Linear motion under a variable force — Paper 3 · A Level Mathematics - Further 9231

5 A particle P is moving along a straight line with acceleration 3ku - kv where v is its velocity at time t, u is its initial velocity and k is a constant. The velocity and acceleration of P are both in the direction of increasing displacement from the initial position. (a) Find the time taken for P to achieve a velocity of 2u. [3] … … … … … … … … … … … … … … … … … … … … … … … … (b) Find an expression for the displacement of P from its initial position when its velocity is 2u. [5] … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 5(a) d d 3 = − v k t u v M1 ( ) ln 3 − − = + u v kt d 0, : ln 2 = = = − t v u d u M1 1 2 : ln 2 = = v u t k A1 3 5(b) d 3 d = − v v ku kv x [  d d 3 = − v v k x u v ] B1 ( ) ( ) 3 3 d d 3 − − + = − u v u v k x u v so ( ) 3 ln 3 −− − = + v u u v kx c M1A1 0, : 3 ln 2 = = = −− x v u c u u u M1 ( ) 2 : 3ln 2 1 = = − u v u x k A1 5

This question in 9231/31 May/June 2020

5 A particle P is moving along a straight line with acceleration 3ku - kv where v is its velocity at time t, u is its initial velocity and k is a constant. The velocity and acceleration of P are both in the direction of increasing displacement from the initial position. (a) Find the time taken for P to achieve a velocity of 2u. [3] … … … … … … … … … … … … … … … … … … … … … … … … (b) Find an expression for the displacement of P from its initial position when its velocity is 2u. [5] … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 5(a) d d 3 = − v k t u v M1 ( ) ln 3 − − = + u v kt d 0, : ln 2 = = = − t v u d u M1 1 2 : ln 2 = = v u t k A1 3 5(b) d 3 d = − v v ku kv x [  d d 3 = − v v k x u v ] B1 ( ) ( ) 3 3 d d 3 − − + = − u v u v k x u v so ( ) 3 ln 3 −− − = + v u u v kx c M1A1 0, : 3 ln 2 = = = −− x v u c u u u M1 ( ) 2 : 3ln 2 1 = = − u v u x k A1 5

This question in 9231/32 May/June 2020

2 A particle Q of mass m kg falls from rest under gravity. The motion of Q is resisted by a force of magnitude mkv N, where v ms -1 is the speed of Q at time t s and k is a positive constant. Find an expression for v in terms of g, k and t. [6] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: 2 d d = − v t g kv M1 ( ) 1 ln − − = + g kv t A k M1 1 0, 0: ln = = = − t v A g k A1 1 ln − = − g kv t k g M1 ( ) 1 − = − kt g v e k M1A1 6

This question in 9231/33 May/June 2020

7 A particle P of mass m kg moves in a horizontal straight line against a resistive force of magnitude mkv2 N , where v ms -1 is the speed of P after it has moved a distance x m and k is a positive constant. The initial speed of P is u ms -1 . 1 1 (a) Show that x = ln 2 when v = 2 u . [4] k … … … … … … … … … … … … … … … … … … … … … … … … … Beginning at the instant when the speed of P is 12 u , an additional force acts on P. This force has 5mmagnitude N and acts in the direction of increasing x. v (b) Show that when the speed of P has increased again to u ms -1 , the total distance travelled by P is given by an expression of the form 1 A - ku 3 ln 3 3k e B - ku o, stating the values of the constants A and B. [7] … … … … … … … … … … … … … … … … … … … … … … Additional Page If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown. … … … … … … … … … … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 7(a) 2 dv mv kmv dx = − B1 N2L, with m lnv kx c = − + M1 Separate variables and integrate 0, : ln x v u c u = = = M1 Use initial condition 1 1 : ln , 2 2 v u kx = = − 1 ln 2 x k = A1 AG 4 7(b) 2 d 5 d v m mv mkv x v = − + B1 N2L (allow missing m in this part) 2 3 d d 5 v v x kv = − ( ) 3 1 ln 5 ( ) 3 kv x d k − − = + M1A1 Separate variables and integrate Using (a) ( ) ( ) 3 3 1 1 1 ln 5 ln 5 ln2 3 3 kv x ku k k k − − = − − − M1M1 Use condition. M0 if 1 , 0 2 v u x = = used unless 1 ln2 k is added on later Rearrange dependent on ln solution 3 3 1 40 ln 3 5 ku x k ku   − =   −   M1A1 Use v u = 7

This question in 9231/32 Oct/Nov 2020

1 A particle P of mass 1 kg is moving along a straight line against a resistive force of magnitude 10 v -1 2 N, where v ms is the speed of P at time t s. When t = 0, v = 25. ( t + 1 ) Find an expression for v in terms of t. [5] … … … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 1 ( ) 2 d 10d 1 v t v t = − + 10 2 1 = + + v A t M1 A1 Attempt to integrate. 0, 25, 0 t v A = = = M1 Use correct initial condition. ( ) 2 25 1 = + v t A1 CAO 5

This question in 9231/31 May/June 2021

1 A particle P of mass 1 kg is moving along a straight line against a resistive force of magnitude 10 v -1 2 N, where v ms is the speed of P at time t s. When t = 0, v = 25. ( t + 1 ) Find an expression for v in terms of t. [5] … … … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 1 ( ) 2 d 10d 1 v t v t = − + 10 2 1 = + + v A t M1 A1 Attempt to integrate. 0, 25, 0 t v A = = = M1 Use correct initial condition. ( ) 2 25 1 = + v t A1 CAO 5

This question in 9231/32 May/June 2021

5 A particle P of mass m kg is projected vertically upwards from a point O, with speed 20ms -1 , and moves under gravity. There is a resistive force of magnitude 2mv N, where v ms -1 is the speed of P at time t s after projection. (a) Find an expression for v in terms of t, while P is moving upwards. [6] … … … … … … … … … … … … … … … … … … … … … … … … … The displacement of P from O is x m at time t s. (b) Find an expression for x in terms of t, while P is moving upwards. [2] … … … … … … … … … … … … (c) Find, correct to 3 significant figures, the greatest height above O reached by P. [2] … … … … … … … … … … … …

10 marks

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

This question in 9231/33 May/June 2021

2 A particle P of mass m kg moves along a horizontal straight line with acceleration ams -2 given by v ( 1 - 2 t 2 ) a = , t where v ms -1 is the velocity of P at time t s. (a) Find an expression for v in terms of t and an arbitrary constant. [3] … … … … … … … … … … … (b) Given that a = 5 when t = 1, find an expression, in terms of m and t, for the horizontal force acting on P at time t. [3] … … … … … … … … … … …

6 marks

Mark scheme: 2(a) Separate variables and integrate: 2 1 2   − =     dv t dt v t so 2 ln ln = − + v t t c M1 A1 2 − = t v Ate , 2 − −= t v Ate , 2 − = − t v Ate A1 CAO. 3 Question Answer Marks Guidance 2(b) ( ) ( ) 2 2 2 2 1 2 1 2 − − − − = = − − t t Ate t a Ae t t M1 Substituting their answer to part (a) into given formula ( ) 1, 5 5 t a A e = = = M1 Use initial condition. Force = ( ) 2 1 2 5 2 1 − − t me t A1 Use N2L, correct work only. Alternative method for question 2(b) ( ) 2 1 2 − = v t a t substitute 1, 5 t a = = so 5 = − v M1 Use initial condition. Use N2L, correct work only. Substituting in their answer to part (a) so ( ) 5 = A e M1 Force = ( ) 2 1 2 5 2 1 − − t me t A1 3

This question in 9231/31 Oct/Nov 2021

6 A particle P of mass 2 kg moves along a horizontal straight line. The point O is a fixed point on this line. At time t s the velocity of P is v ms -1 and the displacement of P from O is x m. 128 A force of magnitude 8x - 3 N acts on P in the direction OP. When t = 0, x = 8 and v = - 15 . e x o 2 2 (a) Show that v = - ( x - 4) . [5] x … … … … … … … … … … … … … … … … … … … … … … … … (b) Find an expression for x in terms of t. [4] … … … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 6(a) 3 128 2 8 = − dv v x dx x 2 2 2 4 64 − = + + v x x c A1 OE. x = 8, v = ‒15 and c = ‒32 M1 Use initial condition. ( ) 2 4 2 2 4 8 16 = − + v x x x or 2 2 64 4 32 − + x x A1 Correct expression for 2v , AEF. ( ) 2 2 2 2 4 4 v x x = − giving ( ) 2 2 4 = − − v x x A1 Convincingly shown, e.g. v is negative initially, AG. 5 6(b) ( ) 2 1 ln( 4) 2 2 − = − + x t A M1 Use = dx v dt and integrate. 1 0, 8, ln60 2 t x A = = = DM1 Use initial condition. 2 1 4 ln 2 2 60   − = −     x t giving 2 4 4 60 − − = t x e M1 Remove log. 4 4 60 − = + t x e A1 CAO 4

This question in 9231/32 Oct/Nov 2021

2 A particle P of mass m kg moves along a horizontal straight line with acceleration ams -2 given by v ( 1 - 2 t 2 ) a = , t where v ms -1 is the velocity of P at time t s. (a) Find an expression for v in terms of t and an arbitrary constant. [3] … … … … … … … … … … … (b) Given that a = 5 when t = 1, find an expression, in terms of m and t, for the horizontal force acting on P at time t. [3] … … … … … … … … … … …

6 marks

Mark scheme: 2(a) Separate variables and integrate: 2 1 2   − =     dv t dt v t so 2 ln ln = − + v t t c M1 A1 2 − = t v Ate , 2 − −= t v Ate , 2 − = − t v Ate A1 CAO. 3 Question Answer Marks Guidance 2(b) ( ) ( ) 2 2 2 2 1 2 1 2 − − − − = = − − t t Ate t a Ae t t M1 Substituting their answer to part (a) into given formula ( ) 1, 5 5 t a A e = = = M1 Use initial condition. Force = ( ) 2 1 2 5 2 1 − − t me t A1 Use N2L, correct work only. Alternative method for question 2(b) ( ) 2 1 2 − = v t a t substitute 1, 5 t a = = so 5 = − v M1 Use initial condition. Use N2L, correct work only. Substituting in their answer to part (a) so ( ) 5 = A e M1 Force = ( ) 2 1 2 5 2 1 − − t me t A1 3

This question in 9231/33 Oct/Nov 2021

3 A particle P is moving in a horizontal straight line. Initially P is at the point O on the line and is moving -1 4000 -2 with velocity 25 m s . At time t s after passing through O, the acceleration of P is 3 m s in ( t5 + 4 ) the direction PO. The displacement of P from O at time t is x m. Find an expression for x in terms of t. [5] … … … … … … … … … … … … … … … … … … … … … … … … …

5 marks

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

This question in 9231/31 May/June 2022

3 A particle P is moving in a horizontal straight line. Initially P is at the point O on the line and is moving -1 4000 -2 with velocity 25 m s . At time t s after passing through O, the acceleration of P is 3 m s in ( t5 + 4 ) the direction PO. The displacement of P from O at time t is x m. Find an expression for x in terms of t. [5] … … … … … … … … … … … … … … … … … … … … … … … … …

5 marks

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

This question in 9231/32 May/June 2022

5 A particle P of mass 4 kg is moving in a horizontal straight line. At time t s the velocity of P is v ms -1 and the displacement of P from a fixed point O on the line is x m. The only force acting on P is a resistive force of magnitude ( 4e -x + 12) e -x N . When t = 0, x = 0 and v = 4 . 1 + 3 e x (a) Show by integration that v = x . [4] e … … … … … … … … … … … … … … … … … … … … … … … … … (b) Find an expression for x in terms of t. [4] … … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

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

This question in 9231/33 May/June 2022

4 A particle of mass 0.5 kg moves along a horizontal straight line. Its velocity is v m s -1 at time t s. The forces acting on the particle are a driving force of magnitude 50 N and a resistance of magnitude 2v 2 N . The initial velocity of the particle is 3 m s -1 . (a) Find an expression for v in terms of t. [7] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … (b) Deduce the limiting value of v. [1] … … … … … … … … … … … … … …

8 marks

Mark scheme: 4(a) dv 2 dv 2 B1 N2L m = 50 − 2v = 4 25 − v ( ) dt dt 1 1 1 M1 Separate variables and use partial fractions.  + dv =  4 dt 10 5 − v 5 + v 1 ( − ln ( 5 − v ) + ln ( 5 + v ) ) = 4t + A M1 A1 Integrate(Note: formulainto logonterms.MF19). 10 1 M1 Use initial condition. Use t = 0, v = 3 to give A = ln 4 10 1 5 + v 5 + v 40 t M1 Rearrange to make v the subject. 4t = ln leading to = e 10 4 ( 5 − v ) 20 − 4v −40 t A1 5 4 − e ( ) v = 4 + e −40 t 7 4(b) As t →, v → 5 B1 1

This question in 9231/31 Oct/Nov 2022

4 A particle P of mass 5 kg moves along a horizontal straight line. At time t s, the velocity of P is v m s -1 and its displacement from a fixed point O on the line is x m. The forces acting on P are a force of 500 1 2 magnitude N in the direction OP and a resistive force of magnitude 2 v N . When t = 0 , x = 0 and v v = 5 . (a) Find an expression for v in terms of x. [6] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … (b) State the value that the speed approaches for large values of x. [1] … … … … … … … … … … … … … … …

7 marks

Mark scheme: 4(a) dv 500 1 2 B1 Sight of m or 5 is required. 5v = − v dx v 2 10 v 2 dv M1 Separate variables and attempt to integrate into a log term. = dx 1000 − v 3 3 A1 1000 − v = x ( + A ) − 10ln ( ) 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 − v 3 1 A1 1 3 : A0 if eln terms. v =  (1000 − 875e −0.3 x  3 v = 5  (8 − 7 e −0.3 x      6 4(b) Maximum value of v is 10 B1 No FT: result can be found from initial equation. 1

This question in 9231/32 Oct/Nov 2022

4 A particle of mass 0.5 kg moves along a horizontal straight line. Its velocity is v m s -1 at time t s. The forces acting on the particle are a driving force of magnitude 50 N and a resistance of magnitude 2v 2 N . The initial velocity of the particle is 3 m s -1 . (a) Find an expression for v in terms of t. [7] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … (b) Deduce the limiting value of v. [1] … … … … … … … … … … … … … …

8 marks

Mark scheme: 4(a) dv 2 dv 2 B1 N2L m = 50 − 2v = 4 25 − v ( ) dt dt 1 1 1 M1 Separate variables and use partial fractions.  + dv =  4 dt 10 5 − v 5 + v 1 ( − ln ( 5 − v ) + ln ( 5 + v ) ) = 4t + A M1 A1 Integrate(Note: formulainto logonterms.MF19). 10 1 M1 Use initial condition. Use t = 0, v = 3 to give A = ln 4 10 1 5 + v 5 + v 40 t M1 Rearrange to make v the subject. 4t = ln leading to = e 10 4 ( 5 − v ) 20 − 4v −40 t A1 5 4 − e ( ) v = 4 + e −40 t 7 4(b) As t →, v → 5 B1 1

This question in 9231/33 Oct/Nov 2022

6 A particle of mass m kg falls vertically under gravity, from rest. At time t s, P has fallen x m and has velocity v m s -1 . The only forces acting on P are its weight and a resistance of magnitude kmgv N, where k is a constant. (a) Find an expression for v in terms of t, g and k. [5] … … … … … … … … … … … … … … … … … … … … … … … … … (b) Given that k = 0.05, find, in metres, how far P has fallen when its speed is 12 m s -1 . [5] … … … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

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

This question in 9231/33 May/June 2023

2 A ball of mass 2 kg is projected vertically downwards with speed 5 ms -1 through a liquid. At time t s after projection, the velocity of the ball is v ms -1 and its displacement from its starting point is x m. The forces acting on the ball are its weight and a resistive force of magnitude 0.2v 2 N . (a) Find an expression for v in terms of t. [6] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … (b) Deduce what happens to v for large values of t. [1] … … … … … … … … … … … … … …

7 marks

Mark scheme: 2(a) dv 2 B1 2 = 2 g − 0.2v dt dv M1 Integrate to a ln term of the correct form. Separate variables and attempt to integrate = dt 0.1 100 − v 2 ( ) 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 2 t 10 + v M1 Rearrange, removing ln. 2t = ln , e = 3 (10 − v ) 3 (10 − v ) 30 − 10e −2 t A1 AEF v = 3 + e −2 t 6 2(b) v →10 B1FT FT from expression of correct form. 1

This question in 9231/31 Oct/Nov 2023

2 A particle P of mass 0.5 kg moves in a straight line. At time t s the velocity of P is v m s -1 and its displacement from a fixed point O on the line is x m. The only forces acting on P are a force of 150 magnitude 2 N in the direction of increasing displacement and a resistive force of magnitude ( x + 1) 450 3 N . When t = 0 , x = 0 and v = 20 . ( x + 1) Ax + B Find v in terms of x, giving your answer in the form v = , where A and B are constants to be ( x + 1) determined. [6] … … … … … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: 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: 0.5v 2 = − + + A x + 1 ( x + 1) 2 x = 0 , v = 20 ; A= 50 M1 Use initial condition. 2 100( x 2 − 4 x + 4) A1 AEF Rearrange: v = ( x + 1) 2 2 100( x − 2) 2 10( x − 2) A1 v = so v =  ( x + 1) 2 ( x + 1) 20 − 10 x From initial condition, sign must be negative, v = Signs dealt with convincingly. x + 1 6

This question in 9231/32 Oct/Nov 2023

2 A ball of mass 2 kg is projected vertically downwards with speed 5 ms -1 through a liquid. At time t s after projection, the velocity of the ball is v ms -1 and its displacement from its starting point is x m. The forces acting on the ball are its weight and a resistive force of magnitude 0.2v 2 N . (a) Find an expression for v in terms of t. [6] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … (b) Deduce what happens to v for large values of t. [1] … … … … … … … … … … … … … …

7 marks

Mark scheme: 2(a) dv 2 B1 2 = 2 g − 0.2v dt dv M1 Integrate to a ln term of the correct form. Separate variables and attempt to integrate = dt 0.1 100 − v 2 ( ) 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 2 t 10 + v M1 Rearrange, removing ln. 2t = ln , e = 3 (10 − v ) 3 (10 − v ) 30 − 10e −2 t A1 AEF v = 3 + e −2 t 6 2(b) v →10 B1FT FT from expression of correct form. 1

This question in 9231/33 Oct/Nov 2023

5 A particle P of mass 2 kg moving on a horizontal straight line has displacement x m from a fixed point O on the line and velocity v m s -1 at time t s. The only horizontal force acting on P is a variable force F N which can be expressed as a function of t. It is given that v 3 - t = x 1 + t and when t = 0, x = 5 . (a) Find an expression for x in terms of t. [4] … … … … … … … … … … … … … … … … … … … … … … … (b) Find the magnitude of F when t = 3. [3] … … … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 5(a) d x  4  M1 Separate variables, obtain RHS in integrable form. =  − 1  d t x  t + 1  ln x = 4ln t + 1 −+t A A1 t = 0, x = 5: A = ln5 M1 4 t A1 x = 5 ( t + 1) e− 4 5(b) 3 t M1 v = ( 3 − t )  5 ( t + 1) e− dv − t 3 2 3 Acceleration = = 5e − ( t + 1) + ( 3 − t ) 3 ( t + 1) − ( 3 − t )( t + 1) ( ) dt − t 2 AEF Acceleration = 5e ( t + 1) ( 5 − t )(1 − t ) − t 2 M1 F = 2  acceleration, so at F = 10e ( t + 1) ( 5 − t )(1 − t ) At t = 3, magnitude of force is 640 e− N3 A1 31.9 N 3

This question in 9231/31 Oct/Nov 2024

7 A particle P of mass m kg is held at rest at a point O and released so that it moves vertically under gravity against a resistive force of magnitude 0.1mv 2 N, where v m s -1 is the velocity of P at time t s. (a) Find an expression for v in terms of t. [6] … … … … … … … … … … … … … … … … … … … … … … … … … … The displacement of P from O at time t s is x m. (b) Find an expression for v2 in terms of x. [5] … … … … … … … … … … … … … … … … … … … … … … … … … … Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. … … … … … … … … … … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 7(a) dv 2 B1 Use of suvat means 0 marks in this part m = mg − 0.1mv Note that no mg term means 0 marks in this part. dt d v Must see m, may be cancelled before a = used d t dv 2 1 2 M1* = 10 − 0.1v = 100 − v ( ) dt 10 dv 1 = dt 100 − v 2 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 2 t A1 −2 t 10 e − 1 10 1 − e ( ) ( ) v = aef v = e 2 t + 1 1 + e −2 t 6 7(b) dv 1 2 M1* Use of suvat means 0 marks in this part v = 100 − v ( ) dx 10 Separate variables and integrate, allow missing + A for M1 only. 1 2 1 − ln 100 − v = 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 v 2 in terms of x M1 x A1 AEF − v 2 = 100(1 − e 5 ) Allow 10g instead of 100. 5

This question in 9231/32 Oct/Nov 2024

5 A particle P of mass 2 kg moving on a horizontal straight line has displacement x m from a fixed point O on the line and velocity v m s -1 at time t s. The only horizontal force acting on P is a variable force F N which can be expressed as a function of t. It is given that v 3 - t = x 1 + t and when t = 0, x = 5 . (a) Find an expression for x in terms of t. [4] … … … … … … … … … … … … … … … … … … … … … … … (b) Find the magnitude of F when t = 3. [3] … … … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 5(a) d x  4  M1 Separate variables, obtain RHS in integrable form. =  − 1  d t x  t + 1  ln x = 4ln t + 1 −+t A A1 t = 0, x = 5: A = ln5 M1 4 t A1 x = 5 ( t + 1) e− 4 5(b) 3 t M1 v = ( 3 − t )  5 ( t + 1) e− dv − t 3 2 3 Acceleration = = 5e − ( t + 1) + ( 3 − t ) 3 ( t + 1) − ( 3 − t )( t + 1) ( ) dt − t 2 AEF Acceleration = 5e ( t + 1) ( 5 − t )(1 − t ) − t 2 M1 F = 2  acceleration, so at F = 10e ( t + 1) ( 5 − t )(1 − t ) At t = 3, magnitude of force is 640 e− N3 A1 31.9 N 3

This question in 9231/33 Oct/Nov 2024

3 A ball of mass m kg is projected vertically upwards with initial speed U m s -1 and moves under gravity. At time t s after projection, the ball has travelled a distance x m and its speed is v m s -1 . There is a resistive force of magnitude mkv 2 N , where k is a positive constant. 1 g + kU 2 (a) Show that the distance travelled by the ball when it is moving upwards is x = ln f p. [4] 2k g + kv 2 … … … … … … … … … … … … … … … … … … … … … … … … It is given that k = 0.025 and that U = 20 . (b) Find the time taken for the ball to reach its maximum height. [4] … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 3(a) 2 dv B1 Correct use of N2L with m seen. − mg − mkv = mv dx v 1 2 M1 Attempt to integrate by separating variables. d x = − ln g + kv d v leading to x = − 2 ( ) + c  . Correct form. Allow missing + c .  2 k  g + kv 1 2 M1 v = U when x = 0 , so c = ln g + kU . ( ) 2 k 1  g + kU 2  A1 AG, shown convincingly. x = ln  2  (metres) 2 k  g + kv  4 3(b) 1 2 dv *M1 Attempt to integrate by separating variables. 10 + 40 v = − dt 40 2  dt = − 400 + v dv 40 −1 1 A1 Correct integration. t = − 20 tan ( 20 v )( +C ) When t = 0 , v = 20 , so C = 12 π DM1 Find C and substitute v = 0 to find value for t . π t = −2tan −1 (0) + 12 t = 1.57 (seconds) A1 CWO, accept final answer of 12 π . 4

This question in 9231/33 May/June 2025

2 A particle P of mass m kg moves along a horizontal straight line against a resistive force of magnitude 2mv 3 N , where v ms -1 is the velocity of P at time t s. When t = 0, v = 1. (a) Find an expression for v in terms of t. [4] … … … … … … … … … … … … (b) Find the displacement of P from its initial position when t = 6 . [3] … … … … … … … … … … … …

7 marks

Mark scheme: 2(a) dv 3 M1 N2L (m must be seen, possibly as F = ma), no extra terms. m = −2 mv Attempt to integrate by separating variables, allow sign errors. dt −2 2dt  −3 Allow errors in coefficient of v . v dv =  − Allow missing +A . − 12 v −2 = −2t + A A1 Any correct equivalent form. When t = 0 , v = 1 leading to A = − 12 . M1 Use correct initial conditions to find value for A. 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 1 M1 FT FT their answer to part 2(a). = so x = 2 4t + 1 [ + B ] Allow missing +B for this mark. dt 4t + 1 When t = 0 , x = 0 leading to B = − 12 . M1 Use correct initial conditions to find value for B. When t = 6 , x = 2 (metres). A1 CAO Alternative method for question 2(b) 6 M1 FT Integration attempt, FT their answer to part 2(a). 6 1 1 x = dt =  4t + 1   2  0 Ignore limits. 0 4t + 1 x = 12 25 − 12 1 M1 Substitute correct limits. x = 2 (metres) A1 CAO 3

This question in 9231/34 May/June 2025

7 A particle P of mass m kg moving along a rough horizontal table has displacement x m from a fixed point O on the table and velocity v m s -1 at time t s. The particle P is subject to a resistive force of magnitude mgkv N, where k is a positive constant, and a frictional force of magnitude nmg . The particle P is initially at O with speed U m s -1 . 1 kU + n (a) Show that t = ln e o. [4] gk kv + n … … … … … … … … … … … … … … … … … … … … … … … … It is given that U = 10 , k = 0.04 and n = .02 . (b) Find the distance P moves before coming to rest. [4] … … … … … … … … … … … … … … … … … … (c) Find the average speed of P over the period it is moving. [2] … … … … … … … Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 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 7(b) 2 dv M1 Three terms in equation of motion, separating − 5 v − 2 = v variables and expressing integrand in a form dx that can be directly integrated. 5 v 5   5  x = − 2  dv = − 2   1 −  dv May be in terms of k ,U and , for example  v + 5   v + 5  1   k  x = − 1 − dv .    gk v + k     Use of suvat in this part cannot be awarded any credit. x = − 52 v + 252 ln ( v + 5 ) + C  M1 Correct form. May be in terms of k ,U and , for example 1    x = − v + ln  v +  + C . 2 gk gk  k  When x = 0, v = 10: C = 25 − 252 ln (15 ) M1 Use initial condition to find constant OR use correct limits in a definite integral. 25 When v = 0, x = 25 − 252 ln (15 ) + 2 ln ( 5 ) = 11.3 [m] A1 Alternative method for question 7(b) 2 M1 Rearrange answer to part 7(a) in the form − t  15  5 − 5 From part (a), t = 52 ln   , v = 15e v = ae bt + c .  v + 5  Use of suvat in this part cannot be awarded any credit. 75 − 52 t M1 Integrate to correct form. x = − 2 e − 5t  + B  75 75 − 52 t 75 M1 Use initial condition to find constant OR use When t = 0, x = 0, B = 2 , x = − 2 e − 5t + 2 correct limits in a definite integral. When v = 0, t = 2.5ln3, x = 25 − 252 ln3 = 11.3 [m] A1 7(b) Alternative method for question 7(b) 2 t M1 Rearrange answer to part 7(a) in the form 5  15  − 5 , v = 15e − 5 From part (a), t = 2 ln   v = ae bt + c .  v + 5  Use of suvat in this part cannot be awarded any credit. 2.5ln3 − 52 t M1 When v = 0, t = 2.5ln3 so x =  0 (15e − 5 ) dt 75 x =  − − 52 t − 5t  2.5ln3 M1 Integrate to correct form. 2 e   0 75 1 75 25 A1 [m] x = − 2 −3 5 12.5  ln3 −−( 2 ) = 25 − 2 ln3 = 11.3 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 [m s–1] A1 4.102, accept values in the range 4.10 – 4.11. 2

This question in 9231/32 Oct/Nov 2025