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UNIVERSITY OF KWAZULU-NATAL FOUNDATION PHYSICS SUPPLEMENTARY EXAM 30th NOVEMBER 2009 UNIVERSITY OF KWAZULU-NATAL SECOND SEMESTER SUPPLEMENTARY MEMORANDUM: 30th November 2009 Subject, Course and Code: Physics Foundation PHYS099 and PHYS199 DURATION: 3 HOURS TOTAL MARKS:180 Student number: __________________________________ Seat number: _______________ INTERNAL EXAMINERS: Ms B. Adams, Ms W. Dlamini, Mrs S. Halstead, Mr R. Webber and Mr E Zhandire EXTERNAL EXAMINER Dr V Couling ANSWER ALL QUESTIONS ON THIS PAPER. IN THEIR OWN INTERESTS STUDENTS ARE REQUESTED TO WRITE LEGIBLY. THIS PAPER CONSISTS OF 14 PAGES. PLEASE SEE THAT YOU HAVE THEM ALL. Throughout paper, take off half a mark for incorrect or missing units in the answer to each section. PROPERTIES OF MATTER Question 1 (8 marks) Give your answers to the following questions with the correct number of significant figures or decimal places. a. 16.02 litres + 8.3649 dm3 + 27.9 cm3 = (3) = 16.02 dm3 + 8.3649 dm3 + 0.0279 dm3 = 24.41 dm3 b. An engineer finds that a metal has a volume of 0.15 cm3 per gram. i) What is the density of the metal? d= 1 g.cm-3 = 6.7 g.cm-3 0.15 ii) What is the mass of a 17.8 cm3 volume of this metal? How many shares of 0.15 cm3 are there in 17.8 cm3? Each share will have a mass of 1 gram Therefore mass = 17.8 = 120 g 0.15 OR m = d x V = 6.7 x 17.8 = 119.26 g = 119 (OR 120 g if you did not round off earlier) Page 1 of 14 (2) (3) UNIVERSITY OF KWAZULU-NATAL 30th NOVEMBER 2009 FOUNDATION PHYSICS SUPPLEMENTARY EXAM velocity (m/s) MECHANICS NOTES: 1. Take the magnitude of acceleration due to gravity as 9.8 ms-2 2. Show the reference frame that you use for each situation. Question 2 (9 marks) The velocity-time graph below shows the motion of an object over 12 seconds. 10.0 D 8.0 C B 6.0 E 4.0 A 2.0 F tim e (s ) 0.0 0 2 4 6 8 10 G 12 -2.0 -4.0 The object has a mass of 10 kg and started off travelling north. Refer to the letters on the graph when answering the following questions. NB Initially traveling NORTH a. Over what section was the object travelling fastest? D b. In what direction is it traveling, over: i) section B? ii) section E? iii) section G? north north south (1) (3) c. Over what section does the object first remain stationary? F (1) d. What section has maximum acceleration/deceleration? (1) e. What is the resultant force on the object in section D? Show reasons. (3) Object has constant velocityT so by Newton I (or show calc with Newton II)T the resultant force is ZeroT Page 2 of 14 E UNIVERSITY OF KWAZULU-NATAL 30th NOVEMBER 2009 FOUNDATION PHYSICS SUPPLEMENTARY EXAM Question 3 (10 marks) Zipho throws a ball vertically upward with an initial speed of 15.0 ms-1 to Ntombi who is leaning over a balcony that is 8.0 m above Zipho’s hand. She catches the ball on its way down. Calculate: a. The velocity of the ball when she catches it. (6) (take UP as +, –½ if not stated) v is + v is – v = – 8.3 ms-1 (or 8.3 ms-1down) NB Must be negative square root b. The time taken to reach Ntombi. v = u + at = – 8.26 + 15.0 + (–9.8)t a is –9.8 ms-2 for whole journey Zipho = (15.0)2 +2(–9.8)(8.0) = 68.2 (NB signs must be correct according to ref frame) ∆x or s = + 8.0 m Ntombi v2 = u2 + 2as (4) (NB signs must be correct according to ref frame) t = 2.4 s Question 4 (6 marks) Sinah stands on a building so she is 11.0 m above the ground. She sees her boyfriend walking towards the building at a steady pace of 1.6 ms-1. She drops her cellphone to him when he is 3.5 m from the building. If he keeps the same pace, will he be able to catch the phone without stretching his arms? (6) Sinah a is +9.8 ms-2 for phone Phone ∆x or s = + 11.0 Boy Boy v is 1.6 ms-1 Boy ∆x or s = 3.5 m a = 0 (const v) NB DO NOT MIX DATA FOR PHONE & BOY Vertical Motion - Sinah’s phone: take down as + (– 12 not stated) Horizontal motion – Boyfriend Take direction of motion as + s = ut + 12 at 2 T 11 = 0 + 12 9.8t2 T t = 1.5sT s = ut + 12 at 2 3.5 = 1.6t + 1/2 (0)t2T t = 1.75s = 1.8sT Which is longer than 1.5 s OOPS he does not get there in time. T Page 3 of 14 UNIVERSITY OF KWAZULU-NATAL 30th NOVEMBER 2009 FOUNDATION PHYSICS SUPPLEMENTARY EXAM Question 5 (12 marks) a) A student gets off a bus at university and walks 105 m on a bearing of 45°, then 87 m on a bearing of 292° and finally 220 m due East. The displacement vectors are represented on the diagram below. How far and in what direction do they need to walk to get back to the bus at the end of the day? (8) It may help to sketch the vectors individually – BE CAREFUL to get the correct angles B = 87 m y 105 m 87 m 450 220 m x 2920 y A =105 m y y By =87 sin 220 m Ay =105 sin 450 m 450 220 C = 220 m 0 45 0 Ax =105 cos 45 m Bx = –87 cos 220 m x x x 2920 x-components (east) 105cos45o =74 m y-components (west) 105sin45o =74 m 87sin22o =33 m C 87cos22o = –- 81 m 220 m Resultant 213 m 107 m A B 0 To find the journey back to the bus, we are looking for the equilibrant – the opposite of resultant – so we make the vectors head-to-tail and come back to the start. s2 = (213)2 + (107)2 therefore If you want marks for θ you MUST show θ on a diagram. 107 tan θ = = 1.99 213 θ = 26.7 N s = 238 m 213 m θ 107 m s B̂ o b) A pilot wishes to fly due North from Airport A to Airport B. The wind is blowing at 30 kmhr -1 due west. The small plane averages a velocity of 200 kmhr-1. What direction should the pilot take in order to reach Airport B? Use a sketch to help solve. (4) The RESULTANT is what must happen overall – the plane must RESULT IN GOING NORTH 30 NOT tan sin θ = = 0.15 θ = 8.60 200 The plane must fly bearing 8.60 OR 8.60 East of North You are not asked to find the resultant speed Page 4 of 14 Vp= 200 kmhr-1 N B̂ Resultant – how plane needs to end up the student must walk 238 m bearing of 2430 OR 26.70south of west OR 63.3owest of south θ vw=30 kmhr-1 UNIVERSITY OF KWAZULU-NATAL FOUNDATION PHYSICS SUPPLEMENTARY EXAM 30th NOVEMBER 2009 Question 6: (16 marks) The drawing shows a truck, with mass 2000 kg, towing a caravan. While the truck accelerates from rest to a maximum speed of 22.5 ms-1 in 25.0 seconds, the engine of the truck exerts an average force on the road of 4000 N, and the average frictional force from the air is 1000 N. a) Calculate the mass of the caravan. (7) Take right as + direction The car has to accelerate itself AND the caravan a= ∆v 22.5 T= 0.900 ms-2T = ∆t 25.0 1000 N 4000 N F res = 4000 N – 1000 N = 3000 N right. TT F res = ma, therefore 3000 N = mtotal x 0.900 T so m total = 3333 kgT (3330 to 3 sig figs) Therefore caravan mass = 3330 – 2000 = 1330 kgT b) When stopping, the maximum braking force on the truck is 9 000 N. Calculate how far the truck will take to stop from maximum speed when it is NOT towing the caravan? (6) We still have friction but NO driving force from the engine, 1000 N we now also have a braking force 9000 N F res = –9000 + (–1000) = –10000N T −10000 T= –5.000 ms-2 T From F = ma we get: a = 2000 The initial speed is 22.5 ms-1and final speed is 0, v 2 = u 2 + 2as so c) 0 = 22.52 + 2 x (– 5.00) s TT so s = 50.6 metresT When the car is towing the caravan, will the stopping distance increase, decrease or stay the same. Give a brief explanation in terms of physics. (3) There will be more mass/inertia with the caravanT, so with same braking force we have smaller acceleration (deceleration) T therefore stopping distance increasesT Page 5 of 14 UNIVERSITY OF KWAZULU-NATAL FOUNDATION PHYSICS SUPPLEMENTARY EXAM 30th NOVEMBER 2009 Question 7: (10 marks) The diagram below shows a woman at the airport pulling her 20 kg suitcase at a constant speed along a horizontal surface, using a strap that makes an angle θ above the horizontal. She pulls on the strap with a 35 N force and the frictional force on the suitcase is 20 N. FNormal Ty = 35 sin θ N T = 35 N θ Ffriction = 20 N Tx = 35 cos θ N Fg = 20 x 9.8 N a) What angle does the strap make with the horizontal? (3) Constant velocity therefore Fresultant = 0 Instead of T or 35 N we use the horizontal and vertical components Tx and Ty Notice that there are now 2 forces vertically upwards Consider horizontal forces: (Take forward as positive) FRes x = 0 this means: Tx + Ffriction = 0 20 cosθ = = 0.5714 35 θ = 55o therefore 35 cos θ +(–20) = 0 b) What normal force does the ground exert on the suitcase? Consider vertical forces: (Take UP as positive) (4) FRes y = 0 FN + Ty + Fg = 0 therefore: FN + 35 sin θ – 20 x 9.8 = 0 FN = 196 – 35sin55o = 167 N (or 170N) c) If the strap breaks, what would be the magnitude and direction of the acceleration of the suitcase? (3) The only horizontal force will be Ffriction or –20N F = ma a= −20 = −1.0ms −2 20 Page 6 of 14 UNIVERSITY OF KWAZULU-NATAL FOUNDATION PHYSICS SUPPLEMENTARY EXAM 30th NOVEMBER 2009 ELECTROSTATICS NOTE: Take the electrostatic constant in air to be k = 9.00 x 109 Nm2C-2 and the charge on an electron to be Qe = 1.602x10-16 C Question 8 a) n= (26 marks) In walking across a carpet, you acquire a net negative charge of 50 µC. How many excess electrons do you have? (2) 50x10−6 = 3.1x1011 electrons −16 1.602x10 b) The diagram below shows four charges located at the corners of a square. q1 = - 10µC q2 = - 10µC F12 F32 F42 q4 = + 5.0µC q3 = + 5.0µC i) Use the diagram to show F12, F32 and F42. (3) ii) What is the magnitude of the force F12? (3) F12 = iii) kQ1Q2 9 x10 9 x(10 x10 −6 ) 2 = = 90 N r2 (0.10) 2 What is the distance separating q2 and q4? (2) r2 = (0.10)2 + (0.10)2 r = 0.14 m iv) F42 = Determine the magnitude of the force F42. kQ 2Q 4 9.00x109 × (10x10−6 ) × (5.0 × 10−6 ) = = 23N r2 (0.14) 2 Page 7 of 14 (3) UNIVERSITY OF KWAZULU-NATAL c) FOUNDATION PHYSICS SUPPLEMENTARY EXAM 30th NOVEMBER 2009 Charge Q acts as a point charge to create an electric field. Its strength, measured a distance of 30 cm away, is 40 NC-1. What would be the electric field strength 30 cm away from charge 2Q ? i) Enew = 2 E1 = 80 NC-1 (in the same direction) 90 cm away from charge 0.5Q? ii) E new = iii) 1 2 2 3 = 1 x 40 = 2.2 NC-1 (in the original direction) 18 (4) Calculate the force on a 40 nC charge 30 cm away from charge Q. F = E1 x q = 40 x 40x10-9 = 1.60x10-6 N d) (2) (in direction of field) (3) A proton is accelerated between two points where the potential difference is10 V. What is the work done on the proton? Qp= + 1.602 x 10-19 C. (4) W = q X Vab = 1.602 X 10-19 X 10 = 1.6 X 10-18 J Page 8 of 14 UNIVERSITY OF KWAZULU-NATAL FOUNDATION PHYSICS SUPPLEMENTARY EXAM 30th NOVEMBER 2009 ELECTRIC CIRCUITS Question 9 (10 marks) Questions a, b and c refer to the circuits below, which contain identical batteries and bulbs. The internal resistance of the batteries can be ignored. A E C F B D a. If bulb C is unscrewed, the brightness of bulb B will ____same___________(2) b. If bulb E is unscrewed, the brightness of bulb D will ____increase________ (2) c. If bulb E is unscrewed, the brightness of bulb F will ____ decrease ______ (2) Questions d and e refer to the circuit on the right. A battery is connected to two bulbs, an ammeter, and a voltmeter as shown. Ignore internal resistance. If a third bulb is added in parallel to the two bulbs shown, the V d. voltmeter reading will ____ same_____ (2) e. ammeter reading will ____ increase ___ (2) Page 9 of 14 A UNIVERSITY OF KWAZULU-NATAL 30th NOVEMBER 2009 FOUNDATION PHYSICS SUPPLEMENTARY EXAM Question 10 (11 marks) In the circuit shown calculate: a) I4Ω = b) the current through each 4.0 Ω resistor,(2) 2 7 0.50Ω x 0.700 = 0.20 A 4.0Ω the total resistance of the circuit, 4.0Ω 8.0Ω (3) 700mA 1 1 1 1 1 7 = + + + = Req 4 4 4 8 8Ω Req = 1.1Ω RT = 0.5 +1.1 = 1.6Ω c) 4.0Ω the battery voltage, and (3) VT = ITxRT = 0.70 x 1.6 = 1.1 V d) the voltage across the 8.0 Ω resistor. V8 = IxR = 0.1 x 8.0 = 0.80 V OR (3) V8=VT – 0.70x0.50 = 0.75 V Question 11 (10 Marks) a) For the circuit on the right, calculate the unknown current I. (2) I + 1.0 = 5.0 I = 4.0 A I Loop 1 5.0Ω X [3] c) Use loop 1, as shown, to calculate the resistance R. 24 – 22.5 = 4.0R + (-1.0)5.0 6.5 R= = 1.6Ω 4.0 Page 10 of 14 Loop 2 3.5Ω 24V b) Use loop 2, as shown, to calculate voltage X. (4) X = 5.0 x 3.5 + 1.0 x 5.0 = 22.5 V = 23 V 5.0A 1.0A R node a (4) UNIVERSITY OF KWAZULU-NATAL FOUNDATION PHYSICS SUPPLEMENTARY EXAM 30th NOVEMBER 2009 Question 12 (16marks) In the circuit on the right, the “lost volts” across the battery is 1.0 V. If a current of 2.5 A flows in the circuit, calculate: a) the internal resistance (r) (3) 2.5A Vl = I.r V 1.0 r= l = = 0.40Ω I 2.5 5.2Ω 2.4Ω emf b) the terminal potential difference (3) r Vtpd = IRext = 2.5 x 7.6 = 19 V c) the emf of the battery (3) emf = I(R + r) = 2.5 x 8.0 = 20 V d) the energy lost in the external circuit in 1 minute, and W = I2 Rt V2 t R 192 (60) = 7.6 OR W = = (2.5) 2 (7.6)(60) = 2.9 ×103 J (or 2850 J) e) (4) OR W = VIt = 19 × 2.5 × 60 = 2.9 ×103 J = 2.9 ×103 J the power loss in the battery P =I V OR P = I2 R = (1.0)( 2.5) = (2.5)2 ( 0.4) = 2.5 W Page 11 of 14 = 2.5 W (3) OR P= V 2 (1.0) 2 = = 2.5W R 0.4 UNIVERSITY OF KWAZULU-NATAL FOUNDATION PHYSICS SUPPLEMENTARY EXAM 30th NOVEMBER 2009 Question 13 (7 marks) a) Complete the following diagrams to show what happens to a ray of light passing through the perspex prisms or lenses: . f O Page 12 of 14 f O (7) UNIVERSITY OF KWAZULU-NATAL Question 14 FOUNDATION PHYSICS SUPPLEMENTARY EXAM 30th NOVEMBER 2009 (16 marks) (5) F I O F (5) 1 v = 1 f − u1 = 1 6.0 − 21.5 = - 0.2334 v = - 4.28 = - 4.3 cm (4) mag = −v u = 4.3 2.5 = 1.72 = 1.7 (2) Any 2 of: Virtual Enlarged upright (same side as object) (can’t be cast on screen) Page 13 of 14 UNIVERSITY OF KWAZULU-NATAL Question 15 FOUNDATION PHYSICS SUPPLEMENTARY EXAM 30th NOVEMBER 2009 (13 marks) (13) φwater y tany = 1.0/1.5 y = 33.7o θair = 56.3o npsinθp = nasinθa sinθp = 1.00 x sin 56.3 = 0.5738 1.45 θp = 35o sinφw = 1.45 x0.5738 = 0.6255 1.33 φw = 39o and θp = φp nwsinφw = npsinφp Page 14 of 14