Linear motion under a variable force

1 dv 10dt M1 Separate variables. =− v ( t+1 )2 10 M1 A1 Attempt to integrate. 2 v = + A t+1 t =0, v=25, A=0 M1 Use correct initial condition. 25 A1 CAO v= ( t+1 )2 5 © UCLES 2021 Page 6 of 13

1 dv 10dt M1 Separate variables. =− v ( t+1 )2 10 M1 A1 Attempt to integrate. 2 v = + A t+1 t =0, v=25, A=0 M1 Use correct initial condition. 25 A1 CAO v= ( t+1 )2 5 © UCLES 2021 Page 6 of 13

5(a) dv B1 Use of SUVAT implies 0 marks. m =−mg−2mv dt N2L, must include . m ( ) ( ) ln 5+v =−2t +A M1 Separate variables and integrate 3-term N2L, condone omission of constant. ( ) ln 5+v =−2t +A A1 FT FT only sign error in N2L. t=0, v=20, A=ln25 M1 Use correct initial condition. 2t =ln   5 2 + 5 v    , e2t = 5 2 + 5 v M1 Remove all logs. v = 25e−2t −5 A1 Alternative method for question 5(a) dv B1 N2L, must include . m =−mg−2mv m dt dv M1 +2v=−g : Integrating factor = e2t dt ( ) d ve2t M1 Integrate both sides, condone omission of constant. =−ge2t, ve2t =− g e2t (+A ) dt 2 g A1 FT FT only sign error in N2L. ve2t =− e2t +A 2 © UCLES 2021 Page 11 of 16 5(a) t =0, v=20, A=25 M1 Use correct initial condition. g A1 ve2t =− e2t +25, v=25e −2t −5 2 6 5(b) x=− 25 e −2t −5t (+B ) M1 Use of SUVAT implies 0 marks. Integrate their expression from part (a). 2 t=0, x=0, B= 25 A1 FT FT only expressions of the form v=Pekt +Q for P, Q non-zero. 2 25( ) x= 1−e −2t −5t 2 2 5(c) 1 M1 Use of SUVAT in part (a) or part (b) implies 0 marks. Greatest height when v = 0, so t = 0.8047…or ln5 Find value of t, may be embedded. 2 x = 5.98 m A1 CWO 2 © UCLES 2021 Page 12 of 16

2(a) Separate variables and integrate: M1 A1 Must include logs. Condone missing modulus. dv 1−2t2  =  dt so ln v =lnt−t2 +c v t   v = Ate −t2 , −v= Ate −t2 , v=−Ate −t2 A1 CAO. 3 © UCLES 2021 Page 6 of 12 2(b) −Ate −t2 1−2t2 ( ) M1 Substituting their answer to part (a) into given formula a= =−Ae −t2 1−2t2 t t =1, a=5 ( A=5e ) M1 Use initial condition. ( ) Force = 5me1−t2 2t2 −1 A1 Use N2L, correct work only. Alternative method for question 2(b) ( ) v 1−2t2 M1 Use initial condition. a= substitute t=1, a=5 so v=−5 Use N2L, correct work only. t Substituting in their answer to part (a) so ( A=5e ) M1 ( ) Force = 5me1−t2 2t2 −1 A1 3 © UCLES 2021 Page 7 of 12

6(a) dv 128 M1 Separate variables and integrate, + c not required for M1. 2v =8x− dx x3 v2 =4x2 +64x −2 +c A1 OE. x = 8, v = ‒15 and c = ‒32 M1 Use initial condition. v2 = 4 ( x4 −8x2 +16 ) or 4x2 −32+ 64 A1 Correct expression for v2, AEF. x2 x2 v2 = 4 ( x2 −4 )2 giving v=− 2( x2 −4 ) A1 Convincingly shown, e.g. v is negative initially, AG. x2 x 5 6(b) 1 ln(x2 −4)=−2t (+A ) M1 Use v= dx and integrate. 2 dt 1 DM1 Use initial condition. t=0, x=8, A= ln60 2 1 x2 −4 x2 −4 M1 Remove log. ln  =−2t giving =e −4t 2  60  60 x= 4+60e −4t A1 CAO 4 © UCLES 2021 Page 11 of 12

2(a) Separate variables and integrate: M1 A1 Must include logs. Condone missing modulus. dv 1−2t2  =  dt so ln v =lnt−t2 +c v t   v = Ate −t2 , −v= Ate −t2 , v=−Ate −t2 A1 CAO. 3 © UCLES 2021 Page 6 of 12 2(b) −Ate −t2 1−2t2 ( ) M1 Substituting their answer to part (a) into given formula a= =−Ae −t2 1−2t2 t t =1, a=5 ( A=5e ) M1 Use initial condition. ( ) Force = 5me1−t2 2t2 −1 A1 Use N2L, correct work only. Alternative method for question 2(b) ( ) v 1−2t2 M1 Use initial condition. a= substitute t=1, a=5 so v=−5 Use N2L, correct work only. t Substituting in their answer to part (a) so ( A=5e ) M1 ( ) Force = 5me1−t2 2t2 −1 A1 3 © UCLES 2021 Page 7 of 12

3 dv 4000 400 M1 A1 Integrate.  ; v A dt 5t43 5t42 Constant of integration needed for A1. t 0, v25 A25250 M1 Find constant. dx M1 Integrate and find constant. v : x805t41 B dt x0, t0 B20 80  100t  A1 x 20   5t4  5t4 5 © UCLES 2022 Page 7 of 14

3 dv 4000 400 M1 A1 Integrate.  ; v A dt 5t43 5t42 Constant of integration needed for A1. t 0, v25 A25250 M1 Find constant. dx M1 Integrate and find constant. v : x805t41 B dt x0, t0 B20 80  100t  A1 x 20   5t4  5t4 5 © UCLES 2022 Page 7 of 14

5(a) 4v dv   4ex 12  ex B1 dx 1 1 M1 Expression of the correct form. v2  e2x 3exA 2 2 9 A1 v4, x0, A 2 v2 e2x 6ex 9  3ex2 A1 AG Must see the factorisation. 13ex Condone lack of justification for taking positive square root. v3ex  ex 4 5(b) dx 13ex ex M1* A1 Integration to obtain ln term  so  dx1 dt Correct answer with constant of integration dt ex 3ex 1 1 ln  3ex 1  tB 3 1 DM1 Find the constant and substitute into their general solution. t 0, x0, B ln4 3 3ex 1 3t ln 4 4 1 A1 OE xln e3t   3 3 4 © UCLES 2022 Page 10 of 13

4(a) m dv =50−2v2 dv =4 ( 25−v2) B1 N2L dt dt 1 1 1 M1 Separate variables and use partial fractions.  + dv=4dt 10 5−v 5+v 1 M1 A1 Integrate into log terms. (−ln(5−v)+ln(5+v))=4t+A (Note: formula on MF19). 10 1 M1 Use initial condition. Use t=0, v=3 to give A= ln4 10 1 5+v 5+v M1 Rearrange to make v the subject. 4t = ln leading to =e40t 10 4(5−v) 20−4v 5 ( 4−e−40t) A1 v= 4+e−40t 7 4(b) As t→, v→5 B1 1 © UCLES 2022 Page 8 of 12

4(a) dv 500 1 B1 Sight of m or 5 is required. 5v = − v2 dx v 2 10v2dv M1 Separate variables and attempt to integrate into a log term. =dx 1000−v3 − 10 ln ( 1000−v3) =x(+A) A1 3 10 M1 Evaluate constant: correct initial condition used. x=0, v=5, A=− ln875 3 10 875 M1 Make v the subject: correct use of logs. x= ln 3 1000−v3 1 A1 1 v=(1000−875e−0.3x3 v=5(8−7e−0.3x3 : A0 if eln terms.     6 4(b) Maximum value of v is 10 B1 No FT: result can be found from initial equation. 1 © UCLES 2022 Page 10 of 15

4(a) m dv =50−2v2 dv =4 ( 25−v2) B1 N2L dt dt 1 1 1 M1 Separate variables and use partial fractions.  + dv=4dt 10 5−v 5+v 1 M1 A1 Integrate into log terms. (−ln(5−v)+ln(5+v))=4t+A (Note: formula on MF19). 10 1 M1 Use initial condition. Use t=0, v=3 to give A= ln4 10 1 5+v 5+v M1 Rearrange to make v the subject. 4t = ln leading to =e40t 10 4(5−v) 20−4v 5 ( 4−e−40t) A1 v= 4+e−40t 7 4(b) As t→, v→5 B1 1 © UCLES 2022 Page 8 of 12

6(a) dv M1 v 6v v9 and attempt to separate variables and integrate dx 2 v9 6xA A1 x2, v72 A6 M1 Use initial condition to find constant. v9x12 9 A1 Correct, AEF. 4 6(b)    d d x t 9  x2 2x  , xx dx 2 9dt    1 2    1 x  x 1 2    dx9dt M1 S   e a pa  rate b va   r d ia x b  le d s t a n d write in the form  x xc 1  x  A1 Integrate, any correct form. ln 9tB 2 x2 1 1 M1 Use initial condition to find constant. t 0, x2 B ln 2 2  2x  2x M1 Take logarithms 18t ln  e18t  x2 x2 2e18t 2 A1 Any correct form x or x 2e18t 2e18t 1 5 © UCLES 2023 Page 13 of 15

6(a) dv M1 v 6v v9 and attempt to separate variables and integrate dx 2 v9 6xA A1 x2, v72 A6 M1 Use initial condition to find constant. v9x12 9 A1 Correct, AEF. 4 6(b)    d d x t 9  x2 2x  , xx dx 2 9dt    1 2    1 x  x 1 2    dx9dt M1 S   e a pa  rate b va   r d ia x b  le d s t a n d write in the form  x xc 1  x  A1 Integrate, any correct form. ln 9tB 2 x2 1 1 M1 Use initial condition to find constant. t 0, x2 B ln 2 2  2x  2x M1 Take logarithms 18t ln  e18t  x2 x2 2e18t 2 A1 Any correct form x or x 2e18t 2e18t 1 5 © UCLES 2023 Page 13 of 15

6(a) dv B1 Mass must be seen at this point or earlier. m mg1kv [SUVAT does not apply.] dt 1 M1 Separate variables and integrate to logarithm.  ln1kvgt A k A1 Correct, with constant of integration. t0, v0 [A0] M1 Use initial condition to evaluate their constant. v 1 1ekgt A1 Any correct form with v as subject. Final A0 if numerical value of g present. k 5 © UCLES 2023 Page 11 of 14 6(b) k 0.05 and so dx 20  1e0.5t *M1 Attempt to integrate if expression contains a term of dt the form bect . Integrate: x20  t2e0.5t B A1 1 1  x t ekgt B k gk  t0, x0 B40 DM1 Use initial condition to evaluate their constant. When v12, from part (a), e0.5t 10.05120.4, t2ln0.4 M1 1.83… x40ln0.4400.440 12.7 A1 5 40ln 24 2 Alternative method for question 6(b) dv  1  *M1 Separate variables and write in integrable form v g1kv leading to  1  dvkgdx dx  1kv 1 DM1 A1 Dependent on previous M1. Attempt to integrate. v ln1kvkgxB k v0, x0 B0 and k 0.05, v12 1 220ln0.40.5x M1 Dependent on both previous M1s. Use initial condition to evaluate their constant and use v12 x12.7 A1 5 40ln 24 2 5 © UCLES 2023 Page 12 of 14

2(a) dv B1 2 =2g−0.2v2 dt dv M1 Integrate to a ln term of the correct form. Separate variables and attempt to integrate =dt 0.1 ( 100−v2) 1 10+v A1 ln   =0.1t+c 20 10−v 1 M1 Use initial condition. t=0, v=5, c= ln3 20 10+v 10+v M1 Rearrange, removing ln. 2t =ln , e2t = 3(10−v) 3(10−v) 30−10e−2t A1 AEF v= 3+e−2t 6 2(b) v→10 B1FT FT from expression of correct form. 1 © UCLES 2023 Page 7 of 13

2 dv 150 450 M1 Allow sign errors. 0.5v = − dx (x+1)2 (x+1)3 300 450 M1A1 Correct powers, allow sign errors. Integrate: 0.5v2 =− + + A x+1 (x+1)2 x=0, v=20; A=50 M1 Use initial condition. 100(x2 −4x+4) A1 AEF Rearrange: v2 = (x+1)2 100(x−2)2 10(x−2) A1 v2 = so v= (x+1)2 (x+1) 20−10x From initial condition, sign must be negative, v= Signs dealt with convincingly. x+1 6 © UCLES 2023 Page 6 of 15

2(a) dv B1 2 =2g−0.2v2 dt dv M1 Integrate to a ln term of the correct form. Separate variables and attempt to integrate =dt 0.1 ( 100−v2) 1 10+v A1 ln   =0.1t+c 20 10−v 1 M1 Use initial condition. t=0, v=5, c= ln3 20 10+v 10+v M1 Rearrange, removing ln. 2t =ln , e2t = 3(10−v) 3(10−v) 30−10e−2t A1 AEF v= 3+e−2t 6 2(b) v→10 B1FT FT from expression of correct form. 1 © UCLES 2023 Page 7 of 13

6(a) dv 1 dv 1 *M1 Separate variables and attempt to integrate both 2  2v12 et so  etdt dt 10 2v12 20 sides. Where p and q are constants. p qet  A 2v1 1 1 A1 AEF   et  A 22v1 20  3  DM1 Substituting the boundary condition and obtain a t0, v3, A    value.  20 1 5et *M1 A1 Find v in terms of t . AEF. v  2 3et 1 5 6(b) Integrate: x ptqln(ret s) B *M1 1 5 A1 AEF x t ln(3et 1) B 2 3 5 DM1 Substituting the boundary condition and obtain a t 0, x1, B1 ln2 value. 3 1 5 (3et 1) A1 AEF x1 t ln 2 3 2 4 © Cambridge University Press & Assessment 2024 Page 11 of 13

6(a) dv 1 dv 1 *M1 Separate variables and attempt to integrate both 2  2v12 et so  etdt dt 10 2v12 20 sides. Where p and q are constants. p qet  A 2v1 1 1 A1 AEF   et  A 22v1 20  3  DM1 Substituting the boundary condition and obtain a t0, v3, A    value.  20 1 5et *M1 A1 Find v in terms of t . AEF. v  2 3et 1 5 6(b) Integrate: x ptqln(ret s) B *M1 1 5 A1 AEF x t ln(3et 1) B 2 3 5 DM1 Substituting the boundary condition and obtain a t 0, x1, B1 ln2 value. 3 1 5 (3et 1) A1 AEF x1 t ln 2 3 2 4 © Cambridge University Press & Assessment 2024 Page 11 of 13

7(a) dv B1 No marks in this part if suvat used. m m10v dt Must have sight of m (for example in F = ma). ln10v t A or lnv10t A *M1 A1 Separate variables and integrate to obtain a ln term. Constant may be omitted. Constant needed for A1 Use t 0, v50: Aln 40 DM1 Find constant, dependent on previous M1. May use limits instead. 0.1vet 4et M1 Remove all logs . v1040et A1 Correct work only . 6 7(b) x10t40et B M1 No marks in this part if suvat used in part (a) or part (b). Integrate their answer to part (a). Constant may be omitted. Use t 0, x0: B40 M1 Use initial condition in their expression for x in terms of t . x10t40et 40 A1 3 © Cambridge University Press & Assessment 2024 Page 14 of 15 7(c) 1 M1 No marks in this part if suvat used in part (a) part When v15, et  , t 2.08 or ln 8 (b) or part (c). 8 Find value of t from their answer to part (a). x 55.8 (metres) A1 Note 35 + 10 ln 8 scores A0. 2 © Cambridge University Press & Assessment 2024 Page 15 of 15

5(a) dx  4  M1 Separate variables, obtain RHS in integrable form. = −1dt x t+1  ln x =4ln t+1−t+ A A1 t=0, x=5: A=ln5 M1 x=5(t+1)4 e−t A1 4 5(b) v=(3−t)5(t+1)3 e−t M1 Acceleration = dv =5e−t ( −(t+1)3 +(3−t)3(t+1)2 −(3−t)(t+1)3) dt Acceleration = 5e−t(t+1)2(5−t)(1−t) AEF F=2acceleration, so at F =10e−t(t+1)2(5−t)(1−t) M1 At t=3, magnitude of force is 640e−3 N A1 31.9 N 3 © Cambridge University Press & Assessment 2024 Page 9 of 12

7(a) dv B1 Use of suvat means 0 marks in this part m =mg−0.1mv2 Note that no mg term means 0 marks in this part. dt dv Must see m, may be cancelled before a = used dt dv =10−0.1v2 = 1 ( 100−v2) M1* dt 10 dv 1 = dt 100−v2 10 Separate variables and integrate. May see partial fractions, but integral is on Formula sheet, v+10 allow missing +A for M1 only ln   =2t+A 10−v A1 Must see modulus sign Use t=0, v=0, A=0 DM1 Remove logs to obtain v in terms of t M1 10 ( e2t −1 ) A1 10 ( 1−e−2t) v= aef v= e2t +1 1+e−2t 6 © Cambridge University Press & Assessment 2024 Page 18 of 19 7(b) v dv = 1 ( 100−v2) M1* Use of suvat means 0 marks in this part dx 10 Separate variables and integrate, allow missing +A for M1 only. 1 ( ) 1 − ln 100−v2 = x+B 2 10 A1 For A1, allow missing modulus sign 1 DM1 Use x=0, v=0, B=− ln100 2 Remove logs to obtain v2 in terms of x M1 x A1 AEF − v2 =100(1−e 5) Allow 10g instead of 100. 5 © Cambridge University Press & Assessment 2024 Page 19 of 19

5(a) dx  4  M1 Separate variables, obtain RHS in integrable form. = −1dt x t+1  ln x =4ln t+1−t+ A A1 t=0, x=5: A=ln5 M1 x=5(t+1)4 e−t A1 4 5(b) v=(3−t)5(t+1)3 e−t M1 Acceleration = dv =5e−t ( −(t+1)3 +(3−t)3(t+1)2 −(3−t)(t+1)3) dt Acceleration = 5e−t(t+1)2(5−t)(1−t) AEF F=2acceleration, so at F =10e−t(t+1)2(5−t)(1−t) M1 At t=3, magnitude of force is 640e−3 N A1 31.9 N 3 © Cambridge University Press & Assessment 2024 Page 9 of 12

1(a) dv B1 Correct use of N2L. x3 +4x=8v dx (x3+4x) dx=8v dv M1 Attempt to integrate by separating variables. dv Must have v . 4v2 = 1x4 +2x2 +c dx 4 Must include +c. v2 = 1 x4 + 1x2 +1  = ( 1x2 +1 )2 A1 Use initial conditions to find c and rearrange to 16 2  4  obtain an expression for v in terms of x. v= 1x2 +1 x2 +4 4 Condone v= . 4 Condone omitted explanation for choosing positive v. 3 1(b) dx  1 M1 Separate variables and integrate totan−1 form. = 1x2 +1 leading to  dx=dt . dt 4  1x2 +1 Only FT their v of the form ax2 +b. 4 t=4   1 dx=2tan−1(1x )+C  x2 +4 2 C=0 so x=2tan (1t ) A1 Must have +C and statement that C=0. 2 2 © Cambridge University Press & Assessment 2025 Page 9 of 17

1(a) dv B1 Correct use of N2L. x3 +4x=8v dx (x3+4x) dx=8v dv M1 Attempt to integrate by separating variables. dv Must have v . 4v2 = 1x4 +2x2 +c dx 4 Must include +c. v2 = 1 x4 + 1x2 +1  = ( 1x2 +1 )2 A1 Use initial conditions to find c and rearrange to 16 2  4  obtain an expression for v in terms of x. v= 1x2 +1 x2 +4 4 Condone v= . 4 Condone omitted explanation for choosing positive v. 3 1(b) dx  1 M1 Separate variables and integrate totan−1 form. = 1x2 +1 leading to  dx=dt . dt 4  1x2 +1 Only FT their v of the form ax2 +b. 4 t=4   1 dx=2tan−1(1x )+C  x2 +4 2 C=0 so x=2tan (1t ) A1 Must have +C and statement that C=0. 2 2 © Cambridge University Press & Assessment 2025 Page 9 of 17

3(a) dv B1 Correct use of N2L with m seen. −mg−mkv2 =mv dx dx=−   v dv leading to x=− 1 ln ( g+kv2)+c. M1 A Co tt r e r m ec p t t f t o o r m in . t e A g l r l a o t w e b m y i s s s e i p n a g r a + ti c ng . variables.  g+kv2 2k v=U when x=0, so c= 1 ln ( g+kU2) . M1 2k 1 g+kU2  A1 AG, shown convincingly. x= ln  (metres) 2k  g+kv2  4 3(b) dv *M1 Attempt to integrate by separating variables. 10+ 1 v2 =− 40 dt  40 dt=− dv 400+v2 t=−40tan−1( 1 v )(+C) A1 Correct integration. 20 20 When t =0,v=20, so C = 1π DM1 Find C and substitute v=0 to find value for t. 2 t =−2tan−1(0)+ 1π 2 t=1.57 (seconds) A1 CWO, accept final answer of 1π. 2 4 © Cambridge University Press & Assessment 2025 Page 11 of 17

2(a) dv M1 N2L (m must be seen, possibly as F = ma), no extra terms. m =−2mv3 Attempt to integrate by separating variables, allow sign errors. dt v−3dv=−2dt Allow errors in coefficient of v−2. Allow missing +A. −1v−2 =−2t+ A A1 Any correct equivalent form. 2 When t =0, v=1 leading to A=−1 . M1 Use correct initial conditions to find value for A. 2 1 A1 1 v= AEF, A0 for  . 4t+1 4t+1 SC B2 for complete solution except for m not seen. 4 2(b) dx 1 M1 FT FT their answer to part 2(a). = so x= 1 4t+1 [+B] dt 4t+1 2 Allow missing +B for this mark. When t =0, x=0 leading to B=−1 . M1 Use correct initial conditions to find value for B. 2 When t =6, x=2(metres). A1 CAO Alternative method for question 2(b)  6 1 6 M1 FT Integration attempt, FT their answer to part 2(a). x= dt=1 4t+1 Ignore limits.  4t+1 2  0 0 x= 1 25−1 1 M1 Substitute correct limits. 2 2 x= 2 (metres) A1 CAO 3 © Cambridge University Press & Assessment 2025 Page 10 of 17

3(a) dv =−1 ( v2 +4 ) *M1 Forming differential equation and integrating to a term dt 2 involving an inverse tangent. Condone omission of +A. 1tan−1(1v ) =−1t +A Allow M1 for integrating dv = 1 ( v2 +4 ) to obtain a term 2 2 2 dt 2 involving an inverse tangent. A1 CWO, condone omission of +A. t=0, v=2 : A=1tan−1(2) =1π DM1 Use initial condition to find constant. 2 2 8 dx DM1 Integrate into term involving ln of cosine or sine. v= =2tan (1π−t ) dt 4 Accept modulus in ln term. OE, for example x=2ln ( cos(t−1π)) +B 4 v=−2tan(t−1π), x=2ln cos(t−1π) +B. 4 4 x=0, t=0: B=ln2 A1 Use initial condition and obtain correct expression. AEF, may see an expression in terms of sec. x=2ln ( cos(t−1π)) +ln2 =ln ( 2cos2(t−1π)) 4  4  Accept modulus in ln term. 5 3(b) 2ln ( cos(t−1π)) +ln2=0 M1 Solve equation to find a value for t. 4 t = 1π A1 CWO 2 2 © Cambridge University Press & Assessment 2025 Page 11 of 17

7(a) dv B1 Must have m. −mgkv−mg =m dt 1 1 *M1 Separate variables and integrate to a natural t=−  dv logarithm term. gkv+ Ignore modulus signs in this part. 1 t=− ln(kv+) +C gk 1 DM1 Use initial condition to find constant OR use When v = u, t = 0: C = ln(kU +) correct limits. gk Dependent on having integrated to obtain a natural logarithm term. 1 1 A1 AG, shown convincingly. t=− ln(kv+)+ ln(kU +) Must see an intermediate step before reaching gk gk given result. 1 kU + Not dependent on B1. t = ln  gk  kv+ 4 © Cambridge University Press & Assessment 2025 Page 15 of 17 7(b) dv M1 Three terms in equation of motion, separating −2v−2=v 5 dx variables and expressing integrand in a form that can be directly integrated.  v  5  x=−5 dv=−51− dv May be in terms of k,U and , for example 2v+5 2 v+5 1    x=− gk      1− v+ k  k    dv. Use of suvat in this part cannot be awarded any credit. x=−5v+ 25ln(v+5) +C M1 Correct form. 2 2 May be in terms of k,U and , for example 1    x=− v+ lnv+ +C . gk gk2  k  When x=0, v=10: C =25− 25ln(15) M1 Use initial condition to find constant OR use 2 correct limits in a definite integral. When v=0, x=25− 25ln(15)+ 25ln(5)=11.3 [m] A1 2 2 Alternative method for question 7(b) 2 M1 Rearrange answer to part 7(a) in the form  15  − t From part (a), t= 5ln   , v=15e 5 −5 v=aebt +c. 2 v+5 Use of suvat in this part cannot be awarded any credit. x=−75e− 5 2t −5t +B M1 Integrate to correct form. 2 When t=0, x=0, B=75, x=−75e− 5 2t −5t+75 M1 Use initial condition to find constant OR use 2 2 2 correct limits in a definite integral. When v=0, t =2.5ln3, x=25− 25ln3=11.3 [m] A1 2 © Cambridge University Press & Assessment 2025 Page 16 of 17 7(b) Alternative method for question 7(b) From part (a), t= 5 2 ln   v 1 + 5 5    , v=15e− 5 2t −5 M1 R v e = ar a r e a b n t g + e c a . n swer to part 7(a) in the form Use of suvat in this part cannot be awarded any credit. When v=0, t=2.5ln3 so x= 2.5ln3( 15e− 5 2t −5 ) dt M1 0 2.5ln3 M1 Integrate to correct form. x=−75e− 5 2t −5t  2  0 x=−751−512.5ln3− ( −75) =25−25ln3=11.3 [m] A1 2 3 2 2 4 7(c) "11.3" "11.3" M1 Divide their x from part 7(b) by the value of t Average speed = = 1 0.0410+0.2 2.5ln3 when v = 0 using the given answer from part ln  7(a). Condone if their x from part 7(b) is 100.04  0.2  obtained using suvat. 4.10 [ms–1] A1 4.102, accept values in the range 4.10 – 4.11. 2 © Cambridge University Press & Assessment 2025 Page 17 of 17

3(a) dv =−1 ( v2 +4 ) *M1 Forming differential equation and integrating to a term dt 2 involving an inverse tangent. Condone omission of +A. 1tan−1(1v ) =−1t +A Allow M1 for integrating dv = 1 ( v2 +4 ) to obtain a term 2 2 2 dt 2 involving an inverse tangent. A1 CWO, condone omission of +A. t=0, v=2 : A=1tan−1(2) =1π DM1 Use initial condition to find constant. 2 2 8 dx DM1 Integrate into term involving ln of cosine or sine. v= =2tan (1π−t ) dt 4 Accept modulus in ln term. OE, for example x=2ln ( cos(t−1π)) +B 4 v=−2tan(t−1π), x=2ln cos(t−1π) +B. 4 4 x=0, t=0: B=ln2 A1 Use initial condition and obtain correct expression. AEF, may see an expression in terms of sec. x=2ln ( cos(t−1π)) +ln2 =ln ( 2cos2(t−1π)) 4  4  Accept modulus in ln term. 5 3(b) 2ln ( cos(t−1π)) +ln2=0 M1 Solve equation to find a value for t. 4 t = 1π A1 CWO 2 2 © Cambridge University Press & Assessment 2025 Page 11 of 17

5(a) dv B1 m must be seen. mv =−0.025v2m−10m dx  v *M1 Separate variables and integrate to a term involving  dv=−dx 10+0.025v2 ln ( av2 +b ) or ln   av+b  .  pv+q −0.025x+c= 1ln ( v2 +400 ) 2 A1 Ignore modulus signs. Condone missing +c. OE, for example −x+c=20ln ( v2 +400 ) . c= 1ln800 DM1 Use initial condition to find constant of integration. 2 A1 AEF v=20 2e− 2 1 0 x −1 5 5(b) dv M1 Equation and integration to a term involving tan−1. m =−0.025v2m−10m dt Accept missing m.  1 −dt= dv 10+0.025v2 A1 Condone missing +k . v −t+k=2tan−1 20 k=2tan−120 =1π M1 v 20 2 Use initial condition with a term involving tan−1  . a v=20tan (1π−1t ) A1 4 2 4 © Cambridge University Press & Assessment 2025 Page 14 of 18

3(a) © Cambridge University Press & Assessment 2026 Page 12 of 22 8 − 0 .5 v 2 = 6 .4 v d d v x *B1 Correct use of Newton’s second law – acceleration must be v d d v x .  6.4v *M1 Separate variables correctly (from a differential x= dv=−6.4ln8−0.5v2 +c 8−0.5v2 equation of the form or  6.4v x=− dv=−6.4ln 0.5v2 −8 +c 0.5v2 −8 2 d d    + v = v v x ) and obtain an equation of the form x = k 1 ln ( k 2 + k 3 v 2 ) where k 1 , k 2 and k 3 are non-zero constants OR x= Aln(k +k v)+Bln(k +k v) where A and B 1 2 3 4 must have the same sign. Condone lack of +c. x=−6.4ln8−0.5v2 +c (*) A1 OE, for example, or x=−6.4ln 0.5v2 −8 +c (**) − 12 ln ( 4 − v ) − 12 ln ( 4 + v ) = 1 x 2 .8 + c . Condone lack of modulus signs in (*). However, if solving x = −  0 6 .5 .4 2 v v − 8 d v then modulus must be included in (**). Condone lack of +c. x = 0 , v = 0  c = 6 .4 ln 8 DM1 Use correct initial conditions to find c. Note that 0 = − 6 .4 ln ( 0 − 8 ) + c leading to a value of c is DM0. 1 .3 7 m s–1 A1 WWW AWRT 1.37 (1.3711489…) – or exact, for example, v = 4 1 − e − 18 . 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/ 3(b) dv B1 Correct use of Newton’s second law – acceleration 8−0.5v2 =6.4 dt must be © Cambridge University Press & Assessment 2026 Page 13 of 22 d d v t .  12.8 4+v M1 Separate variables correctly (from a differential t= dv=1.6ln +c 16−v2 4−v equation of the form 2 d d    + v = v t ) and obtain an equation of the form t = A ln ( k 1 v + k 2 ) + B ln ( k 3 v + k 4 ) where A , B , k 1 , k 2 , k 3 and k are non-zero constants but A 4 and B must have different signs. Allow a r t a n h ( )   t = v for any , 0   where 1 .    4+v t =1.6ln +c(*) 4−v or t = − 1 .6 ( ln v − 4 − ln v + 4 ) + c A1 Allow unsimplified and (**) t = 3 .2 a r t a n h ( 14 v ) + c . Condone lack of +c. Condone lack of modulus signs in (*), but in (**) must see modulus sign on at least the ln ( v − 4 ) term. Find c: t = 0 when v = 0  c = 0 t = 1 .6 ln 4 4 + − v v (*) or t = 1 .6 ln 4 v + − v 4 A1 As a minimum must state, ‘ (**) t = 0 , v = 0  c = 0 ’. ISW if re-arranged to make v the subject. Allow as two separate log terms. Allow t = 3 .2 a r t a n h ( 14 v ) . Condone lack of modulus sign for (*) but required in (**) in at least the ln ( v − 4 ) term. If answer is in terms of x, for example 4+x t =1.6ln , then maximum B1 M1 A0 A0 4−x possible. Allow recovery for full marks. 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/

3(a) © Cambridge University Press & Assessment 2026 Page 12 of 22 8 − 0 .5 v 2 = 6 .4 v d d v x *B1 Correct use of Newton’s second law – acceleration must be v d d v x . x =  8 6 − .4 v 0 .5 v 2 d v = − 6 .4 ln 8 − 0 .5 v 2 + c or x = −  0 6 .5 .4 2 v v − 8 d v = − 6 .4 ln 0 .5 v 2 − 8 + c *M1 Separate variables correctly (from a differential equation of the form 2 d d    + v = v v x ) and obtain an equation of the form x = k 1 ln ( k 2 + k 3 v 2 ) where k 1 , k 2 and k 3 are non-zero constants OR x = A ln ( k 1 + k 2 v ) + B ln ( k 3 + k 4 v ) where A and B must have the same sign. Condone lack of +c. x = − 6 .4 ln 8 − 0 .5 v 2 + c (*) or x = − 6 .4 ln 0 .5 v 2 − 8 + c A1 OE, for example, (**) − 12 ln ( 4 − v ) − 12 ln ( 4 + v ) = 1 x 2 .8 + c . Condone lack of modulus signs in (*). However, if solving x = −  0 6 .5 .4 2 v v − 8 d v then modulus must be included in (**). Condone lack of +c. x=0,v=0c=6.4ln8 DM1 Use correct initial conditions to find c. Note that 0=−6.4ln(0−8)+c leading to a value of c is DM0. 1 .3 7 m s–1 A1 WWW AWRT 1.37 (1.3711489…) – or exact, for example, v = 4 1 − e − 18 . 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/ 3(b) © Cambridge University Press & Assessment 2026 Page 13 of 22 8 − 0 .5 v 2 = 6 .4 d d v t B1 Correct use of Newton’s second law – acceleration must be d d v t .  12.8 4+v M1 Separate variables correctly (from a differential t= dv=1.6ln +c 16−v2 4−v equation of the form 2 d d    + v = v t ) and obtain an equation of the form t = A ln ( k 1 v + k 2 ) + B ln ( k 3 v + k 4 ) where A , B , k 1 , k 2 , k 3 and k 4 are non-zero constants but A and B must have different signs. Allow a r t a n h ( )   t = v for any ,0 where 1 .    4+v t =1.6ln +c(*) 4−v or t = − 1 .6 ( ln v − 4 − ln v + 4 ) + c A1 Allow unsimplified and t=3.2artanh (1v ) +c. 4 Condone lack of +c. Condone lack of modulus signs in (*), but in (**) must see modulus sign on at least the (**) ln ( v − 4 ) term. Find c: t = 0 when v = 0  c = 0 t = 1 .6 ln 4 4 + − v v A1 As a minimum must state, ‘t=0, (*) or 4+v t =1.6ln (**) v−4 v = 0  c = 0 ’. ISW if re-arranged to make v the subject. Allow as two separate log terms. Allow t = 3 .2 a r t a n h ( 14 v ) . Condone lack of modulus sign for (*) but required in (**) in at least the ln ( v − 4 ) term. If answer is in terms of x, for example t = 1 .6 ln 4 4 + − x x , then maximum B1 M1 A0 A0 possible. Allow recovery for full marks. 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/

2 EITHER: © Cambridge University Press & Assessment 2026 Page 10 of 20 m v d d v x = − k 1 x 2 (*B1 N2L to form a differential equation in terms of v d d v x . Accept k. B0 for m v d d v x = − k 1 d 2 .  m v d v = −  k x 2 d x leading to m v 2 2 = k x + c *M1 Separate variables correctly and integrate to obtain an equation of the form m v 2 x ( c )  = + where 0. Condone omission of +c.  1212 m m v u 2 2 = 1 4  *B1 Use kinetic energy loss to find v when x= 1d. v2 =1u2 or v=1u 2 4 2 Implied by x = 12 d and v= 1u as a boundary condition. 2 x=d, v=u  m u 2 2 = k d + c DM1 Use both sets of boundary conditions correctly to obtain two equations in c, k and d using their mu2 2k x= 1d, v= 1u  = +c 2 2 8 d m v 2 2 = k x + c . Be aware k can be negative here, k =  38 m d u 2 . c may already have been found, c =  78 m u 2 . 1 4  k d + c  = 2 d k + c implies B1M1. −3mdu2 7mu2 DM1 Dependent on all marks so far. 0= 8 + Substitute x 8 v = 0 . c = − 7 3 k d k 7k 0= − d 3d 3d A1) SC B3 for correct answer if m is not present throughout. 7 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/ 2 OR: Alternative method with x measured from A. © Cambridge University Press & Assessment 2026 Page 11 of 20 m v d d v x = − k ( d 1 − x ) 2 (*B1 N2L to form a differential equation in terms of v d d v x . Accept k.  m v d v = −  ( d k − x ) 2 d x leading to m v 2 2 = − ( d k − x ) + c *M1 Separate variables correctly and integrate to obtain an  equation of the form mv2 = (+c) where 0. (d −x) Condone omission of +c.  1212 m m v u 2 2 = 1 4  *B1 Use kinetic energy loss to find v when v2 =1u2 or v=1u 4 2 x = 12 d . Implied by x = 12 d and v= 1u as a boundary condition. 2 x = 0 , v = u mu2 k  =− +c 2 d x= 1d, v= 1u 2 2  m u 8 2 = − ( d − k 12 d ) + c DM1 Use both sets of boundary conditions correctly to obtain two mv2 k equations in c, k and d using their =− +c. 2 (d −x) Be aware k can be negative here, k =  38 m d u 2 . c may already have been found, c=7mu2. 8 3mdu2 7mu2 M1 Dependent on all marks so far. 0= 8 + Substitute v=0. (d−x) 8 x = 47 d , minimum distance = d − 47 d = 37 d A1) SC B3 for correct answer if m is not present throughout. Available marks 6 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/

5(a) The initial velocity u > 0 and the initial acceleration © Cambridge University Press & Assessment 2026 Page 15 of 25 u 2 − k u 2  0 . B1 The following three parts are all required: • u > 0 or initial velocity > 0 or in words (for example, initial velocity is positive). However, ‘velocity is > 0’ only is B0. • The expression u2 −ku2explicitly seen (OE, for example, − u 2 ( k − 1 ) not with m). • a < 0 OE (for example, u 2 − k u 2 < 0, ‘the acceleration is negative’ condone lack of ‘initial’ as this is covered by seeing the correct expression). 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/ 5(b) mv dv =m ( v2 −ku2) or v dv =v2 −ku2 B1 Acceleration must be v dv . dx dx dx  v x= dvx= 1ln v2 −ku2 +c v2 −ku2 2 or © Cambridge University Press & Assessment 2026 Page 16 of 25 x = −  k u 2 v − v 2 d v  x = 12 ln k u 2 − v 2 + c *M1 Separate variables correctly and obtain an equation of the form x = A ln B v 2 + C k u 2 where A, B and C are non-zero constants. Or x = A ln ( B v + C k u ) + D ln ( E v + F k u ) where A, B, C, D, E and F are non-zero constants, but A and D must have the same sign. Condone lack of modulus and condone no +c. x  = c 0 , = v − = 12 u ln  u 0 2 = − k 12 u ln 2 u 2 o − r k u c 2 = + − c 12 ln ( k u 2 − u 2 ) o r c = − 12 ln k u 2 − u 2 DM1 Use correct initial conditions of x = 0 and v = u correctly to find c. Note that c=−1ln ( u2 −ku2) is M0 (as this 2 gives an undefined value for c). ISW after seeing a correct c. Note that these values of c are dependent on the side in which the candidate added +c (so may see c being the opposite sign).  x = 1 2 ln k u u 2 2 − − k v u 2 2 , v = 0   x = 1 2 ln  k k − 1  1 k A1 1 k or x= ln WWW, OE, for example, ln , 2 k−1 2 1−k 1  k  ln  etc.   2 1−k   1  k  Note: ln  is A0. 2 1−k 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/ 5(c) m dv =m ( v2 −ku2) or dv =v2 −ku2 B1 Acceleration must be dv . dt dt dt © Cambridge University Press & Assessment 2026 Page 17 of 25 t =  v 2 − 1 k u 2 d v = 2 u 1 k ln v v − + u u k k + c M1 Separate variables correctly and obtain an equation of the form or  1 1 u k +v t=− dv=− ln +c  ku2 −v2 2u k u k −v t = u K k ln   v v   u u k k where K is a non-zero constant (or equivalent as two separate log terms). Condone lack of modulus and condone no +c. [When t = 0, v = u  ] 1 1− k c=− ln 2u k 1+ k or c = 2 u 1 k ln k k + − 1 1 A1 OE, for example, c = − 2 u 1 k ln  k k − + 1 1  or − 2 u 1 k ln  1 1 − + k k  but − 2 u 1 k ln  1 1 − + k k  (or any other undefined expression) is A0. Award even if u appears in all terms in the fraction in the argument of the natural logarithm. Note that these values of c are dependent on the side in which the candidate added +c (so may see c being the opposite sign).  t = 2 u 1 k ln v v − + u u k k − 2 u 1 k ln 1 1 − + k k s o w h e n v = 0   [ t = ] 2 u 1 k ln  k k + − 1 1  A1 1− k  1 WWW OE, for example, − ln  2u k 1+ k    1  k −1 1  k +k or − ln  or ln  1 1− k  2u k   k +1   2u k  k− k   but A0 for − ln    etc. 2u k 1+ k  Must have scored the previous A1 mark. Allow u in all terms in the fraction in the argument of the natural logarithm. 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/