Cambridge A Level Mathematics 9709 — 2017 Oct/Nov Paper 4 · Variant 3

1 20 N 60Å P F N 60Å 30 N Three coplanar forces of magnitudes F N, 20 N and 30 N act at a point P, as shown in the diagram. The resultant of the three forces acts in a direction perpendicular to the force of magnitude F N. Find the value of F. [3] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1 (X=) 20 cos 60 + 30 cos 60 – F B1 [F = 20 cos 60 + 30 cos 60] M1 Use of horizontal component of resultant = 0 F = 25 A1 3

More questions on Forces and equilibrium

2 A lorry of mass 7850 kg travels on a straight hill which is inclined at an angle of 3Å to the horizontal. There is a constant resistance to motion of 1480 N. (i) Find the power of the lorry’s engine when the lorry is going up the hill at a constant speed of 10 m s−1. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Find the power of the lorry’s engine at an instant when the lorry is going down the hill at a speed of 15 m s−1 with an acceleration of 0.8 m s−2. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 2(i) [F = 1480 + 7850g sin 3] ( = 5588) M1 P M1 Using P = Fv and solving for P [10 = 1480 + 7850g sin 3] →P = … Power = 55 900 W A1 3 2(ii) [F + 7850g sin 3 – 1480 = 7850 × 0.8] M1 Use of Newton’s Second Law (F = 3652) P M1 Using P = Fv and solving for P [15 + 7850g sin 3 – 1480 = 7850 × 0.8] →P = … Power = 54800 W A1 3

More questions on Forces and equilibrium

4 Two particles A and B have masses 0.35 kg and 0.45 kg respectively. The particles are attached to the ends of a light inextensible string which passes over a small fixed smooth pulley which is 1 m above horizontal ground. Initially particle A is held at rest on the ground vertically below the pulley, with the string taut. Particle B hangs vertically below the pulley at a height of 0.64 m above the ground. Particle A is released. (i) Find the speed of A at the instant that B reaches the ground. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Assuming that B does not bounce after it reaches the ground, find the total distance travelled by A between the instant that B reaches the ground and the instant when the string becomes taut again. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 4(i) EITHER: (M1 Applies Newton’s Second Law to one of [T – 0.35g = 0.35a the particles or forms system equation in or 0.45g – T = 0.45a a (mBg – mAg = (mA + mB)a) or 0.45g – 0.35g = 0.8a] [0.45g – T = 0.45a M1 Applies Newton’s Second Law to form or T – 0.35g = 0.35a] →a = … second equation in T and a and solves for a or solves system equation for a a = 1.25 m s–2 A1 [v2 = 2 × 1.25 × 0.64] (= 1.6) M1 Using v2 = u2 + 2as Velocity = 1.26 ms–1 A1) OR: (M1 Attempts PE loss [PE loss = 0.45g × 0.64 – 0.35g × 0.64] [KE gain = ½ (0.35 + 0.45) v2] M1 Attempts KE gain PE loss = 0.45g × 0.64 – 0.35g × 0.64 A1 and KE gain = ½ (0.35 + 0.45) v2 [½ (0.8) v2 = 0.1g × 0.64] (v2 = 1.6) M1 Using PE loss = KE gain Velocity = 1.26 ms–1 A1) 5 4(ii) EITHER: (M1 Using v2 = u2 + 2as [0 = 1.6 – 2 gs] (s = 0.08) Distance = 0.16 m A1) OR: (M1 Using PE gain = KE loss for particle A [0.35gh = ½ (0.35) × 1.6] (h = 0.08) Distance = 0.16 m A1) 2

More questions on Kinematics of motion in a straight line

5 A particle starts from a fixed origin with velocity 0.4 m s−1 and moves in a straight line. The acceleration a m s−2 of the particle t s after it leaves the origin is given by a = k 3t2 −12t + 2 , where k is a constant. When t = 1, the velocity of P is 0.1 m s−1. (i) Show that the value of k is 0.1. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Find an expression for the displacement of the particle from the origin in terms of t. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (iii) Hence verify that the particle is again at the origin at t = 2. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 5(i) v = ∫ k(3t2 – 12t + 2) dt *M1 Use of v = ∫ a dt = k(3t3/3 – 12t2/2 + 2t) + C v = k t 3 − 6t 2 + 2t + C A1 Condone C missing ( ) C = 0.4 B1 0.1 = k(1 – 6 +2) + 0.4 [–0.3 = –3k] DM1 Substitutes t = 1, v = 0.1 k = 0.1 A1 AG 5 5(ii) [s = ∫ 0.1(t3 – 6t2 +2t) + 0.4 dt M1 Use of s = ∫ v dt = 0.1(t4/4 – 6t3/3 + 2t2/2) + 0.4t + C] s = 0.025t4 – 0.2t3 + 0.1t2 + 0.4t A1 C = 0 seen or implied 2 5(iii) Substitutes t = 2 to show s = 0 B1 AG 1

More questions on Kinematics of motion in a straight line

6 v (m s−1) V Q t (s) 0 2 6 10 12 T −6 P The diagram shows the velocity-time graphs for two particles, P and Q, which are moving in the same straight line. The graph for P consists of four straight line segments. The graph for Q consists of three straight line segments. Both particles start from the same initial position O on the line. Q starts 2 seconds after P and both particles come to rest at time t = T. The greatest velocity of Q is V m s−1. (i) Find the displacement of P from O at t = 10. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Find the velocity of P at t = 12. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (iii) Given that the total distance covered by P during the T seconds of its motion is 49.5 m, find the value of T. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (iv) Given also that the acceleration of Q from t = 2 to t = 6 is 1.75 m s−2, find the value of V and hence find the distance between the two particles when they both come to rest at t = T. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 6(i) [Area = ½ (10 + 4) × 6 = 42 m] B1 Displacement = 42 m 1 6(ii) v 6 M1 Using similar triangles or using = acceleration = gradient and v = u + at 2 4 or [gradient =1.5, v = 6 + 1.5 × 6] v = 3 ms–1 A1 2 6(iii) Total distance travelled B1 FT Area found with FT distance from (i) and = 42 + ½ (T – 10) × 3 FT speed from (ii) [42 + ½ (T – 10) × 3 = 49.5] →T= … M1 For equation and solving for T T = 15 s A1 3 6(iv) V = 1.75 × 4 = 7 ms-1 B1 Q travels [½ (13 + 6) × 7 = 66.5 m] M1 Finding area for Q and interpreting total Distance apart = [66.5 + 42 – 7.5] distance between particles Distance between P and Q = 101 m A1 3

More questions on Kinematics of motion in a straight line

7 T N 15Å P 30Å A particle P of mass 0.2 kg rests on a rough plane inclined at 30Å to the horizontal. The coefficient of friction between the particle and the plane is 0.3. A force of magnitude T N acts upwards on P at 15Å above a line of greatest slope of the plane (see diagram). (i) Find the least value of T for which the particle remains at rest. [6] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ The force of magnitude T N is now removed. A new force of magnitude 0.25 N acts on P up the plane, parallel to a line of greatest slope of the plane. Starting from rest, P slides down the plane. After moving a distance of 3 m, P passes through the point A. (ii) Use an energy method to find the speed of P at A. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 7(i) R = 0.2g cos 30 – T sin 15 B1 [F = 0.3 × (0.2g cos 30 – T sin 15)] M1 Use of F = µR M1 For resolving along the plane T cos15 + 0.3 × (0.2g cos30 – T sin15) A1 = 0.2gsin30 M1 For solving a 4 term equation for T T = 0.541 A1 6 7(ii) 0.3 × 0.2g cos 30 × 3 [= 1.5588 J] B1 WD against F = friction × distance WD = 0.25 × 3 [= 0.75 J] B1 WD against 0.25 force 0.2g × 3 sin 30 [= 3 J] B1 PE loss = mgh [½ (0.2) v2 = 3 – 1.5588 – 0.75] M1 Work/Energy equation Speed = 2.63 ms-1 A1 5

More questions on Forces and equilibrium