Cambridge A Level Mathematics 9709 — 2022 May/June Paper 4 · Variant 1

1 A car starts from rest and moves in a straight line with constant acceleration for a distance of 200m, reaching a speed of 25ms−1. The car then travels at this speed for 400m, before decelerating uniformly to rest over a period of 5s. (a) Find the time for which the car is accelerating. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Sketch the velocity–time graph for the motion of the car, showing the key points. [2] (c) Find the average speed of the car during its motion. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 1(a) 0 25 200 2   t For use of 2  u v s t or other complete method to find t e.g. 2 2 2   v u as followed by   v u at N.B. a = 1.5625 t = 16s A1 2 1(b) Trapezium B1 Through (0, 0) and positive t-axis All correct through (0, 0), (16, 25), (32, 25), (37, 0) B1 FT their value of t from part (a) (t +16 + 5) 2 1(c) Total distance = 1 200 400 25 5 2     [= 662.5] M1 Or trapezium    1 25 16 37 662.5 2    Or 25 0 200 400 5 2     Allow their value of t from part (a) in calculating 200 and their time from the constant speed section from part (b) in calculating 400 Allow their a from part (a) if used in calculating 200 Average speed = 662.5 37 = 17.9 m s−1 (3 s.f.) A1 Allow 1325 67 17 74 74  2

More questions on Kinematics of motion in a straight line

2 Two particles P and Q, of masses 0.5kg and 0.3kg respectively, are connected by a light inextensible string. The string is taut and P is vertically above Q. A force of magnitude 10N is applied to P vertically upwards. Find the acceleration of the particles and the tension in the string connecting them. [5] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 Use of Newton’s second law for P or Q or system M1 Allow sign errors. Do not allow g missing For P: 10 – 0.5g – T = 0.5a, For Q: T – 0.3g = 0.3a, For system: 10 – 0.5g – 0.3g = 0.8a A1 For any two correct For attempt to solve for either a or T M1 From equation(s) with no missing or extra terms. Allow g missing. a = 2.5 ms–2 A1 T = 3.75 N A1 5

More questions on Kinematics of motion in a straight line

3 A crate of mass 300kg is at rest on rough horizontal ground. The coefficient of friction between the crate and the ground is 0.5. A force of magnitude X N, acting at an angle ! above the horizontal, is applied to the crate, where sin ! = 0.28. Find the greatest value of X for which the crate remains at rest. [5] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 3 For attempt at resolving horizontally or vertically M1 Allow sin/cos mix. Allow sign error. Allow g missing. Correct number of terms. R = 300g – 0.28X or 300 sin16.3   R g X A1 α = 16.26… 0.96X – F = 0 or 0.96 0.5(300 sin ) 0     X g X Or X cos 16.3 – F = 0 or X cos 16.3 0.5(300 sin ) 0     g X A1 Or using their F Use of F = 0.5R M1 Use to get an equation in X only. Allow sin/cos mix. Allow sign error. Allow g missing. Must be from 2 term R, which is a linear combination of 300(g) and a component of X X = 1360 [1363.63…] A1 5

More questions on Forces and equilibrium

4 Three coplanar forces of magnitudes 20N, 100N and F N act at a point. The directions of these forces are shown in the diagram. Given that the three forces are in equilibrium, find F and !. [6] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 4 Attempt to resolve in any direction M1 For resolving. Allow sin/cos mix. Allow sign error. Correct number of terms. F cos α – 20 cos 40 – 100 sin 20 = 0 [F cos α = 15.320…+ 34.202… = 49.5229…] A1 F sin α + 20 sin 40 – 100 cos 20 = 0 [F sin α = 93.969… − 12.855… = 81.1135…] A1 F     2 2 49.5229 81.1135     M1 OE; Attempt to solve for F; one term missing in total α 1 81.1135 tan 49.5229           M1 OE; Attempt to solve for α; one term missing in total F = 95(.0), α = 58.6 A1 F = 95.0364… and α = 58.5943… Alternative mark scheme for question 4: For candidates who use cosine and/or sine rule Attempt at cosine rule from triangle of forces M1 Must use lengths 100 and 20 with a suitable angle 2 2 2 100 20 2 100 20cos70      F A1 Correct F = 95[.0] A1 95.0364 20 95.0364 100 OR sin70 sin sin70 sin     M1 Attempt at sin rule A1 where   70     where   40     α = 58.6 A1 α = 58.5943… Question Answer Marks Guidance 4 Alternative mark scheme for question 4: For candidates who resolve in other directions Attempt to resolve (e.g. parallel or perpendicular to 100 N) M1 For resolving. Allow sin/cos mix. Allow sign error. Correct number of terms.     sin 20 20sin20 100 0 sin 20 93.159             F F A1     cos 20 20cos20 0 cos 20 18.793            F F A1 2 2 93.159 18.793   F M1 OE; Attempt to solve for F; one term missing in total 1 93.159 tan 20 18.793             M1 OE; Attempt to solve for α; one term missing in total F = 95[.0], α = 58.6 A1 F = 95.0364… and α = 58.5943… 6

More questions on Forces and equilibrium

5 Two racing cars A and B are at rest alongside each other at a point O on a straight horizontal test track. The mass of A is 1200kg. The engine of A produces a constant driving force of 4500N. When A arrives at a point P its speed is 25ms−1. The distance OP is d m. The work done against the resistance force experienced by A between O and P is 75000J. (a) Show that d = 100. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ Car B starts offat the same instant as car A. The two cars arrive at P simultaneously and with the same speed. The engine of B produces a driving force of 3200N and the car experiences a constant resistance to motion of 1200N. (b) Find the mass of B. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) Find the steady speed which B can maintain when its engine is working at the same rate as it is at P. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 5(a) For attempt at work energy equation M1 3 terms. Allow sign errors. M0 for (constant) acceleration method 4500d – 75 000 = 2 1 1200 25 2   [= 375 000] A1 Correct equation d = 100 A1 AG Accept verification with d substituted in above line to show LHS = 375 000 or LHS −RHS = 0 If no marks scored allow SCB1 for 2 1 1200 25 2   3 5(b) 2 25 0 2 100    a [leading to a = 3.125] B1 Allow B1 if acceleration found in part (a) as 3.125 and used or stated here 3200 – 1200 = m  3.125 M1 Newton’s second law with 3 terms. Allow sign errors and their a. Mass of car B = 640 kg A1 Alternative mark scheme for question 5(b) For attempt at work energy equation M1 3 terms. Allow sign errors. (3200 – 1200)  100 = 2 1 25 2   m A1 Correct equation Mass of car B = 640 kg A1 3 Question Answer Marks Guidance 5(c) At P power = 3200  25 [= 80 000] B1 For use of power = Fv 80000 1200 0   v M1 Attempt Newton’s second law for car B with a = 0 Allow their 80 000 (dimensionally correct) Steady speed = 66.7 m s−1 A1 Allow 200 2 66 3 3  3

More questions on Energy, work and power

6 A particle starts from a point O and moves in a straight line. The velocity vms−1 of the particle at time t s after leaving O is given by v = k 3t2 −2t3 , where k is a constant. (a) Verify that the particle returns to O when t = 2. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) It is given that the acceleration of the particle is −13.5ms−2 for the positive value of t at which v = 0. Find k and hence find the total distance travelled in the first two seconds of motion. [6] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 6(a) For attempt at integration *M1 The power of t must increase by 1 with a change of coefficient in at least 1 term. Allow if k is omitted.   3 4 1 2         k t t C A1 Allow unsimplified   1 8 16 0 2          k DM1 For use of limits 0 and 2 or substituting t = 2 OR equate to 0, then solve a quartic equation in t. Allow if k is omitted. 0 A1 AG, CWO including stating C = 0 if not using limits 4 Question Answer Marks Guidance 6(b) v = 0 when   2 3 3 2 0   k t t M1 For solving for t Leading to t = 1.5 [or t = 0] A1 A0 for other solutions not discarded   2 6 6   a k t t M1 For differentiation, the power of t must decrease by 1 with a change of coefficient in at least 1 term. Allow if k is omitted   2 [ 13.5 6 1.5 6 1.5      k ⇒   13.5 9 13.5    k ⇒] k = 3 A1 Distance from t = 0 to t = 1.5 is 1.5 3 4 3 4 0 1 1 3 3 1.5 1.5 0 2 2                        t t DM1 For use of limits. 1.5 0 2   or 2 15 2  or both integrals. Dependent on first M in part (a), unless they restart [= 2.53125] So total distance = 2  2.53125 = 5.06 m A1 Allow distance 81 1 5 16 16   If DM0 then SCB1 for 5.06 without working 6

More questions on Kinematics of motion in a straight line

7 Two particles A and B, of masses 0.4kg and 0.2kg respectively, are moving down the same line of greatest slope of a smooth plane. The plane is inclined at 30Å to the horizontal, and A is higher up the plane than B. When the particles collide, the speeds of A and B are 3ms−1 and 2ms−1 respectively. In the collision between the particles, the speed of A is reduced to 2.5ms−1. (a) Find the speed of B immediately after the collision. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ After the collision, when B has moved 1.6m down the plane from the point of collision, it hits a barrier and returns back up the same line of greatest slope. B hits the barrier 0.4s after the collision, and when it hits the barrier, its speed is reduced by 90%. The two particles collide again 0.44s after their previous collision, and they then coalesce on impact. (b) Show that the speed of B immediately after it hits the barrier is 0.5ms−1. Hence find the speed of the combined particle immediately after the second collision between A and B. [7] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 7(a) v = 3 ms−1 A1 Allow M1A0 if g included with the masses. 2 7(b) For A 0.4 sin30 0.4   g a or for B 0.2 sin30 0.2   g a or sin30   mg ma M1 For either. Allow sin/cos mix. a =  5 or  sin30 g A1 Allow sin30 g without working for M1A1 For B when hits barrier 2 2 3 2 5 1.6    v [⇒ v = 5] OR   3 5 0.4 5        v u at v v M1 Using their  a g and their v from part (a) OR: use of 2  u v s t   3 1.6 0.4 5 2      v v OR 2 2 1 1 0.2 0.2 3 0.2 1.6 sin30 2 2         v g Speed after hitting barrier = 0.1  5 = 0.5 A1 AG vA = 2.5 + 5  0.44 [= 4.7] vB = −0.5 + 5  0.04 [= −0.3] or vB = 0.5 + (–5)  0.04 [= 0.3] *M1 Use of   v u at for either with correct t-value, with initial speeds 2.5 or 0.5   their  a g 0.4  4.7 + 0.2  (−0.3) = 0.6 vcomb DM1 Use of   v u at for BOTH with correct t-values, initial speeds 2.5, 0.5   and  their acceleration (same for both) and use of conservation of momentum with correct number of terms. Allow sign errors. vcomb = 3.03 ms−1 A1 Allow 91 1 3 30 30   v Allow DM1A0 if g included with the masses. 7

More questions on Momentum