Further Mechanics Exam Pack MS

8/10/2019 Further Mechanics Exam Pack MS

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Further Mechanics Exam Pack

Mark Scheme

1. C[1]

2. B

[1]

3. D[1]

4. A[1]

5. B[1]

6. A satellite orbits the Earth once every 120 minutes. Calculate the satellites angular speed.Correct substitution into angle/time (1)

Answer with correct unit (1)

r.p.m. etc. not allowed

Angular speed = e.g. 0.052 rad min11!0"h1(2 mar#s)

Dra a !ree"body !orce diagram !or the satellite.

(1 mar#)($% the &arth is shown' then the direction must be correct)

#he satellite is in a state o! !ree !all. $hat is meant by the termfree fall? %o can the heighto! the satellite stay constant i! the satellite is in !ree !all&

ree %all when gra*itational %orce is the onl+ %orce acting on an ob,ect (1)

-eight (1) %or each clear and rele*ant ph+sics statement (1) (1)

( mar#s)[otal mar#s]

Feltham Community College 1


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7. '(etch o! to graphs)

a

0 t

'inusoidal*egative start

a

dO

+inear through 0,0*egative gradient -

Amplitude o! tide = .1 m

*e/t mid"tide at 12.00 noon

*e/t lo tide at 1.00 pm

Calculation o! time at hich !alling ater levels reaches ring 3)

x=x0sin

12

2 t

4Allo cosine5

1.6 m = .1 m sin h12

2 t

4Error carried !orard !or their amplitude above7 not 1.2 ml

t= 1.28 h ort= -.2 h i! cosine used

#ime at 3 = 12.00 h 9 1.28 h = 1.28 h 1.18 pm -[11]



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8. Demonstration that ater must be thron bac(ards at about 1 m s:1)

;orce = t

waterofmomentum

(1)

< 10

* = 8 10-

V(1)

V=-

108

10 108$ ?3 .1 @$ (1)

4Allo . @$ i! 1. m s:lused5 2

?verall e!!iciency)

oer in = 1.8 sl

.- 10> lJ

= .-- 10>$ (1)

4ntermediate value not e/plicitly needed5

oer out = 18.0 1089 .- 108= 21.- 108 (1)

E!!iciency = .--.21

= 0.6 6 (1) [10]

9. De!inition o! linear momentum)

@ass velocity 4$ords or de!ined symbols7 *?#ft5 (1) 1*etons second la)

+ine only (1) 1

*etons third la)

+ine 2 ?3 1F 2 (1) 1



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+abelled a/es and line shoing E increasing ith time'inusoidal shape (1)60 s, . 10:rad s:12(1)

= 1.6 *(1)

4*o e.c.! here unless in rad s:15



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Calculation o! value o! !orce B)

;orce B = 6* : 1.6*(1)

= > *(1) 2

4e.c.!. e/cept here 3.; = 05

;orce)

'cales read > * same as B 4allo e.c.!.5

*etons rd la!orce student e/erts on scales(1) 1[12]

12. @omentum) mass velocity 4accept de!ined symbols5 1

hysical Guantity)

*et !orce (1)

on lorry (1) 2

4M3ate o! change o! momentumN scores one only5

@agnitude)

s-0

81*s

= 1. 10-*

Iradient measurements (1)

Correct calculation (1) 2

4Accept 1.- : 1.> to 2 s.!.5

E/planation o! shape)

;orce decreases as speed increases (1)

4Allo Mrate o! change o! momentumN5

Any one o!)

Air resistance increases

#ransmission !riction increases

Engine !orce reduces 2[3]



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13. @omentum o! driver)

Correct use o!p = m4?3 ith numbers5 (1)

= 100 * s ?3 100 (g m s:1(1) 2

Average resultant !orce)Correct choice o!F t = p ?3 F = ma(1)

F 0.0> s = 100 * s F= 0 -260 00.0> (1)

= 21 (* = 21 (* (1)

4gnore sign o! anser5

$hy resultant !orce is not the same as !orce e/erted on driver by seatbelt)

Air bags !loor!rictionseatsteering heel (1)4*amed !orce other than eightreaction5 1

[]

14. rinciple o! conservation o! linear momentum)

4*ot Oust eGuation : symbols must be de!ined5

'um o! momentatotalmomentum remains constant (1)

4EGuation can indicate5

4*ot MconservedN5

! no resultant e/ternal !orce acts (1)

4NotMclosedisolated systemN52

+as o! @otion)

2ndand rdlas (1)1



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2= 2gx *ote) derived !rom 2= u2+ 2as = 0

4;or 2= 2gx9 1 (1)speed at bottom5

2gx/r gat top

xr2x0.2 m

4or =5 (1)[5]

16. Calculation o! total momentum)

n 1 s,p= 1-00 J 10 10:8

* s (g m s:1

4*o u.e.5 2

;orce e/erted on hole sail)

@omentum changetime 4?3 symbols5 (1)

= -.> 10:8 1. 108

= >.0 >.1 * (1) 2

E/planation o! hy !orce is doubled)hotons bounce bac( (1)

'o their change o! momentum is doubled (1) 2

Calculation o! ma/imum increase in speed)

a = F/m =>1200 m s:2 4allo e.c.!.5 (1)

= 0.008 m s:2

= a t= 0.008 m s:2 80-


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17. Discussion)*o eGuilibrium or there is a resultant !orce (1)Direction changing or otherise ould move in a straight lineor o!! at an tangent (1)acceleration or velocity changing (1)

;orce toards centre or centripetal (1)#he tension provides this !orce 4conseGuent5 (1)4?3 !or last 2 mar(s) eight o! ball acts donards (1)vertical component o! tension balances it (1)5

;ree"body diagram)

$eightmggravitational Pattraction 4not Pgravity5 (1) 1[]

18. E/periment2 light gates (1)Iate gives time trolley ta(es to pass 4 not Oust Pthe time5 (1)'peed = length o! Pinterruptertime ta(en (1)

?3

2 tic(er timers (1)Dots at (non time intervals (1)'peed = length o! tape sectiontime ta(en (1) 4ruler 9 cloc( could obtain third mar( only, speci!ying a lengthtime5

#otal momentum o! trolleysQero (1)t as Rero initially ormomentum is conserved 4conseGuent5 (1) 2

'peed v o! ASse o! momentum = mass velocity (1)Sse o! mass speed A = mass speed B (1)

1.< m s14ignore "ve signs5 (1)

[!]



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19. Diagrams shoing !orces)

D i a g r a m 1 D i a g r a m 2

(1) (1) 2

4Each diagram, one arro only5

Discussion re arti!icial gravityAny !our !rom)

centripetal !orce, in conte/t

!orces !elt by astronaut both MupardN

so !eels li(e eightgravity

a2r F= m2 r a=2rF= m2rF= m

=2

2

2

22

10

210--2 )/(

t

rr/

t

r =

=

4Any one5 or calculate or

= 2.0 m s:24*o u.e.5

arti!icial !ield varies ith radius (1) (1) (1) (1) -

1.8 * (g:14to one signi!icant !igure5 ?3 other Ousti!ication 4e.c.!5 (1) 1[3]

20. %o ions are accelerated

Electric !ield e/ists beteen 9, : electrodes (1)

!orce on ions !orce acceleration (1) 2

'peed o! /enon atom

eT = m2eT =((1)

= mBeT2 (1)

=

1

2

16

ms102.2

221080108.12

+

(1)

= -. 10-m s:14*o u.e.5 (1) -



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#hrust on space probe

;orce = rate o! change o! momentum (1)

= 2.1 10:8 - 000 * (1)

= 0.060 * (1)

4Ssing - 10-m s:1givesF= 0.00 (m to 210 (m 8


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22. E/planation o! elton heel

Vuote o!F = mt4or in ords5(1)

*egative momentvelocity a!ter(1)

ncreased tice momentum velocity changem" mucomparedith !alling o!! plane paddle(1)

dea o! doubled(1) @a/

ercentage e!!iciency o! station

Energy available = mgh(1)

oer input = 2>0 6.< 20(1)

E!!iciency = 00 000 100 881 00 = >8 (1)

?ther desirable properties

;or e/ample)

%ard : does not earscratchdent(2)

#ough : can ithstand dynamic loadsplastic de!ormation(2)

'trong : high brea(ing stress!orce(2)

'mooth : lo !riction sur!ace(2)

Durable : properties do not orsen ith time(2) @a/ -

#ime to repay initial investment

Each hour orth 00 0.0->- = W2.>0 ?3 total no o! ($ hreGuired 1.0000000.0->- = 2.1 10>

*o. o! hours to repay = 1 000 000 2.>0 ?3 2.1 10>00

2- 8 = -.< years to repay debt(1)

Assumption

Constant poer production no interest charges no repair costs

no ages !or employees(1) - [14]

23. $hy person moving in a circle must have an acceleration

Acceleration due to changing direction?3! not it ould continue in straight line(1) 1



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Centripetal acceleration

a=

22

?3rwr

v

(1)

=

= ==

srad10.>

?3

ms-880802- 10-.822

1

1

8

w

!#rv

(1)

a = 0.0- m s"2 4 no u.e.5(1)

$hich !orce is the larger

mg is larger than$/$ is smaller than mg(1)

mg % $ centripetalacceleratingresultant !orce acts toards centre(1) 2

Di!!ering apparent !ield strength

0.0- L 6.


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25. @omentum and its unit

@omentum = mass velocity 1

(g m s:1or * s 1

@omentum o! thorium nucleus be!ore the decay

Qero 1

'peed o! alpha particleradium nucleus and directions o! travel

Alpha particle because its mass is smallerlighter 1

'o higher speed !or the same magnitude o! momentum?3 * argument 1

?pposite directionsalong a line 1[]

26. Deceleration o! Earth

F=m t= > 10-* 4EGuation or substitution5 (1)=2.1 10* (1)a=F/m=2.1 108.0 102- 4EGuation or substitution5 (1)= . 10:20m s:2(1) -

?3in 1 second) !or debris = 10-m s:1

a= 10-

m s

:2

F=ma=> (g 10-ms:2

= 2.1 10* (1) (1)

Fon Earth =Fon debris

aearth=F/m=(g108

*101.22-

= . 10:20m s:2(1) (1)?3 Conservation o! momentum X (1)

or m1u1+m22= metc X

8 102- 10-9 > 0 = 8 102-9 > (1)?3 An anser . 10:20m s:2 calculated !rom any speci!ic (1) (1) (1)arbitrary time5 (1)Comment) negligible (1) 14?3 energy loss negligible5

[5]



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27. Direction o! centripetal acceleration

#oards centredonardsinards (1) 1

E/planation

F=ma

aandFin same direction (1) 1

3esultant !orce

= !r2

= .-


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E!!ectiveness o! mallet ith rubber head

tgoes uppgoes up (1)

less!orce'lesse!!ectivemore!orce, moree!!ective 4conseGuent5 (1) 2[11]

30. 'peed o! rim o! drum

v= r or v= 2r!4either used5 (1)

= s80minrev


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Calculation o! change in direction

sintan = 0002102.1

00010> 10:8

(1)

= 8.< 10:8rad ?3 .6 10:-degrees ?3 1.- seconds (1) 2

;ormula !or net !orce in terms o! momentum

F = d/dtm ?3 ords (1) 1

Calculation o! number o! roc(ets reGuired

NFt= 2 1011?3N= 2 1011 > 10810 (1)

N= 220 4must be a hole number5 (1) 2[]

32. 3esultant !orce

Direction o! travel changing (1)

Telocity changingaccelerating (1)

;orce is toards centre o! circle (1)

$hy no sharp bends

3elate sharpness o! bend to r(1)

3elate values o! , randF(1) 2

4e.g. i! rlarge, can be large ithout !orce being too largei! rsmall,must be small to prevent !orce being too large5

Bobsleigh

Ncos= mg(1)

Nsin(1)

= m2ror ma(1)

roo! success!ully completed 4conseGuent on using correct !ormula5 (1) -

Calculation o! angle

>> >


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33. @omentum o! neutron

Sse o!p= m(1)

p= .0 10:20* s(g m s:1(1) 2

'peed o! nucleus#otal mass attempted to be !ound (1)Conservation o! momentum used (1)= 2.01 108m s:14ec! !rompabove only5 (1)

$hether collision as elastic

Sse o! (.e. = m2(1)

(e = >.- 10:1J .08 10:1-J ec! (1)

A correct comment based on their to values o! (e. (1) [!]

34. 'peed o! car 2

2[ u2= 2as

[u2= [ 2 .- 2.6 4substitution or rearrange5 m s:1 (1)

u= 12.< m s:1 4value5 (1) 2

@agnitude o! the momentum o! car 2p= m(1)

= 1-0 12.< 1 (g m s:1

= 1< 00 1< 800 * s or (g m s:1(1) 2

Calculation o! easterly component o! momentum

Component = momentum cos (1)

Car 1) 2


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E/planation o! ho investigator could use conservation la

Any to !rom)

@omentum conservation A!ter collision there is signi!icant northerly momentum

Be!ore collision car 1 had no northerly momentumonly car 2had northerly momentum (1)(1) 2

[12]

35. a 3esultant !orce reGuired

#he direction o! speed ?3 velocity is changing (1)#here is an accelerationrate o! change in momentum (1) 2

b i Angular speedSse o! an angle divided by a time (1)

>. 10:rad s:1?3 0.28 rad h:1?3 -.2 10:os:1?3 1oh:1(1) 2

ii 3esultant !orce on student

Sse o!F= mr]2?3 v= r] ithF= rmv 2

(1)2.0 * (1) 2

iii 'cale reading

Evidence o! contact !orce = mg [ resultant !orce (1)$eight o! girl =


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iii at B resultant provided by !orce V being greater than (1) 1[10]

38. a ;rom hat height&

Sse o! mghand mv2(1)4ignore poer o! 10 errors5

mg h= mv2(1)

4shon as !ormulae ithout substitution, or as numbers substitutedinto !ormulae5

Anser 40.


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c #ime !or truc(s to stop

4Do not penalise candidates !or using a total !rictional !orce o! 0.8 *. possible5

Either

Correct use o! poer =f v and mv2(1)

4Do not penalise poer o! 10 errors or not dividing by 2 in ! T eGuation5

Sse o! energy divided by poer(1)

Anser in range 2.8 s to 2.> s (1)4ec! their value !or u*

= 0.12 * 2.1

m s:1= 0.0 s(1)

:0.12 * =


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OR

'electFt= p(1)'ubstitution :0.12t = : s(1) [10]

39. 'mo(e detectors3ecognition that mass alpha = - (1)dea o! " - to !ind resulting nucleus mass 42>5 (1)

-udaughter product= mvalphaor momentum eGuations in conte/t (1)

valpha= 6.2 udaughter product4allo ec! incorrect masses eg - and 2-1 ) 80.25 (1)

Sse o! mv2#o give ratio = 6.2 4allo ec! as long as rounds (1)to 80 7 must have speeds sub 7 valid use o! p22m5

Energy)

= m.2 energy must have come !rom mass (1)

#otal mass a!ter is a little less than be!oremass lossmass de!ectbinding energy (1) ma/ 8

[]

40. 'peed o! sphere

@omentum conserved 4stated or implied5 _(1)

Correct subs +.%.' or 3.%.' o! conservation o! momentum eGuation (1)

Correct anser 4` = 1.-m s:15 (1)

E/ample o! calculation)

- 2.> 9 0 = - ` 9 26 2.12 g m s:11< = - ` 9 81.-m5 (1) 2

E/ample o! calculation)

r

= 8.8> 10:11

* m2

(g:2

8.0 102-

(g 2- 80 80 s2

-2

= >.8 1022m

'o r= -.2 10>m

e i 'atellite ith greater mass)Kes : because, in geostationary orbit, rconstant soacceleration remains the same, regardless o! mass (1)

ii 'atellite ith greater speed)

*o 9 suitable argument (1) 24e.g. !or geostationary orbit, !and r are !i/ed, so vcannotincrease v= 2#r!5

! $hy satellite must be over eGuator)dea that centre o! satellites orbit must be the centre o! theEarth can be shon on diagram (1)there must be a common a/is o! rotation !or the satellite andthe Earth the satellites orbit must be at right angles to thespin a/is o! the Earth (1) 2

[11]

47. i to correct arros 4ignore labelling5 (1) 1



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ii 'ome use o! mv2/rith vcorrectly subbed ?3mr]2ith vcorrectly used (1)4subbing may happen later in anser5

!cos= mg mg

?3!sin = mv2/r4either gains (1)5 (1)

tan = v2/rg(1)

r= v2gtan

= 0 06.


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b %eight above Earths sur!ace

'tatement use o!2

2

2?3

r

m1-

r

mv

r

m1- == mr2(1)

4EGuations may be given in terms o! accelerations rather than !orces5

4#hird mar( !rom belo may also be aarded here i! r9h is used !or r5Correct value !or r, i.e. -.2 10>m (1)Sse o! h= their r % $E(1)

Correct anser = .< : .80 10>m 4no ec!5 (1)

E/ample o! anser)

r

mv

r

m1-2

2 =

rr

r

r

v

r

1- 222

2

===

1-= 2

r

21

2-2211

2

s102>.>

(g106.8

==

1-r

= -.2 10>mh= -.2 10>m : 8.< 108m= .6 10>m -

[3]

49. a #able

4gnore crosses. ! more than one tic( in a line, no mar(.5#op line) #o the le!t (1)Bottom line) Donards (1) 2

b Calculation o! rotation period

Sse o! ! = 2#r/v or ! = 2#/ and = v/r (1)Correct anser 40.0


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ii 'peedometer reading

'peedometer reading is too high because !reGuency (1)

?3 ?3 revs per second ?3 rate o! rotation o! heelis increased

4Allo ec! !rom MdecreasedN in c i5 1 []

50. a 1 eGuating E and E (1)2 recall o! mv2r (1) !ind centripetal !orce = 2mg (1)- !orce on rider = centripetal !orce 9 eight ?3 !orce = mg (1) hence Mg"!orceN = (1)

b height not a !actor, so B is correct (1)some ill reach this conclusion via much longer routes 1

[]

51. a recall o! p = mv 4eGn or sub5 (1)anser (1)

p = mv

= 2 0.02- 0.


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d 2 points !rom(inetic energy is lost as soundthrough de!ormationto heat (1)?3 collisions not per!ectly elastic@omentum still conserved (1)as the total (e decreases column D more balls are in motion (1) @a/ 2

[12]

52. a Calculation o! angular speed

Sse o! = 2!(1)>.2> 10:42 sig !ig minimum5 (1) 2

22-h 800 s h:1 = >.2> 10rad s:1

b i Calculation o! accelerationSse o! a = r2?3 v =rand a = v2r(1)

0.0-01 m s:2(1) 2

8-00 10Jm>.2> 10:rad s:12

= 0.0- m s:2

ii Direction o! accelerationArro to the le!t (1)4*o label needed on arro. ! more than one arro shon, no

mar( unless correct arro labelled acceleration5 1

iii ;ree"body diagramArro to le!t labelled $eight&mgpull o! Earthgravitational!orce (1)Arro to right labelled *ormal reactionN$push o! Earth ?3groundnormalcontact !orce (1)

4Dont accept MgravityN as label54@ore than to !orces ma/ 154Diagram correct e/cept rotated gets 1 out o! 25 2

iv %o the acceleration is produced

N is less than &(1)3esultant ?3 net ?3 unbalanced !orce toards centre (1)

4Accept donard centripetal !or toards the centre, butnot as an alternative to MresultantN5 2

[]



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53. a denti!y particle

Alpha particle %elium nucleus %e-2

-2%e -2

-2

-2

-2 aalphaBalphB 1

b @omentum o! particle

@omentum eGuation 4n symbols or ith numbers5 (1)

Either

Correct substitution into 2

1

mv2= energy (1)

Sse the relationship to determine the mass 48.8 10:2>(g5 (1)Anser 46. 10:20(g m s:1 @ust be given to 2 sig !ig. *ounit error5 (1)

Or

3earrangement o! E(= 21

mv2to give momentum ie vE2 ^

(1)Correct substitution (1)

Anser 46. 10:20(g m s:1. @ust be given to 2 sig !ig. *ounit error5 (1)

Eg 2

1

m1.-1 10>m s:12= 8.< 10:1J

m =21>

1

sm10-1.1

J10(gmomentum = 8.8 10:2>(g 1.-1 10>m s:1

= 6. 10:20(g m s:1

?r

@omentum =1>

1

sm10-1.1

J10


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54. Any < mar(s !rom)3ecall o!p =mv(1)Sse o! momentum be!ore collision = momentum a!ter collision (1)Correct value !or speed (1)

E/ample)120 2


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56. a i Sse o!3 = r2leading to 0. m (1) 1

ii Correct use o! = 2t leading to 82.< rad s:1 (1) 1

iii Correct use o! v = r= m s:14allo use o! sho that value5 (1) 1

b i 'ubstitution intop = 3v(1)

0-> $ (1) 2

ii Air is hitting at an angleall air not stopped by blades (1)Energy changes to heat and sound (1) 2

c i Attempts to !ind volume per second 3 v (1)

-- (g s:1(1) 2

ii Sse o!F = mvt(1)

F = 810 * (1) 2

d 3ecognises that 100 $ is produced over 2- hours (1)

Estimates i! this ould !ul!il lighting needs !or a day(1)Estimates energy used by lo energy bulbs in day(1)Conclusion (2)#he anser must be clear and be organised in a logical seGuence

E/ample)

#he 100 $ is an average over the hole day. @ost households ould uselight bulbs !or 8 hours a day in no more than - rooms, so this ould meanno other energy as needed !or lighting.- lo energy bulbs ould be -- $ !or 8 each hours so ould reGuire energy!rom the *ational grid.

4Accept an argument based on more light bulbslonger hours that leads tothe opposite conclusion5 [1]

57. a #otal linear momentum o! a system is constant, (1)

provided no resultant e/ternal !orce acts on the system (1) 2



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b #he anser must be clear, use an appropriate style and be organised in alogical seGuence

Sse o! a light gate (1)Sse o! second light gate (1)

Connected to timer or inter!ace 9 computer accept Plog"it (1)Cards on gliders (1)

@easure length o! cards (1)Telocity = length L time (1) 8

c @ultiplies mass velocity to !ind at least one momentum (1)

180 g cm s:10.018 (g m s:1 be!ore and a!ter (1) 2[10]

58. a #hey act on the same body or do not act on di!!erent bodies (1)#hey are di!!erent types o!, or they are not the same type o!, !orce(1) 2

b As the passenger or capsule or heel has constant speed (1)there is *o resultant tangential !orce acting on the passenger (1) 2

c ;riction beteen seat F person or push o! capsule all on person (1) 1[5]

59. a i measured thic(ness o! lead -" mm (1)measured radius 2 " < mm (1)Talue beteen < : > mm (1)

Eg actual radius = mm 8 mm-. mm

ii Sse o!p=45r4 any to values sub5 (1)

Anser range 6.1 10:21" 1.- 10:20* s or (g m s:1

4allo ec!5(1) 2

b #rac( gets more curved above lead r smaller abovelead (1)@ust be sloing don less momentum losesenergy (1)Sp 4dependent on either anser above5 (1)

c nto page (1)

4ec! out o! page i! don in b5 1



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d i Division by 6.11 10:1(g (1)

Anser range 1.0 : 1.8 1010m s:1(1) 2

ii greater than speed o! light (1)

impossible so mass must have increased (1) 2[1]

60. a ud identi!ied (1) 1

b Conversion o! I (1)Conversion o! either eT or divided by c2 (1)2. 10:2


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ii Sse o! v = %o d to sho6o = t1

(1)

Correct anser (1) 2

E+a',-e o# ca-c"-ation

6o = t1

= s18101.12

1

= 2.810:1


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Further Mechanics Exam Pack MS