Physics Department Fisika Departement Semester Test 1 Semestertoets 1 FSK116 16 March 011 / 16 Maart 011 Surname and initials Van en voorletters Signature Handtekening MEMO Student number Studentenommer Time ½ hours Ma. Total 85 Marks Ma Tyd ½ ure Maks. Totaal 85 punte Maks Internal Eaminer: Interne Eksaminator Dr J M Nel Eternal Eaminer Eksterne Eksaminator Mr R Q Odendaal Answer all questions. Cleary show all calculations. Do not just write down the answer. Clearly state all assumptions, rules and laws used. Where applicable draw the free-body diagrams. Demonstrate clearly that you understand the physics that you use to answer the question. This means that a numerically correct answer will not necessarily be awarded full marks. Programmable calculators may NOT be used. No tetbooks allowed. A formula sheet is attached. All test and eamination regulations of the University of Pretoria are applicable. eantwoord alle vrae. Wys al jou berekeninge. Moenie net n antwoord neerskryf nie. Stel duidelik alle aannames, reëls en wette wat u gebruik. Waar van toepassing, teken die vryliggaamsdiagram (of kragte diagramme). Wys dat jy die fisika wat jy gebruik om die vraag te beantwoord verstaan. Dit beteken dat volpunte sal nie noodwendig vir n numeries-korrekte antwoord gegee word nie. Programmeerbare sakrekenaars MAG NIE gebruik word nie. Geen handboeke word toegelaat nie. n Formuleblad is aangeheg. Alle toets- en eksamenregulasies van die Universiteit van Pretoria is van toepassing. ---------------------------------------------------------- ------------------------------------------------ --- Semester Test 1 FSK 116 Semestertoets 1 16 March 011 / 16 Maart 011 Surname and initials Van en voorletters Signature Handtekening Student number Studentenommer Venue Lokaal
Formula Sheet Formuleblad Circle: Circumference πr omtrek Area πr Sphere: Area 4πr Volume 4 / 3 πr 3 Cylinder / Silinder Area (πr ) + πrh Volume πr h Trapesium: Area ½(a+b)h sin θ + cos θ 1 sin θ sinθ cosθ sin ( α ± β ) sinα cos β ± cosα sin β cos ( α ± β ) cosα cos β sinα sin β sinα ± sin β sin ½ α ± β cos½ α β ( ) ( ) a b ab cosφ a b a b + a b + a b a b a b b a iˆ ± y y z z ( y z y z ) ( ) + a b b a ˆj z z ( y y ) + a b b a kˆ b b 4ac a T T 73.15 c K 9 TF TC + 3 5 L Lα T V V β T V f Vi ( f i ) ( f i ) Q C T T Q cm T T Q Lm W pdv Question 1 / Vraag Calculate the following to 3 significant figures: ereken die volgende tot 3 beduidende syfers: [7] 1.1 1 1 1.08 10 4 8.00 10 + 4 300 1. 1 -.00 ln e 1.3 log ( 1.34 ).18 1.4 ( ) log1.34 1.19 1.5 Cos ( 48 ) 0.669 1.6 Tan-1( 57.5) 1.55 rad in radians / in radiale 1.7 Convert 5.88 radians to degrees Skakel 5.88 radiale na grade toe 337 1.8 A circular race track has an internal circumference of km and a width of 6 m. Calculate the area of the paving on the track in square metres. n Sirkelvormige resiesbaan het n binneomtrek van km en n padwydte van 6 meter. ereken die oppervlakte van die plaveisel van die pad in vierkante meter. [5] Determine the inner radius of the track: c π r c 000 m ri 318.31 m π π Outer radius: ro ri + 6 m 318.31+ 6 m 34 m Area of paving:
Paved Area Total area inner area π r π r o i ( ro ri ) π (( 34) ( 318) ) π 1101 m 1.9 Iron has a density of 7.87 g/cm 3, and the mass of an iron atom is 9.7 10-6 kg. If the atoms are spherical and tightly packed, what is the volume of an iron atom? Yster het n digtheid van 7.87 g/cm 3, en die massa van n yster-atoom is 9.7 10-6 kg. Indien die atome sferies is en styf gepak is, wat is die volume van ʼn yster atoom? [5] Mass of iron atom: 9.7 10-6 kg 9.7 10-3 g mass n mass of atom Density volume volume 3 3 density volume 7.87 g/cm 1cm n 3 mass of atom 9.7 10 g 8.49 10 atoms Volume of 1 atom (assuming no space in between) Volume with n atoms volume of 1atom n atoms 3 1 cm 8.49 10 atoms 1.1779 10 cm 1.18 10 m 3 3 9 3 Question / Vraag.1 Determine the first derivative of the function y ( 3). epaal die eerste afgeleide van die funksie y ( 3) ( 3) y u dy dy du d du d ( )( ) 3 4 3 16 4 or ( ). [] y + 4 3 4 1 9 dy d 3 16 4. Determine the first derivative of the function. y cos( 3) epaal die eerste afgeleide van die funksie y cos( 3).. [3]
( ) y cos 3 y cos Then : ( u) dy dy du d du d 4sin 3 sin ( 3) [ 4] ( ).3 An eisting stone wall on a farm is about 00 m long. The farmer has enough material to erect a fence of length 100 m. He wishes to use a portion of the eisting wall as the one side of a rectangular pen for his cattle and the other three sides are to be completed by using the available fencing material. Calculate the dimensions and the area of the pen which is constructed in this way and will have a maimum area. Remember to draw a sketch. n Lang reguit klipmuur wat 00 m lank is, is geleë op n plaas. Die boer besit genoeg kampmateriaal om n draadheining van 100 m lank te maak. Hy gebruik die bestaande klipmuur as een sy en voltooi die ander drie sye van n reghoekige kamp met sy beskikbare materiaal. ereken die afmetings en oppervlakte van die kamp wat so gebou word dat die oppervlakte daarvan n maksimum sal wees. Onthou om ʼn skets te teken. [9] 00 m Wall/muur b l A l b l 00 100 b + l ( l) b ½ 100 ( 00 )( ½ ( 100 00 )) ( 00 )( 50 ½) A + + 10 000 + 150 ½ da d 0 150 m for a ma or min d 10 000 150 ½ d + 150 l 00 50 m ( l) b ½ 100 5 m A l b 50 5 150 m
.4 (a) Calculate the area bounded by the t-ais and the graph of the function v 3 + 4t, between t 1 and t 5. The units of v are meters per second. ereken die area tussen die t-as en die grafiek van die funksie v 3 + 4t, tussen die waardes t 1 en t 5. Die eenhede vir v is meter per sekonde. [4] v 3 + 4t 5 1 ( 3 4 ) Area under curve + 5 3t ½( 4) t 1 ( 3( 5) ( 5) ) 3( 1) ( 1) + 60 m t dt ( ) + + (b) What does this area in (a) represent and what are the units of this quantity? Wat word verteenwoordig deur die area in (a), en wat is die eenhede van hierdie hoeveelheid? [] Area represents the displacement Units are in meters Question 3 / Vraag 3 3.1 A has the magnitude 1.0 m and is angled 60.0 counter-clockwise from the positive direction of 1.0 m iˆ + 8.00 m ˆj on the same coordinate the ais of an y coordinate system. Also, ( ) ( ) system. We now rotate the system counter-clockwise about the origin by 0.0 to form the y system. On this new system, what are A and in unit-vector notation? Sketch the two situations. A het n grootte van 1.0 m en wys in n rigting wat n hoek van 60.0 anti-kloksgewys met die 1.0 m iˆ + 8.00 m ˆj en is ook op dieselfde positiewe as maak in n y koördinaatstelsel. ( ) ( ) koördinaatstelsel. Ons roteer nou die koördinaatstelsel deur n hoek van 0.0 anti-kloksgewys om die oorsprong om die y stelsel te vorm. In hierdie nuwe koördinaatstelsel, wat is A en in eenheidsvektor notasie? Skets die twee situasies. [+10] y A 60 1.0 8.0 m y A Original aes Rotated aes If we convert to the magnitude-angle notation (as A already is) we have
14.4 m ( ) 1 θ tan 8 33.7 1 A' 1.0 m ' 14.4 m After rotating: and θ ' A 60 0 40.0 θ ' 33.7 0.0 13.7 Converting these results to (, y) representations, we obtain A' Acos θ ' 9.19 m (a) A' y Asin θ ' 7.71 m A' (9.19 m) ˆi + (7.71 m) ˆj. ' cos θ ' 13.99 14.0 m (b ' y sin θ ' 3.4 m. (14.0 m) ˆi + (3.41 m) ˆj 3. Consider the following vectors: eskou die volgende vektore: A (.00 m) iˆ + ( 3.00 m) ˆj C 4.00 m iˆ + 6.00 m ˆj ( ) ( ) : 4.00 m, at +65.0 D : 5.00 m, at 35 (a) Determine the sum of the four vectors in unit-vector notation. epaal die som van hierdie vier vektore in eenheidsvektor-notasie. [6] Epressing (1.69i ˆ + 3.63j) ˆ m and D (.87i ˆ + 4.10j) ˆ m in unit-vector notation, we have and, respectively. (a) Adding corresponding components, we obtain the resultant R ( A )i ˆ ( ) ˆ + + C + D + Ay + y + Cy + Dy j. ( 3.18 m)i ˆ + ( 4.73 m) ˆj (b) Determine the magnitude and direction of the resulting vector in (a) epaal die grootte en rigting van die resultante vector in (a). [4] The magnitude is R ( R ) + ( R ) y The angle is + ( 3.18 m) (4.7 m) 5.69 m.
1 y θ tan 1 4.7 m tan 3.18 m 56.0 above ais (in nd quadrant) 3.3 Consider the following vectors a 3.0iˆ + 3.0 ˆj + 3.0kˆ and b.0iˆ + 1.0 ˆj + 3.0kˆ. eskou die volgende vektore a 3.0iˆ + 3.0 ˆj + 3.0kˆ en b.0iˆ + 1.0 ˆj + 3.0kˆ. (a) Determine the angle between the two vectors. epaal die hoek tussen die twee vektore. [4] a a 3 + 3 + 3 7 5. b b + 1 + 3 14 3.7 a b abcosθ a b + a b + a b a b + a b + a b cosθ ab y y z z ( 3)( ) + ( 3)( 1) + ( 3)( 3) ( 5.)( 3.7) 0.936 y y z z θ 0.7 This final answer depends on rounding. If it has been marked incorrect, and you have a final answer around 0, you can bring it to me for re-marking. (b) Determine a vector which is at right angles to both these vectors. epaal n vektor wat n 90 hoek maak met altwee hierdie vektore. [4] a b a b b a iˆ + a b b a ˆj + a b b a kˆ ( y z y z ) ( z z ) ( y y ) (( 3)( 3) ( 1)( 3) ) ˆ (( 3)( ) ( 3)( 3) ) ˆ (( 3)( 1) ( )( 3) ) i + j + kˆ i + j + kˆ ( 6) ˆ ( 3) ˆ ( 3) Question 4 / Vraag 4 4.1 When the temperature of a copper coin is raised by 100 C, its diameter increases by 0.0%. To two significant figures, answer the following (given reasons for your answers): Die diameter van ʼn koper muntstuk verander met 0.0% wanneer die temperatuur met 100 C styg. Antwoord die volgende, en gee jou antwoord tot beduidende syfers (Gee ook redes vir jou antwoorde). (a) Determine the percentage increase in the area of the face (top side) of the coin. epaal die persentasie toename in die area van die bokant van die muntstuk. [5] If D 0.0D 0, then r 0.0r 0
A π r First derivative gives the small change in A when there is a change in r da π r dr da π rdr A π r r ( r) A π r r r 0. A π r r r Therefore A r (0.0)0.40% of A Do NOT use actual values to determine the change in area. You can also get the same answer by using: A π r ( ) A' π r + r A A' A A % change in A 100 A (b) Determine the percentage increase in the thickness of the coin. epaal die persentasie toename in die dikte van die muntstuk. [] Epansion is the same in all directions and therefore the increase in thickness is 0.0% (c) Calculate the coefficient of linear epansion of the coin. ereken die linieêre uitsettings-koëffisiënt van die muntstuk. [3] D D α T 0 0 0 ( 0.0 10 ) D D0 5 α.0 10 / C D T D ( 100 C) 4. What mass of steam at 100 C must be mied with 150 g of ice at its melting point, in a thermally insulated container, to produce liquid water at 50.0 C? Water massa stoom by 100 C moet met 150 g ys by dié se smelt punt gemeng word in ʼn termies geïsoleerde houer, om vloeistof water by 50.0 C te kry? [6] c water 4180J/kgK L fwater 333kJ/kg L v,water 56kJ/kg Let the mass of the steam be m s and that of the ice be m i. Then Q 0 Q + Q Q Q ice/ water steam/ water ice/ water steam/ water L m + c m ( T 0.0 C) m L m c ( T 100 C) F c w c f s s s w f, where T f 60 C is the final temperature. We solve for m s :
L m + c m ( T T ) m L m c ( T T ) F c w c f i, ice s s s w f i, steam m s LF mc + cwmc ( Tf 0.0 C) L c ( T 100 C) s w f 3 3 3 ( )( ) ( )( ) 3 ( ) ( ) 333 10 kj/kg 150 10 kg + 4180 J/kg.K 150 10 kg (50 C 0.0 C) 3 3.98 10 kg 33.0g 56 10 kj/kg 4180 J/kg.K (50 C 100 C) 4.3 A lab sample of gas is taken through a cycle abca shown in the p-v diagram. The net work done is +1.5 J. Along path ab, the change in the internal energy is +3.0 J and the magnitude of the work done is 5.0 J. Along path ca, the energy transferred to the gas as heat is +.5 J. ʼn Laboratorium gasmonster word deur die siklus abca geneem, soos aangedui in die p-v diagram. Die netto arbeid verrig is +1.5 J. Langs pad ab, is die verandering in interne energie +3.0 J en die grootte van die arbeid verrig is 5.0 J. Langs pad ca word +.5 J energie oorgedra na die gas. (a) How much energy is transferred as heat along path ab? Hoeveel energie word oorgedra as warmte langs die pad ab? [3] We note that process a to b is an epansion, so W > 0 for it. Thus, W ab +5.0 J. We are told that the change in internal energy during ab process is +3.0 J, so application of the first law of thermodynamics for that ab yields E Q W ab,int ab ab Q E + W ab int + 3.0 J + 5.0 J Q ab +8.0 J.