Free until 14 Nov · 80% off for the first 100 · Pricing
Cambridge AS & A Level Mathematics (9709), Paper 4 Mechanics, October/November 2022. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
9709_w22_qp_419709_w22_ms_419709/41/O/N/22
Each question as it's printed, and the mark scheme's points for it, as text. Diagrams, tables and graphs are in the PDFs above.
Question
Coplanar forces of magnitudes P N, Q N, 16 N and 22 N 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
1 Attempt at resolving horizontally or vertically M1 Allow sign errors, allow sin/cos mix. 3 terms. P cos 25 = 22 + 16cos55 A1 Q + 16sin 55 = 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
Question
Small smooth spheres A and B, of equal radii and of masses 6 kg and 2 kg respectively, lie on a smooth horizontal plane. Initially A is moving towards B with speed 5 m s−1 and B is moving towards A with speed 3 m s−1 . After the spheres collide, both A and B move in the same direction and the difference in the speeds of the spheres is 2 m s−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) = 6vA + 2vB different. Use vB = v A + 2 or v A = vB − 2 with their momentum equation and solve for v A or v B DM1 Allow vB = v A ± 2 or v A = vB ± 2 . v A = 2.5 or vB = 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 1 × 6 × 52 + × 2 × ( −3) [ = 84] 2 Initial KE = 2 2 1 1 Final KE = × 6 × ( their 2.5 ) + × 2 × ( their 4.5 ) 2 2 2 2 1 1 2 Change in KE for A = ± × 6 × 52 − × 6 × ( their 2.5) 2 2 1 1 2 Change in KE for B = ± × 2 × ( −3) − × 2 × ( their 4.5) 2 2 2 Loss of KE = 45 J A1 Allow –45 J. Allow if mgv used in momentum equation. 5
Question
A constant resistance of magnitude 1400 N acts on a car of mass 1250 kg. (a) The car is moving along a straight level road at a constant speed of 28 m s−1 . Find, in kW, the rate at which the engine of the car is working. [2] (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 43.5 kW. Find this speed. [3] (c) On another occasion, the car pulls a trailer of mass 600 kg 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 5000 N and the resistance to the motion of the trailer is 300 N. The resistance to the motion of the car remains 1400 N. Find the acceleration of the system and the tension in the cable. [4]
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 × sin 6.89 = 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 = 1250a Allow use of 6.89º or 6.9º. Trailer: T − 300 − 600 g × 0.12 = 600a Allow with g missing. System: 5000 −1400 − 300 −1250g × 0.12 − 600g × 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
Question
A block of mass 8 kg 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 0.2 m s−2 . Find T . [7] (b) The block is initially at rest. Find the distance travelled by the block during the fourth second of motion. [2]
Mark scheme
4(a) Attempt at N2L parallel to the plane *M1 4 terms. Allow sign errors, sin/cos mix, allow g missing. T cos 26 − 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 = 8g 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 1 1 1 s = 2 × 0.2 × 4 − 2 × 0.2 × 3 OR s = 2 ( 0 + 0.2 × 4 ) × 4 − 2 ( 0 + 0.2 × 3) × 3 2 2 Distance = 0.7 m A1 If 0 marks scored then 1 SCB1 for s = × 0.2 × 42 = 1.6 2 2
Question
A particle P moves on the x-axis from the origin O with an initial velocity of −20 m s−1 . The acceleration a m s−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. [5] (b) Find the total distance travelled by P in the interval 0 ≤ t ≤ 12. [5]
Mark scheme
5(a) Attempt to integrate 12 − 2t 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. 2t 2 A1 No +c required for this mark. v = 12t − ( +c ) = 12t − t 2 ( +c ) Allow unsimplified. 2 Use boundary conditions to get c = −20 B1 Solve 12t − t 2 − 20 = 0 to get t = 2 and t = 10 B1 soi Correct graph inverted quadratic starting at ( 0, −20) and ending at (12, −20) B1 Ignore anything outside 0 t 12 . 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 1 1 A1ft ft their +c ≠ 0 . Allow unsimplified. s = t 2 − t 3 − 20t ( +d ) = 6t 2 − t 3 − 20t ( +d ) 2 3 3 1 DM1 Correct use of correct limits for one time Attempt to evaluate their 6t 2 − t 3 − 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 1 DM1 Correct use of correct limits for all 3 time Attempt to evaluate their 6t 2 − t 3 − 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) B1 20 2 1 3 2 56 Either s = 0 6t − 3 t − 20t dt = 3 = 18.7 0.25t − 8t + 60 dt = 26 . 2 Allow 10 10 1 256 Or s = 6t 2 − t 3 − 20t dt = 2 3 3 = 85.3 12 1 3 56 Or s = 6t − 3 t − 20t dt = 3 = 18.7 2 10 56 256 56 368 B1 Awrt 123 s = + + = ≈ 123 m 12 3 3 1 3 368 6t − 3 t − 20t dt = 3 ≈ 123 3 3 Allow s = 2 0 m for B2. 5
Question
A 4 kg B 3 kg 30Å Fig. 6.1 Fig. 6.1 shows particles A and B, of masses 4 kg and 3 kg 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. [6] (b) A 4 kg 1m 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 1 m (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. [6] 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.
Mark scheme
6(a) T = 4g B1 soi R = 3g cos 30 B1 Attempt to resolve parallel to the plane M1 3 terms, allow g missing. Allow sign errors, sin/cos mix. F = T − 3g sin 30 *A1 May see F = 25 . Eliminate T and use F = μ R to get an equation in μ 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 sin 30 y OR B gains y m in height and A loses . sin 30 2 A1 EITHER x + x sin 30 = 1 x = 3 y 1 OR y + =1 y = sin 30 3 1 1 1 B1 Change in KE = × 4 × v 2 + × 3 × v 2 = × 7 × v 2 2 2 2 y B1 x or y need not be substituted. Change in PE ± ( 4 gx − 3gx sin 30) or ± 4 g − 3gy OR ± ( 4 gx − 3gy ) sin 30 Conservation of energy M1 4 terms. 1 1 x or y need not be substituted. 4 gx − 3gx sin 30 = × 4 × v 2 + × 3 × v 2 Must be same v for both particles. 2 2 y 1 1 OR 4 g − 3gy = × 4 × v 2 + × 3 × v 2 sin 30 2 2 1 1 OR 4 gx − 3gy = × 4 × v 2 + × 3 × v 2 2 2 100 10 21 A1 2.182178902 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 − 3g sin 30 = 3a M1 Attempt at 2 equations from N2L on either particle or the system. Allow sign errors. 4 g − T = 4a Allow sin/cos mix. Correct number of terms. 4g − 3g sin 30 = ( 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 1 M1 Attempt at work energy using their T× = × 3 × v 2 + 3gy OR 4 gx = Tx + × 4 × v 2 sin 30 2 2 T ( ≠ 4g or 3g sin 30) . May be in terms of x and/or y. 100 10 21 A1 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 sin 30 y OR B gains y m in height and A loses . sin 30 2 A1 EITHER x + x sin 30 = 1 x = 3 y 1 OR y + =1 y = sin 30 3 25 B1 T − 3g sin 30 = 3a and 4 g − T = 4a a = = 3.57 7 25 OR 4 g − 3g sin 30 = ( 4 + 3) a a = = 3.57 7 100 10 21 B1 Uses constant acceleration to get speed = = = 2.18 m s–1 21 21 6
Checked your answers against these? Let AI mark the whole paper, on your own working.
Mark my 9709/41Examiners follow the generic marking principles at the front of the scheme, then each question's points: method (M), accuracy (A) and independent (B) marks, error carried forward where the scheme allows it, and positive marking, so you're credited for what you got right.
Done this paper? Upload your script and CIE Buddy's AI marker marks it question by question against this exact scheme, puts the ticks on your own working and says what each lost mark needed.
Question papers and mark schemes © Cambridge University Press & Assessment. CIE Buddy is independent and not connected to Cambridge.
On this page, free: the mark scheme for AS & A Level Mathematics 9709/41, October/November 2022 (9709_w22_ms_41), and the question paper (9709_w22_qp_41), to read or download, with each question's marking points written out underneath.
Mathematics 9709/41 for October/November 2022 is out of 50 marks, with 1 hour 15 minutes to answer it.
Yes. Upload your answers to 9709/41/O/N/22 as a PDF or photos, and CIE Buddy marks every question against this paper's official mark scheme, using Cambridge's marking rules. You get your score, each lost mark explained on your own script, and the topics to revise. It's a guide, not an official Cambridge grade.