AQA Maths M2. Topic Questions from Papers. Circular Motion. Answers

Size: px

Start display at page:

Download "AQA Maths M2. Topic Questions from Papers. Circular Motion. Answers"

Arabella Armstrong
6 years ago
Views:

1 AQA Mths M Topic Questions fom Ppes Cicul Motion Answes PhysicsAndMthsTuto.com

2 PhysicsAndMthsTuto.com Totl 6 () T cos30 = 9.8 Resolving veticlly with two tems Coect eqution 9.8 T = cos30 T =.6 N AG 3 Coect T fom coect woking (b) v T cos60 = 0.6 3()(i) d - v =.84 ms 4 Coect v Totl 7 Resolving hoizontlly. Coect eqution Solving fo v (Q, Jn 006) 6() = m + mg(3 3cos θ ) v = 4+ 6 g( cos θ ) AG Totl 7 Thee tem enegy eqution Coect eqution d 4 Solving fo v. Coect esult fom coect woking (b) v mg cosθ = m 3 Resolving towds the cente Coect eqution 3gcosθ = 4+ 6g 6gcosθ d Solving fo cosθ 4+ 6g cosθ = 9g Coect cosθ θ = 44.6 Coect ngle Totl 9 (Q6, Jn 006) 4() 3 mu = + mgl( cos 60 ) U = v + gl = v U gl Totl d 4 thee/fou tem enegy eqution with tig tem coect eqution solving fo v o v coect v in simplified fom (b) v T mgcos60 = m l U gl g U g T = m + = m l l d d Totl 9 esolving towds the cente of the cicle with thee tems substituting fo v coect eqution mking T the subject coect expession fo T. Simplifiction not necessy. (Q4, June 006)

3 PhysicsAndMthsTuto.com 4 () 4 = = F = = 4704 N AG d 4 finding cceletion coect cceletion use of F = m coect foce fom coect woking (b) R = = 760 B noml ection 4704 µ µ 760 µ 0.4 dv AG 3 Totl 7 pplying F µ R coect esult fom coect woking (Q, June 006) 3() mg = Enegy eqution v= g 3 (b) All tems fo, no component T ñ mg= T = 3mg F 3 ft if T > 0 Totl 6 (Q3, Jn 007)

4 PhysicsAndMthsTuto.com 6 6() 40 π 4π = (d/sec) 60 3 (b) 4π = ω = π 4 (c)(i) = Accept 0.36 π ( 3sf ) B (ii) Veticlly No cceletion, foces blnce mg = T cosθ B (iii) Hoizontlly 6π Tsinθ = m 4 F ft cceletion T cosθ = mg m ω SC tnθ = g st 3 mks fo quoting nd using coectly 6π tnθ = 4g o tnθ = 0.38( 08) F ft povided θ = 0! F ened in (b) Totl (Q6, Jn 007)

5 PhysicsAndMthsTuto.com 7() Using consevtion of enegy (lowest nd highest points): m(7v) = + mg fo 7v nd v 48 v = g Needs 48 o 4 v = g AG (b) Velocity t A is g Resolving veticlly t A: 3 tems v m + R = mg, coect 3 tems, coect signs m g R = mg ñ ñ mg A = mg 4 Condone mg λ Totl 9 (Q, June 007) 8 8() Q is in equilibium E Q t est, o not moving T = g = 49 N B AG (b) Resolving veticlly fo P: T cos θ = 3g cos θ = 3 θ = cos = 3. 3 Do not condone 3 3 (c) 4 sin θ = Resolving hoizontlly fo P: B = T sin θ tems: tem coect, othe tem includes sin o cos 3v 4 = g 3 4 = 4g = 48 4g =. 4 SC3 3 Totl 9 TOTAL 7 (Q8, June 007)

6 PhysicsAndMthsTuto.com () Acceletion is v 9 = 0. = 0 m s (b) θ = 30 B Resolve veticlly: T cos θ = mg T cos θ = 4g T = 4.3 N 4 AG (c) Resolve hoizontlly: T sin θ + T = fo 3 tems, coect 4.3sin θ + T = 4 0 T = 7.4 N 3 Condone 7.3 N Totl 9 λ (Q, Jn 008) Mks Totl Comments 0 7() Consevtion of enegy: ( ) m 3 g + mg= fo 3 tems: KE nd PE 9 mg + mg = v= 3g 4 (b) At A, conside veticl foces: T mg = fo 3 tems, coect T = mg + 3mg m T = 4mg ft 4 ft fom () Totl 8 (Q7, Jn 008)

7 Totl 8 7() At top, fo complete evolutions: = mg whee v is speed t top v = g Consevtion of enegy fom B to top : 3 tems, KE nd PE + mg = mu u = 4g + v = g u = g AG PhysicsAndMthsTuto.com (b) At C, speed of pticle is 3g B Resolving hoizontlly t C: T = T = m 3g T = 3mg 3 Needs coect tems (c) No i esistnce B Bed is pticle Totl 9 (Q7, June 008) () 40 evolutions pe minute = 80π dins pe minute B o 3 ev pe second = 4π 3 dins pe second B AG (b) (c) Resolve veticlly: T cos 30 = 6g tem ech side, coect T = 67.9 N 3 AG Resolve hoizontlly: sin 30 = T mω 67.9 sin 30 = 6 ( 4π ) tem ech side, coect T sin 30 RHS 3 = 0.3 m 4 Condone 0.33 (using π s 3.4) Totl 9 (Q, Jn 009)

8 PhysicsAndMthsTuto.com 37() = m mg 8 Totl 7 3 tems, KE nd PE v = = 4.8 v = Accept 4.8 (b) Using F = m dilly: R= mgcos 60 + = 6g cos = 66.6 N 4 Totl 7 3 coect tems (not necessily coect signs) B B fo 60 (Q7, Jn 009) Q Solution Mks Totl Comments 44() Resolving veticlly: T cos 60 + T cos 40 = mg.66 T = 6g T = 46.4 N 4 AG no mks if g deleted (b) Rdius of cicle is 0.6 tn 60 B =.039 o.04 Hoizontlly: = Tcos0 + Tcos30 Accept sin insted of cos fo 6v 46.4cos cos 30 =.039 o 70.0 v =.3 Speed is 3.48 m s 4 Totl 8 θ (Q4, June 009)

9 PhysicsAndMthsTuto.com Q Solution Mks Totl Comments 7() By consevtion of enegy to point whee QP mkes n ngle θ with upwd veticl: = mu mg( + sinθ fo 3 tems, KE nd PE ) mg ( + sinθ ) tem v = u g( + sinθ ) Resolve dilly R= mgsinθ fo 3 tems, include sinθ o cosθ mu = 3mg sinθ mg 6 AG (b) When pticle leves the tck, R = 0 0= 3mg 3mg sinθ mg sinθ = SC3 sin 3 3 θ = 9. 4 ccept 9.4 o θ = c Totl 0 (Q7, June 009) Q Solution Mks Totl Comments 66() =.sinθ B. cosθ 0 mks (b) Resolve hoiz: Tsinθ = mω T sinθ = 4.sinθ T = 0 T cosθ = mω etc (+ second ) Resolve vet: Tcosθ = 4g cosθ = fo θ = 70.9 o c.4 6 Totl 7 30sinθ tnθ = g (Q6, Jn 00)

10 PhysicsAndMthsTuto.com 77() Using consevtion of enegy: fo 3 tems, KE nd PE mu = mgh o 4 tems, KE nd PE mu = mg ( cosθ ) fo finding h v = u + g cosθ ( ) ( [ cos ]) v= u + g θ AG (b) Using F = m dilly, mg cosθ N = Pticle leves sufce of hemisphee when N = 0 B m mg cosθ = ( u + g[ cosθ] ) u cosθ = + cosθ g u cosθ = + 3 g Totl 0 Coect 3 tems Coect signs ( N o+n ) (Q7, Jn 00) Q Solution Mks Totl Comments 88() Using consevtion of enegy: 3 mg( cos θ ) = mgh = v = 6 g( cos) m v= (6 g[ cos]) =.4 4 SC3:.4 (b) When pticle is t est, esolve dilly T = mg cos T mgcos= o T = mgsin = mg cos m =.3 3 Totl 7 (Q8, June 00)

11 PhysicsAndMthsTuto.com 9 o 9 As pticle moves, T = using unknown s extension: If dius is, extension is. B If extension is x, dius is. + x B λx λx Using T = : Using T = : l l 9(.) T =. T = 9 x. = 60(.) = 60x 8 3 T = 60(.) = T = x =. + x 60 9 = 7 ( o = 86.4 ) 9x+ 60x = = 0 0x + 4x 9 = 0 (0+ 3)( 3) = 0 (0x 3)( x+ 3) = 0 =. o 0.3 x = 0.3 o. Rdius is. 8 Rdius is. Totl 8 (Q9, June 00) 0() Resolve veticlly R = mg If the pticle is on the point of sliding, F = µr F = 0.3R = 0.3mg Mks Totl Comments Ignoe ll inequlities Resolving dilly: F = mω 0.3mg = mω 0.8 ω = 0.3 g 0.8 ω =.9 4 (b)(i) 4 evolutions pe minute = 90π 60 = 3π o 4.7 dins pe second (ii) Resolving dilly: F = mω 3π mµ g = m 0. 3π 0. µ = g eithe side coect second side coect µ = CAO (ccept 0.339) Totl 0 (Q, Jn 0)

12 Totl 6() By consevtion of enegy m( v) = m(3 v) + mg 8v = g g v = o 4 g 4 fo 3 tems, KE nd PE PhysicsAndMthsTuto.com (b) Getest nd lest vlues of tension e t the highest nd lowest points of its pth m(3 v) At top, T = mg = 4 mg ft ft - must be positive tension m( v) At B, T = + mg = 9 4 mg ft Rtio is 9 : CAO Condone : 9 o :.8 Totl 9 (Q6, Jn 0) Q Solution Mks Totl Comments 7() Resolving veticlly T cos cos 0 = 4g : Thee tems, which must include 4g, Tcosθ o Tsinθ nd 0cosθ o 0sinθ, whee θ = 30, 40, 0 o 60. : Coect tems : Coect eqution T cos 30 = T = 30.4 N 4 : Coect finl nswe. Accept 30.4 o AWRT Accept 30.4 o 30. o AWRT 30.4 fom g = 9.8. (b) Hoizontlly: 4 = 30.3 = 0cos 40 + T cos 60 F d : Thee tems, which must include 4 o, Tcosθ o Tsinθ nd 0cosθ o 0sinθ, whee θ = 30, 40, 0 o 60. F: Coect eqution. My include T, m nd v. d: Substitution of vlues fo T, m nd 4 v. Eqution of fom = numbe = = : Coect nswe. Accept 3.7 o 3.8 o AWRT 3.8. Accept 3.7 o AWRT 3.7 fom g = 9.8. Note: Do not ccept = 30.4 o simil. Totl 8 (Q7, June 0)

13 Q Solution Mks Totl Comments 38() Using consevtion of enegy (lowest nd highest points) mu = +mg() : Eqution fo consevtion of enegy with two KE tems nd one o two PE tems. My see m o 0.3. : Coect eqution. u = v + 4g Fo complete evolutions, v > 0 u > 4g PhysicsAndMthsTuto.com u > g AG 3 : Coect esult with sttement of v > 0 nd some intemedite woking including 4g tem. O Use of PE t top nd KE t B () Coect PE nd KE () Coect deduction including inequlity () (b)(i) C of Enegy mu = + mg( + sinθ) v = 9 g g( + sinθ) = g g sinθ Resolve dilly ±R = mg sinθ + = mg sinθ + mg mg sinθ = 3mg sinθ + mg = 3 9 sin θ g OE (must include g) 4 0 : Eqution fo consevtion of enegy with two KE tems nd one o two PE tems including sinθ. My see m o 0.3. : Coect eqution. : Thee tem eqution fom esolving dilly. Coect thee tems, but condone signs nd eplcement of sin by cos. : Coect eqution. My see m o 0.3. : Simplified coect finl nswe. 9 3 Condone sin 0 θ 4 g (ii) When this ection is zeo, 3 9 sin θ g = : Putting thei ection equl to zeo. sinθ = 6 θ is 6.4 bove hoizontl : Coect ngle. Accept AWRT Totl 0 (Q8, June 0)

14 PhysicsAndMthsTuto.com 4 R = mg F = 0.8 mg = 0.8 mg condone v = g m dependent on both s = 83. v = 6.8 m s 6 = 0.8R (fo ) Totl 6 (Q, Jn 0) Q Solution Mks Totl Comments 7() by consevtion of enegy: mu ( ) = ( ) + mg v = u 4g fo 3 tems, KE nd PE; not v = u + s (b)(i) (ii) t point A; T = t point B; T = T T = m ( v ) m ( u ) mg + mg () mu ( ) mg = + mg ( 4 ) m u g mg ( ) mu = + mg B u 0g g = u + g m 3u = 7g both signs incoect eithe coect o T A = T B o T = T, T = T CAO u = 3 g 7 condone 9g u = v + 4g v = g B condone v fom tio : o : nd one tension eqution coect = g tio u : v = 3 : B ccept.34 : o : 0.74 Totl (Q7, Jn 0)

15 Q Solution Mks Totl Comments 6 () Fo pticle B, tension in sting =.g N B PhysicsAndMthsTuto.com Resolve hoizontlly fo pticle A: m = T =.g = 49 Angul velocity is 7 d/sec 4 O m mg o (condone lck of nd ) mg (b) Using v = ω: speed = =. m s Pt (b) mks cn be wded in () (c) Time tken is π / ω = π 7 = sec 6() Using consevtion of enegy: Totl 8 O π. Accept π 7 (0.89 A0) (Q, June 0) 7 6() Using consevtion of enegy: mgh fo o 3 tems, KE nd o PE = mg.4( cos8) m m fo finding h v = 4.8g( cos8) =.30 v =. m s 4 Condone. (b) Resolving veticlly: T = mg + = g Coect 3 tems Coect signs = 36.7 N 3 = 37 N Accept 36 N Totl 7 (Q6, June 0)

16 Totl 7 8 6() Resolve veticlly: T cos θ = mg fo T cos o T sin nd mg 34 cos θ = 9.8 cos θ = θ = (b) Resolve hoizontlly fo pticle: = T sin θ fo T cos o T sin v = 34sin ft fom () v =.3 Speed is 3.33 m s 3 Accept 3.34 PhysicsAndMthsTuto.com (c) Time tken is π / v =. sec ft Totl 8 O find nd use (Q6, Jn 03) Q Solution Mks Totl Comments 9 7() Using consevtion of enegy: mu mgh fo 3 tems, KE nd PE 3 v g. cos fo finding h [ fo.( coso sin ) ] v 4.4 g cos v v 3.7ms 4 Accept 3.7, 3.70, 3.7 (b) Resolving dilly: ccept cos o sin, + o sign T mg cos nd fully coect nd substituted 6. N 3 Accept 6.0 o 6 Totl 7 (Q7, Jn 03)

17 PhysicsAndMthsTuto.com 30 In limiting equilibium, using F = μr Fictionl foce is 0. mg Resolve hoizontlly m = 0. mg = 0. g = = Totl 4 (Q, June 03) Q Solution Mks Totl Comments 38() Using consevtion of enegy: fo 3 [o 4] tems: KE m( u) = m( u) + mg nd [o ] PE u g fo finding h u = 4g 4 (b) Using consevtion of enegy with speed t point S to be V: m( u) = m( V ) + mg( + cos60) mv = m ( u ) mg 4g V = 3g 37g V = Resolving dilly t point S: mv ( ) R= mgcos mg = mg + 3 = 4 mg o.6mg Totl 9 O 4g m( V ) = mg( cos60 ) + m (Q8, June 03)

Mark Scheme (Results) January 2008

Mark Scheme (Results) January 2008 Mk Scheme (Results) Jnuy 00 GCE GCE Mthemtics (6679/0) Edecel Limited. Registeed in Englnd nd Wles No. 4496750 Registeed Office: One90 High Holbon, London WCV 7BH Jnuy 00 6679 Mechnics M Mk Scheme Question