Provincial Examination with Answers 2004

PHYSICS 12 JUNE 2004 PROVINCIAL EXAMINATION PART A: MULTIPLE CHOICE 1. Which of the following is a scalar quantity? A. work B. force C. velocity D. momentum 2. An astronaut on the moon throws a 5.0 kg wrench vertically upwards with an initial speed of 15 m/s. The acceleration due to gravity on the surface of the moon is one-sixth that on the surface of the earth. What is the maximum height reached by the wrench? A. 25 m B. 46 m C. 69 m D. 75 m 3. A ball rolls off a horizontal roof at 15 m/s. v = 15 m/s 24 m/s 4. The units N/kg are used for A. net force. C. electric field strength. How far will the ball have fallen vertically when it reaches a speed of 24 m/s ? A. 4.1 m B. 18 m C. 29 m D. 37 m B. gravitational force. D. gravitational field strength. 5. A 1200 kg trailer is accelerated from rest to 15 m/s in 5.0 s. The average force of friction acting on the trailer is 800 N. What is the pulling force applied to the trailer through the hitch? A. 800 N B. 2800 N hitch C. 3600 N D. 4400 N 6. According to the work-energy theorem, which of the following is equal to the net work done on an object? DELETED A. ΔEK B. ΔEP C. ΔEP + ΔEK D. ΔEP - ΔEK hitch 7. Which of the following is a graph of EK vs. time for a ball dropped off a roof? (Ignore air friction.) A B C. D EK EK EK EK t t 8. A 2.0 kg wood block slides down an incline from rest as shown to the right. Determine the force of friction on the block during the slide if its final velocity at the bottom is 10 m/s. A. 4.0 N B. 6.7 N C. 11 N D. 17 N t t vI = 0 15 m vF = 10 m/s 33° 9. You are pushing a crate across a smooth wood floor with a constant pushing force. The crate is accelerating. Which of the following describes your power output? A. increasing B. decreasing C. increasing then constant D. decreasing then constant 10. A 25 kg object is moving due north at 33 m/s. If an impulse of 330 Ns is applied at 45˚ N of W to this object, what is the final velocity of this object? SPEED DIRECTION A. 33 m/s 45˚ N of W m B. 33 /s 78˚ N of W m C. 43 /s 45˚ N of W m D. 43 /s 78˚ N of W 11. The sum of the torques on an object is zero about a certain point. The object is in rotational equilibrium. Which of the following statements is correct? DELETED A. The sum of the torques is zero about no other point. B. The sum of the torques is zero about all other points. C. The sum of the forces must be zero at all other points. D. The sum of the forces must be zero at that certain point. 12. A 5.0 kg mass suspended by a rope is pulled to the left by a horizontal force of 18 N. What is angle Θ ? A. 16˚ B. 20˚ C. 70˚ D. 74˚ 13. A uniform 2.0 m long beam is to be kept horizontal by a cord. In which of the four situations shown will the tension in the cord be one-half of the beam’s weight? B. C. 45˚ Θ 18 N A. D. 45˚ 14. A 5.0 x 104 kg moonlet travels in a circular path around a planet. The moonlet’s orbital radius is 2.5 x 107 m and the orbital period is 3.7 x 105 s. What is the mass of the planet? A. 1.1 x 108 kg B. 6.8 x 1022 kg C. 3.4 x 1027 kg D. 2.5 x 1028 kg 15. A 0.15 kg toy airplane is suspended as shown. It travels in a horizontal circle at a constant speed. What is the period of the motion of this airplane? A. 0.84 s B. 1.6 s C. 1.8 s D. 2.0 s 16. A small object of mass m is suspended from a fixed point by a light cord. The object is raised to an angle of 60˚ and released from rest. The object moves in an arc of a circle as shown. When the object passes through its lowest position at point P, what is the tension in the cord in terms of the object’s weight (mg)? A. 0.5 mg B. 1.0 mg C. 1.5 mg D. 2.0 mg 33˚ 0.55 m 0.15 kg 60˚ 0.5 r r 17. The graph shows the gravitational force between the earth and an object as a function of the distance of separation, r, from the centre of the earth. As the object is moved from r1 to r2, what is the work done? A. Q B. Q + S C. P+Q D. Q + S +T Fg P Q S r1 T r2 r 18. A 200 kg object is released from rest at an altitude of 1.0 x 107 m. What is its impact speed with the earth? Assume no air resistance. A. 7.0 x 103 m/s B. 8.7 x 103 m/s C. 1.1 x 104 m/s D. 1.4 x 104 m/s 19. Which of the following is equal to the work required per coulomb to move a positive charge from one position to another in an electric field? A. EP, electric potential energy B. ΔEP, change in electric potential energy C. V, electric potential D. ΔV, electric potential difference 20. The diagram below shows a positive charge located near a smaller (in magnitude) negative charge. In which region is there a point where the electric field due to the two charges is equal to zero? IV • I II • III • + A. I B. II C. III D. IV 21. Which of the following is always true for resistors in parallel or in series? RESISTORS IN RESISTORS IN PARALLEL SERIES A. equal currents equal currents B. equal currents equal voltage drops C. equal voltage drops equal currents D. equal voltage drops equal voltage drops 22. In the circuit shown below the voltage loss due to the battery’s internal resistance is 2.0 V. Determine the value of the load resistance R. A. 12.0 Ω B. 13.3 Ω C. 13.5 Ω D. 15.0 Ω Є = 20.0 V r = 1.50 Ω R 23. A resistor is added in parallel to the 4.0 Ω resistor shown in the diagram to the right. What happens to the power dissipated by the 8.0 Ω resistor and by the 4.0 Ω resistor? P8..0Ω P4..0Ω A. decreases increases B. decreases decreases C. increases increases D. increases decreases V 10 Ω 4.0 Ω 8.0 Ω 24. Protons and electrons travelling at high speeds enter a magnetic field parallel to their direction of travel. Which of the following is correct? A. Only protons are deflected. B. Only electrons are deflected. C. Both protons and electrons are deflected. D. Neither protons nor electrons are deflected. 25. Which of the following diagrams best shows the orientation for a set of four compasses placed around a current-carrying wire? A. I B. I C. I D. I 26. An electron travels at a speed of 1.4 x 104 m/s while following a circular path of radius 0.020 m perpendicular to a magnetic field. What is the strength of the magnetic field? A. 0.0 T B. 4.0 x 10-6 T C. 6.3 x 10-3 T D. 4.1 x 10-2 T 27. A potential difference of 22 V is placed across a 570-turn solenoid that has a resistance of 4.9 Ω. The solenoid has a diameter of 0.052 m and is 0.37 m long. What is the magnetic field strength in the centre of this solenoid? A. 8.7 x 10-3 T B. 6.2 x 10-2 T C. 1.2 x 10-1 T D. 3.0 x 10-1 T 28. A 200-turn copper coil of radius 0.050 m is perpendicular to a 0.087 T magnetic field. The magnetic field collapses to zero in 0.63 s. What was the average induced emf in the coil during the collapse? A. 0.0 V B. 1.1 x 10-3 V C. 0.22 V D. 1.4 V 29. The coil of a motor has a resistance of 4.1 Ω. The motor is plugged into a 120 V outlet, and the coil develops a back emf of 118 V when rotating at normal speeds. Find the current when the motor first starts up and the current when it is operating at normal speeds. CURRENT AT START UP CURRENT AT NORMAL SPEEDS A. 0.49 A 0.49 A B. 0.49 A 29 A C. 29 A 0.49 A D. 29 A 29 A 30. A 0.034 kg copper rod is hung by two wires and placed in a constant magnetic field. A current of 14 A runs through the 0.050 m long copper rod, making it hang at an angle of 34˚ B from the vertical as B shown to the left. What is the magnetic 34˚ m = 0.034 field strength I holding the copper 34˚ rod in this position? A. 2.7 x 10-1 T 0.050 m B. 3.2 x 10-1 T C. 3.9 x 10-1 T oblique view side view D. 4.8 x 10-1 T a = 1.5 m/s2 PART B: WRITTEN RESPONSE 1. Two masses are connected by a light cord passing over frictionless pulleys as shown in the diagram below. a) What is m2 if the system accelerates as shown? (5 marks) A block is then added to the system. b) Adding the block decreases the acceleration of the system. Identify and explain two reasons for this decrease. (4 marks) m1 = 4.0 kg m2 Block μ>0 m1 = 4.0 kg m2 2. A 4300 kg truck travelling at 21 m/s in the direction of 31˚ north of east collides with a stationary 1500 kg car. After the collision, the car has a speed of 15 m/s due east. What is the resulting speed of the truck? (7 marks) 3. restraining cable 1.0 m 40˚ A sign is suspended from the end of a 6.0 m long uniform pole of mass 25 kg as shown. If the mass of the sign is 36 kg, what is the tension in the horizontal restraining cable? (7 marks) Highway Toll: $13 4. A 4.20 x 104 kg satellite orbits the earth at an altitude of 700 km (7.00 x 105 m). a) What is the satellite’s orbital speed at this altitude? (4 marks) b) What is the satellite’s total energy at this altitude? (3 marks) 700 km Earth’s surface 5. a) Determine the electric potential, relative to zero at infinity, at point P, midway between the two 5.0 m charges, shown to the right. •P + (5 marks) 50 μC -20 μC b) How much work would it take to move a -15 μC charge 10.0 m from point P to a position infinitely far away? (2 marks) 6. The 8.0 Ω resistor in the circuit shown to the right dissipates 45 J of heat energy in 5.0 s. Determine the value of the resistor R. (7 marks) 20 V 8.0 Ω 5.0 Ω R 7. A 0.13 kg aluminum rod maintains contact with two vertical metal rails. A voltage is applied across the metal rails and a horizontal magnetic field of 0.65 T exists across the whole apparatus as shown. What current must flow through the aluminum rod to have it remain stationary? (7 marks) 0.090 m Aluminum rod B = 0.65 T + - 8. A student uses a wrench to apply a constant force to turn a nut. He applies the force at various angles and measures the F amount of torque produced at each of the Θ angles. The torque data collected by the student along with the sine of the angles is shown 0.60 m below: TORQUE (N•m) SIN Θ 4.0 0.14 11 0.42 17 0.66 22 0.86 25 0.98 a) Plot a graph of torque versus sin Θ on a graph. (2 marks) b) Calculate the slope of your line including units. (1 mark) c) Use the slope of your graph to determine the amount of constant force the student used throughout his experiment. (2 marks) 9. A small plastic ball carrying a positive charge is held near a fixed large positively charged sphere as shown to the right. It is then released. Explain how and why the acceleration of the plastic ball changes as it moves away from the sphere. (4 marks) + + June 2004 Provincial Examination ANSWER KEY PART A: Multiple Choice (each question worth TWO marks) 1. 2. 3. 4. 5. A C B D D 6. delete 7. B 8. A 9. A 10. D 11. delete 12. C 13. D 14. B 15. C 16. 17. 18. 19. 20. D B B D D 21. 22. 23. 24. 25. C C D D A 26. 27. 28. 29. 30. B A C C B PART B: Written Response = 60 marks 1. a) FNet = mTotala m2g – m1g = (m1 + m2)a 3 marks m 2 m 2 m2(9.8 /s ) – (4.0 kg)(9.8 /s ) = (4.0 kg + m2)(1.5 m/s2) m2(9.8 m/s2) – 39.6 N = 6.0 N + (m2)(1.5 m/s2) (9.8 m/s2)(m2) – (1.5 m/s2)(m2) = 6.0 N + 39.2 N (8.3 m/s2)(m2) = 45.2 N m2 = 5.4 kg 2 marks b) Since the system mass has increased, the acceleration must decrease (a=FNet /mtotal). Since there is friction, the net force will decrease and the acceleration will be smaller yet. 4 marks 2. Before: PI = PITruck + PICar = mTvIT + mCvIC 90 300 kgm/s = (4300 kg)(21 m/s) + (1500 kg)(0 m/s) PFT = 90 300 kgm/s After: PF = PFTruck + PFCar = mTvFT + mCvFC 31˚ = (4300 kg)(vFT) + (1500 kg)(15 m/s) 3 marks 22 500 kgm/s = (4300 kg)(vFT) + 22 500 kgm/s By cosine law: PFT = (22 500 kgm/s)2 + (90 300 kgm/s)2 -2(22 500 kgm/s)( 90 300 kgm/s)(cos 31˚) 2 marks 4 kgm PFT = 7.18 x 10 /s 4 kgm vFT = 7.18 x 10 /s = 16.7 m/s = 17 m/s 2 marks 4300 kg 3. ΣτC = ΣτCC 1 mark FPdPsinΘP + FSdSsinΘS = FcdcsinΘc 3 marks FPdPsinΘP + FSdSsinΘS Fc = dcsinΘc m 2 /s )(3.0 m)(sin 50˚) + (36 kg)(9.8 m/s2)(6.0 m)(sin 50˚) = (25 kg)(9.8 = (5.0 m)(sin 40˚) 2 marks = 563.0 N + 1621.6 N 3.21 m = 680 N 1 mark 4. Fg = FC GmM = mv2 R2 R v= GM R = 6.67 x 10-11 N/m2kg2 x 5.98 x 1024 kg 7.00 x 105 m + 6.38 x 106 m v = 7.51 x 103 m/s 4 marks 1 mark b) ET = EP + EK 2 = - GmEm + ½mv 1 mark R = -(6.67x10-11 Nm2/kg2)(5.98x1024 kg)(4.20x104 kg) + ½(4.20x104 kg)(7.51x103 m/s)2 7.00 x 105 m + 6.38 x 106 m = -2.37 x 1012 J + 1.18 x 1012 J = -1.18 x 1012 J 1 mark 5. a) V = V50μC + V-20μC V = 9.0 x 109 Nm2/C2(50 x 10-6 C) + 9.0 x 109 Nm2/C2(-20 x 10-6 C) 5.0 m 5.0 m 4 V = 5.4 x 10 V b) W = ΔEP = EPF - EPI W = 0 – (54 x 103 J)(-15 x 10-6 J ) W = 0.81 J 6. R8.0 Ω = 45 J = 9.0 W 5.0 s 1 mark 1 mark 1 mark 1 mark 3 marks 1 mark P = I2R 9.0 W = (I8.0 Ω)2(8.0Ω) I8.0 Ω = 1.06 A VR = V 8.0 Ω = IR = 1.06 A x 8.0 Ω = 8.48 V IR = Icircuit - I8.0 Ω IR = 20 V – 8.48 V - 1.06 A = 1.24 A 5.0 Ω R = V = 8.48 V = 6.8 Ω I 1.24 A 7. Fg = FB mg = BIl = 0.13 kg x 9.8 m/s2 0.65 T x 0.090 m = 22 A 8. 1 mark 1 mark 2 marks 2 mark 2 marks 2 marks 2 marks 1 mark 28 26 24 22 Torque (Nm) 20 18 16 14 12 10 8 6 4 2 0 0.00 0.20 0.40 0.60 2 marks for graph: 1 mark for line, 1 mark for dots - ½ mark for wayward dots 0.80 sin Θ 1.00 b) slope ≈ 25 Nm 1 mark ½ mark for numeric answer ½ mark for units c) τ = Fl F = slope = τ = 25 nm l l 0.60 m 2 marks ≈ 42 Nm 1 mark for equation and algebra 1 mark for substitution and answer 1 mark for using 25 and it is not slope found in b) 9. The size of the electrostatic force will be decreasing with separation from the sphere (1½ marks), so the acceleration of the plastic ball will be decreasing (1½ marks). According to Coulomb’s Law, where F = kq1q2 (1 mark). r2