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VICTOttlA JUNIOR COLLtrGESUGGf,STtrD SOLUTIONS TO 2OO9PHYSICS H2 P3 PRELIM EXAMSl(a) Newton's Second Law: The rate ofchangc ol monentun ofa body ispropoflional 10 the net force acting on thebod), and taks place in the dircclion oltheNewton's Third l,alv: For every forc(action) which body A exerts on body B,body B willexert an equal and oppositeforce Geaction) on body A.

WN: normal reaction lbrcc exeded by sledItl: \\,cighl ol'boxl: liiclional force bes-een sled and box(bXii) Lel the horizonlal lbrce exerted bythe horse on thc slcd be F. and thefriclional lbrce behveen the sled and thesround be ^. By N2LF-. .f, = (M,,,, + M,t")a4 = (32X2.0) + 64: 128 N(b)(iii) With the horse increasiDg ilsacceleration significanlly, ihe box willdrop offthe sled allcr some time.This is because the friciional forccbei$een the box and thc slcd will not besulTicient to acceleralc thc sled and box asone syslcnr once the acceleration reachos1oo high a vahre.

2(a)

(bXi) Consrruclivc nxerlerence isproduced u,hen the wavos lron Y and Xarrive at P in phase.The wavc displacements ofX and YrgiDlbrce each other to producc alnaxinum wav ampliludc al P.(b)(ii) 6.0 cnl ofpath difTercnccconsponds io 60' ofphase diference.Hence the wavclcnglh ).:6 x 6.0: i6 cll1.From thc glaph. period 1ofrhe lvave: L06 ms.Hcnce, the velociry of lYavc- 0.16(1.06 x l0 r) =340 m s1(b)(iii) The velocity oflhc waves indicateslhat bodr X and Y arc sources ofsound

i(a)(i) Distancc lion] each cornet ol'squarc 10 centre ofsquare is, =lJe o' + 0.0' - +.:+ m2Elecrric licld slrength at O duc 10 charge A

is given byE= 4 = 6.0x10'" 4'6,s' 4(3. r12X8.85 x r oi'? ) (4.242 )tr = J.0 N C'

(b)(i)

8/14/2019 VJC-phy-pap3 answers

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.. wavelength of laser light:,:6.91 x l0? m =694 nm.(iii) The prnping power supplied by ihexenon lanp musr be high enough to causethe number ofchromium atoms in lhcmcta stablc slatc t710 be more than thenumbcr olaloms in the ground stale tr.'l his condition is called populationOnly when population invelsion isachieved willihe passase of694 nm laserphorons through the ruby crystalcausenrore de-excitations (rr + r;) thancxcilations (]?, J trr, resuhing in the netproduction and hence alnplification oflaser light.(b) Mass ofeach deuterhrm nucleus=2 t 1.6'7 tlcl1kc:3.34' 10 2r kg(!wicc thc mass olhydrcgen atoln)... no. ofdeuierium nucleipresent in rbct.ou^ t0roellet = _:2.99 l0' I t4^10 :'Assunring thal the vaporised deuteriumnuclei behave like ideal gas aroms,the average kinetic energy per Duclcustl: &1 .t8^ r0 '') .00 10')))= 2.07 ' l0

5JHerce, energy needed to raise temperaturcofdeuterium to 103 K:2.99 ' 10" ' 2.07 ' l0'5 J:619\107J

6.19 x l0?po$er needed:1 = 1.00x l0 ''z

A geostaiionary satcllile musl satisry ihelbllowing conditions:L lr musi have a period of2,1 houts2. It must movc in a direction similar to!hal ol the Eath's rotation3. Ii must revolve in a plane that includesthe Eadh's equator.

(ii) Angular vclociq/ z) =27ra=__7.21\24(60X60)

energy

(iii) For the circul motion oflhc salellile,

G'e?^lo"X6J"lo")I=:l-1 17.27\ro')'/=4.23x107m GMnt(i\)CPE U= ronlyE s =9!!!! - -!Lr212

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!=I- 6.19 x l0r'w-Section B(a) Newton's Law ol Gravilalion stalesihat any two particles in the universe willexert a force ofaftraction towards cachother proportional to thc product ofthemasses ofthe two particles and invenelyproportionalto rhe square oftheir

(b)(i) A geostationary salcllilc is onewhich revolves around the Ealth in such away thai it always remains stationaryrclative to a point on the surface ofthe

(c) As the satellitc spiraled lowads theEadh, it lvould lose gravitational PE.Pat ofthis loss in PE would be manifestedas work done against air fricdon,and pad olit as kinetic energy ofthcsatcllilc which would consequently

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