Farr High School NATIONAL 5 PHYSICS. Unit 3 Dynamics and Space. Exam Questions

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Transcription:

Farr High School NATIONAL 5 PHYSICS Uni Dynamics and Space Exam Quesions

VELOCITY AND DISPLACEMENT D B D 4 E 5 B 6 E 7 E 8 C VELOCITY TIME GRAPHS (a) I is acceleraing Speeding up (NOT going down he flume ) disance = area under graph () = (½ x 7.5 x 5) + (0 x 5) () = 8.75 + 00 = 8.75 m () (c) a = (v-u)/ () a = 0/5 () a =.0 ms - () d = area under graph () d = (0 5 9 ) + (0 9) + (0 5 9 ) () d = 9 + 90 + 4 5 d = 0 5 m () (a) DO NOT accep goes faser (0) marks for d = v Can award () for implied relaionship if addiion of areas is aemped No significan figure penaly (exac answer) disance = area under graph () = (8 ) + (½ 5 8) () = 4m () 4 disance = area under graph () = ( ½ x 4 x 5 ) + (8 x 5) + (½ 6 x 4.5) () = 6.5m () 5 (a) disance = area under graph () = ( ½ x 40 x 6) + (480 x 6) + (½ 480 x 6) () = 440 m () v = d/ () = 440/00 () =. ms - () 6 (a) 0.6 s disance = area under graph () = (8 x 0.6) + (½ 8 x.) () =.6 m ()

(a) a = (v-u)/ () = (8-0)/5 () =. m s - () ACCELERATION s = area under graph () = (/ x 5 x 8) + (50 x 8) () = 05 m () (c) (wear) igh fiing clohes Crouch (wear) sreamlined helme Sreamlined shoes Solid wheels (ii) Tyres (handle) grips Brakes Shoes on pedals Saddle B B 4 a = v u () = 0 8 ().5 = -. m s - () 5 (a) a = v u () = 0 () 5.8 =.9 m s - () displacemen = area under graph () = ½ x ( x 5 8) + ( x 6) () = 9 + 66 = 97 9 m () If wrong values exraced from graph hen () MAX for equaion deduc () for wrong/missing uni Do no accep a = v/ as wrong equaion- sop marking and award (0) marks If wrong subsiuion hen () MAX Deduc () for wrong/missing uni. Quesion refers o he cyclis in picure so answer should refer o his No: Pushes forward No: Wheels If wrong values exraced from graph hen () MAX for equaion. If = 6 hen wrong subsiuion award () mark max Deduc () for wrong/missing uni. Do no accep a = v/ as his is he wrong equaion - sop marking and award (0) sig. fig. range 4 9, 90, 897, Any aemp o use s = v applied o he graph is wrong physics (0) marks. If firs ime 5 8 hen () mark max for implied equaion. sig. fig. range 97 9, 98, 00

6 (a) a = v u () = 0 () 5 = 0.6 m s - () There is an unbalanced force/fricion, () his acs agains he moion. () (mus have some menion of opposing he moion) Ignore menion of componen of weigh 7 a = v u () = 70 0 () = 5 m s - () 8 a = v u () = 5 0 () 4 =.5 m s - ()

NEWTON S LAWS (a) If A exers a force on B, B exers an equal bu opposie force on A. Mus show a good aemp a saing Law Or To every acion (force) here is an equal and opposie reacion (force) (ii) Engine/exhaus gases pushed down(a on B); gases push rocke up (B on A) Mus refer o engine/exhaus gases D A 4 (a) Mus have correc label and direcion. Accep: Uphrus Upward hrus Upwards force Force of graviy on he rocke Force of graviy Do no accep: Graviy alone, W = mg () = 08 x 0 5 x 9.8 () = 0 x 0 6 N ()

(ii) Fun = 5 000 00 000 = 000 (N) () F = ma (½) () 000 =.08 x 0 5 x a () a =.08 m s - () 4 (c) I moves wih consan speed in he horizonal direcion () while acceleraing due o he force of graviy in he verical direcion () or consisen wih answer in If arihmeic error can be seen in subracion o ge Fun hen deduc () mark. Candidae can sill ge nex () marks. If no subracion is aemped and 5000 or answer from is used in calculaion for acceleraion hen () MAX for equaion. sig. fig. range 5.,.08,.078,.0779 Answer should be based on he following wo poins: saemen relaing o horizonal moion: eg ISS moves forward curvaure of he earh, surface curves away. () saemen relaing o verical moion eg falling (owards he Earh), or force of graviy () Accep: pull of graviy NOT graviy alone (d) The asronau is falling (owards Earh) a he same rae as he ISS The asronau is in freefall Answer mus refer o he asronau. Do no accep: a sraigh saemen of Newon III (ii) The asronau exers a force agains he wall and he wall exers an equal and opposie force agains he asronau (causing him o move) Answer mus refer o he asronau. Do no accep: a sraigh saemen of Newon III 5 E 6 D 7 C

8 (a) BC and DE Boh pars needed Do no accep single leers e.g. B and D NOT where graph is fla/horizonal NOT where speed is consan (ii) W = m g () = 90 0 0 () = 900 N () F res = F u - F d () = 958 5 900 = 58 5 N () a = Fres/m () = 58.5/90.0 () = 0 65 ms - () 5 9 (a) Scale diagram () N () Resulan = (800 + 800 ) () = N () (ii) W = mg () = 80 0 () = 800 N () resulan = 700-800 = 900 N () a = F/m () = 900/80 () = 5 ms - () 4 0 Ew = F d () Ew = 00 5 () Ew = 450 J () a = F/m () = 50/75 () = ms - () (a) a = (v-u)/ () = (70 0)/ () = 5 ms - () F = ma () = 8 000 5 () = 980 000N () (c) addiional force required = oal force aircraf hrus() = 980 000 40 000 = 740 000N () B 4 wo forces: air resisance and weigh () balanced () Accep g = 9 8:W = 88 N 5 sig fig {880, 88, 88 9, 88 90} Accep g = 9 8: W = 88 N 4 sig fig {900, 880, 88, 88 0} Accep answer for weigh from b for W = 880 a = 0 87 ms - for W = 88 a = 0 85 ms - for W = 88 9 a = 0 84 ms - for W = 88 a = 0 89 ms -

PROJECTILE MOTION (a) () mark or zero. I moves wih consan velociy in he horizonal direcion () while acceleraing due o he force of graviy in he verical direcion () Skech should show a reasonable curve from he package (o ground level). Sraigh line zero marks. Answer should be based on he following wo poins: Saemen relaing o horizonal moion, eg package moves forward, or package coninues a a consan velociy () Saemen relaing o verical moion eg package falls owards he road/earh, or force of graviy acs/pulls downwards () g = 9.8 (m s - ) () daa a = v - u () 9.8 = v ( 0) () 0.55 No an answer referring o graviy alone. () daa mark for correc selecion of g from able. Accep 5.9 v = 5.4 m s - () If incorrec relaionship saed (eg a = v/, v = a or v = g) sop marking and award (0) marks bu can sill ge () for daa. 4 Candidaes who sar wih v = 0 55 x 9.8 have no shown an incorrec relaionship so should no be penalised eg v = 0 55 9.8 () for implied formula, () for subsiuion & daa mark v = 5 4 m s - ()

(a) s = v () = 0 x () = 0 55 s Accep 0 6 s () a = v u () 9.8 = v 0 () 0.55 v = 5.4 m s - () (c) 5.4 velociy (m s - ) Figures on axis mus be consisen wih pars (a) and above s vs No marks (d) s = area under graph or () Mus be s = ½ x 0.55 x 5.4 s = (5.4) x 0.55 () s = 5 m s = 5 m () No oher symbols (a) s= v () s = 0 76 () s = 5 m () (ii) a = v - u () 9.8 = v - 0 () 0.76 v = 7.45 m s - ()

Same Mus have explanaion o ge firs mark (ii) All objecs fall wih he same (verical) acceleraion. Will ake he same ime o reach he waer 4 C 5 D 6 a = (v - u)/ () 0 = (v - 0)/0. () v = ms - () 7 (a) Answer should be based on he following wo poins: saemen relaing o verical moion, eg falling (owards he moon), or force (of graviy) () saemen relaing o horizonal moion, eg probe moves forward, or curvaure of Ganymede, eg surface curves NOT graviy alone away () Newon III: The hrusers force gas one way () So he gas exers an equal and opposie force on he probe.() 8 A 9 (a) d = v () = 0 6 () = 80m () a = v u () 4 = v (-0) () 6 v = 4ms - () 0 (a) d = v x () = 4 8 0 65 () = m () a = v u () 0 = v 0 () 0.65 v = 6 5 m/s () C SPACE EXPLATION (a) Eyepiece magnify () Boh correc Objecive collec ligh () Boh correc Explanaions in erms of Newon I and Newon II are also accepable. Newon I: menion of balanced forces () would no slow down/indicaion of consan speed () Newon II: unbalanced force () in he opposie direcion/opposing moion () Quoing NI, NII or NIII alone is insufficien, he answer mus relae o he hrusers/probe

COSMOLOGY (a) s = v () = x 0 8 x ( x 65 x 4 x 60 x 60) () + () = 9.4608 x 0 5 m Differen frequencies/ wavelenghs /signals require differen deecors/elescopes Cerain deecors/elescopes canno pick up cerain frequencies/wavelenghs/signals Differen signals have differen frequencies/ wavelenghs (a) (visible) Ligh NOT a sandard ()marker () for iniial equaion () for speed of ligh () for correc subsiuion of ime No marks for final answer-given Uni no required bu deduc () if wrong Accep he number 56000s in place of ( x 65x 4 x 60 x 60) Accep: Differen elescopes deec differen signals Do no accep: Differen ypes of signals - unless menioned along wih differen wavelenghs/frequencies/elesc opes/ deecors. ypes of radiaion ambiguous could be α or β Any menion of sound ypes of wave or wave oo vague Differen ypes of radiaion have differen properies/wavelenghs/frequencies (c) A he focus Accep diagram wih ligh rays shown (d) Arecibo () Larger dish can deec more waves () (a) disance ravelled by ligh in one year differen deecors () are required for differen radiaions/ frequencies/wavelenghs () C A B () or (0) (ii) accep: X-ray Visible Infrared G-M ube phoographic film (iii) 4 (a) (Radio waves are) longer/greaer/bigger/larger NOT wider ligh has a shorer wavelengh Aerial Radio elescope NOT saellie dish NOT elescope (on is own) (c) Differen frequencies/wavelenghs/signals require differen deecors/elescopes Cerain deecors/elescopes can pick up cerain frequencies/wavelenghs/signals eg differen elescopes deec differen signals NOT ypes of wave or waves oo vague NOT ypes of radiaion ambiguous could be alpha or bea Any menion of sound 0 marks 5 Cadmium and mercury Boh required 6 Helium and nirogen Boh required 7 informaion abou aoms (or elemens) presen age of sar disance o sar speed of sar ype of sar emperaure of sar Accep: gases (idenified) chemicals (idenified) wha i is made of Do no accep: posiion radiaion

VARIOUS (a) Acceleraion is he change of velociy (no speed) in uni ime (ii) Direcion of saellie is (coninually) changing Velociy of saellie is (coninually) changing There is an unbalanced (no resulan ) force on he saellie F = - = 0 N () F = ma () 0 = 50 x a () a = 0 m s - direcion is righ () (a) a = (v-u)/ () a = (6-0)/60 () a = 0. ms - () (a) (a) (ii) s = area under graph () = (0 5 60 6) + (40 6) () = 40 m () (c) v = s/ () = 40/00 () = 4. ms - () (d) W = mg () = 400 0 () = 4000 N () (e) F = ma () = 400 0 () = 40 N () (a) (iii) Upward force = 4000 + 40 = 4040 N () 800N a = (v u)/ () = (7 5)/ () = 0.7 ms - () F = ma () = 400 x 0.7 () = 67 (N) () Need o be indicaion of ime requiremen. 4 4 No aemp o calculae F => () for formula accep 9.8 and 9.8 for g which give 90 N and 94 N mus be consisen wih (a) and (d) Accep 0., 0.67 (a) (iv) (ii) Forward force = 800 + 67 () = 867 N () 5 Disance = area under graph () = ½ x 8 x 7 () = 8m () Time will be longer Sraigh line NOT accepable Ignore any values on graph No labels on axes accepable bu wrong labels is no