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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_439709_w22_ms_439709/43/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
A particle P is projected vertically upwards with speed u m s−1 from a point on the ground. P reaches its greatest height after 3 s. (a) Find u. [1] (b) Find the greatest height of P above the ground. [2]
Mark scheme
1(a) u = 30 B1 From v = u + at or equivalent. 1 1(b) 0 = 302 − 2 ×10 × s M1 Use of suvat formulae. Greatest height = 45 m A1 FT FT u2/2g with their u from (a). 2
Question
A box of mass 5 kg is pulled at a constant speed of 1.8 m s−1 for 15 s up a rough plane inclined at an angle of 20Å to the horizontal. The box moves along a line of greatest slope against a frictional force of 40 N. The force pulling the box is parallel to the line of greatest slope. (a) Find the change in gravitational potential energy of the box. [2] (b) Find the work done by the pulling force. [2]
Mark scheme
2(a) PE = 5 g ×15 ×1.8 sin 20 M1 Attempt to find PE gain. PE = 462 J A1 From 461.727… 2 2(b) WD = 5 g ×15 ×1.8 sin 20 + 40 ×15 ×1.8 M1 Uses WD by pulling force = PE gain + WD against or friction or WD = Fs. WD = (5gsin 20 + 40) ×15 ×1.8 WD = 1540 J A1 FT From 1541.727… FT ‘1080 + PE from (a)’. 2
Question
A ring of mass 4 kg is threaded on a smooth circular rigid wire with centre C. The wire is fixed in a vertical plane and the ring is kept at rest by a light string connected to A, the highest point of the circle. The string makes an angle of 25Å to the vertical (see diagram). Find the tension in the string and the magnitude of the normal reaction of the wire on the ring. [6]
Mark scheme
3 T cos 25 = 40 + R cos 50 M1 Resolving in any direction e.g. horizontal, vertical, along radius or tangent. R sin 50 = T sin 25 M1 Resolving in a second direction. Radially: T cos 25 = R + 40 cos 50 A1 Two correct equations. Tangentially: T sin 25 = 40 sin 50 Parallel to T: T = R cos 25 + 40 cos 25 Perpendicular to T: R sin 25 = 40 sin 25 Vertically: T cos 25 = 40 + R cos 50 Horizontally: R sin 50 = T sin 25 Solving equation(s) to find either T or R M1 T = 72.5 N A1 From 72.504…. R = 40 N A1 6
Question
A particle P travels in the positive direction along a straight line with constant acceleration. P travels a distance of 52 m during the 2nd second of its motion and a distance of 64 m during the 4th second of its motion. (a) Find the initial speed and the acceleration of P. [5] (b) Find the distance travelled by P during the first 10 seconds of its motion. [2]
Mark scheme
4(a) ( 52 = u ( 2 ) + 0.5a ( 2 ) - u (1) + 0.5a (1) 2 2 ) M1 Use of s = ut + ½ at2 or equivalent to form equation for 2nd or 4th second . or 64 = u ( 4 ) + 0.5a ( 4 ) - (u ( 3) + 0.5a ( 3) 2 2 M1 Second equation in u and a. 52 = u +1.5a and 64 = u + 3.5a A1 Two correct equations in u and a. 12 = 2a leading to a = 6 M1 Solves simultaneous equations to find either u or a. Initial speed = 43 ms −1 and acceleration = 6 ms −2 A1 5 4(b) s = 43 × 10 + 0.5 × 6 × 102 M1 Use of s = ut + ½ at2 or equivalent. Distance = 730m A1 FT FT 10u + 50a with u and a from part (a). 2
Question
Particles X and Y move in a straight line through points A and B. Particle X starts from rest at A and moves towards B. At the same instant, Y starts from rest at B. At time t seconds after the particles start moving ³ the acceleration of X in the direction AB is given by 12t + 12 m s−2 , ³ the acceleration of Y in the direction AB is given by 24t − 8 m s−2 . (a) It is given that the velocities of X and Y are equal when they collide. Calculate the distance AB. [6] (b) It is given instead that AB = 36 m. Verify that X and Y collide after 3 s. [2]
Mark scheme
5(a) v = 12t + 12 dt , v X = 6t 2 + 12t or v = 24t − 8 dt , vY = 12t 2 − 8t M1 Uses v = a dt . v X = 6t 2 + 12t vY = 12t 2 − 8t A1 All correct. 10 B1 Solves v X = vY . 6t 2 + 12t = 12t 2 − 8t leading to t = 3 s X = 6t 2 + 12t dt s X = 2t 3 + 6t 2 *M1 Uses s = v dt . s = 12t − 8t dt s = 4t − 4t Y 2 Y 3 2 10 3 10 2 3 10 10 2 DM1 10 sX = 2 + 6 sY = 4 − 4 Evaluates s for each particle. 3 3 3 3 3 3800 2800 1000 A1 AB = − = Distance AB = 37 m 27 27 27 6 5(b) AB = −2t 3 + 10t 2 = −2 × 33 + 10 × 32 M1 Calculates distance AB. [ t = 3 ], AB = 36 m A1 AG 2
Question
A car of mass 1750 kg is pulling a caravan of mass 500 kg. The car and the caravan are connected by a light rigid tow-bar. The resistances to the motion of the car and caravan are 650 N and 150 N respectively. (a) The car and caravan are moving along a straight horizontal road at a constant speed of 24 m s−1 . (i) Find the power of the car’s engine. [2] (ii) The engine’s power is now suddenly increased to 40 kW. Find the instantaneous acceleration of the car and caravan and find the tension in the tow-bar. [5] (b) The car and caravan now travel up a straight hill, inclined at an angle sin−1 0.14 to the horizontal, at a constant speed of v m s−1 . The car’s engine is working at 31 kW. The resistances to the motion of the car and caravan are unchanged. Find v. [3]
Mark scheme
6(a)(i) P = (650 + 150) × 24 M1 Use of P = DF × v. 19 200 W or 19.2 kW A1 2 6(a)(ii) 40 000 = DF × 24 B1 Correct use of P = DF × v. 40000 M1 Use of Newton’s Second Law for the system or for − 800 = 2250a the caravan or for the car. 24 T −150 = 500a A1 Two correct equations. 40000 M1 Solves for a or for T. a= − 800 ÷ 2250 leading to a = 24 −2 Acceleration = 0.385 ms and Tension = 343 N A1 52 From a = = 0.38518…. and 135 9250 T= = 342.59… 27 5 6(b) DF = 800 + 2250 × g × 0.14 M1 Resolving up hill using DF = Total resistances. 31000 M1 Use of P = DF × v to form equation in v . = 800 + 2250 × g × 0.14 v v = 7.85 A1 620 From v = = 7.848… 79 3
Question
Particles of masses 1.5 kg and 3 kg lie on a plane which is inclined at an angle of ! to the horizontal, where tan ! = 34 . The section of the plane from A to B is smooth and the section of the plane from B to C is rough. The 1.5 kg particle is held at rest at A and the 3 kg particle is in limiting equilibrium at B. The distance AB is x m and the distance BC is 4 m (see diagram). (a) Show that the coefficient of friction between the particle at B and the plane is 0.75. [3] The 1.5 kg particle is released from rest. In the subsequent motion the two particles collide and coalesce. The time taken for the combined particle to travel from B to C is 2 s. The coefficient of friction between the combined particle and the plane is still 0.75. (b) Find x. [6] (c) Find the total loss of energy of the particles from the time the 1.5 kg particle is released until the combined particle reaches C. [3] 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
7(a) R = 3g cos α = 3 × 10 × 0.8 B1 F = 3g sin α = 3 × 10 × 0.6 M1 Resolving parallel to plane. 18 3 g sin α A1 F μ= = 0.75 or μ = = tan α = 0.75 Uses μ = AG. 24 3 g cos α R 3 7(b) a = g sin α or PE loss = 1.5 gx sin α for AB and a = 0 for BC B1 Accelerations for AB and BC. 4.5g × sin α − 0.75 × 4.5 g cos α = 4.5a leading to a = 0 v12 = 2 × g sin α ×x ] or [ 1.5g ×x sin α = 0.5 ×1.5 × v12 M1 Uses ’suvat’ or PE loss = KE gain for AB. v12 = 20x sin α = 12x leading to v1 = 12 x A1 1 M1 Conservation of momentum. 1.5 × 12x + 0 = 4.5 × v2 leading to v2 = 12 x 3 2 M1 Use of s = vt on BC since a = 0. 4= × 12 x 3 x=3 A1 7(b) Alternative Method for 7(b) a = g sin α or PE loss = 1.5gx sin α for AB and a = 0 for BC B1 Accelerations for AB and BC. 4.5g × sin α − 0.75 × 4.5g cos α = 4.5a leading to a = 0 4 = 2v2 leading to v2 = 2 M1 Uses s = vt on BC since a = 0. 1.5 × v1 + 0 = 4.5 × 2 M1 Conservation of momentum. v1 = 6 A1 Velocity before collision. 62 = 2 × g sin α × x or 1.5 g × x sin α = 0.5 ×1.5 × 62 M1 Uses suvat or PE loss = KE gain for AB. x=3 A1 6 7(c) KE = 0.5 × 4.5 × 22 = 9J B1 KE gain for AC. 3 3 M1 Evaluates PE loss for AC. PE loss = 15× ( 4 + 3) × + 30 × 4 × = 135 J 5 5 Loss of energy = 126 J A1 3
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