Circular motion

2 ↑ Rcosθ=mg B1 mv2 B1 → Rsinθ= r r =asinθ B1 ( ) 8cosθ=3 1−( cosθ)2 M1 Quadratic equation in cosθ . 1 A1 cosθ= 3 2 A1 x= a 3 6 © UCLES 2021 Page 7 of 13

5 mu2 B1 T −mgcosθ= A a mag B1 T +mgcosθ= B a mu2  mag M1 Use given relationship and combine. T A = 7T B so mgcosθ+ =7  −mgcosθ+  a  a  u2 =ag ( 7−8cosθ) Energy: 1 mu2 − 1 mag =mg ( acosθ+acosθ) M1 A1 Energy equation. 2 2 ( ) So u2=ag 4cosθ+1 Equate expressions for u2 M1 1 A1 CAO cosθ= , θ =60° 2 u= 3ga A1 CAO 8 © UCLES 2021 Page 10 of 13

2 ↑ Rcosθ=mg B1 mv2 B1 → Rsinθ= r r =asinθ B1 ( ) 8cosθ=3 1−( cosθ)2 M1 Quadratic equation in cosθ . 1 A1 cosθ= 3 2 A1 x= a 3 6 © UCLES 2021 Page 7 of 13

5 mu2 B1 T −mgcosθ= A a mag B1 T +mgcosθ= B a mu2  mag M1 Use given relationship and combine. T A = 7T B so mgcosθ+ =7  −mgcosθ+  a  a  u2 =ag ( 7−8cosθ) Energy: 1 mu2 − 1 mag =mg ( acosθ+acosθ) M1 A1 Energy equation. 2 2 ( ) So u2=ag 4cosθ+1 Equate expressions for u2 M1 1 A1 CAO cosθ= , θ =60° 2 u= 3ga A1 CAO 8 © UCLES 2021 Page 10 of 13

3(a) For B: T =mg B1 May be embedded. For A: R+Tcosθ=3mg M1 All 3 terms, allow sign errors, allow sin/cos mix. 3 A1 Use given R to obtain cosθ= 5 3 3(b) 3mv2 M1 May be seen in part (a), allow sin/cos mix. Tsinθ= r r= ARsinθ B1 Or equivalent. 3a 1 A1 [Combine to give AR= , so] BR= a 4 4 3 © UCLES 2021 Page 8 of 16

4 2 *M1 Energy equation, with 2 KE terms and a two-part GPE term, 1 mu2 − 1 m   u  =mg ( acosθ+asinθ) allow cos/sin mix. 2 2 2 3 ( ) A1 u2 =2ag cosθ+sinθ 4 2 B1 N2L u m   At B, tension in string is zero, so mgsinθ= 2 a (u2=4agsinθ ) u2 DM1 Eliminate tanθ=2 OE A1 mu2 B1 N2L At A, T −mgcosθ= a 9 5 M1 A1 Substitute to find T. T = mg ( = 4.02mg) 5 Alternative method for question 4 2 *M1 Energy equation, with 2 KE terms and a two-part GPE term, 1 mu2 − 1 m   u  =mg ( acosθ+asinθ) allow cos/sin mix. 2 2 2 3 u2 =2ag ( cosθ+sinθ) A1 4 © UCLES 2021 Page 9 of 16 4 2 B1 N2L u m   At B, tension in string is zero, so mgsinθ= 2 a (u2=4agsinθ) Eliminate θ : u2 = 8ag 5 DM1 A1 5 mu2 B1 N2L At A, T −mgcosθ= a 9 5 M1 A1 Substitute to find T. T = mg ( = 4.02mg) 5 8 © UCLES 2021 Page 10 of 16

1 3mgx B1 Their tensions equated to obtain a quadratic equation, T = CAO. a 4mga B1 T = 3 ( a+x ) 9x2 +9ax−4a2 =0 leading to ( 3x−a )( 3x+4a )=0 M1 1 A1 x= a 3 4

6(a) 5ag B1 N2L At A T −mgcosθ=m× A a v2 B1 N2L At B T +mgcosθ=m× B a 1 1 M1 Energy equation with correct number of terms. m×5ag− mv2 =mga×2cosθ 2 2 v2 =5ag−4gacosθ A1 Accept multiplied by m and/or divided by a. Use ratio of tensions = 9 : 5 M1 Use ratio and simplify to an expression in cosθ. 2 A1 CAO cosθ= 5 6 6(b) Greatest speed at lowest point M1 Energy equation including lowest point, correct number − 1 m×5ag+ 1 mV2 =mga×( 1−cosθ) of terms. 2 2 31ag A1 FT Ft their cosθ from part (a). V = 5 2 © UCLES 2021 Page 11 of 12

3(a) T =3mg and Tcosθ=mg M1 Must see both of these separately. 1 A1 At least one step of working, AG. Combining, cosθ= 3 2 3(b) (cosθ= a−x ,where x = AR) B1  sinθ= 8  x  3    3 1 a AR= a or BR= a or radius = 4 4 2 mv2 M1 Tsinθ= r Combining to find an equation in v2, a and g only. DM1 v2 =2ga, v= 2ga A1 4 © UCLES 2021 Page 8 of 12

7(a) Coordinates of A: x=asin60, y=a−acos60 B1 ( ) 2 M1 Substitute their (x, y) into correct trajectory equation. a 3 g a = a 3 3− 22 2 2 1 2V2. 4 Rearrange to find V2. M1 3 3 A1 V2 = ag, V = ag 2 2 4 7(b) 1 mu2 − 1 mV2 =mga ( 1+cos60 ) M1 Energy equation. 2 2 9 A1 u is the speed at P. u2 = ag 2 m M1 N2L T −mg = u2 a 11 A1 T = mg 2 4 © UCLES 2021 Page 12 of 12

1 3mgx B1 Their tensions equated to obtain a quadratic equation, T = CAO. a 4mga B1 T = 3 ( a+x ) 9x2 +9ax−4a2 =0 leading to ( 3x−a )( 3x+4a )=0 M1 1 A1 x= a 3 4

6(a) 5ag B1 N2L At A T −mgcosθ=m× A a v2 B1 N2L At B T +mgcosθ=m× B a 1 1 M1 Energy equation with correct number of terms. m×5ag− mv2 =mga×2cosθ 2 2 v2 =5ag−4gacosθ A1 Accept multiplied by m and/or divided by a. Use ratio of tensions = 9 : 5 M1 Use ratio and simplify to an expression in cosθ. 2 A1 CAO cosθ= 5 6 6(b) Greatest speed at lowest point M1 Energy equation including lowest point, correct number − 1 m×5ag+ 1 mV2 =mga×( 1−cosθ) of terms. 2 2 31ag A1 FT Ft their cosθ from part (a). V = 5 2 © UCLES 2021 Page 11 of 12

2 1 1 M1 A1 Energy equation.  mv2  m3gamgacoscos 2 2 m B1 N2L T mgcos v2 a 4 M1 Combine to find cos. agcos3ga2gacos2ga 5 7 A1 cos 15 5 © UCLES 2022 Page 6 of 14

5 For A: Tsinmr2 M1 N2L horizontal. r3asin B1 Correct expression for radius. T m3a2 A1 3 M1 N2L horizontal Similarly, for B: Tcos mrk22 4 3 A1 T  mak22 4 3 M1 Equate expressions for T. m3a2  mak22 4 k2 4, k 2 A1 © UCLES 2022 Page 9 of 14 5 Alternative method for question 5 For A: Tcosmg, Tsinmr2 M1 N2L horizontal and vertical. r3asin B1 Correct expression for radius. 5 5 g A1 Combine to obtain expression for 2. T m3a2  mg,  2   4 12 a 3 M1 N2L horizontal. Similarly, for B: Tcos mrk22 4 3 A1 T  mak22 4 5 3 5g M1 Substitute for T and . mg  mak2 4 4 12a k2 4, k 2 A1 7 © UCLES 2022 Page 10 of 14

2 1 1 M1 A1 Energy equation.  mv2  m3gamgacoscos 2 2 m B1 N2L T mgcos v2 a 4 M1 Combine to find cos. agcos3ga2gacos2ga 5 7 A1 cos 15 5 © UCLES 2022 Page 6 of 14

5 For A: Tsinmr2 M1 N2L horizontal. r3asin B1 Correct expression for radius. T m3a2 A1 3 M1 N2L horizontal Similarly, for B: Tcos mrk22 4 3 A1 T  mak22 4 3 M1 Equate expressions for T. m3a2  mak22 4 k2 4, k 2 A1 © UCLES 2022 Page 9 of 14 5 Alternative method for question 5 For A: Tcosmg, Tsinmr2 M1 N2L horizontal and vertical. r3asin B1 Correct expression for radius. 5 5 g A1 Combine to obtain expression for 2. T m3a2  mg,  2   4 12 a 3 M1 N2L horizontal. Similarly, for B: Tcos mrk22 4 3 A1 T  mak22 4 5 3 5g M1 Substitute for T and . mg  mak2 4 4 12a k2 4, k 2 A1 7 © UCLES 2022 Page 10 of 14

4(a) 1 1 M1 Energy equation with all necessary terms, GPE terms must be mv2  mu2 mgacoscos resolved, allow sin/cos mix, allow sign error. 2 2 1 1 A1 2u may be substituted later. m2u2  mu2 mgacoscos 3 2 2 2 Implied by  ag gacoscos 2 3 m B1 N2L At A, T mgcos u2 a m B1 N2L and use of tension (10T). Also, 10T mgcos 4u2 a Use all three (two N2L and energy) equations to find T in terms of M1 Might see m and g only. 3m 2 9T mgcoscos  ga a 3 coscos1 10coscos4 1 A1 T  mg 3 6 4(b) 1 4m 2 M1 Any appropriate method to obtain cos. Substitute back, 10 mgmgcos  ga 3 a 3 2 A1 cos 3 2 © UCLES 2022 Page 9 of 13

1 2v2 M1 Use F = ma: 20= OR 20=20.62 0.6 50 A1 v2 =6 OR 2 = 3 60v 60 A1 FT 38.9848…. Number of revolutions per min = OR 0.62π 2π so 39(.0) revolutions 3 Question Answer Marks Guidance

5(a) 1 1 B1 Energy equation. mv2 − mu2 =mgacos 2 2  1  kag = ag+2agcos    3  m B1 N2L at B. T −mgcos= v2 a 11 m 11 So mg−mgcos= .kag, −cos=k 6 a 6 Solve simultaneously. M1 4 1 A1 Both. k = , cos= 3 2 4 5(b) Initial speed  = kagsin B1 Use v2 =u2 +2as: 0= ( kagsin )2 −2gs M1 1 A1 s= a 2 1 1 A1 FT Height above lowest point = s+a−acos = a+a− a=a 2 2 4 © UCLES 2022 Page 9 of 12

1 m B1 N2L When string goes slack, mgcos= v2, v2 =agcos May include T, but B1 not awarded until T = 0. a 1 1 B1 Energy equation. m.3ag− mv2 =mg(acos+acos) 2 2  2 M1 Combine. So u2 −agcos=2agcos+   3 4 A1 u2 − ag 3 5 cos= = 3ag 9 4 © UCLES 2022 Page 6 of 15

6(a) 6 B1 May be implied. T = mg 7 Tsin=mr2 =mhtan2 B1 Allow r for radius. Radius of circle = htan B1 6g [So 2 = cos] 7h 9 6g M1 Second scenario, equivalent result . In second scenario, 2 = cos 4 7h 6g 4 6g A1 Combine convincingly to obtain given result. Equate, cos=  cos giving 7h 9 7h 4 cos= cos AG 9 5 © UCLES 2022 Page 12 of 15 6(b) First scenario: N+Tcos=mg 1 B1 Both. Second scenario, N +Tcos=mg 2 6 12 M1 12cos−6cos=7 Equate: mg− mgcos=2mg− mgcos 7 7 3 A1 cos= 4 5 A1 N = mg 7 4 © UCLES 2022 Page 13 of 15

1 2v2 M1 Use F = ma: 20= OR 20=20.62 0.6 50 A1 v2 =6 OR 2 = 3 60v 60 A1 FT 38.9848…. Number of revolutions per min = OR 0.62π 2π so 39(.0) revolutions 3 Question Answer Marks Guidance

5(a) 1 1 B1 Energy equation. mv2 − mu2 =mgacos 2 2  1  kag = ag+2agcos    3  m B1 N2L at B. T −mgcos= v2 a 11 m 11 So mg−mgcos= .kag, −cos=k 6 a 6 Solve simultaneously. M1 4 1 A1 Both. k = , cos= 3 2 4 5(b) Initial speed  = kagsin B1 Use v2 =u2 +2as: 0= ( kagsin )2 −2gs M1 1 A1 s= a 2 1 1 A1 FT Height above lowest point = s+a−acos = a+a− a=a 2 2 4 © UCLES 2022 Page 9 of 12

3(a) m4ag M1 Allow cos instead of sin for M1. Do not award At B, mgsin until tension 0 used. Mass must be seen. No sign 5a error. 4 A1 sin 5 2 3(b) mu2 B1 At A, T mgcos a 1 1 4ag M1 A1 Energy equation with 4 terms, dimensionally Energy mu2  m mgacossin correct. Mass must be present, allow sign errors. 2 2 5 1 Must see in the KE terms. 2 Solve to find T M1 Complete method leading to an expression in mg for T . 21 A1 CWO T  mg 5 5 © UCLES 2023 Page 9 of 15

5(a) mg mg3  B1 Hooke’s law, T 1  a xa or T 2  a  4 xa  mv2 mv2 mg M1 gxxa Also, T = and equate  xa v2  1 x x a a Dimensionally correct terms. mg   3x a   2m   1 v   2 M1 v2  3gx  3 xa    4  2  2a 4  Similarly:  a 3 1 3x x Must have v and on the RHS. 4 2 4 Their dimensionally correct T . 2 Equate expressions for v2 and solve for x in terms of a. M1 x4a A1 WWW 5 SC B3 for answer of 4a using  instead of mg . 5(b) a 2a M1 FT their expression for x.  v2 or  v2 and substitute x4a, v 12ag xgxa 3  3xg xa 4  1 A1 CAO 2 © UCLES 2023 Page 12 of 15

3(a) m4ag M1 Allow cos instead of sin for M1. Do not award At B, mgsin until tension 0 used. Mass must be seen. No sign 5a error. 4 A1 sin 5 2 3(b) mu2 B1 At A, T mgcos a 1 1 4ag M1 A1 Energy equation with 4 terms, dimensionally Energy mu2  m mgacossin correct. Mass must be present, allow sign errors. 2 2 5 1 Must see in the KE terms. 2 Solve to find T M1 Complete method leading to an expression in mg for T . 21 A1 CWO T  mg 5 5 © UCLES 2023 Page 9 of 15

5(a) mg mg3  B1 Hooke’s law, T 1  a xa or T 2  a  4 xa  mv2 mv2 mg M1 gxxa Also, T = and equate  xa v2  1 x x a a Dimensionally correct terms. mg   3x a   2m   1 v   2 M1 v2  3gx  3 xa    4  2  2a 4  Similarly:  a 3 1 3x x Must have v and on the RHS. 4 2 4 Their dimensionally correct T . 2 Equate expressions for v2 and solve for x in terms of a. M1 x4a A1 WWW 5 SC B3 for answer of 4a using  instead of mg . 5(b) a 2a M1 FT their expression for x.  v2 or  v2 and substitute x4a, v 12ag xgxa 3  3xg xa 4  1 A1 CAO 2 © UCLES 2023 Page 12 of 15

1 1 1 M1 Energy equation, 4 terms, dimensionally correct, m.3ag mv2 mgacoscos mass must be present, allow sign errors, allow sin in 2 2 both terms on RHS mv2 B1 N2L, may include tension initially but not awarded mgcos until tension = 0 used a 3 1 4  M1 Dependent on tension = 0 and on an energy 2 mag 2 m.agcosmga  5 cos   equation, eliminate v2. 3 7 cos 2 10 7 A1 If no m in energy equation and no further errors, cos award SCB2 for correct final answer 15 4 © UCLES 2023 Page 4 of 14

5(a) Tcosmg B1 mv2  27ag B1 Tsin  m r 4r   r 12atan used M1 27 3 M1 Finds value for tan OE. Divide: tan , so tan 412tan 4 Reduces to equation in  or x, no k. r 9a, extension of string = 3a A1 Alternative method for question 5(a) Let L be stretched length of string. B1 12a Or T mg Tcosmg L mv2  27ag B1 Tsin  m r 4r   r  Lsin used M1 12a  L2 144a20.5 M1 Use cos and sin and eliminate T. L L [L2 144a2 81a2] L15a, extension of string = 3a A1 5 © UCLES 2023 Page 10 of 14 5(b) kmgL12a B1 Hooke’s law: T  12a kmgL12a mgL M1 Eliminate T:  12a 12a L A1 k  5 L12a 3 Question Answer Marks Guidance

6(a) Angular speeds of P and Q are equal, M1 v v so Q = P x 3a−x A1 Shown convincingly: angular speeds equal stated. 2x ag v = AG Q 3a−x 2 6(b) m4ag B1 For P: T +mgcos60= 3a−x mv2 B1 For Q: T −mgcos60= Q x m.4ag mv2 M1 Eliminate T: −mgcos60+ =mgcos60+ Q 3a−x x m4ag mx4ag M1 Solve to find x. =1+ 3a−x (3a−x)2 Obtain 3-term quadratic equation. 4a(3a−x)=(3a−x)2 +4ax, x2 +2ax−3a2 =0 (x−a)(x+3a)=0, x=a A1 5 © UCLES 2023 Page 12 of 13 6(c) Energy changes from initial position: B1 KEs correct. Gain in KE of P: 1 m ( 4ag−u2) B1FT GPEs correct. 2 1  u 2  Loss in KE of Q: 2 m   2   −v Q 2     Loss in GPE of P = mg(3a−x)(1−cos60) (=mga)  1  Gain in GPE of Q = mgx(1−cos60)  = mga   2  1 2 m ( 4ag−u2) − 1 2 m       u 2    2 −v Q 2    =−mgx(1−cos60)+mg(3a−x)(1−cos60) M1 Energy equation.   5 A1 AEF Simplify: 4ag− u2 +ag =ag 4 16 4 u2 = ag, u= 5ag 5 5 4 © UCLES 2023 Page 13 of 13

1 ↑Tcosθ=mg B1 mv2 B1 → Tsinθ= asinθ Eliminate T and substitute for 𝜃 M1 A1 v= 5ag 6 4 Question Answer Marks Guidance

5(a) mv2 B1 At A: R −mgcos= A A a mv2 B1 At B: −R +mgcos= B B a 1 1 M1 All terms present, allow sign errors, cos/sin mix. Energy: mv2 − mv2 =mga(cos+cos) 2 A 2 B A1 AEF 1 M1 Any equivalent working, leading to a dimensionally Eliminate velocities, use R = R and substitute angle values B 6 A correct equation in R and mg. B 1 1 a( R −mgcos)− a(−R +mgcos)=mga(cos+cos) 2 A 2 B 3 4 7 6R − mg+R − mg=2mg B 5 B 5 5 3 A1 R = mg B 5 6 5(b) 3 3 mv2 M1 From first equation in (a), 6 mg− mg = A 5 5 a v2 =3ag, so A1 A CAO k=3 2 © UCLES 2023 Page 10 of 15

6(a) Angular speeds of P and Q are equal, M1 v v so Q = P x 3a−x A1 Shown convincingly: angular speeds equal stated. 2x ag v = AG Q 3a−x 2 6(b) m4ag B1 For P: T +mgcos60= 3a−x mv2 B1 For Q: T −mgcos60= Q x m.4ag mv2 M1 Eliminate T: −mgcos60+ =mgcos60+ Q 3a−x x m4ag mx4ag M1 Solve to find x. =1+ 3a−x (3a−x)2 Obtain 3-term quadratic equation. 4a(3a−x)=(3a−x)2 +4ax, x2 +2ax−3a2 =0 (x−a)(x+3a)=0, x=a A1 5 © UCLES 2023 Page 12 of 13 6(c) Energy changes from initial position: B1 KEs correct. Gain in KE of P: 1 m ( 4ag−u2) B1FT GPEs correct. 2 1  u 2  Loss in KE of Q: 2 m   2   −v Q 2     Loss in GPE of P = mg(3a−x)(1−cos60) (=mga)  1  Gain in GPE of Q = mgx(1−cos60)  = mga   2  1 2 m ( 4ag−u2) − 1 2 m       u 2    2 −v Q 2    =−mgx(1−cos60)+mg(3a−x)(1−cos60) M1 Energy equation.   5 A1 AEF Simplify: 4ag− u2 +ag =ag 4 16 4 u2 = ag, u= 5ag 5 5 4 © UCLES 2023 Page 13 of 13

5(a) For A: F T ma2 M1 Only allow sign errors. A F mg  1mg B1 Accept with g replaced by 10. A 5 Combine: T  1mg 4 mg M1 To reach an equation in T and mg only. 5 25 Accept with g replaced by 10. T  1 mg A1 CAO 25 4 5(b) For B: F T km2a2 M1 Only allow sign errors. B F kmg 1kmg and combine to find k M1 To reach an equation in k only. B 5 k  1 A1 3 3 © Cambridge University Press & Assessment 2024 Page 10 of 13

5(a) For A: F T ma2 M1 Only allow sign errors. A F mg  1mg B1 Accept with g replaced by 10. A 5 Combine: T  1mg 4 mg M1 To reach an equation in T and mg only. 5 25 Accept with g replaced by 10. T  1 mg A1 CAO 25 4 5(b) For B: F T km2a2 M1 Only allow sign errors. B F kmg 1kmg and combine to find k M1 To reach an equation in k only. B 5 k  1 A1 3 3 © Cambridge University Press & Assessment 2024 Page 10 of 13

2(a) 2mgx B1 2mgx Hooke’s law: T  seen anywhere. a a mga B1 RHS seen anywhere. 2 May be in terms of radius or extended length, for  T  a mga mga a x 4 example 2 , 2 . 1 r l a 4 mga M1 Equate two expressions for T and obtain a Equate: 2mgx  2 , 4x2 3axa2 0 simplified homogeneous quadratic equation a 3a 4l2 5al0, x 4 4r2 3ara2 0, 2k2 3k20 a A1 Single correct answer only. x 4 4 2(b)  T kmg B1 kmg seen anywhere in an equation. This may be seen in part (a). Note that no response in part (b) can earn B1 if kmg seen in part (a). 2mgx 1 B1 CWO. Part (a) needs to be correct. T  , k  a 2 2 © Cambridge University Press & Assessment 2024 Page 7 of 15

3(a) mu2 B1 Condone r used consistently instead of a At lowest point, T mg  throughout this question. a mv2 B1 When string makes angle  with upward vertical, Smgcos a 1 1 M1 Must include m. Allow sin instead of cos for Energy: mu2  mv2 mga1cos 2 2 this mark, allow sign errors. Eliminate u2 and v2 M1 Need to see at least one line of working. S T 3mg1cos A1 AG 5 3(b) When string goes slack, S = 0 so T 3mg1cos M1 May use v2 agcos substituted into energy equation. mu2 2 A1 But T mg mg4mg 5mg, so cos a 3 2 © Cambridge University Press & Assessment 2024 Page 8 of 15

2 mv2 B1 At B, T+mgsin= a 1 1 M1A1 Energy A to B: mu2 − mv2 =mga(cos+sin) 2 2 Substitute for u and  to find T: M1  4 3 T =mg5−2 −3   5 5 8 A1 T = mg 5 5 © Cambridge University Press & Assessment 2024 Page 6 of 12

6(a) At P:  T cos=T cos+0.05g B1 1 2 At Q:  T cos=0.04g 2 B1 OR: whole system: T cos=0.09g 1 T =0.15g =1.5 N B1 1 3 6(b) T sin−T sin=0.050.82 M1 Allow sin/cos mix 1 2 T 2 sin=0.041.42 M1 T sin=0.050.82 +0.041.42 A1 1 5 2 =12.5,  = 2 2 3 6(c) T cos=0.04g and T sin=0.041.42 M1 From part (a) and part (b) 2 2 7 Divide: tan= 4 =60.3 A1 2 © Cambridge University Press & Assessment 2024 Page 10 of 12

1 100 B1 100 Hooke’s law: T = x T = (r−0.8) 0.8 0.8 N2L: T =2(0.8+x)52 B1 T =2r52 Equate and solve: 50(0.8+x)=125x, x=0.533 B1 8 15 3 © Cambridge University Press & Assessment 2024 Page 6 of 19

6(a) For P to lowest point L: M1* 1 1 Dimensionally correct, all terms present, allow Energy: mv2 = mu2 +mga(1−cos) sign errors, allow cos/sin error. 2 2 28 [v2 = ag−2agcos] 9 From L to string goes slack: M1* 4a 1 1 4a Energy: mv2 = mw2 + mg(1+cos) Dimensionally correct, all terms present, with , 9 2 2 9 allow sign errors, allow cos/sin error.  20 26  w2 = ag− agcos    9 9  Both equations correct, allow unsimplified. A1 mw2 B1 4a When string goes slack: mgcos= 9 Equate expressions for w2 to find a value for cos DM1 2 A1 cos= 3 6 © Cambridge University Press & Assessment 2024 Page 15 of 19 9231/32 Cambridge International AS & A Level – Mark Scheme October/November 2024 PUBLISHED Question Answer Marks Guidance Alternative method for question 6(a) For P from start to string goes slack: M2 4 Dimensionally correct, all terms present, with a 9 1 1  4  Energy: mw2 = mu2 +mga (1−cos)− (1+cos) Allow sign errors, allow cos/sin error   2 2  9  5  4 RHS may appear with a−  1+  acos 9  9 4 Allow M1 if is missing or if an attempt at both 9 heights, but all other conditions are met.  10 5 13  20 26  A1 Correct, allow unsimplified.  w2 = ag+2ag  − cos  = ag− agcos   9 9 9  9 9  mw2 B1 4a When string goes slack: mgcos= 9 Equate expressions for w2 to find a value for cos DM1 2 A1 cos= 3 6 © Cambridge University Press & Assessment 2024 Page 16 of 19 6(b) mv2 M1 EITHER equation, dimensionally correct, allow Tension before: T −mg = sign error only 1 a mv2 Tension after: T −mg = 2 4 a 9  16  25 A1 EITHER tension correct v2 = ag, T = mg T =5mg   9   1 9 2 Find the other tension (from a valid equation) and find ratio of tensions M1 mv2 Equation must be of the form T −mg = r Ratio is 5 : 9 A1 5 Any equivalent ratio, allow 9 4 © Cambridge University Press & Assessment 2024 Page 17 of 19

2 mv2 B1 At B, T+mgsin= a 1 1 M1A1 Energy A to B: mu2 − mv2 =mga(cos+sin) 2 2 Substitute for u and  to find T: M1  4 3 T =mg5−2 −3   5 5 8 A1 T = mg 5 5 © Cambridge University Press & Assessment 2024 Page 6 of 12

6(a) At P:  T cos=T cos+0.05g B1 1 2 At Q:  T cos=0.04g 2 B1 OR: whole system: T cos=0.09g 1 T =0.15g =1.5 N B1 1 3 6(b) T sin−T sin=0.050.82 M1 Allow sin/cos mix 1 2 T sin=0.041.42 M1 2 T sin=0.050.82 +0.041.42 A1 1 5 2 =12.5,  = 2 2 3 6(c) T cos=0.04g and T sin=0.041.42 M1 From part (a) and part (b) 2 2 7 Divide: tan= 4 =60.3 A1 2 © Cambridge University Press & Assessment 2024 Page 10 of 12

3(a) 32x M1 λx 8= leading to x=0.5. Use T = to find x. May be implied by 2 l correctly derived value for sin=0.6. Hence, PA =2.5 and sin=1.5 =0.6. A1 AG, shown convincingly. 2.5 2 3(b) ( ) *M1 For either equation (resolving horizontally or  R=1.6g−8cos vertically). Allow sin/cos wrong way round. () mr2 =8sin+R Correct number of terms and correct signs. Dimensionally correct. Numbers or algebraic. A1 Both equations correct. 8(3)+0.5 ( 1.6g−8(4)) =1.61.52 DM1 Eliminate R to obtain equation involving 2 5 5 with no other unknown quantities. 2 =4 leading to =2 A1 60 A1 19.0985… Number of revolutions per minute = =19.1 . Accept final answer of 19. π 60 Allow final answer of . π 5 © Cambridge University Press & Assessment 2025 Page 11 of 17

7(a) mu2 B1 Both equations correct. R = Use of N2L towards centre at A and B. A a May see u2 = 6ag already substituted. mv2 5 R B = a −mgcos May see k or cos. mu2 mv2 *M1 Use of R = 1R . = −mgcos leading to v2 = 1u2 +agk. B 2 A 2a a 2 Condone use of R = 1R for this mark. A 2 B Two terms required in their expression for R . B May see u2 = 6ag already substituted. 5 May see k or cos. 1mu2 = 1mv2 +magk leading to v2 =u2 −2agk . *M1 Energy equation, dimensionally correct. 2 2 6ag−2agk = 3ag+agk DM1 Dependent on both previous M marks. 5 5 Combine N2L and energy equations to obtain an equation that simplifies to involve k only. k = 1 A1 5 5 7(b) v2 =0 leading to u2 = 2ag u= 2ag B1 FT FT their 2agk .  5  5 1 © Cambridge University Press & Assessment 2025 Page 15 of 17 7(c) 2vsin M1 Complete method to find v2. Vertically B to C: t = g Must use 2asin and not 2a. May quote range formula. 2vsin Horizontally B to C: 2asin=vcost =vcos FT their k from part 7(a). g cos=their k, sin= 1−their k2 ag A1 ag v2 = May seev2 = or v2 =5ag . cos k ag M1 Use their energy equation from part 7(a) u2 = +2agk cos ( v2 =u2 −2agk ) . Must substitute for v2. u2 = 27ag, u= 27ag A1 CAO 5 5 Alternative method for question 7(c) Equation of trajectory: (M1) Complete method to find v2.  ( )2 Must use 2asin and not 2a. 0=2asin (tan)− g(2asin)2  0= 4 6a ( 2 6 ) − g 5 4 6  FT their k from part 7(a). 2(vcos)2   5 2 2 5 v2   cos=their k, sin= 1−their k2   ag A1 ag v2 = May seev2 = or v2 =5ag . cos k ag M1 Use their energy equation from part 7(a) u2 = +2agk cos ( v2 =u2 −2agk ) . Must substitute for v2. u2 = 27ag, u= 27ag A1 CAO 5 5 © Cambridge University Press & Assessment 2025 Page 16 of 17 7(c) Alternative method for question 7(c) gx2 dy gx M1 Complete method to find v2. y=xtan− leading to =tan− 2v2cos2 dx v2cos2 Must use asin and not a. FT their k from part 7(a). gasin cos=their k, sin= 1−their k2 Reaches maximum when x=asin: 0=tan− v2cos2 ag A1 ag v2 = May seev2 = or v2 =5ag . cos k ag M1 Use their energy equation from part 7(a) u2 = +2agk cos ( v2 =u2 −2agk ) . Must substitute for v2. u2 = 27ag, u= 27ag A1 CAO 5 5 4 © Cambridge University Press & Assessment 2025 Page 17 of 17

3(a) 32x M1 λx 8= leading to x=0.5. Use T = to find x. May be implied by 2 l correctly derived value for sin=0.6. Hence, PA =2.5 and sin=1.5 =0.6. A1 AG, shown convincingly. 2.5 2 3(b) ( ) *M1 For either equation (resolving horizontally or  R=1.6g−8cos vertically). Allow sin/cos wrong way round. () mr2 =8sin+R Correct number of terms and correct signs. Dimensionally correct. Numbers or algebraic. A1 Both equations correct. 8(3)+0.5 ( 1.6g−8(4)) =1.61.52 DM1 Eliminate R to obtain equation involving 2 5 5 with no other unknown quantities. 2 =4 leading to =2 A1 60 A1 19.0985… Number of revolutions per minute = =19.1 . Accept final answer of 19. π 60 Allow final answer of . π 5 © Cambridge University Press & Assessment 2025 Page 11 of 17

7(a) mu2 B1 Both equations correct. R = Use of N2L towards centre at A and B. A a May see u2 = 6ag already substituted. mv2 5 R B = a −mgcos May see k or cos. mu2 mv2 *M1 Use of R = 1R . = −mgcos leading to v2 = 1u2 +agk. B 2 A 2a a 2 Condone use of R = 1R for this mark. A 2 B Two terms required in their expression for R . B May see u2 = 6ag already substituted. 5 May see k or cos. 1mu2 = 1mv2 +magk leading to v2 =u2 −2agk . *M1 Energy equation, dimensionally correct. 2 2 6ag−2agk = 3ag+agk DM1 Dependent on both previous M marks. 5 5 Combine N2L and energy equations to obtain an equation that simplifies to involve k only. k = 1 A1 5 5 7(b) v2 =0 leading to u2 = 2ag u= 2ag B1 FT FT their 2agk .  5  5 1 © Cambridge University Press & Assessment 2025 Page 15 of 17 7(c) 2vsin M1 Complete method to find v2. Vertically B to C: t = g Must use 2asin and not 2a. May quote range formula. 2vsin Horizontally B to C: 2asin=vcost =vcos FT their k from part 7(a). g cos=their k, sin= 1−their k2 ag A1 ag v2 = May seev2 = or v2 =5ag . cos k ag M1 Use their energy equation from part 7(a) u2 = +2agk cos ( v2 =u2 −2agk ) . Must substitute for v2. u2 = 27ag, u= 27ag A1 CAO 5 5 Alternative method for question 7(c) Equation of trajectory: (M1) Complete method to find v2.  ( )2 Must use 2asin and not 2a. 0=2asin (tan)− g(2asin)2  0= 4 6a ( 2 6 ) − g 5 4 6  FT their k from part 7(a). 2(vcos)2   5 2 2 5 v2   cos=their k, sin= 1−their k2   ag A1 ag v2 = May seev2 = or v2 =5ag . cos k ag M1 Use their energy equation from part 7(a) u2 = +2agk cos ( v2 =u2 −2agk ) . Must substitute for v2. u2 = 27ag, u= 27ag A1 CAO 5 5 © Cambridge University Press & Assessment 2025 Page 16 of 17 7(c) Alternative method for question 7(c) gx2 dy gx M1 Complete method to find v2. y=xtan− leading to =tan− 2v2cos2 dx v2cos2 Must use asin and not a. FT their k from part 7(a). gasin cos=their k, sin= 1−their k2 Reaches maximum when x=asin: 0=tan− v2cos2 ag A1 ag v2 = May seev2 = or v2 =5ag . cos k ag M1 Use their energy equation from part 7(a) u2 = +2agk cos ( v2 =u2 −2agk ) . Must substitute for v2. u2 = 27ag, u= 27ag A1 CAO 5 5 4 © Cambridge University Press & Assessment 2025 Page 17 of 17

1 Tcos=mg B1 B1 for both equations, SOI. Tsin=m2r '5mg' sin=m2asin M1 Combining and using r=asin. 3 Allow sin/cos mix in resolving. Must have a trigonometric ratio on both sides. 5g A1 OE, CAO = 3a 3 © Cambridge University Press & Assessment 2025 Page 9 of 17

5(a) 1mv2 +mgr cos=1m (3gr ) v2 = 3gr−2grcos M1 Energy equation. All 3 terms. Allow sin/cos error. 2 2 2  2  Allow sign errors. mv2 B1 N2L at B. mgcos= r mgcos=m (3g−2gcos ) M1 Combining to eliminate v. 2 cos= 1,  =60 A1 2 4 5(b) v2 = 1gr B1 FT FT their 60. 2 ( )2 M1 Allow sign error. 0= sin'60' 'v2' −2gh h= 3 r A1 FT FT their 60and their v2. 16 Total height = 3 r+rcos60=11r A1 16 16 4 © Cambridge University Press & Assessment 2025 Page 13 of 17

4 For B: T =mg B1 Must be seen explicitly. ( ) *M1 Allow sin/cos mix, allow sign error. For A:  Rsin=Tcos+mg Allow this mark with mg instead of T. (→) Rcos+Tsin=mr2 *M1 Allow sin/cos mix, allow sign error. Allow this mark with mg instead of T. r=htan B1 Seen anywhere. Combine equations to find an expression for 2 in terms of g and DM1 mg(1+cos) May see R=3mg or R= . h only. sin g A1 If T =mg is not seen explicitly, maximum possible mark is =2 h B0 M1 M1 B1 DM1 A0. 6 © Cambridge University Press & Assessment 2025 Page 12 of 17

7 Energy, highest to lowest: M1 Note that assuming the same speed at highest and lowest 1mu2 −1mv2 =2amg v2 =u2 −4ag points could give the correct final answer but will not score 2 2   full marks. Energy, start to lowest: One energy equation, all terms present, dimensionally correct, 1mu2 −1m(4ag)=mg(a−acos) u2 =ag(6−2cos) allow sign error. 2 2   M1 A second energy equation, all terms present, dimensionally Energy, start to highest, correct, allow sign error 1m(4ag)−1mv2 =mg(a+acos) v2 =2ag(1−cos) 2 2   A1 One fully correct energy equation. mu2 B1 Use of N2L at the lowest point, may see T =mg(7−2cos). At lowest point, T −mg = L L a mv2 B1 Use of N2L at the highest point, may see T =mg(1−2cos). At highest point, T +mg = H H a Must use a different speed. T =11T B1 mu2 L H +mg a SOI, must be correct way around. For example, =11. mv2 −mg a Combine to find equation involving cos only. M1 Must see an energy equation involving cos only. cos= 1 A1 5 8 © Cambridge University Press & Assessment 2025 Page 16 of 17

2 N cos =mg B1 N2L horizontally and vertically for the first case. 1 1 →N sin=m(rsin) 2 5 4 g=r 1 2 1 1 1 1 N cos =mg B1 N2L horizontally and vertically for the second case. 2 2 →N sin =m(rsin) 2 5 3 g=r 2 2 2 2 2 2 5g r2 M1 Combine to find ratio. 4 = 1 5g r2 3 2  A1 1 = 1 3  2 2 4 © Cambridge University Press & Assessment 2025 Page 10 of 17

6(a) Let v be the speed of P before the collision. M1 Dimensionally correct energy equation with a GPE term and Energy conservation before the collision: two KE terms. 1m(17ag)+3mag=1mv2 2 2 2 A1 v= 20ag 2 6(b) Let w be speed of P after the collision and V be its speed at the top. B1 Dimensionally correct equation. P T = 0 must be seen or implied. mV2 N2L: =(T)+mg V = ag   a Energy equation: 1mw2 =1mV2+2mga M1 Dimensionally correct energy equation with a GPE term and 2 P 2 two KE terms. A1 w = 5ag P 3 6(c) Let w the speed of Q after the collision. B1 Momentum conserved (must see masses). Q mv=kmw −mw Q P w +w =ev =v B1 NEL with consistent signs. P Q OR OR Use conservation of kinetic energy. 1mv2 =1mw2 +1kmw2 2 2 P 2 Q k=3 B1 Must have e=1 seen or implied. 3 © Cambridge University Press & Assessment 2025 Page 15 of 17

1 T cos45+T sin45=m2lsin45 B1 Correct horizontal equation. A B Allow sin/cos mix. B0 B0 M0 A0 for using the same tension in each string. T sin45−T cos45=mg B1 Correct vertical equation. A B Allow sin/cos mix. 2T =m2l+ 2 mg M1 Combine, isolating term in T . A A T = 1m ( 2l+ 2 g ) A1 AEF A 2 4 Question Answer Marks Guidance

6 A to ground: 1mu2 +2mga=1m(3u)2 u2 = 1ga *M1 One energy equation, all terms present. 2 2  2  *M1 Second energy equation, all terms present. A to B: 1mv2+mgacos=1mu2+mga v2 =u2 +2ag(1−cos) 2 2   A1 At least one correct energy equation involving cos . B to ground: 1mv2 +mgacos=1m(3u)2 −mga v2 =9u2 −2ag(1+cos) 2 2   mv2 B1 N2L at B. mgcos= v2 =agcos   a gacos= 1ga+2ag(1−cos) DM1 Eliminate u and v. 2 cos= 5 A1 6 Horizontally at ground: vcos=3ucos. M1 A complete method to find . 5ga (5) cos= 6 6 = ( 5 15 ) 3 1ga 54 2 OR v =vcos= 5ga (5) , v2 =(vsin)2 +2ga(1+cos)=847ga x 6 6 y 216 847ga tan= 216 5ga (5) 6 6 =69.0 A1 8 © Cambridge University Press & Assessment 2025 Page 14 of 17

2 N cos =mg B1 N2L horizontally and vertically for the first case. 1 1 →N sin=m(rsin) 2 5 4 g=r 1 2 1 1 1 1 N cos =mg B1 N2L horizontally and vertically for the second case. 2 2 →N sin =m(rsin) 2 5 3 g=r 2 2 2 2 2 2 5g r2 M1 Combine to find ratio. 4 = 1 5g r2 3 2  A1 1 = 1 3  2 2 4 © Cambridge University Press & Assessment 2025 Page 10 of 17

6(a) Let v be the speed of P before the collision. M1 Dimensionally correct energy equation with a GPE term and Energy conservation before the collision: two KE terms. 1m(17ag)+3mag=1mv2 2 2 2 A1 v= 20ag 2 6(b) Let w be speed of P after the collision and V be its speed at the top. B1 Dimensionally correct equation. P T = 0 must be seen or implied. mV2 N2L: =(T)+mg V = ag   a Energy equation: 1mw2 =1mV2+2mga M1 Dimensionally correct energy equation with a GPE term and 2 P 2 two KE terms. A1 w = 5ag P 3 6(c) Let w the speed of Q after the collision. B1 Momentum conserved (must see masses). Q mv=kmw −mw Q P w +w =ev =v B1 NEL with consistent signs. P Q OR OR Use conservation of kinetic energy. 1mv2 =1mw2 +1kmw2 2 2 P 2 Q k=3 B1 Must have e=1 seen or implied. 3 © Cambridge University Press & Assessment 2025 Page 15 of 17

1 For A: T =mg M1 Both equations. For B vertically: Tcos=0.5mg Allow sin/cos mix. =60 A1 Equate and solve for , SOI. For B horizontally: Tsin=0.5m(BRsin)2 M1 Allow sin/cos mix. Allow their stated BR. BRcos=AR BR=8a B1 SOI 1 g A1 AEF in terms of g and a. = 2 a 5 © Cambridge University Press & Assessment 2025 Page 9 of 18

4(a) mw2 B1 Accept equality. At top R≥0, so ≥mg. a 1mv2−1mw2≥mga(1−cos) M1 Energy inequality with correct number of terms and 2 2 correct dimensions, accept equality. A1 Fully correct energy inequality, accept equality. 1mv2≥13mga M1 Substitute for cos and w2. 2 9 v≥1 26ag A1 AG, shown convincingly. 3 If equality used throughout, introduction of inequality must be convincing. 5 4(b) Maximum reaction is at bottom: M1 Energy equation with correct number of terms and 1mv2+mga(1+cos)=1mV2 V2 =5ag correct dimensions and correct signs. 2 2   Allow sin/cos mix. mV2 B1 R −mg = max a R =6mg A1 max 3 © Cambridge University Press & Assessment 2025 Page 13 of 18

2 (F =)71242 [=1344] B1 Correct unsimplified expression for the radial force. radial Attempt to apply N2L either horizontally or vertically to form an equation. M1 Correct number of terms, dimensionally correct, allow sign errors and sin/cos mix only. Values of sin/cos must be substituted. Allow 37 or better (36.8698…) for the angle between AP and the horizontal and 23 or better (22.6198…) for the angle between BP and the horizontal. 3T = 5T +7g 5 A 13 B © Cambridge University Press & Assessment 2026 Page 11 of 22 (  3 9 T A = 2 5 T B + 4 5 5 0 ) A1 Allow 37 or better (36.8698…) for the angle between AP and the horizontal and 23 or better (22.6198…) for the angle between BP and the horizontal, for example, T A s i n ( 3 7 ) = T B s i n ( 2 3 ) + 7 g . 45 T A + 11 23 T B = 7  1 2  4 2 (  1 3 T A + 1 5 T B = 2 1 8 4 0 ) A1 Allow 36 or better (36.8698…) for the angle between AP and the horizontal and 23 or better (22.6198…) for the angle between BP and the horizontal, for example, T cos(37)+T cos(23)=71242. A B T =675 and T =871 A1 AWRT 675 and 871 only. A B 5 Trace ID: PC-76076FFBDF Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/3a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/31 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/

5(a) 1m ( 2 ag )2 =mga(1−cos)+ 1mv2 2 2 or © Cambridge University Press & Assessment 2026 Page 16 of 22 m g a 12 m ( 2 a g ) 2 m g a c o s 12 m v 2  − +  = − + *M1 Three term energy equation (OE). Must be dimensionally correct. Allow sign errors and sin/cos mix only. Mass must be present. v2 =2agcos+2ag   mv2 B1 N2L radially. T −mgcos= a c o s ( 2 c o s 2 )   T − m g = m a  a g + a g DM1 Combine 2 equations (N2L radially and energy). Their N2L radially can have sign errors and sin/cos mix only. Dimensionally correct. [T =]mg(3cos+2) A1 OE 4 Trace ID: PC-76076FFBDF Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/3a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/31 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 5(b) mg(3cos+2)=0 © Cambridge University Press & Assessment 2026 Page 17 of 22 c o s 23    = −  *M1 Setting their dimensionally correct expression for T from 5(a) to zero (OE from a restart, so may see c o s 23  = being derived). Their expression for T must be two terms only with only one term containing either sine or cosine. v2 =2ag(−2)+2ag v2 = 2ag DM1 Find an expression (possibly unsimplified) for 3  3  v 2 or v when the string goes slack in terms of a (and possibly g). Their expression for v 2 must be two terms only with only one term containing either sine or cosine. DM1 For v = 2ag 1− ( −2)2 1 30ag y 3 3  9  s i n  v following through their v and their value of c o s . If correct, might see (1.9245…)a. Dependent on both previous method marks. 0= ( 1 30ag )2 +2(−g)s 9 y   s y = 52 7 a  DM1 For use of v 2 = u 2 + 2 a s (or equivalent complete method) with v = 0 , a = − g and their v y for u. If correct, might see (0.1851…)a. Dependent on all previous method marks. 50a A1 Condone 1.85a or better (1.85185…)a or 27 52 7 a g . Dependent on full marks in part (a) (or a fully correct restart in this part). 5 Trace ID: PC-76076FFBDF Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/3a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/31 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/

2 © Cambridge University Press & Assessment 2026 Page 11 of 22 ( F ra d ia l = ) 7  1 2  4 2 [ = 1 3 4 4 ] B1 Correct unsimplified expression for the radial force. Attempt to apply N2L either horizontally or vertically to form an equation. M1 Correct number of terms, dimensionally correct, allow sign errors and sin/cos mix only. Values of sin/cos must be substituted. Allow 37 or better (36.8698…) for the angle between AP and the horizontal and 23 or better (22.6198…) for the angle between BP and the horizontal. 35 T A = 51 3 T B + 7 g (  3 9 T A = 2 5 T B + 4 5 5 0 ) A1 Allow 37 or better (36.8698…) for the angle between AP and the horizontal and 23 or better (22.6198…) for the angle between BP and the horizontal, for example, T sin(37)=T sin(23)+7g. A B 45 T A + 11 23 T B = 7  1 2  4 2 (  1 3 T A + 1 5 T B = 2 1 8 4 0 ) A1 Allow 36 or better (36.8698…) for the angle between AP and the horizontal and 23 or better (22.6198…) for the angle between BP and the horizontal, for example, T cos(37)+T cos(23)=71242. A B T A = 6 7 5 and T =871 A1 AWRT 675 and 871 only. B 5 Trace ID: PC-852B15CDF8 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/3a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/32 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/

5(a) © Cambridge University Press & Assessment 2026 Page 16 of 22 12 m ( 2 a g ) 2 m g a (1 c o s ) 12 m v 2   = − + *M1 Three term energy equation (OE). Must be dimensionally correct. or −mga+ 1m ( 2 ag )2 =−mgacos+ 1mv2 Allow sign errors and sin/cos mix only. 2 2 Mass must be present. v2 =2agcos+2ag   c o s 2  T − m g = m v a B1 N2L radially. c o s ( 2 c o s 2 )   T − m g = m a  a g + a g DM1 Combine 2 equations (N2L radially and energy). Their N2L radially can have sign errors and sin/cos mix only. Dimensionally correct. [ ] ( 3 c o s 2 )  T = m g + A1 OE 4 Trace ID: PC-852B15CDF8 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/3a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/32 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 5(b) © Cambridge University Press & Assessment 2026 Page 17 of 22 ( 3 c o s 2 ) 0  m g + = c o s 23    = −  *M1 Setting their dimensionally correct expression for T from 5(a) to zero (OE from a restart, so may see c o s 23  = being derived). Their expression for T must be two terms only with only one term containing either sine or cosine. v 2 = 2 a g  ( − 23 ) + 2 a g   v 2 = 23 a g  DM1 Find an expression (possibly unsimplified) for v 2 or v when the string goes slack in terms of a (and possibly g). Their expression for v2 must be two terms only with only one term containing either sine or cosine. v y = 23 a g  1 − ( − 23 ) 2   19 3 0 a g  DM1 For s i n  v following through their v and their value of c o s . If correct, might see (1.9245…)a. Dependent on both previous method marks. 0 = ( 19 3 0 a g ) 2 + 2 ( − g ) s y   s y = 52 7 a  DM1 For use of v 2 = u 2 + 2 a s (or equivalent complete method) with v = 0 , a = − g and their v y for u. If correct, might see (0.1851…)a. Dependent on all previous method marks. 50a A1 Condone 1.85a or better (1.85185…)a or 5 ag. 27 27 Dependent on full marks in part (a) (or a fully correct restart in this part). 5 Trace ID: PC-852B15CDF8 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/3a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/32 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/

1  Rcos=mg B1 Resolve vertically. © Cambridge University Press & Assessment 2026 Page 9 of 20 R s i n m ( a s i n ) 2    = B1 Resolve horizontally. B1 for 2  R = m a (with s i n  cancelled). B0 for s i n 2   R = m r unless r replaced with s i n  a later in working. B1 for s i n 3 5 2   R = m a . Sight of t a n ( s i n ) 2   m g = m a earns B1 B1. g ( a c o s ) 2  = g or 2 = or a cos c o s a g 2   = M1 OE, eliminate R and m to get an expression in 2  . Must be obtained from equations with sin/cos mix only. t a n 13 5 g a 2   = 13 a a a c o s  = − , c o s 23  = B1 12 5 13 5 g a 2  = 3 2  = g a A1 OE, for example, 1 .2 2 g a . 5 Trace ID: PC-202C151F9E Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/3a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/33 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/

3 At B, © Cambridge University Press & Assessment 2026 Page 12 of 20 c o s 2  m g = m v a B1 N2L radially. If their equation contains R, only award this mark when R is set to 0. 1mv2−1mu2 =mga(cos−cos) *M1 Energy equation with 4 terms and correct dimensions (mass 2 2 must be present). Allow sign errors and sin/cos mix only. A1 Fully correct energy equation. a g c o s ( 23 ) 2 a g c o s 2 a g ( c o s c o s )     − = − DM1 Combine their two equations and substitute u = 23 v to obtain an equation in, for example, c o s  and c o s  only. Allow sin/cos mix only in their N2L equation radially. May, instead, see the following: c o s ( 23 ) 2 c o s 2 ( c o s c o s )     − = − c o s 49 c o s 2 ( c o s c o s )     − = − c o s : c o s 2 3 : 1 8   = A1 OE 5 Trace ID: PC-202C151F9E Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/3a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/33 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/

1 © Cambridge University Press & Assessment 2026 Page 9 of 25 T c o s m g  = B1 m(2 ag)2 B1 m(2 ag)2 Tsin= Do not accept Tsin= , r=asin asin r until both combined with r eliminated. c m o g s s i n m ( 2 a s i a n g ) 2     = M1 Eliminating T from their dimensionally correct horizontal and vertical equations. Their horizontal and vertical equations can contain sin/cos mix only (therefore their r must be either asin or a c o s  only). 1−cos2=4cos cos=−2+ 5 ONLY A1 WWW OE, accept AWRT 0.236 (0.2360679…).   No working is required to solve the quadratic equation, but if present then it must be correct. If c o s 2 5  = −  , then c o s 2 5  = − − must be rejected (but no reasoning is required for rejection). 4 Trace ID: PC-051B44EFC7 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/3a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/34 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/

3(a) © Cambridge University Press & Assessment 2026 Page 12 of 25 T A + m g c o s 6 0 = m u a 2 M1 N2L radially at A (radius must be a) with correct number of dimensionally correct terms. Allow sin/cos mix and sign errors only. Angle must be substituted. m(2u)2 B1 Correct application of N2L radially at B. T −mg = B a v = 2 u must be substituted. Radius must be a. 12 m  ( 2 u ) 2 − 12 m u 2 = m g a (1 + c o s 6 0 ) *M1 Three-term energy equation (OE) for the motion at A and at B. Must be dimensionally correct.   3 2 u2 = 3 2 gau2 =ga  Allow sign errors and sin/cos mix only. Mass must be present. Angle must be substituted. T A + m g c o s 6 0 = m a  g a   T A = 12 m g  or T B − m g = m  ( 2 a g a ) 2   T B = 5 m g  DM1 Substitute their expression for u (which must be in terms of a and possibly g) into either of their N2L equations to obtain an equation containing T or T (not both). Their N2L must contain the A B correct number of dimensionally correct terms (with sin/cos mix and sign errors only). T A = 12 m g or T A = 5 m or T B = 5 m g or T B = 5 0 m A1 For either tension correct in terms of m (and possibly g) or, for example, 2 T A = 1 0 m . k = 1 0 A1 WWW, accept T B = 1 0 T A OE. 6 3(b) [Loss in KE =] 2mga B1 WWW, if KE terms seen then K E b o tto m = 2 m g a and K E to p = 0 . 1 Trace ID: PC-051B44EFC7 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/3a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/34 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/