PHYSICS 025 FINAL EXAMINATION Friday, 2003 April 11
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1 Print Name: Student No. A 1-10 A 11-0 A 1-30 B1 B B3 B4 C1 C C3 Total PHYSICS 05 FINAL EXAMINATION Friday, 003 April 11 Time: 7:00 10:00 PM 1. This is a closed book test. You may use only a calculator, and the list of formulas provided.. Except for multiple-choice questions, for maximum credit you should supply complete solutions, not just the answers. Don t forget units! Present all work that you wish to be marked on the page where the problem is stated. You may use the reverse side for rough work. 3. Feel free to use sketches or diagrams in solving the problems. 4. a. Do all the multiple choice questions in Part A (30 Marks) b. Do any 3 problems from Part B (30 Marks) c. Do any problems from Part C (0 Marks)
2 PART A 1. A car goes from 0 m/s to rest in 10 s. Its average acceleration is A).0 m/s B).0 m/s C) 0 m/s D) 0 m/s E) 00 m/s. Which of the curves on the graph below best represents the vertical component v y versus t for a projectile fired at an angle of 45 above the horizontal? A) OC B) DE C) AB D) AE E) AF 3. An object moving at constant velocity in an inertial frame must: A) have a net force acting on it B) eventually stop due to gravity C) not have any force of gravity acting on it D) have zero net force acting on it E) have no frictional force acting on it 4. The "reaction" force does not cancel the "action" force because: A) the action force is greater than the reaction force B) they act on different bodies C) they are in the same direction D) the reaction force exists only after the action force is removed E) the reaction force is greater than the action force 5. A car of mass m is being driven at speed v on a level, unbanked curve of radius R. If the car does not skid, the friction force must be equal to A) µ s mg B) mg C) mgr D) mv R E) µ mv R s 6. A force with a given magnitude is to be applied to a wheel. The torque can be maximized by: A) applying the force at the rim, tangent to the rim B) applying the force near the rim, radially outward from the axle C) applying the force near the axle, parallel to a tangent to the wheel D) applying the force near the axle, radially outward from the axle E) applying the force at the rim, at 45 to the tangent 7. The following point-like weights are placed on a rod of negligible weight: 10 N at x = 1.0 m 0 N at x = 3.0 m 40 N at x = 4.0 m. The x-coordinate of center of gravity is A).71 m B) 3.00 m C) 3.17 m D) 3.9 m E) 3.45 m 8. A force acting on a particle is conservative if: A) it is not a frictional force B) it obeys Newton's second law C) it obeys Newton's third law D) its work equals the change in the kinetic energy of the particle E) its work depends on the end points of the motion, not the path between
3 9. A passenger elevator lifts a given load from the first floor to the tenth in less time than it takes for a freight elevator to do the same lift. A) The freight elevator does less work. B) The passenger elevator does less work. C) The freight elevator operates at a lower power level. D) The passenger elevator operates at a lower power level. E) Both elevators did the same amount of work and operate at the same power level. 10. Planet X has the same mass as the Earth but four times the radius. If the escape velocity from the Earth is ve, the escape velocity from planet X is A) ve/4 B) ve/ C) ve D) ve E) 4ve 11. An inelastic collision within an isolated system is one in which: A) momentum is conserved but kinetic energy is not conserved B) momentum is not conserved but kinetic energy is conserved C) total mass is not conserved but momentum is conserved D) neither kinetic energy nor momentum is conserved E) the total impulse is equal to the change in kinetic energy 1. A man, with his arms at his sides, is spinning on a light frictionless turntable. When he extends his arms: A) his angular velocity increases B) his angular velocity remains the same C) his moment of inertia decreases D) his rotational kinetic energy increases E) his angular momentum remains the same 13. A 4.0-kg bone has a moment of inertia of 0.30 kg m about an axis perpendicular to the bone and through its centre of mass. What is its moment of inertia about a parallel axis through the joint at the end of the bone, 0.40 m from the centre of mass? A) 0.30 kg m B) 0.34 kg m C) 0.64 kg m D) 0.94 kg m E) 1.9 kg m 14. For simple harmonic motion, there must be a restoring force proportional to the: A) amplitude B) frequency C) velocity D) displacement E) displacement squared 15. A particle is in simple harmonic motion along the x axis. The amplitude of the motion is A. At one point in its motion its kinetic energy is K = 4J and its potential energy (measured with U = 0 at x = 0) is U = 3J. When it is at x = A, the kinetic and potential energies are: A) K = 4J and U = 3J B) K = 4J and U = 3J C) K = 7J and U = 0 D) K = 0 and U = 7J E) K = 0 and U = 7J 16. Young s Modulus for bone is N/m. If the cross-sectional area of the bones in each lower leg is m in a human weighing 600 N, what is the strain on the bone in each leg when standing with the weight evenly distributed? A) B) C) D) E)
4 17. A tin can has a volume of 1000 cm 3 and a mass of 100 g. What is largest number of grams of lead shot can it carry without sinking in water? (Density of water = 1.00 g/cm 3 ) A) 100 B) 900 C) 980 D) 1000 E) A needle 3.0 cm long is carefully laid on the surface of a liquid so that it does not sink. The surface tension of the liquid is N/m and the surface tension force is directed upwards. The weight of the needle is A) N B) N C) N D) N E) N 19. Choose the correct statement concerning electric field lines: A) field lines point away from negative charge B) field lines may cross C) field lines are far apart where the field is small D) a point charge released from rest moves along a field line E) none of these is correct 0. The diagram shows two negative charges Q with the same magnitude. The electric field at point P on the perpendicular bisector of the line joining them is: A) Q B) P C) D) E) zero Q 1. In a certain region of space the electric potential increases uniformly from south to north and does not vary in any other direction. The electric field: A) points north and varies with position B) points north and does not vary with position C) points south and varies with position D) points south and does not vary with position E) points east and does not vary with position. Two equal and opposite charges are placed equidistant from the origin as shown. A test charge q (not shown) is brought from a very great distance to the origin. The work done on the test charge by the electric field of the source charges is A) zero B) k q Q a C) k q Q a D) kq a E) kq a Q a a Q 3. The equipotential surfaces associated with an infinite plane of charge are: A) radially outward from the plane B) planes parallel to the charge plane and non-uniformly spaced C) planes perpendicular to the charge plane and non-uniformly spaced D) planes parallel to the charge plane and uniformly spaced E) planes perpendicular to the charge plane and uniformly spaced
5 4. A battery is connected across a parallel combination of two identical resistors. If the potential difference across the terminals is V and the current in the battery is i, then: A) the potential difference across each resistor is V/ and the current in each resistor is i B) the potential difference across each resistor is V and the current in each resistor is i/ C) the potential difference across each resistor is V/ and the current in each resistor is i/ D) the potential difference across each resistor is V and the current in each resistor is i E) none of the above is true 5. Ammeters A) have a low resistance and are used in parallel in a circuit B) have a low resistance and are used in series in a circuit C) have a high resistance and are used in parallel in a circuit D) have a high resistance and are used in series in a circuit E) none of the above 6. A proton (charge e), traveling perpendicular to a magnetic field, experiences the same force as an alpha particle (charge e) which is also traveling perpendicular to the same field. The ratio of their speeds, v proton /v alpha is: A) 0.5 B) 1 C) D) 4 E) 8 7. The diagrams show five possible orientations of a magnetic dipole ± in a uniform magnetic field 4. For which of these does the magnetic torque on the dipole have the greatest magnitude? A) I B) II C) III D) IV E) V 8. A solenoid is 6.0 cm long and has a radius of 0.50 cm. It is wrapped with 500 turns of wire carrying a current of.0 A. The magnetic field in Tesla at the center of the solenoid is: A) B) C).1 10 D) 8.0 E) none of these 9. A camper covers herself with a down sleeping bag 5.0 cm thick. It is in contact with 0.70 m of her clothing, which is 30 K warmer than the air above the bag. Thermal conductivity of down is W/m/K. What is the heat flow through sleeping bag? A) 0.00 W B) 5.6 W C) 8.0 W D) 16 W E) 55 W 30. How many Joules are required to change one gram of ice at 0 C ice to steam at 100 C? The latent heat of fusion is 333 J/g and the latent heat of vaporization is 55 J/g. The specific heat of water is J/g K. A) 417 B) C) D) E)
6 PART B 1. Calculate the following quantities for the circuit shown in terms of given symbols: a) the equivalent resistance of the upper pair of resistors R R R R = R+ R= 3R eq + E b) the equivalent resistance of the combination of all three resistors 3 1 R = 1 R+ 1 3R. Therefore R = R eq eq 4 c) the current through the battery The potential difference across the equivalent resistor of part b is E. Thus I battery = E /R eq = (4/3) E /R d) the current through the upper pair of resistors The potential difference across the equivalent resistance of part a is E. Thus I = E /R eq = E /3R e) the power input to the circuit from the battery and the total power dissipated as heat in all the resistors Pin = Ibattery Vbattery = (4/3) (E /R) E = (4/3) (E /R). This must be equal to the total power dissipated as heat in the resistors because of energy conservation. Accept also an explicit calculation of the power dissipated in each resistor and then summed.
7 . Two parallel metal plates are spaced by.00 cm and a 90-V battery is attached to them as shown. a. What is the direction of the electric field at points not too close to ends of the plates? Since the top plate is positively charged, a test charge would move away from it. Thus E is down. + b. What is the magnitude of the electric field at points not too close to ends of the plates? E V d 3 = = = V/m (d is the plate spacing) c. If the area of each plate is 100 cm, how much charge is on each plate? Q = EA πk= π = C ( ) ( ) d. If the potential of the bottom plate is chosen to be 0 V, what is the potential at a point exactly midway between the plates? Draw the equipotential surface passing through this point. Assuming the field is nearly uniform, it follows that the rate of change of the potential with distance is constant, i.e. the equipotentials are equally spaced. That means the potential must be 45 V at this midway point. The surface is a plane parallel to the plates. A positive ion of mass 40 u and charge +e (the gray sphere) is formed at the top plate at rest. e. What is the kinetic energy and velocity of the ion just as it reaches the bottom plate? 17 From energy conservation K = U = e V = e( ) = J Since the ion starts from rest, this is also the final kinetic energy K f. K 1 4 Since K = mv then v = f = m/s
8 3. The figure below shows a long straight wire carrying a current of 4.00 A into the page. a) Draw the (closed) magnetic field line that passes through P, indicating the direction of the field at P with an arrow. P Place thumb of right hand along current, and fingers follow field lines. b) What is the magnitude of the magnetic field B at the point P, which is 5.00 cm away from the centre of the wire? Since B = ki r, B = T Suppose a second long straight wire carrying a current of 4.00 A out of the page is placed parallel to the first wire at a distance of 5.00 cm from the first wire, passing through P. A 3 c) Find the net magnetic field, magnitude and direction, at a point A midway between the two wires. Using the same RH rule as in b, the field lines for the LH wire are clockwise, while those for the RH wire are counterclockwise. Thus both fields at A are down and they are equal in magnitude as well 7 5 B = B = = T because of the symmetry. The net field is thus net 1 ( ) 3 d) The two wires are electrically neutral (i.e. uncharged) and yet there is a force between them. Giving your reasoning, find the direction of the force on the right hand wire. The force between them is not electrical because the wires are uncharged. The LH wire produces a downward field at the position of the RH wire. Using F= I l B with l coming out of the page for the RH wire tells us that F on the RH wire is to the right: the wires repel each other.
9 4. a. Water enters a house.0 m below ground at a speed of.0 m/s through a 4.0-cm diameter pipe. The pressure in this pipe is Pa. A smaller.0-cm diameter pipe carries water to a sink 5.0 m above ground, on the second floor. You may neglect the viscosity of the water and assume steady flow. (ρ water = kg/m 3 ) i. Find the speed of the water in the smaller pipe. va = va so v ( ) ( ) = A1 A v1 = = 8.0 m/s 1 1 ii. Find the pressure of the water in the smaller pipe at the second-floor level. Since is constant in the flow, P P ρ( v v ) ρg( y y ) 1 P ρv ρ gy that y y ( ) 1 = + 5 = 7 m, we get = + +. Remembering P = Pa b. The average velocity of blood in an aorta is m/s. The radius of the aorta is 1. cm. The density of whole blood is kg/m 3 and its viscosity at 37 o C is Pa s 1 i. What is the flow rate (m 3 /s) of blood through the aorta? ( ) Q = Av = π = m /s ii. Using a numerical test, show that the blood flow is laminar, i.e. not turbulent. NR = ρvr η = 113. Since this number is less than 000, the flow is laminar. iii. What is the pressure drop due to viscosity along 0 cm of the aorta? Since Q = PπR 4 8ηl, we can solve as P = lq R = 4 8η π 0.15 Pa 1 iv. What is the power required to pump blood through this portion of the aorta? P = Q P = W
10 PART C 1. A 5.0-kg load of bricks is fastened to a rope, which passes over a frictionless pulley and is attached to a 0.0-kg sandbag on the ground. The pulley has rotational inertia of kg-m and a radius of 10.0 cm. The brick load released when it is 3.40 m above the ground and the rope is taut. When the system moves, the rope does not slip on the pulley. a. At some instant during the motion of the system, the bricks and sandbag have some speed v and the pulley has angular speed ω. Explain physically why these two quantities are related, and give in symbols the mathematical relation that connects them. If the rope doesn t slip on the pulley, it must have the same speed as the rim of the pulley. This linear velocity of the rim is connected to the angular velocity by v= ωr where r = 0.10 m b. Find the change in the gravitational potential energy U g of the system from the time of release until just before the bricks reach the ground. U g =Since Ug = mg y for a single mass m, here = ( ) 9.8 = 167 J U g c. What is the relationship between U g and the change in kinetic energy K of the system over the same interval? Since there are no frictional losses in the system, mechanical energy is conserved. Thus K = U g 4 d. Use energy ideas to find the speed of the bricks just before they hit the ground. ( ) K = m + m v + I ω. Substitute ω = vr, equate to U g and solve for v: pulley ( v ) 45v = 167 gives v =.46 m/s. 1 1
11 . a. A kg hockey puck is moving on an icy, frictionless, horizontal surface. At t = 0, when the puck is moving to the right at 3.00 m/s, a hockey stick applies an average force of 5.0 N directed to the right for s. Find the final velocity (magnitude and direction) of the puck. Since av t = mv v = F t. F p p and the force and initial momentum are in the same direction, ( ) av Solving for the final velocity gives v = 10.8 m/s, moving to the right. b. Hockey player Mark (85.0 kg) is skating at 11.0 m/s toward a defender (110 kg), who is skating toward him at 5.00 m/s. Immediately after a collision, Mark is moving at 1.50 m/s in his original direction. Neglect external horizontal forces exerted by the ice on the skaters during the collision. i. What is the velocity (magnitude and direction) of the defender immediately after the collision? Since external forces are negligible, the total linear momentum of the skaters is conserved: If we take Mark s direction as positive, then ( 5.00) = v d. Solving gives v d =.34 m/s. The positive sign implies the defender is now moving in Mark s direction. ii. Would you expect the total kinetic energy of the two players to change as a result of their collision? Give a physical explanation of your answer. A collision like this must conserve momentum but not necessarily energy. Some energy is lost to heat and sound, so the KE decreases. c. A door (shown in a TOP view) with moment of inertia 5.0 kg-m is hinged to swing without friction A police officer fires a bullet of mass 10 g into the exact centre of the door in a direction normal to the plane of the door. The bullet sticks in the door. i. Find the angular momentum of the bullet before it hits the door, using an origin at the door hinge. For a particle, L = r p. In magnitude, o L = sin 90 =.00 kg m /s bullet v=400 m/s 0.50 m ii. Find the angular velocity of the door as it swings after being hit. Because there is no frictional torque on the door, the angular momentum of the bullet+door must = Idoor + mbulletr ω. Solving for ω gives rad/s. be conserved. Thus ( )
12 3. a. A cart with mass kg is placed on an air track and attached to a horizontal spring of force constant 450 N/m. At t = 0 it is pulled m from its equilibrium position and released. i. Find the oscillation period of the cart. Since T = π ω and ω = km, the period is 0.09 s. ii. Find the magnitude of the maximum acceleration of the cart. What is the velocity of the cart when it has maximum acceleration? = ω, the acceleration is maximum when the displacement is maximum: amax = k m = 36.0 m/s. When the cart is at is maximum displacement, its instantaneous velocity is zero. Since a ( k m) x x magnitude ( ) iii. Write an algebraic equation x(t) for the cart with all constants expressed numerically. (Equilibrium position is x = 0.) Equation is x = Acosωt where A = m and ω = km= 30.0 rad/s (Using a sin would make x=0 at t=0, whereas x=a at t=0.) b. After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 50.0 cm. She finds that the pendulum makes 100 complete swings in 181 s. i. What is the value of g on this planet? The period T of this pendulum is 1.81 s and we know that f 1 T g = π T = 6.03 m/s. solve for g as ( ) =. Since f 1 ( π ) g we can ii. If she can throw a ball 1 m straight up on Earth, how high could she throw the same ball on this planet? (You may assume she gives the ball the same initial speed.) If a ball is thrown vertically upward with speed v 0, its maximum height above the release point 1 can be found from conservation of energy: mv = mg y. Thus y 1 g and ( ) 19.5 m y = g g y = X Earth X Earth 0
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