Physics 221. Final Exam Spring 2003

Transcription

1 Physics 1. Final Exam Spring 003 The situation below refers to the next two questions: A block of mass m = 0. kg starts from rest at point A and slides down a frictionless curved surface to point B, where its speed is v = 8 m/s. Then, it slides along a rough horizontal surface for d =10 m and comes to rest at point C. A h B C 1. What is the initial height h of the block? a. h = 1. m b. h =.5 m *c. h = 3.3 m d. h = 4.6 m e. h = 5.1 m d Mechanical energy is conserved while no friction is present: 1 mgh + 0= 0+ mv v h= = 3.3m g. Find the coefficient of kinetic friction between the block and the horizontal surface between B and C. a. µ k = 0.1 b. µ k = 0. *c. µ k = 0.3 d. µ k = 0.4 e. µ k = 0.5 The missing mechanical energy must be equal to the work done by friction: E = W non conservative 1 mv 0 = fkd v where fk = µ kn = µ kmg µ k = = 0.3 gd Page 1 of 8

2 Physics 1. Final Exam Spring 003 The situation below refers to the next two questions: The graph below shows the potential energy U and the total mechanical energy E of a particle as a function of position x. The only force acting on this particle is the force associated to this potential energy. U (J) E = 5.0 J x (m) 3. At which position does the particle have the greatest kinetic energy? a. At x = 1 m b. At x = m *c. At x = 4 m d. At x = 5 m e. At x = 6 m Max KE Min U At x = 4 m 4. Which of the following statements about the force on the particle at x = 3 m is true? a. It is zero b. It points in the x direction *c. It points in the +x direction d. The particle can never be at x = 3 m, so there is no force to be discussed. e. The direction of the force depends on the direction of the velocity of the particle. du Fx = The force is minus the slope of the U(x) curve, dx. At x = 3 m, the curve has a negative slope, so the force must point in the +x direction. Page of 8

3 Physics 1. Final Exam Spring 003 The situation below refers to the next two questions: A rope is attached to the handle of a bucket with a stone of mass m inside. A person swirls the whole system in vertical circles as shown in the figure below. The distance between the stone and the center of the circle is L = 1.5 m. L 5. What is the minimum speed that the system must have at the highest point of the trajectory if the stone is to stay in contact with the bottom of the bucket? *a. v min = 3.8 m/s b. v min = 4. m/s c. v min = 5.7 m/s d. v min = 6.1 m/s e. v min = 7.8 m/s The condition for minimum speed is that the normal force exerted by the bottom of the bucket on the stone is equal to zero when the system is at the highest point of the trajectory. This means that the only force providing the required centripetal acceleration to the stone is its weight mg. v mg = m L min vmin = gl = 3.8 m/ s 6. As the bucket moves from the highest to the lowest point in its trajectory, the work done by the tension on the rope is and the work done by gravity is. a. negative, negative b. zero, negative c. positive, negative d. negative, positive *e. zero, positive The tension on the rope does zero work along any portion of the circle because it is always perpendicular to the displacement. Gravity does a positive work along this displacement, because the force points down and the y-component of all the differential displacement vectors dl that this displacement is Page 3 of 8

4 Physics 1. Final Exam Spring 003 made of also point down. In other words, the angle between the force and the displacement is always acute so dw = Fg dl = Fgdy > 0 Or: W g = - U g, where U=mgh. As the stone moves from top to bottom of the trajectory, its gravitational potential energy decreases. Therefore U<0, so W>0. The situation below refers to the next two questions: A firecracker rocket is flying vertically up when it explodes in two fragments of masses m and M (> m). The bigger mass is ejected to the left and the small mass to the right.: 7. During this process, it is reasonable to assume that: a. total kinetic energy is conserved, but total linear momentum is not conserved *b. total linear momentum is conserved, but total kinetic energy is not conserved c. both total kinetic energy and total linear momentum are conserved d. total linear momentum is conserved and the kinetic energy of each part is conserved. e. both kinetic energy and linear momentum of each part is conserved The external forces acting on this system (gravity, air resistance) are small compared to the forces involved in the explosion (or the change in momentum of each part due to the external forces p ext =F ext t is negligible compared to the change in momentum due to the internal forces p int =F int t), so F ext = 0 is a reasonable approximation. The, total linear momentum is conserved. Kinetic energy is not conserved because some energy is used to restructure the system (one object goes to two). An explosion is an inverted perfectly inelastic collision. 8. Immediately after the explosion, the center of mass of the system moves: *a. Up b. Down c. Up and to the left d. To the left e. It does not move. F ext = ma cm, so a cm = 0. Therefore, v cm is constant. (Or: p total is constant, but p total = m total v cm, so v cm = constant). Before the explosion, the center of mass was moving up. Thus after the explosion it is also moving up. Page 4 of 8

5 Physics 1. Final Exam Spring A marble of mass m = 8.0 g is fired into a block of soft wood hung from two wires as shown below.the block has a mass M =.0 kg and the marble imbeds itself in the block causing it to rise to a maximum height h = 0.1 m. Find the speed v of the marble just before impact. (Ignore the mass of the wires) m v e. v = 440 m/s M h a. v = 90 m/s *b. v = 385 m/s c. v = 400 m/s d. v = 45 m/s When the marble hits the block, we have a perfectly inelastic collision: m+ M v = v' mv= (m+m)v, so m After the collision, the system moves as a pendulum and the total mechanical energy is conserved: 1 ( ) ' M + mv = ( M+ mgh ) v' = gh Thus, m+ M v= gh = 385 m/ s m Page 5 of 8

6 Physics 1. Final Exam Spring A car must achieve an average speed of 50 km/h on a track of total length of 1500 m. The car travels the first half of the track at an average speed of 00 km/h. Therefore, the average speed needed in the second half of the track has to be: *a. 333 km/h b. 300 km/h c. 75 km/h d. 65 km/h e. 60 km/h. If the average speed must be 50 km/h, the time to cover the complete track must be: xtotal 1.5km 3 ttotal = = = 6 10 h v 50 km/ h ave, total The time used in the first half is x1 0.75km 3 t1 = = = h vave,1 00 km/ h So the second half must be covered in 3 t = ttotal t1 =.5 10 h Thus, x1 0.75km vave,1 = = = 333 km/ h 3 t.5 10 h Two balls are thrown from the roof of a building with an initial speed of 10 m/s. Ball number 1 is thrown vertically downward and ball number vertically upward. Ball 1 hits the street below the building in 4 seconds. How much later does the second ball hit the ground? (Take g=10 m/s ). a. t = 1s *b. t = s c. t = 3s d. t = 4s e. t = 5s Using the first ball, we can figure out the initial height for both: 1 0 = h (10 m/ s)(4 s) (10 m/ s )(4 s) h= 10m For the second ball, 1 0 = 10 m+ (10 m/ s) t (10 m/ s ) t The (positive) solution to this quadratic equation is t = 6 s, so ball hits the ground seconds after ball 1. Page 6 of 8

7 Physics 1. Final Exam Spring An object is thrown at ground level with an initial speed of 0 m/s. It hits the ground seconds later. At what angle above the horizontal was the object thrown? a. 10 o b. 0 o *c. 30 o d. 40 o e. 45 o Consider the y(t) equation for this motion: 1 0= 0+ v0 yt gt 1 0= v0 sinθt gt gt sinθ = = 0.5 θ = 30 v A point on the rim of a 5-cm-radius wheel has a constant centripetal acceleration of 4.0 m/s. The tangential acceleration of that point is a. 6.0 m/s b. 4.0 m/s c..0 m/s d. 1.0 m/s *e. None of the above. If the centripetal acceleration in a circular motion is constant, the speed must be constant as well (a c = v /R), so the tangential acceleration is zero. Page 7 of 8

8 Physics 1. Final Exam Spring When a man holding weights and spinning on a frictionless rotating stool extends his arms horizontally and thereby doubles its moment of inertia, the rotational kinetic energy is: a. twice as much as before *b. half as much as before c. the same as before d. four times as much as before e. one fourth as much as before. There is no sizable external torque on this system (friction on the axis of the stool and air resistance are quite small), so angular momentum is conserved. Iω = I ω i i f f Ii Ii ωi ωf = ωi = ωi = I f Ii Compare the kinetic energies: 1 KEi = Iiωi 1 1 ω i 1 1 KE f = I fω f = ( Ii) = Iiωi = KEi 4 Page 8 of 8

9 Physics 1. Final Exam Spring 003 The situation below refers to the next two questions: A uniform solid sphere rolls down an incline. The magnitude of the acceleration of the g center of mass is observed to be What is the angle θ between the incline plane and the horizontal? a. θ = 4 *b. θ = 8 c. θ = 15 d. θ = 0 e. θ = 33 We need a free-body diagram: N f s mg translation of the cm along the incline: mg sinθ fs = macm rotation about the cm: fsr= Icmα rolling without slipping: acm = Rα 3 equations with 3 unknowns (a,α,f S ) Icmα Icmacm fs = = R R Icmacm mg sinθ = ma cm R mg sinθ g sinθ acm = = Icm Icm m + 1+ R mr For a uniform solid sphere, g sinθ 5 acm = = g sin θ If this is to be equal to g/10, Icm = 5 mr, so Page 9 of 8

10 Physics 1. Final Exam Spring g g sin θ = sinθ = 50 o θ = If a frictionless block were to slide down the same incline, its acceleration a would be: g a > *a. 10 g a = b. 10 g a < c. 10 d. The answer depends on the ratio of the block s and sphere s masses e. The answer depends on the radius of the sphere and the height of the block Mass never matters when the only force doing work (i.e., speeding up or slowing down the object) is gravity. The size of an object does not matter either for sliding objects. For rolling objects, only the shape matters, not the actual dimensions: We always encounter the ratio I/mR, where I mr, so the R always cancels out. Therefore, d and e are false. When an object slides down a frictionless incline, all the gravitational potential energy becomes translational kinetic energy. When an object rolls down an incline, the gravitational potential energy becomes both translational kinetic energy and rotational kinetic energy. So the translational kinetic energy is less, and therefore the final speed is less than for the sliding object. In terms of acceleration, this means that the acceleration is greater for the sliding object (it attains a larger speed). Page 10 of 8

11 Physics 1. Final Exam Spring A uniform rod is free to rotate about its center of mass, that we shall take as the origin, as shown in the figure. Its initial position is such that the end labeled P is at r P = (i 3j) m. A force F = (i + 5j) N is applied at that point. y x P What is the torque due to this force on the rod about the origin? a. τ = ( 7 k) Nm b. τ = ( 11 k) Nm c. τ = ( 3 k) Nm *d. τ = (16 k) Nm e. τ = (1 k) Nm iˆ ˆj kˆ τ = r F = 3 0 = 16ˆ k 5 0 Or τ = r F = (iˆ 3 ˆj) (iˆ+ 5 ˆj) = 4iˆ iˆ 6 ˆj iˆ+ 10iˆ ˆj 15 ˆj ˆj = 0 + 6kˆ+ 10kˆ+ 0 = 16kˆ Page 11 of 8

12 Physics 1. Final Exam Spring The mobile shown below is made of four identical objects of mass m. The rods and the strings are massless. The upper rod has length L and is supposed to be horizontal. Find the distance d between the left end of the rod and the point from which the mobile should be hung from the ceiling. L d = *a. 4 L d = b. 3 L d = c. 3 3L d = d. 4 L d = e. 3mg d T L mg Net force = 0 T mg 3mg = 0 T = 4mg Net torque (about the left end) = 0 mgl mgl L Td mgl = 0 d = T = 4mg = 4 Or: Take torque about the point where T is applied. Then net torque equal zero means: mgl L 3 mgd mg( L d) = 0 d = = 4mg 4 Page 1 of 8

13 Physics 1. Final Exam Spring A small dust particle with mass m = 10 9 g and electrostatic charge q = C has been suspended between the plates of a horizontal parallel plate capacitor so that the electrostatic force compensates the gravitational force. The magnitude of the electric field inside the capacitor is: a. E = 10 5 V/m *b. E = V/m c. E = V/m d. E = V/m e. E = V/m qe mg E The net force must be zero: qe mg = 0 E mg q 5 = = 3 10 N/ C Page 13 of 8

14 Physics 1. Final Exam Spring Two point charges Q 1 and Q are placed at (x = a, y = 0) and (x = 0, y = a), respectively, as shown in the figure below. Find the work done by the electric field when a third charge q is brought from infinity to the origin. y (x = 0, y = a) q Q Q 1 (x = a, y = 0) x Q 1 = 3.0 µc Q = Q 1 = 6 µc q =.0 µc a = 5.0 cm *a. W =. J b. W = 1.1 J c. W = 0.31 J d. W = 1.3 J e. W = 5.4 J ( f i) ( f 0) W = U U = U = U f (where the initial point is infinity, so U i = 0 ) U f = qvorigin = q( k Q e + k Q e ) = q( k Q Q e + ke ) = k qq e =.J a a a a a Page 14 of 8

15 Physics 1. Final Exam Spring A pear-shaped conductor (gray shape in the figures) has a certain net charge Q 0. Which of the figures is a reasonable representation of the electric field lines in the region around the conductor? (a) (b) (c) *(d) (e) In a isolated conductor in equilibrium, there cannot be positive charges (E-lines out) on one side and negative charges (E-lines in)on the other side (they are free to move and they attract each other), (a) and (e) are bad options. The electric field (and therefore the electric field lines) must be perpendicular to the surface at all points, so (c) is a bad option. The electric field is stronger near the regions with higher curvature (or smaller curvature radius), which means that there must be a higher density of lines in the region with higher curvature (the right-hand side). So (b) is a bad option. Option (d), instead, verifies all these criteria. Page 15 of 8

16 Physics 1. Final Exam Spring 003. Charge is distributed uniformly with the volume charge density ρ throughout the volume of an infinitely long cylinder of radius R. Find the magnitude of the electric field at a distance r>r from the cylinder axis. ρr E = a. ε 0r ρr E = b. πε 0r ρ E = c. ε 0r ρr E = *d. ε 0r ρ E = e. 4πε 0r r L Use a Gaussian cylinder of length L and radius r: Φ = E ds = E ds + E ds = E ds cylinder side caps side (no flux through the caps because E and ds are perpendicular there). So: Φ= E ds = Eds= E ds= Eπ rl side side side Using Gauss s law, qenclosed ρπ Φ= = ε0 ε0 Therefore, ρπ R L Eπ rl= ε ρr E = ε r 0 0 R L Page 16 of 8

17 Physics 1. Final Exam Spring Three capacitors with the same shape and dimensions are made of three different conductors: gold, copper and aluminum. The conductivity of gold is greater than the conductivity of copper, and the conductivity of copper is greater than the conductivity of aluminum. Rank the capacitances of these capacitors. a. C gold > C copper > C aluminum b. C gold < C copper < C aluminum c. C gold = C copper = C aluminum (if they are parallel plate capacitors only) *d. C gold = C copper = C aluminum (always) e. Not enough information to answer the question. Capacitance only depends on the geometry of the capacitor (same for all of them) and on the dielectric between the plates, but not on the material of the plates (it just needs to be a conductor). 4. If you stretch a cylindrical wire and it remains cylindrical, how does this affect the resistance of the wire (measured end to end along its length)? Before After *a. It increases. b. It decreases. c. It remains the same. d. The result could be different for different materials. e. The result could be different for different cylinders and different stretches. Resistivity is the same for both, since that depends only on the material used. But l R = ρ resistance is A, so increasing l (length) and decreasing A (cross section area) both increase the resistance. Page 17 of 8

18 Physics 1. Final Exam Spring 003 The situation below refers to the next two questions: A copper wire has cross-sectional area m and length 4.0 m. The resistivity of copper is ρ = Ωm. A current of.0 A is uniformly distributed across the crosssection. 5. What is the magnitude of the electric field along the wire? *a. E = V/m b. E =.7 10 V/m c. E = V/m d. E = V/m e. E = V/m Electric field and current density J are related through resistivity: I E = ρj = ρ = V / m A 6. How much electric energy is transferred to thermal energy in 30 min? a. 185 J *b. 45 J c. 375 J d. 435 J e. 55 J Power is energy per unit time, so the total energy for an interval t is U=P t. l P = I R = I ρ A l = = A U I ρ t 45J (Note: Minutes must be converted to seconds) Page 18 of 8

19 Physics 1. Final Exam Spring The figure shows a circuit containing one ideal battery of emf ε =1 V, and four resistances with the following values: R 1 = 0 Ω, R = 0 Ω, R 3 = 30 Ω, R 4 = 8.0 Ω. What is the magnitude of the current through the battery? ε R 1 R 3 R 4 R a. I = 0.10 A b. I = 0.0 A *c. I = 0.30 A d. I = 0.40 A e. I = 0.50 A This circuit can be reshaped as follows: R R 1 R 4 R 3 ε The equivalent resistance for the two resistors in parallel is = + R 3 = 1Ω R R R 3 3 So the equivalent resistance of the whole circuit is: R eq = R 1 + R -3 + R 4 = 40 Ω Current through the battery: I ε = = 0.30A R eq Page 19 of 8

20 Physics 1. Final Exam Spring In the circuit shown below, ε =1.0 V, R = 1.40 MΩ, and C =.00 µf. The capacitor is initially uncharged and the switch is open. The switch is closed at t = 0. How long does it take for the charge to build up to 16.0 µc? ε R a. t = 1.1 s b. t = 1.4 s c. t =.1 s d. t =.8 s *e. t = 3.1 s C The charge as a function of time for this circuit is: / ( 1 t ) Q = Q e τ so Q Q t τ ln 1 ln Q τ = = Q Q The time constant is τ = RC =.8 s The final charge is determined by the maximum voltage across the capacitor. When the current is zero, the plates are connected to the battery, so V C = ε. Therefore, Q = εc=4 µc. If Q = 16 µc, Q t = τ ln = 3.1s Q Q Page 0 of 8

21 Physics 1. Final Exam Spring 003 lab table motion detector 135 cm clamp X force probe spring plate mesh 9. In lab, you studied the vertical oscillation of a plate. Position data collected by the computer is referenced to the X axis shown in the figure at the right. Assume that the plate hangs at its equilibrium position of X = 50 cm. Consider two trials of its motion which differ only in the way in which they are begun, namely, 1. In trial #1, the plate is displaced downward to X = 60 cm and released at rest;. In trial #, the plate is displaced downward to X = 70 cm and released at rest. For these two trials, compare: 1. the time required for the plate to first return to its release point;. the maximum acceleration experienced by the plate. For trial # (as compared to trial #1), a. the time to return is twice as long, while the maximum acceleration is the same; b. the time to return is twice as long, while the maximum acceleration is twice as large; c. the time to return is twice as long, while the maximum acceleration is four times as large; d. the time to return is the same, and the maximum acceleration is the same; *e. the time to return is the same, while the maximum acceleration is twice as large. The period of the oscillations is independent of the amplitude (10 cm and 0 cm, respectively), so the time to return must be the same. The maximum acceleration is proportional to the maximum displacement from equilibrium or amplitude. Since the amplitude is twice as large in trial, the maximum acceleration will also be twice as large. Page 1 of 8

22 Physics 1. Final Exam Spring X As you did in lab, consider the motion of a cart which is given an initial velocity up an inclined track toward a "motion detector". Assume the cart goes part of the way up, and then returns back down the track. (As was done in lab, assume the sensor gives positions relative to the X axis illustrated above). 30. Which of the following best graphs illustrates the velocity, V X, of the cart (after leaving your hand) versus time? V x V x V x V x V x 0 t t t t t A B C D E The cart moves in the negative-x direction, so we must start with a negative velocity. This already leaves only d! Also, the cart turns around. This means that the speed must be 0 at some point. This happens in d and e only. 31. Which of the following best illustrates the acceleration, a X, of the cart versus time (during the same period)? a x a x a x a x a x 0 t t t t t A B C D E The acceleration points down the ramp all the time and is constant (a=g sin θ). The positive x is taken down along the ramp, so the acceleration must be positive. Page of 8

23 Physics 1. Final Exam Spring Consider the collision of two identical hockey pucks on a smooth ice rink. Assume that by analyzing the videotape of the collision, various but not all components of the momenta of the two pucks before and after collision have been determined and are recorded in the table below. Assume that there is negligible friction with the ice, but that the pucks are soft and that considerable kinetic energy is lost in the collision. Under these assumptions, what is the value (if it can be determined) of the entry marked with the X? Before Collision After Collision Puck #1 Puck # Puck #1 Puck # Px X Py a) -0.0 *b) -0.4 c) +0.6 d) e) Cannot be determined from the information given. Conservation of linear momentum in the x direction: = X X = = 0.4 Y 33. The trajectory of a puck sliding freely on a level air table is recorded (as done in 1 lab) on a piece of (metric) graph paper, shown at the right. Assume that the spark generator was operating at a rate of 60 Hertz, and that the figure at the right is approximately full scale. *b) v x =84 cm/s c) v x =60 cm/s d) v x = 0.0 cm/s X Estimate the magnitude of the X component of the velocity of the puck. a) v x = 103 cm/s e) It is not possible to calculate an estimate without a protractor to measure angles. The horizontal distance between the first and the fourth points is 4. cm. Page 3 of 8

24 Physics 1. Final Exam Spring 003 The time elapsed between the first and the fourth points is 3 time 1/60 = 0.05 s. 4.cm vx = = 84 cm/ s So the x-component of velocity is 0.05s. Page 4 of 8

25 Physics 1. Final Exam Spring In the rotational motion experiment, the following graph was obtained when the wheel of the apparatus which you used was rotated by hand. Provide the t likely explanation for this particular graph. a. The wheel was rotated rapidly then stopped for a second or two, then rotated rapidly, then stopped again, and so on. b. The wheel was rotated at different velocities, increasing in steps as time went on. c. The wheel was rotated slowly, and the angular encoder on the wheel axis has a resolution of 1/0 of a revolution. *d. The wheel was rotated slowly, and the angular encoder on the wheel axis has a resolution of 1/00 of a revolution. e. The apparatus likely was malfunctioning. Any rapid rotation would produce fluctuations in the graph, so a and b not good options. You saw this graph when the wheel was rotated slowly. The angle jumps correspond to the angular precision of the apparatus. Between 0.1 rad and 0.3 rad, there are 6 jumps. So each jump corresponds to 0./6=0.03 rad. 1turn rad = turns = turns π rad 188 This is closer to answer (d) than to answer (c). Page 5 of 8

26 Physics 1. Final Exam Spring 003 #1 # 35. Consider two conducting spheres mounted on insulating rods and located (with the relative positions shown) in a region of large electric field, near a negatively charged rod as shown. Assume that both of the spheres are initially uncharged. Then one of the two spheres is displaced slightly so that the two spheres touch each other briefly. Assume that neither touches the rod, and that their relative positions always are as shown. Which of following are possible values for the resulting charge on each sphere, #1 and #, respectively, after the sequence of events described? #1 # A +3.0 nc 3.0 nc *B 3.0 nc +3.0 nc C 3.0 nc 3.0 nc D +3.0 nc +3.0 nc E 0 nc 0 nc When the spheres touch, the negative charges in sphere # flow to sphere #1 (repelled by the negatively charged rod). When the spheres are separated (in the presence of the rod), sphere #1 remains therefore negatively charged and sphere # remains positively charged. The total charge must be zero, because both spheres were initially uncharged. Page 6 of 8

27 Physics 1. Final Exam Spring As done in lab, a student, seat, chain, and two load cells are at rest and supported by cables attached to the upper (sensitive) fittings of the two load cells. The cell readings and those of the large protractors are protractors given in the table below. load cells 0 0 chain Y seat X LEFT RIGHT Cell reading 531 N 600 N Protractor reading Use this information to estimate the X component of the force exerted by the left cell upon the end of the left chain. a. +66 N b. 66 N *c. 460 N d N e. 0 (The force must be zero, since nothing is accelerating.) L R θ 1 W θ The x-component of force L is negative according to the x-y axis given in the figure. L x = L sin θ 1 = (531 N) (sin 60 ) = 460N Page 7 of 8

28 Physics 1. Final Exam Spring 003 Did you bubble in your exam version? Page 8 of 8