Semester Test 1 Semestertoets 1 FSK March 2013 / 15 Maart Time 2½ hours Max. Total 65 Marks Max Tyd 2½ ure Maks. Totaal 65 punte Maks

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1 Physics Department Fisika Departement Semester Test 1 Semestertoets 1 FSK March 13 / 15 Maart 13 Surname and initials Van en voorletters Signature Handtekening MEMO Student number Studentenommer Time ½ hours Ma. Total 65 Marks Ma Tyd ½ ure Maks. Totaal 65 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. Beantwoord 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 15 March 13 / 15 Maart 13 Surname and initials Van en voorletters Signature Handtekening Student number Studentenommer Venue Lokaal 8

2 Formula Sheet 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 ( ± ) = sin cos ± cos sin cos( ± ) = cos cos sin sin sin ± sin = sin ½ ± cos½ a b = ab cosφ α β α β α β α β α β α β α β α β α β a b = a b + a b + a b y y z z ( ) a b = a b b a iˆ ( y z y z ) + a b b a ˆj z z ( y y ) + a b b a kˆ ± = a L = Lα T b b 4ac V = V β T V f Vi ( f i ) ( f i ) Q = C T T Q = cm T T Q = Lm W = pdv c Formuleblad T = T K 9 TF = TC v = ½ v + v v = v + at = + v t + at ½ = + v t o ( ) v = v + a = ½ v + v t = vt ½at Question 1 6 Convert the following to SI units: (a) 1. day [3] 1. day 4 hours 6 min 6s 1. day = 1day 1hour 1min = 864s = s (b) 8 km/h [3] 8 m 1h 1min 8 km / h = h 6 min 6s =. m / s 8 km/h IT IS ESSENTIAL THAT YOU INCLUDE YOUR UNITS IN YOUR MULTIPLICATION. YOU CANNOT JUST MULTIPLY NUMBERS. 9

3 Question 7 Vector A is directed along the + ais and vector B has a magnitude of 6. m. The sum of these two vectors is a third vector that is directed along the +y ais, with a magnitude 3. times that of A. What is that magnitude of A? Vektor A is langs die +-as gerig en vektor B het n grootte van 6. m. Die som van die twee vektore is n derde vektor wat langs die +y-as gerig is met n grootte van 3. keer dié van A. Wat is die grootte van A? [7] Given information: A = a î B = b ˆi ˆ + by j B = 6. m = b + by C = A + B = a + b ˆi + b ˆj = by ĵ A + B = 3. A y Therefore a and by = 3a C = b = 3. a therefore : = b ( ) y b = 3 a = 3 b y B = 6. m = b + b b b y 3 1 = b + b = b 36 = = = 1.9 m = a A =1.9 m 1

4 Question 3 7 In the product F = qv B, take q = 3. C, v =.î + 4.ˆj + 6. kˆ m.s -1 and F = 4.î +.ˆj + 1 kˆ N. Determine B (in tesla (T)) in unit vector notation if B = B y. In die produk F = qv B, aanvaar q = 3. C, v =.î + 4.ˆj + 6. kˆ m.s -1 en F = 4.î +.ˆj + 1 kˆ N. Bepaal B (in tesla (T)) in eenheid-vektor notasie as B = B y. [7] F = qv Bq v + v + v B + B + B ( ˆ i ˆ j k ˆ ) ( ˆ i ˆ j k ˆ y z y z ) ( ˆ i ˆ j k ˆ y z z y z z y y ) = q v B v B + v B v B + v B v B = 4.î +.ˆj + 1 kˆ Therefore: F = q v B v B = 4. y z z y F = q v B v B =. y z z F = q v B v B = 1 z y y If B = B y, then F = 3..B 4.B = 1N z 6.B = 1 B =.T = B y And due to an error in the question, the answer was marked depending of whether you used Fy or F to solve for Bz. Final answer had to be in unit vector notation: B =.i ˆ.ˆj ( B value) kˆ z You need to multiply the cross product out so that you have the three components of the force vector in terms of the magnetic field components. 11

5 Question 4 6 An aluminium-alloy rod has a length of 1. cm at. C and a length of 1. cm at the boiling point of water. n Aluminium-allooi staaf het n lengte van 1. cm by. C en n lengte van 1. cm by die kookpunt van water. (a) What is the length of the rod at 5. C? Wat is die lengte van die staaf by 5. C? [3] Given: T = 8. C L o = 1. cm L =. cm From the definition of epansion: L =. cm = Lα T L. cm α = = L T 1.cm 8 K / C = L = Lα T cm 6 ( C ) = 1. cm 5. 1 / 3 K = L = 1.75 cm at 5 C You need to first determine α (b) What is the temperature of the rod if its length is 1.3 cm? Wat is die temperatuur van die staaf indien die lengte 1.3 cm is? [3] Given: T = 8. C L o = 1. cm L =.3 cm L =.3cm = Lα T L.3cm T = = Lα 1. cm 5. 1 / C = 1 C T = 3 C 6 ( ) please use α from part (a) 1

6 Question 5 9 When a system is taken from state i to state f along path iaf in the figure shown, Q = 1 J and W = 84 J. Along path ibf Q = 15 J. Wanneer n stelsel van toestand i na toestand f geneem word langs die pad iaf soos aangedui in die figuur is Q = 1 J en W = 84 J. Langs pad ibf is Q = 15 J. (a) Determine W along path ibf. Bepaal W langs pad ibf. [3] For path iaf: Eint = Q W = 1J 84 J = 16 J This is the same for all paths i-f.for path ibf: E = Q W int W = Q E = 15 J 16 J = 4 J int (b) If W = 54 J for the return path fi, determine Q for this path. Indien W = 54 J vir die pad fi, bepaal Q vir hierdie pad. [3] For path fi: E = Q W int Q = Eint + W = 16 J+ 54 J = 18 J negative E!and W (c) If E int,i = 41 J, what is E int,i? Indien E int,i = 41 J, wat is E int,i? [3] From (a) and path iaf: Eint = Q W = 1J 84 J = 16 J Thus E = E E int int, f int, i E = E + E int, f int int, i = 16 J+ 41J = 167 J 13

7 Question 6 7 When the acceleration is constant, the average and instantaneous acceleration are equal: v v a = a =, where a is the acceleration, v is the velocity at time t and v o is the initial velocity at t t = s. We can also write the average velocity as v =, where is the position at time t and t o is the initial position at t = s. Using these two equations show that when a particle is accelerating v = v + a. Show all steps and state all in the direction, its velocity is given by o reasoning clearly. Wanneer die versnelling konstant is, is die gemiddelde en oombliklike versnelling dieselfde: v v a = agem = waar a die versnelling is, v die snelheid by tyd t en v o die aanvanklike snelheid by t t = s is. Ons kan ook die gemiddelde snelheid as vgem = skryf, waar die posisie by tyd t en o t die aanvanklike posisie by t = s is. Deur gebruik te maak van hierdie twee vergelykings, bewys dat v = v + a. Wys al die stappe en verduidelik jou aannames. o [7] v v a = a = t v = v + at v = t = + v t But v ½ ( v v ) = +. 1 Substituting this for v in : = + ½( v + v ) t From the first equation, we substitute for v and rearrange the terms: = + ½( v + v ) t = ½vt + ½v t = ½ v + at t + ½v t = vt + ½at v v From 1: t = a Substituting into above: 14

8 = v t + ½at ( ) a = v ( v v ) + ( v v ) v v v v = v + ½a a a a = v v v + v + v v v v v a t v = t = + v t v = v + a = a = 1 v v From 1: t = 3 a Substituting 3 in : ( ) a = ½ ( v + v ) v v ( ) a = ( v + v )( v v ) = v v v = v + a Shorter method Question 7 9 A motorcyclist who is moving along an ais directed toward the east has an acceleration given by a = t m.s - for t 6. s. At t = s, the velocity and position of the cyclist are.47 m.s -1 and 7.3 m respectively. n Motorfietsryer beweeg oos langs n -as gerig en het 'n versnelling gegee deur a ( t ) [3] = ms - vir t 6. s. By t = s, is die die snelheid en posisie van die fietsryer,47 ms -1 en 7,3 m onderskeidelik. (b) What is the maimum speed achieved by the cyclist? Wat is die maksimum spoed wat die motorfietsryer bereik? [5] ( t ) a = The velocity will be maimum when the first derivative of v ie. acceleration = m/s. ( t) a = = 6.1 t = = s 1. The velocity is obtained by the integrating the acceleration epression 15

9 ( t ) ( ) a = v = adt = t dt = t t c Since v =.47 m/s at t = s, then c =.47 m/s Need to now determine the maimum speed at 5.1 s v = = m.s 1 (b) What total distance does the cyclist travel between t = s and 6. s. Wat is die totale afstand afgelê deur die ryer tussen t = s en 6. s. [4] Total distance will be the area under the v-t curve (integral of the v(t) function between and 6 s. v = t t ( ) = vdt = t t + dt 3 = 3.5t.t +.47t = 81.4 m 6 You need to determine the value (total distance) between the limits of t = s and t = 6s, since the starting point at t = s is at = 7.3 m. The following is not necessary, but could have been used to determine distance travelled. 6 ( ) = vdt = t t + dt t t t c 3 = Since = 7.3 m/s at t = s, then c = 7.3 m 3 f = = = 88.7 m = 7.3 m i = f i = = 81.4 m 16

10 Question 8 8 A small rocket, such as those used for meteorological measurements of the atmosphere, is launched vertically with an acceleration of 3 m/s. It runs out of fuel after 3 s. What is the maimum altitude of the rocket? n Klein vuurpyl, soos dié wat gebruik word vir weerkundige metings van die atmosfeer, word vertikaal gelanseer met n versnelling van 3 m/s. Die brandstof raak op na 3 s. Wat is die maksimum hoogte van die vuurpyl? [8] There are two stages to this problem. Firstly, when the rocket is firing and using fuel, secondly when the fuel is finished, the rocket is then in free-fall. Phase 1 (before fuel runs out): v o = m/s a 1 = +3m/s t = 3 s y 1 =? v 1 =? v = v + a t 1 1 = ( 3 m.s )( 3s) = 9 m.s 1 y = y + v t + ½a t = m+ ( m.s )( 3s) + ½ ( 3 m.s )( 3s) y = 135 m 1 The rocket is moving upwards at a velocity of 9 m/s. It has not yet reached its maimum height. Phase takes care of the distance travelled till vy = m/s. Phase (after fuel runs out): v 1 = 9 m/s a 1 = - 9.8m/s y =? v = m/s v = v + a y 1 v v y = a = ( m.s ) ( 9 m.s ) ( 9.8 m.s ) = 4136 m m m m y = y + y = + = ENJOY THE WEEKEND / GENIET DIE NAWEEK 17