Cambridge A Level Mathematics - Further 9231 — 2020 Oct/Nov Paper 3 · Variant 1

1 A particle P of mass m is placed on a fixed smooth plane which is inclined at an angle i to the horizontal. A light spring, of natural length a and modulus of elasticity 3mg, has one end attached to P and the other end attached to a fixed point O at the top of the plane. The spring lies along a line of greatest slope of the plane. The system is released from rest with the spring at its natural length. Find, in terms of a and i, an expression for the greatest extension of the spring in the subsequent motion. [3] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 1 Gain in EPE = 2 1 3 . 2 mgx a Loss in GPE = sin mgx θ Equate M1 Equate energies 2 sin 3 x a θ = A1 Using forces scores B0M0A0 3

More questions on Hooke's law

2 O a P i 4 5 5ag A particle P is attached to one end of a light inextensible string of length a. The other end of the string is attached to a fixed point O. The particle P is held with the string taut and making an angle i with the downward vertical. The particle P is then projected with speed 45 5ag perpendicular to the string and just completes a vertical circle (see diagram). Find the value of cosi. [5] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 2 At top, tension = 0, so 2 mv mg a = ( 2 ) v ag = B1 ( ) 2 2 1 1 1 cos 2 2 mv mu mga θ = − + M1 A1 Energy equation Substitute for u and v : ( ) 16 .5 2 1 cos 25 ag ag ag θ = − + M1 Eliminate 1 cos 10 θ = A1 5

More questions on Circular motion

3 One end of a light elastic string, of natural length a and modulus of elasticity 4mg, is attached to a fixed point O. The other end of the string is attached to a particle of mass m. The particle moves in g a horizontal circle with a constant angular speed with the string inclined at an angle i to the a downward vertical through O. The length of the string during this motion is ( k + 1) a . (a) Find the value of k. [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find the value of cosi. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 3(a) 4 .ka T mg a = B1 Use Hooke’s law ( ) sin 1 sin . mrg g T m k a a a θ θ   = = +     M1 N2L horizontally. Must see T and k . ( )1 T mg k = + A1 Equate: 1 3 k = A1 4 3(b) cos T mg θ ↑ = M1 ( 4 ) 3 T mg = 3 cos 4 4 3 mg mg θ = = A1 2

More questions on Circular motion

4 E D 2r C h A B 6r The diagram shows the cross-section ABCD of a uniform solid object which is formed by removing a cone with cross-section DCE from the top of a larger cone with cross-section ABE. The perpendicular distance between AB and DC is h, the diameter AB is 6r and the diameter DC is 2r. (a) Find an expression, in terms of h, for the distance of the centre of mass of the solid object from AB. [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ The object is freely suspended from the point B and hangs in equilibrium. The angle between AB and the downward vertical through B is i. (b) Given that h = 134 r , find the value of tani. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 4(a) Volume Centre of mass from AB Small cone 2 1 . 3 2 h r π 1 9 . 4 2 8 h h h   + =     Large cone ( ) 2 1 3 3 . 3 2 h r π 1 3 3 . 4 2 8 h h   =     Object ( ) 2 26 6 r h π x B1 For 9h/8 or 3h/8 (unsimplified) Take moments about AB 2 2 2 13 27 3 1 9 . . . 3 6 8 6 8 h h r h x r h r h π π π = − M1 A1 Moments equation: Allow use of relative masses 1, 26, 27 9 26 h x = A1 4 4(b) tan 3 x r θ = M1 (= 3 ) 26 h r Use 13 4 h r = 3 tan 8 θ = A1 2

More questions on Equilibrium of a rigid body

5 A particle P is projected with speed u at an angle a above the horizontal from a point O on a horizontal plane and moves freely under gravity. The horizontal and vertical displacements of P from O at a subsequent time t are denoted by x and y respectively. (a) Derive the equation of the trajectory of P in the form gx 2 2 a . [3] y = x tan a - 2 sec 2u ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ The point Q is the highest point on the trajectory of P in the case where a = 45° . u 2 (b) Show that the x-coordinate of Q is . [3] 2g ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (c) Find the other value of a for which P would pass through the point Q. [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 5(a) α → = 2 1 sin 2 y u t gt α ↑ = − B1 Both Eliminate t: 2 1 x sin .ucos 2 ucosα x y u g α α   = −     M1 Eliminate 2 2 2 tan sec 2 gx y x u α α = − A1 AG 3 5(b) Greatest height = ( ) 2 sin 2 u g α = 2 4 u g M1 A1 Accept alternative methods, for example differentiate expression in (a) and equate to 0. sin 45 t u = /g so cos45. sin 45 d u u = /g = 2 2 u g A1 AG 3 Question Answer Marks Guidance 5(c) Use greatest height displacements in trajectory equation 2 2 4 2 2 2 tan sec 4 2 2 4 u u gu g g u g α α = − M1 Use equation of trajectory (substitute coordinates of Q 2 2 2 2 2 tan (1 tan ) 2 u u u α α = − + M1 Use of ( ) 2 2 sec 1 tan α α = + 2 tan 4tan 3 0 α α − + = M1 Obtain a three-term quadratic in tanα tan 1, 3 α = so 71.6 α = ° A1 Both solutions needed 4

More questions on Motion of a projectile

6 Two smooth spheres A and B have equal radii and masses m and 2m respectively. Sphere B is at rest on a smooth horizontal floor. Sphere A is moving on the floor with velocity u and collides directly with B. The coefficient of restitution between the spheres is e. (a) Find, in terms of u and e, the velocities of A and B after the collision. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ Subsequently, B collides with a fixed vertical wall which makes an angle i with the direction of motion of B, where tan i = 34 . The coefficient of restitution between B and the wall is 2.3 Immediately after B collides with the wall, the kinetic energy of A is 325 of the kinetic energy of B. (b) Find the possible values of e. [7] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 6(a) = + B1 Momentum equation (with m) v w eu − = B1 Restitution with consistent signs ( )1 3 u v e = + ( ) 1 2 3 u w e = − B1 Both correct. 3 Question Answer Marks Guidance 6(b) Perpendicular to plane: sin y ev θ = Parallel to plane: cos x v θ = B1 Both Speed of B = 2 2 x y + = 2 2 2 4 2 3 ( ) . 5 3 5 v     +           (= 2 5 v ) M1 Speed of B KE of B = ( ) 2 2 1 4 .2 . 1 2 5 9 u m e + M1 KE of B in terms of u . 1 2 and 2m needed KE of A = ( ) 2 2 1. . 1 2 2 9 u m e − So ( ) ( ) 2 2 2 2 1 5 1 4 . . 1 2 . .2 . 1 2 9 32 2 5 9 u u m e m e − = + M1 A1 Relate the two KEs ( ) ( ) 2 2 4 1 2 1 e e − = + or 2 15 18 3 0 e e − + = M1 Rearrange and simplify to quadratic ( ) 1 2 1 2 e e + = ± − 1, 1 5 e = A1 Both values 7

More questions on Momentum

7 A particle P moving in a straight line has displacement x m from a fixed point O on the line at time t s. -2 200 100 The acceleration of P, in ms , is given by 2 - 3 for x 2 0 . When t = 0 , x = 1 and P has velocity x x -1 10 ms directed towards O. -1 10 ( 1 - 2 x) (a) Show that the velocity v ms of P is given by v = . [5] x ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Show that x and t are related by the equation e -40 t = ( 2x - 1) e 2 x -2 and deduce what happens to x as t becomes large. [5] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ 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) 3 2 d 100 200 d v v x x x = − + 2 2 50 200 2 v A x x = − + M1 A1 Correct equation and attempt to integrate Correct 1, 10: 200 x v A = = − = M1 Use initial condition ( ) 2 2 2 100 2 1 x v x − = M1 Rearrange to find 2v ( ) 10 2 1 x v x − = ± and take negative sign to meet initial condition, so ( ) 10 1 2x v x − = A1 Convincingly shown (no mention of ± scores A0) AG 5 Question Answer Marks Guidance 7(b) d 10d 1 2 x x t x = − 1 1 1 d 10d 2 1 2 x t x   − =   −   1 ln 1 2 10 4 2 x x t B − − − = + M1 A1 Rearrange and attempt to integrate 1 0, 1: 2 t x B = = = − M1 Use initial condition 2 2 40 ln(1 2 ) x t x − = − − − so ( ) 40 2 2 2 1 t x e x e − − = − A1 Convincingly shown, working required AG For large values of t, 1 2 x → B1 CAO 5

More questions on Differential equations