscillation of a 1.00 kg mass connected to a Hooke s law spring with a spring constant of 400 N/m is rad/s. A. 5 B. 10 C. 15 D. 0 E. 5 = k m = 400 N/m 1.00 kg = 0.0 rad / s. Problems 46 through 60 are worth 4 points each 46. A 0.55 kg block of wood slides on a horizontal wood table where the coefficient of kinetic friction between the block and the table is µ k = The block goes once around a closed square path of sidelength L = m as shown in the figure. The net work done by the kinetic friction force on the block by the table as the block goes around the closed path is J. A. 0 B..5 C..5 D. 5.0 E. 5.0 W = f k D = (µ k mg)(4l) = 0.500(0.55 kg)(9.80 m/s )4(0.500 m) =.50 J. 5
6 47. A toy gun has an ideal Hooke s law spring in it that is compressed by.00 cm when the gun is cocked. The spring constant of the spring is 490 N/m. The gun fires a 5.00 g rubber bullet straight up into the air from ground level. The bullet goes up to a height of m before stopping. (Ignore any influence of air friction on the motion of the bullet) A. B. 3 C. 4 D. 5 E. 6 Conservation of mechanical energy gives (1 / )kx = mgy, so y = kx (490 N/m)(0.000 m) = mg ( kg)(9.80 m/s ) =.00 m. 48. A car with mass 1000 kg and speed 30.0 m/s heading east on 4 th Street collides with a car with mass 000 kg and speed 0.0 m/s heading north on Grand Avenue. If the two cars stick together after the collision, the speed of the wreck just after the collision is m/s. (Ignore the friction of the wheels of the cars with the road during and after the collision) A. 17 B. 19 C. 1 D. 3 E. 5 Let east be the +x direction and north be the +y direction. The initial total momentum is P = m A v A + m B vb = (1.00 Mg)(30.0 m/s)î + (.00 Mg)(0.0 m/s)ĵ = (30.0 Mg m/s)î + (40.0 Mg m/s)ĵ. Conservation of total momentum of the two cars gives the final velocity of the wreck as v = P (30.0 Mg m/s)î + (40.0 Mg m/s)ĵ = = (10.0 m/s)î + (13.3 m/s)ĵ. m A + m B 3.00 Mg Then v = v x + v y = (10.0 m/s) + (13.3 m/s) = 16.6 m / s. 49. Two particles A and B have masses and positions given, respectively, by m A =.00 kg, r A = (.00 m)î (3.00 m)ĵ m B = 8.00 kg, r B = (3.00 m)î + (.00 m)ĵ. The position of the center of mass of the two particles is. A. (0 m)î (10 m)ĵ B. (4 m)î (1 m)ĵ C. (1 m)î + (8 m)ĵ D. ( m)î + (1 m)ĵ E. ( m)î + ( m)ĵ r cm = m Ar A + m B rb m A + m B = (.00 kg)[(.00 m)î (3.00 m)ĵ] + (8.00 kg)[(3.00 m)î + (.00 m)ĵ] kg = (.00 m)î + (1.00 m)ĵ. 6
7 50. Two 1.00 kg particles A and B collide. The initial velocities of the two particles before the collision are, respectively, v A1 = (.00 m/s)î (7.00 m/s)ĵ v B1 = (6.00 m/s)î + (3.00 m/s)ĵ. The velocity of the center of mass of the two particles after the collision is. A. (4 m/s)î ( m/s)ĵ B. (4 m/s)î + (4 m/s)ĵ C. (8 m/s)î (10 m/s)ĵ D. (8 m/s)î (4 m/s)ĵ E. 0 The velocity of the center of mass does not change due to the collision. Thus we have v cm = m v A1 + m v B1 = 1 m + m ( v A1 + v B1 ) = 1 [(8.00 m/s)î (4.00 m/s)ĵ] = (4.00 m/s)î (.00 m/s)ĵ. 51. A 10.0 kg solid uniform sphere of radius 0.50 m is rotating about an axis through its center with an angular speed of 0.0 rad/s. The rotational kinetic energy of the sphere is J. [The moment of inertia of a solid uniform sphere rotating about an axis through its center is (/5)MR ] A. 50 B. 100 C. 150 D. 00 E. 50 K = 1 I cm = 1 " % 5 MR & ' = 1 5 M (R) = 1 5 (10.0 kg)[(0.50 m)(0.0 rad/s)] = 50.0 J. 5. A square flat plate of side length L = m has three equal forces exerted on it at the same time in the plane of the plate as shown. The magnitude of each force is 0.0 N. The point O is at the lower left corner of the plate. The magnitude of the net torque on the plate about the point O due to the three forces is N m. A. 10 B. 15 C. 0 D. 5 E. 30 All the forces have the same magnitude F, and the torque direction is the same for F and F 3 (the torque due to F 1 is zero) so we can just add the magnitudes. The moment arms of the forces about O are 1 = 0, = L / and 3 = L. Thus the magnitude of the net torque is net = F 1 + F + F 3 = F( ) = 3FL / = 3(0.0 N)(0.500 m)/ = 15.0 N " m. 7
8 53. When a grinding wheel is in use, the power required by an electric motor to keep the grinding wheel rotating at 1000 revolutions per minute is 157 W. The magnitude of the torque exerted by the electric motor on the grinding wheel is N m. A. 0.5 B. 1.0 C. 1.5 D..0 E..5 " = 1000 rev % " min & ' ( rad rev = P " = 157 W 105 rad/s = 1.50 N m. % " & ' 1 min % 60 s & ' = 105 rad/s. 54. A uniform solid cylinder of radius 0.18 m rolls down an inclined plane without slipping as shown in the figure. The angle of the inclined plane with respect to the horizontal is θ = 30.0 degrees. The magnitude of the angular acceleration of the cylinder is rad/s. [The moment of inertia of a uniform solid cylinder about its continuous rotational symmetry axis is (1/)MR ] A. 5 B. 10 C. 15 D. 0 E. 5 Using the traditional method of solving this problem, we need 3 equations to solve for the 3 unknowns: the magnitudes of the acceleration a of the center of mass of the cylinder directed down the slope, the angular acceleration α of the cylinder directed into the page, and the static friction force f s directed up the inclined plane. Using Newton s nd law, the net force directed down the slope due to gravity and f s is f s + Mgsin" = Ma. (1) The magnitude of the torque about the axis of rotation of the cylinder is = Rf s (the weight of the cylinder and the normal force of the inclined plane on the cylinder contribute no torque on the cylinder about this axis). Using the rotational analogue of Newton s nd law gives = Rf s = I cm ". () The condition for rolling without slipping is a = R. (3) gsin" Solving Eqs. (1) (3) for α gives = R 1+ I cm & % MR ' (. Using the given I cm = 1 MR yields = gsin" 3R = (9.80 m/s )sin(30.0 ) = 15.0 rad / s. 3(0.18 m) 8
9 An alternate, faster and conceptually simpler way to solve this problem is to put reference point O at the contact point between the cylinder and the inclined plane instead on the continuous rotational symmetry axis of the cylinder. This approach was discussed in lecture and was applied to understanding the behavior of the Giant Yoyo demonstration. Then the torque on the cylinder about the point O is just due to the xcomponent of the weight acting on the center of mass, since the static friction force, the ycomponent of the weight acting on the center of mass and the normal force on the cylinder produce no torque on the cylinder about the point O. One then obtains the magnitude of the torque about the point O as = w x R = MgRsin" = I, where for the righthand equality we used the rotational analogue of Newton's nd law. Note that the angular acceleration of the cylinder about the point O is the same as that about the continuous rotational symmetry axis of the cylinder. Thus, the magnitude of the angular acceleration is = MgRsin". I The moment of inertia I about the point O is found using the Parallel Axis Theorem to be I = I cm + Md = 1 MR + MR = 3 MR. Substituting this into the expression for gives = MgRsin" = MgRsin" = I 3 MR g sin" 3R, which is, of course, the same result as obtained above using the traditional method of solution. 55. An ice skater is spinning on the ice about a vertical axis at an angular speed of 30.0 rad/s. She has her arms held against her body and has a moment of inertia about her vertical spinning axis of 50.0 kg m. She then extends her arms directly outward and her new moment of inertia is 75.0 kg m. The angular speed at which she now spins is rad/s. A. 5 B. 10 C. 15 D. 0 E. 5 Angular momentum is conserved in this process: L 1 = I 1 1 = L = I, yielding = I 1 I 1 = 50.0 kg " m (30.0 rad/s) = 0.0 rad / s kg " m 9
10 56. A horizontal uniform plank of mass 10.0 kg and length L =.00 m is attached at its left end to a hinge that is attached to a vertical wall, as shown in the figure. The plank is at rest and has a massless rope attached to its right end at an angle θ = 37.8 to the horizontal. The other end of the rope is attached to the vertical wall. The tension in the rope is N. (Hint: set the net torque on the plank about the hinge to be zero) A. 60 B. 65 C. 70 D. 75 E. 80 Let the zaxis point into the page. The weight of the plank acts through its center of mass at a distance L/ from the hinge, so the torque about the hinge due to the weight of the plank is plank z = w = MgL /. The tension T in the rope produces a torque about the hinge pointing out of the page given by rope z = "T = "TL sin. Since the angular acceleration of the plank about the hinge is zero, we set the total zcomponent of the torque to zero: MgL TL sin" = 0. Then solving for T gives T = Mg sin = (10.0 kg)(9.80 m/s ) = 79.9 N. sin(37.8 ) 57. A small projectile is launched from the surface of a large uniform solid spherical body that is at rest, in a direction directly away from the center of the body. The body has no atmosphere, has a mass M = kg and a radius R = m. In order for the projectile to reach a distance of R from the center of the body before momentarily stopping, the launch speed of the projectile has to be km/s. A. 7.9 B. 8.7 C. 9.5 D E. 11. Let m be the mass of the projectile. Conservation of mechanical energy gives K 1 + U 1 = K + U 1 mv GmM R = 0 GmM R. Solving for the launch speed v gives v = GM R = ( "11 N m /kg )( kg) m 10 = 7.90 km / s.
11 58. A planet of mass m = kg is in a circular orbit of radius r = m about a star with mass M = kg. The period of the orbit is days. A. 400 B. 500 C. 600 D. 700 E. 800 T = r 3/ GM = (.08 " m) 3/ % 1 h ( % (6.67 " N m /kg )(1.99 " kg) & ' 3600 s) * 1 day ( & ' 4 h ) * = 599 days. 59. A 4.00 kg block is oscillating along the xaxis of a frictionless horizontal table on the end of a massless Hooke s law spring with spring constant 400 N/m. The equilibrium position of the block is at x = 0. When the position of the block is x = m, the speed of the block is 5.00 m/s. The timeindependent mechanical energy of oscillation of the blockspring system is J. A. 0 B. 40 C. 60 D. 80 E. 100 E = K + U = 1 mv + 1 kx = 1 " (4.00 kg)(5.00 m/s) + (400 N/m)(0.500 m) = 100 J. 60. A simple pendulum on the surface of the earth has a period of oscillation of 5.00 s for smallamplitude oscillations. On the surface of another planet, the period of oscillation of the same pendulum for smallamplitude oscillations is 3.50 s. The magnitude of the freefall acceleration on the surface of the other planet is m/s. A. 5 B. 10 C. 15 D. 0 E. 5 T g = g 1 1 " % & T = (9.80 m/s 5.00 s ) " 3.50 s% & = 0.0 m / s. 11
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