ZNOTES.ORG UPDATED TO 2020-22 SYLLABUS CAIE A2 LEVEL FURTHER MATHS (9231) SUMMARIZED NOTES ON THE FURTHER MECHANICS SYLLABUS CAIE A2 LEVEL FURTHER MATHS (9231) Final velocity, according to the SUVAT equations, is also dependent on the initial velocity and any acceleration acting on the projectile: 1. Projectiles v = u + at 1.1. Initial Velocity Above is a scalar equation which can be written in each component as: vx = ucosθ vy = usinθ − gt o Note that only vy has a −gt term because gravity only affects the y -component, and it is negative due to gravity Consider the following diagram: going downwards. The final speed of the projectile can be achieved by taking the magnitude: v= vx2 + vy2 = (ucosθ)2 + (usinθ − gt)2 1.4. Displacement This shows that an initial speed can be defined in components as: In this notation: θ defines the angle of projection from the horizontal. u x defines the horizontal velocity component. u y defines the vertical velocity component The distance the particle has moved from the origin. Displacement is dependent on initial velocity and acceleration, or final velocity and acceleration: sx = ucos(θ)t sy = usin(θ)t − 12 gt2 These can be derived from one of the SUVAT equations: s = ut + 12 at2 1.2. Gravitational Acceleration Gravity affects all projectiles. Gravity only affects the vertical component of speed. g = 10ms−2 for both Mechanics syllabus in CAIE. Note: this assumes that air resistance is negligible 1.3. Final Velocity Sample Question Question: A ball was projected at an angle of 60° to the horizontal. One second later another ball was projected from the same point at an angle of 30° to the horizontal. One second after the second ball was released, the two balls collided. Show that the velocities of the balls were 12.99ms−1 and 15ms−1 . Take the value of g to be 10ms−2 . Solution: Visualise the scenario: At zero time: At t This defines the velocity of the projectile after seconds of travel. WWW.ZNOTES.ORG =1 CAIE A2 LEVEL FURTHER MATHS (9231) It can also be written in terms of & only as: u2 sin2θ g x max = Maximum Horizontal Range: the maximum possible horizontal distance the projectile can cover at a given initial velocity. u2 sin2θ , the maximum distance can g Knowing x max = be achieved when θ = 45° Cartesian Equation of a Trajectory: For any projectile, the trajectory can be modelled by the equation: y = tanθ − At t =2 gx2 2u2 sec 2 θ o θ is the angle of projection. o u is the initial speed. Therefore, when given a cartesian equation as: y = ax − bx 2 Then, a = tanθ b = Let velocity for first ball be U1 and second ball be U2 Displacement from origin is equal for both at impact. ∴ r1 = r2 r 1 = (U1 (2)cos60)i + (U1 (2)sin60 − 5(2)2 )j r 2 = (U2 (1)cos30)i + (U2 (1)sin30 − 5(1)2 )j Equate horizontal components: 2 vx = ucos30° vy = usin30° − gt ∴ 2U1 sin60 − 20 = U2 sin30 − 5 (2( 23 )( 23 ) − 12 )U2 = 15 U2 = 15ms−1 Find U1 U1 = 23 (15) = 12.99ms−1 to 0: Equate vertical components and substitute info above: 1.5. Special Events Time at Maximum Vertical Height: the time at which the projectile reaches its max height, at a given initial velocity and angle of elevation. This occurs when vertical component of velocity is : U sinθ − gt = 0 ∴ t = usineθ g Total Flight Time: knowing the time taken to reach max vertical height, its total flight time is just twice of it. Hence: t= 2usinθ g Horizontal Range: the horizontal distance the projectile covers at a given initial velocity and angle of elevation. Occurs when the vertical displacement is : utsinθ − 12 gt2 = 0 sec 2 θ Find, in terms of u, the speed of P at time 3 T after projection. Solution Get the projectile’s horizontal & vertical component: Simplifying: 9231/32/M/J/20 - Question 1 A particle is projected with speed u at an angle of 30° above the horizontal from a point O on a horizontal plane and moves freely under gravity. The particle reaches its greatest height at time T after projection. ∴ 2U1 cos60 = U2 cos30 U1 = 23 U2 g 2u2 vx = 23 u vy = 12 u − gt For P to be in max height, vertical compoment must be equal 0 = 12 u − gT u → T = 20 We used g = 10 . The above can be instantly found if you remember the formula for the time taken to reach max height. u and put it into our components: We can then let t = 23 T = 30 3 2 u vx = vy = Simplifying vy : 1 2u u g( 30 ) − vy = 12 u − = u6 u 3 We can take the magnitude of these components to find the speed: v= = ( 3 2 2 u) + ( u6 )2 \n = u x max = utcosθ WWW.ZNOTES.ORG 7 3 u 2. Equilibrium of a Rigid Body 2.1. Moment of a Force Find and substitute into the equation below: ( 3 2 )2 + ( u6 )2 Moment of a force = ∣F ∣ × ∣F ∣: magnitude of the force d CAIE A2 LEVEL FURTHER MATHS (9231) d : perpendicular distance from pivot to point of force Units: Newton-meter (N m) Moments are vector quantities and act clockwise or anticlockwise around pivot. Clockwise is generally considered as positive direction. Principle of Moments: when a system is in equilibrium the sum of anticlockwise moments is equal to the sum of clockwise moments. i. How far out from the cliff can a man of mass 75kg safely walk? ii. The man wishes to walk to the end of the plank. What is the minimum mass he should place on the other end of the plank to do this? Solution: Part (i) Draw up diagram of given scenario: anti − clockwise moments = clockwise moments Sample Question: The diagram shows an aerial view of a revolving door. Four people are exerting forces of 40N , 60N , 80N and 90N as shown. Find the distance x if the total moment of the forces about O is 12N m Use the principle of moments and solve for x : anti-clockwise moments = clockwise moments 100g × 2 = 75g × x ∴ x = 100g×2 = 2.67 75g Part (ii) Draw up diagram of given scenario: Solution: Use the sum of turning forces equation: clockwise moments + anticlockwise moments = 12N m Find clockwise moments: (80 × 1.6) + (60 × 0.8) = 128 + 48 = 176 Find anticlockwise moments: −((90 × 1.6) + (40 × x)) = −(144 + 40x) Substitute back into formula and solve for x : 176 + (−(144 + 40x)) = 12 176 − 144 − 40x = 12 x = 0.5m 2.2. Unknown Forces Magnitude of forces may not always be given. Eliminate an unknown/unwanted force by making the point on which it acts the pivot Sample Question: A uniform plank is m long and has mass kg. It is placed on horizontal ground at the edge of a cliff, with m of the plank projecting over the edge. WWW.ZNOTES.ORG Minimum mass therefore maximum distance from pivot Use the principle of moments and solve for x : anti-clockwise moments = clockwise moments (100g × 2) + (mg × 8) = 75g × 4 ∴ m = (75g×4)−(100g×2) = 12.5 m g×8 2.3. Forces in Different Directions Forces may act at an angle to the plane Equilibrium maintained using components of forces Sample Question A uniform ladder of mass 20kg and length 3m rests against a smooth wall with the bottom of the ladder on smooth horizontal ground and attached by means of a light inextensible string, 1m long, to the base of the wall i. Find the tension in the string ii. If the breaking strain of the string is 250N, find how far up the ladder a man of mass 80kg can safely ascend. Solution: Part (i) Daw up diagram of given scenario: CAIE A2 LEVEL FURTHER MATHS (9231) ∴ (m1 + m2 + …) (xy)   = m1 ( x1 x ) + m2 ( 2 ) y1  y2  +… We can solve for x & y separately instead of using vectors: (m1 + m2 + …)(x ) = m1 x 1 + m2 x 2 + … And (m1 + m2 + …)y = m1 y1 + m2 y2 + … 2.6. Centre of Mass of Rigid Bodies 1-Dimensional Objects: Uniform rod: centre of mass lies at midpoint of the rod 2-Dimensional Objects: Uniform Rectangular Lamina: centre of mass is at the intersection of diagonals The following and more informational can be found in the formula sheet List MF19 Find the angle of elevation, θ , from the horizontal: cosθ = 13 θ = 70.5° G= Use the principle of moments and solve for T : ( 1.5 × 20g × cos70.5 = 3 × T × cos19.5 T = 34.7 Part (ii) In above scenario assume that the man can take any position on the ladder, call it x Use the principle of moments and solve for x : anti-clockwise moments = clockwise moments  idpoint of m midpoint of AB ) C D Uniform Circular Lamina: centre of mass is at the centre of the circle ((1.5)20g + (x)80g)cos70.5 = (3)(250)cos19.5 x = 2.32 2.4. Centre of Mass Centre of Mass: centre of gravity of the system when it is placed in a gravitational field such that each part of system is subject to the same gravitational acceleration Centroid: geometrical centre; coincides with the centre of mass when the object is made of a uniformly dense material 2.5. Finding Centre of Mass We can find the centre of mass by taking moments If each mass mi has position vector, r i then the position vector of the centre of mass r is: r WWW.ZNOTES.ORG = Σmi r i Σmi ( ) ( ) Uniform Triangular Lamina: CAIE A2 LEVEL FURTHER MATHS (9231) 1 x x x ), B = ( 2 ), C = ( 3 ) y1  y2  y 3  1  (x + x 2 + x 3 ) ∴ G = ( 31 1 ) 3 (y1 + y2 + y3  A=( Uniform Circular sector with angle 2θ : C entre of mass : 14 h from base or 34 h from vertex Uniform Hemisphere: centre of mass lies in the centre of the base and height 38 of the radius h= 2rsinθ 2θ 3- Dimensional Objects: Uniform Solid Prism: centre of mass lies at the centre of mass of the cross-section and in the midpoint of length h= Uniform Cone: centre of mass lies in the centre of the base and height in ratio 3:1 to its height 3r 8 2.7. Composite Bodies Split composite body into simple geometrical shapes Find the centre of mass of each shape individually By taking moments with vectors, find the centre of mass of the composite body If the separate geometrical shapes have different densities, use V × ρ instead of just V Therefore, in general: x x x (v1 ρ 1 + v2 ρ 2 + …) ( ) = v1 ρ 1 ( 1 ) +v2 ρ 2 ( 2 ) y  y1  y2 +… WWW.ZNOTES.ORG CAIE A2 LEVEL FURTHER MATHS (9231) For a lamina with a hole in it, find the centre of mass of the lamina, then find centre of mass of the hole and use moments to find centre of mass of the shaded part In these cases: 9231/32/M/J/20 - Question 4: A uniform sqaure lamina ABCD has sides of length 10cm. The point E is on BC = 7.5cm, and the point F is on DC with CF = xcm. The triangle EFC is removed from ABC D (see diagram). The centre of mass of the resulting shape ABEF D is a distance x cm from CB and a distance y cm from C D. 400−x2 a) Show that x = 80−3x and find a corresponding expression for y The shape ABEF D is in equilibrium in a vertical plane with the edge DF resting on a smooth horizontal surface b) Find the greatest possible value of x , giving your answer in the fomr of a + b 2 , where a and b are constants to be determined Solution: Part a) 2.8. Sliding and Toppling Sliding: when the resultant force on the object parallel to the plane of contact becomes non-zero, that is, the limiting friction force is exceeded by the other forces, the object will slide Toppling: when total moment of the forces acting on the object becomes non-zero, the object will topple over Solution: First find the angle needed to slide: Draw diagram at hypothetical angle Resolve forces horizontal to slope: F − mgsinθ = 0 ∴ F = mgsinθ Resolve forces vertical to slope: R − mgcosθ = 0 WWW.ZNOTES.ORG CAIE A2 LEVEL FURTHER MATHS (9231) ∴ R = mgcosθ Form a relationship by dividing the unknowns: Particle P moves in a circular path (the red line) Angular Displacement: θrad F R Angular Velocity: = tanθ = 0.4R into the equation Substitute F tanθ = 0.4 θ = tan−1 0.4 = 21.8 Now find the angle needed to topple: Find centre of mass by using composite bodies rule:  x 0 .15 0 .1 m ( ) = 23 m ( ) + 13 m ( ) y  0 .15 0 .4 x 0 .13 ( )=( ) y  0 .23 v is the velocity r is the radius From here we can deduce that the velocity is Centripetal Acceleration: Prism topples when centre of mass is vertically above point A Calculate this required angle using Pythogaras: Centripetal Force: Applying F = ma to centripetal acceleration, we get, Time taken for each complete revolution: Linear Displacement: tanϕ = 0.13 0.23 ϕ = 29.7 From this we can see θ < ϕ, thus when the equilibrium is broken, the object will slide 3. Circular Motion 3.1. Motions in a circular path Sample Question: A particle of mass 3kg is placed on a rough, horizontal turntable and is connected to its centre by a light, inextensible string of length 0.8m. The coefficient of friction between the particle and the turntable is 0.4. The turntable is made to rotate at a uniform speed. If the tension in the string in 50N , find the angular speed of the turntable. Solution: Consider the following diagram: 3.2. Motion in Horizontal Circles WWW.ZNOTES.ORG CAIE A2 LEVEL FURTHER MATHS (9231) In this scenario, a body is attached to a fixed point by string and travels in a horizontal circle below that point Resolve forces horizontally and vertically with respect to the rotating body Resolve forces vertically: Rcos25 = mg Resolve forces horizontally: Rsin25 = mv 2 r Form a relationship by dividing the two: 2 v tan25 = 150g v = 26.2ms−1 Part (b): At max speed, car about to slip upwards ∴ frictional force would act towards the centre of the circle 3.3. Banked Curves Curved sections of public roads/railway tracks banked, enabling cars/trains to travel more quickly around curves Normal reaction between the vehicle and the track has a horizontal component when track is banked This component helps to provide central force needed to keep vehicle travelling in a circle Resolve forces vertically (direction of gravity): mg + 0.4Rsin25 = Rcos25 Resolve forces horizontally (direction of centripetal): mv 2 r = 0.4Rcos25 + Rsin25 Pull out common factors from both sides: m(g) = R(cos25 − 0.4sin25) v2 m( 150 ) = R(0.4cos25 + sin25) Sample Question: A car is travelling round a curve of radius 150m banked at 25° to the horizontal. The coefficient of friction between the wheels and the road is 0.4 a) What is the ideal speed of the car round the curve: that is, no lateral frictional force? b) What is the maximum safe speed of the car? c) What is the minimum safe speed of the car? Solution: Part (a): Draw diagram of scenario: WWW.ZNOTES.ORG Form a relationship by dividing these equations: v2 150g = 0.4cos25+sin25 cos25−0.4sin25 Part (c): At min speed, car about to slip downwards ∴ frictional force would act away from the centre of the circle CAIE A2 LEVEL FURTHER MATHS (9231) Resolve forces vertically (direction of gravity): mg = Rcos25 + 0.4Rsin25 Resolve forces horizontally (direction of centripetal): mv 2 r = Rsin25 − 0.4Rcos25 Pull out common factors from both sides: m(g) = R(cos25 + 0.4sin25) v2 m( 150 ) = R(sin25 − 0.4Rcos25) Form a relationship by dividing these equations: v2 150g = sin25−0.4cos25 cos25+0.4sin25 v = 9.06ms−1 3.4. Motion in a Vertical Circle When moving in a vertical circle a body is in circular motion with non-uniform speed If the tension or reaction force is 0 during the motion, then the object has left the circular path and is now a projectile The condition for a particle to complete vertical circles with radius r , starting from the lowest point, its initial speed must satisfy: u≥ 5gr R≥0 Apply the conservation of energy on solving vertical circles questions: + P EA = 12 mv 2 + P EB Where A is usually the starting point, and B is the general point of the motion Apply ΣF = ma towards the centre of the circle Once the two equations have been found, it can be simultaneously solved For motion on hemispheres Using ΣF = ma towards the centre gives: mgcosθ − R = mv 2 r When a particle is projected from the top of a hemisphere, provided the initial speed 0 < θ cos−1 23 Springs can be compressed and stretched Elastic strings can only stretch; they go slack Although it is not mandatory to memorize the above, another way to deduce if a particle completes vertical circles is that its reaction force must not be 0 at the highest point of the circle, that is: 1 2 2 mu 4. Hooke’s Law WWW.ZNOTES.ORG < 4.2. Hooke’s Law The tension in an elastic string or spring is directly proportional to the extension of the string/spring T = λx l T is the magnitude of tension x is the extension or compression λ is the modulus of elasticity l is the string/spring’s natural length When a system is resting in equilibrium, the tension is constant and proportional to the extension 4.3. Scenarios If a mass is hanging at one end of a spring with the other attached to a fixed point, the tension in the spring must be equal to the weight of the object CAIE A2 LEVEL FURTHER MATHS (9231) = λx2 2l Used in conjunction with kinetic and potential energy (work-energy principle) If two springs are attached between two fixed points and both are extended, the tension in both must be the same in order for there to be no net force overall If two springs are attached to fixed points and an object in the middle, tensions and frictional force must act in such way that overall force on mass = 0 If a spring is attached to an object on a rough surface, the frictional force acts in the direction opposing tension in the spring (preventing it to return to its original shape) If a mass is on an incline and held at rest by a spring, the tension in the spring must be equal to the component of the weight parallel to the slope 4.4. Elastic Potential Energy Work done in stretching a string/spring is given by: W = λx2 2l Also gives work done in compressing a spring. Derivation: The case when and is arbitrary WWW.ZNOTES.ORG CAIE A2 LEVEL FURTHER MATHS (9231) Then apply boundary conditions (usually called initial conditions) 5. Linear Motion under a Variable Force In this chapter, we denote: x as displacement v as velocity a as acceleration t as time 5.2. Terminology 5.4. Acceleration with respect to displacement Not all the time we can solve differential equations with respect to t If the question gives an acceleration function in terms of x or v , then we need to use an alternative form of Velocity is the rate of change of displacement with respect to time. Acceleration is the rate of change of velocity with respect to time. acceleration: dv a = v dx 5.3. Acceleration with respect to time We can denote velocity and acceleration as a derivative v= dx dt a= dv dt = 2 d x dt 2 If given velocity, we can find its displacement function by integrating the velocity function x = ∫ vdt Same goes for finding velocity if given acceleration v = ∫ adt WWW.ZNOTES.ORG Derivation: a= dv dt By the chain rule: dv dt dv = dx × But v = dx dt dv ∴ a = v dx dx dt CAIE A2 LEVEL FURTHER MATHS (9231) 5.5. Variable Forces Bodies undergo variable acceleration due to the effect of variable forces e.g., gravitational fields. Important to put negative sign if decelerating e.g., when a body falls in resistive medium. Deriving an expression for force in terms of velocity and displacement: dv F = ma a = v dx dv F = mv dx Deriving an expression for force in terms of work done: For an object moving from x 1 to x 2 the change in kinetic energy or work done can be defined as: x W = ∫x12 F dx ∴ dW dx = F 6. Momentum Deriving an expression of power in terms of roce and velocity P = dW dt = dW dx × dx dt = Fv 6.1. Momentum M omentum = mv Conservation of linear momentum: The total momentum of a system in a particular direction remains constant unless an external force is applied. Momentum Before Collision = Momentum After Collision Momentum before = (6 × 4) + Momentum after = 6 × v = 6v (−4 × 2) = 16 16 = 6v v = 2.67ms−1 6.3. Newton’s Law of Restitution When two objects collide directly, e= Separation Speed Approach Speed = v2 −v1 u1 −u2 Separation speed = The positive difference in speed of the two bodies after they collide Approach speed = the positive different in speed of the bodies before they collide The constant e is called the coefficient of restitution for the objects and takes a value between 0 and 1 e = 0 ; totally inelastic impact, no rebounding e = 1 ; perfectly elastic impact, speeds unchanged In realty, perfectly elastcitiy does not occur hence, 0 6.4. Oblique Impacts WWW.ZNOTES.ORG ≤e≤1 CAIE A2 LEVEL FURTHER MATHS (9231) Impact between a smooth sphere and a fixed surface: The law of conservation of momentum applies along line of impact Solution: Sketch a diagram of what is happening: The impulse on the sphere acts perpendicular to the surface, along line of impact Newton’s law of restitution applied to the component of the velocity of the sphere along line of impact The component of the velocity of the sphere along plane of impact is unchanged Impact between two spheres: Impulse affecting each sphere also acts along line of impact Components of velocities of spheres along plane of impact unchanged Newton’s law of restitution applied to components of the velocities of the spheres along line of impact WWW.ZNOTES.ORG CAIE A2 LEVEL Further Maths (9231) Copyright 2022 by ZNotes These notes have been created by Devandhira Wijaya Wangsa for the 2020-22 syllabus This website and its content is copyright of ZNotes Foundation - © ZNotes Foundation 2022. 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