Cambridge A Level Mathematics 9709 — 2022 Oct/Nov Paper 4 · Variant 1

1 Coplanar forces of magnitudes PN, QN, 16N and 22N act at a point in the directions shown in the diagram. The forces are in equilibrium. Find the values of P and Q. [5] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1 Attempt at resolving horizontally or vertically M1 Allow sign errors, allow sin/cos mix. 3 terms. P cos25 = 22 + 16cos55 A1 Q + 16sin55 = P sin 25 A1 Allow their P. Attempt to solve for P or Q M1 No missing/extra terms. P = 34.4 Q = 1.43 A1 P = 34.40025941 , Q = 1.431745128 . 5

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2 Small smooth spheres A and B, of equal radii and of masses 6kg and 2kg respectively, lie on a smooth horizontal plane. Initially A is moving towards B with speed 5ms−1 and B is moving towards A with speed 3ms−1. After the spheres collide, both A and B move in the same direction and the difference in the speeds of the spheres is 2ms−1. Find the loss of kinetic energy of the system due to the collision. [5] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 Use conservation of momentum *M1 4 dimensionally correct terms. Allow sign errors, v A and v B must be 6 +5 2 −( 3) = 6v A + 2vB different. Use v B = v A + 2 or v A = v B − 2 with their momentum equation and solve for v A or v B DM1 Allow v B = v A  2 or v A = v B  2 . v A = 2.5 or v B = 4.5 A1 Attempt at initial KE, or final KE, or change in KE for A , or change in KE for B M1 Allow use or their v A and/or v B . Allow if 2 KE equations seen. 1 2 1 2 Initial KE = 6 5 + −2 ( 3 )  = 84  2 2 1 2 1 2 Final KE = 6 ( their 2.5 ) + 2 ( their 4.5 ) 2 2  1 2 1 2  Change in KE for A =   6 5 − 6 ( their 2.5 )   2 2   1 2 1 2  Change in KE for B =   −2 ( 3) − 2 ( their 4.5 )   2 2  Loss of KE = 45 J A1 Allow –45 J. Allow if mgv used in momentum equation. 5

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3 A constant resistance of magnitude 1400N acts on a car of mass 1250kg. (a) The car is moving along a straight level road at a constant speed of 28ms−1. Find, in kW, the rate at which the engine of the car is working. 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(b) The car now travels at a constant speed up a hill inclined at an angle of 1 to the horizontal, where sin 1 = 0.12, with the engine working at 43.5kW. Find this speed. 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(c) On another occasion, the car pulls a trailer of mass 600kg up the same hill. The system of the car and the trailer is modelled as particles connected by a light inextensible cable. The car’s engine produces a driving force of 5000N and the resistance to the motion of the trailer is 300N. The resistance to the motion of the car remains 1400N. Find the acceleration of the system and the tension in the cable. 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Mark scheme: 3(a) Power = 1400  28 B1 Power = 39.3 kW B1 2 3(b) 43500 B1 oe DF = v Attempt to resolve parallel to the hill M1 3 terms, no need for DF in terms of v . Allow sign errors, sin/cos mix.  DF = 1400 + 1250 g  0.12 = 2900  Allow use of 6.89º or 6.9º.    or DF = 1400 + 1250 g  sin6.89 = 2899.544602  Speed = 15 m s–1 A1 Awrt 15.0 3 3(c) Attempt at N2L on either car, trailer or the system M1 Allow sign errors, sin/cos mix. Correct number of relevant terms. Car: 5000 − 1400 − 1250 g  0.12 − T = 1250 a Allow use of 6.89º or 6.9º. Allow with g missing. Trailer: T − 300 − 600 g  0.12 = 600a System: 5000 − 1400 − 300 − 1250 g  0.12 − 600 g  0.12 = (1250 + 600 ) a A1 For any 2 equations correct. Solve for a or T M1 From equation(s) with at most 1 term. missing/extra in total. Allow with g missing. 108 50700 A1 Awrt 0.584 and 1370. Acceleration = = 0.584 ms-2, Tension = = 1370 N a = 0.583787838 , T = 1370.27027 . 185 37 4

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4 A block of mass 8kg is placed on a rough plane which is inclined at an angle of 18Å to the horizontal. The block is pulled up the plane by a light string that makes an angle of 26Å above a line of greatest slope. The tension in the string is T N (see diagram). The coefficient of friction between the block and plane is 0.65. (a) The acceleration of the block is 0.2ms−2. Find T. 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(b) The block is initially at rest. Find the distance travelled by the block during the fourth second of motion. 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Mark scheme: 4(a) Attempt at N2L parallel to the plane *M1 4 terms. Allow sign errors, sin/cos mix, allow g missing. T cos26 − 8 g sin18 − F = 8  0.2 A1 Allow with their F . Attempt at resolving perpendicular to the plane *M1 3 terms Allow sign errors, sin/cos mix, allow g missing. R + T sin 26 = 8 g cos18 A1 Use of F = 0.65R to get an equation in T only DM1 R is a linear combination of a component of T and a component of weight. Using equations with no missing terms. Solve for T M1 Dependent on all 3 previous M marks. T = 64 (.0 ) N A1 7 4(b) Complete method to find s using constant acceleration formula(e) M1 Finding distance moved between t = 3 and t = 4 , must be using a = 0.2  1 2 1 2 1 1  s =  0.2  4 −  0.2  3 OR s = ( 0 + 0.2  4 ) −4 ( 0 + 0.2  3 )  3    2 2 2 2  Distance = 0.7 m A1 If 0 marks scored then  1 2  SCB1 for s =  0.2  4 = 1.6    2  2

More questions on Kinematics of motion in a straight line

5 A particle P moves on the x-axis from the origin O with an initial velocity of −20ms−1. The acceleration ams−2 at time t s after leaving O is given by a = 12 −2t. (a) Sketch a velocity-time graph for 0 ≤t ≤12, indicating the times when P is at rest. 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(b) Find the total distance travelled by P in the interval 0 ≤t ≤12. 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Mark scheme: 5(a) Attempt to integrate 12 −t2 M1 For integration, the power of t must increase by 1 in at least 1 term with a change of coefficient in the same term. No +c required for this mark. s = vt is M0. 2 A1 No +c required for this mark.  2t  2 = 12t − t ( + c ) Allow unsimplified. v  = 12t − ( + c )   2  Use boundary conditions to get c = −20 B1 Solve 12t − t 2 − 20 = 0 to get t = 2 and t = 10 B1 soi Ignore anything outside 0 t 12 . Correct graph inverted quadratic starting at ( 0, −20 ) and ending at (12, −20 ) B1 t = 2 and t = 10 need not be shown. 5 5(b) Attempt to integrate their 12t − t 2 − 20 *M1 Integrating their 2 or 3 term expression for v from (a) which has come from integration. For integration, the power of t must increase by 1 in at least 1 term with a change of coefficient in the same term. 12 2 1 3  2 1 3 A1ft ft their +c  0 . Allow unsimplified. s = t − t − 20t ( + d ) = 6t − t − 20t ( + d )    2 3  3  2 1 3  DM1 Correct use of correct limits for one time Attempt to evaluate their  6t − t − 20t  for any of t = 0 to t = their 2 or interval.  3  t = their 2 to t = their 10 or t = their 10 to t = 12  2 1 3  DM1 Correct use of correct limits for all 3 time Attempt to evaluate their  6t − t − 20t  for all of t = 0 to t = their 2 or intervals, ignore signs here.  3  t = their 2 to t = their 10 or t = their 10 to t = 12  56   200  56    200  368 A1 Awrt 123 s = −− + −− −   −0    48 −  =  123 m      3   3  3    3  3 5(b) 2 B1 20  2 1 3  56 2 Either s =   6t − t − 20t  dt = = 18.7 Allow  0.25t − 8t + 60 dt = 26 . 0  3  3 10 10  2 1 3  256 Or s =   6t − t − 20t  dt = = 85.3 2  3  3 12  2 1 3  56 Or s =   6t − t − 20t  dt = = 18.7 10  3  3  56 256 56  368 B1 Awrt 123 s = + + =  123 m 12    3 3 3  3 2 1 3 368 6t − t − 20t dt =  123 Allow s =  0 3 3 m for B2. 5

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6 A 4 kg B 3 kg 30Å Fig. 6.1 Fig. 6.1 shows particles A and B, of masses 4kg and 3kg respectively, attached to the ends of a light inextensible string that passes over a small smooth pulley. The pulley is fixed at the top of a plane which is inclined at an angle of 30Å to the horizontal. A hangs freely below the pulley and B is on the inclined plane. The string is taut and the section of the string between B and the pulley is parallel to a line of greatest slope of the plane. (a) It is given that the plane is rough and the particles are in limiting equilibrium. Find the coefficient of friction between B and the plane. 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(b) A 4 kg 1 m B 3 kg 30Å Fig. 6.2 It is given instead that the plane is smooth and the particles are released from rest when the difference in the vertical heights of the particles is 1m (see Fig. 6.2). Use an energy method to find the speed of the particles at the instant when the particles are at the same horizontal level. 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Mark scheme: 6(a) T = 4 g B1 soi R = 3 g cos30 B1 Attempt to resolve parallel to the plane M1 3 terms, allow g missing. Allow sign errors, sin/cos mix. F = T − 3 g sin30 *A1 May see F = 25 . Eliminate T and use F = R to get an equation in  only DM1 Where R is a component of their weight. Coefficient of friction = 0.962 A1 5 3 allow . 9 allow 0.96. If F negative must say why using positive for this mark. 6 6(b) Find height gained by B relative to height lost by A M1 A loses x m in height, B gains x sin30 y OR B gains y m in height and A loses . sin30 2 A1 EITHER x + x sin30 = 1  x = 3 y 1 OR y + = 1  y = sin30 3 1 2 1 2  1 2  B1 Change in KE = 4 v + 3 v = 7 v   2 2  2   y  B1 x or y need not be substituted. 4 g − 3 gy 4 gx − 3 gy ) Change in PE  ( 4 gx − 3 gx sin30 ) or    OR  (  sin30  Conservation of energy M1 4 terms. 1 2 1 2 x or y need not be substituted. 4 gx − 3 gx sin30 = 4 v + 3 v Must be same v for both particles. 2 2 y 1 2 1 2 OR 4 g − 3 gy = 4 v + 3 v sin30 2 2 1 2 1 2 OR 4 gx − 3 gy = 4 v + 3 v 2 2 A1 2.182178902 100 10 21 Speed = = = 2.18 ms-1 SC B1 B1 M1 3/6 max for using x = y = 0.5 21 21 6(b) Alternative method 1 for final 4 marks of question 6(b) T − 3 g sin30 = 3a M1 Attempt at 2 equations from N2L on either particle or the system. Allow sign errors. 4 g − T = 4 a Allow sin/cos mix. Correct number of terms. 4 g − 3 g sin30 = ( 4 + 3) a 18 A1 5 25 Solve to get T = g  25.7 May see a = g =  3.57 7 14 7 y 1 2 1 2 M1 Attempt at work energy using their T  = 3 v + 3 gy OR 4 gx = Tx + 4 v sin30 2 2 T (  4 g or 3g sin30 ) . May be in terms of x and/or y. A1 100 10 21 Speed = = = 2.18 ms-1 21 21 6(b) Alternative method 2 for final 4 marks of question 6(b): Special case where constant acceleration assumed. Score maximum 4/6 Find height gained by B relative to height lost by A M1 A loses x m in height, B gains x sin30 y OR B gains y m in height and A loses . sin30 2 A1 EITHER x + x sin30 = 1  x = 3 y 1 OR y + = 1  y = sin30 3 25 B1 T − 3 g sin30 = 3a and 4 g − T = 4 a  a = = 3.57 7 25 OR 4 g − 3 g sin30 = ( 4 + 3 ) a  a = = 3.57 7 B1 100 10 21 Uses constant acceleration to get speed = = = 2.18 m s–1 21 21 6

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