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CHAPTER 2 : FORCE & MOTIONANSWER (d) (i) the first tape2.1 Linear motionQuestion 1(a) Scalar quantity (ii) the last tape(b) Arrow: (e) acceleration (f)(c) 700(d) Zero 2.2 Motion GraphQuestion 2 Question 4(a) A.c current (a)(i) velocity(b) (i) 0.02 s (ii) constant velocity (ii) (iii) (b)(i) acceleration (b)(ii)Question 3(a) Ac current (b)(iii)(b) Tape chart Question 5 (a) Displacement Time (b) R : from AB (c) (ii)(c) Constant acceleration FIZIKMOZAC 2010
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Question 6(a) 15 m (e) mPuP = (mP + mQ)v(b) Velocity (2)(1) = (2 + m) (0.4)(c) (i) constant velocity m = (2 – 0.8)/ 0.4 (ii) at rest = 3 kgQuestion 7 Question 10(a)(i) acceleration (a) quantity of matter in object(a)(ii) displacement (b) (i) equal but opposite direction(b) (ii) 0 = MV + mv MV = - mv Section of Type of motion (iii) principle of conservation of the graph of the car momentum (c) (1200)v = (4)(60) = 240 Constant OA v = 240/1200 = 0.2 m s-1 acceleration Constant velocity AB Question 11 (a) Principle of conservation of momentumQuestion 8 (b) Momentum of the man (forward)(a)(i) constant velocity is equal with the momentum of(a)(ii) constant acceleration and the boat (backward) but in constant velocity after 2.01 pm opposite direction(b)(i) zero (c) 0 = (50 x 2) + (20)v(b)(ii)the car moves with constant v = - (100)/20 = - 5 m s-1 velocity (d) Rocket(c) Question 12 (a) Mass x velocity (b) Momentum = 0.08 x 100 = 8.0 kg ms-1 (c) Decrease (d) To lengthen the time of impact / to reduce the impulsive force Question 13 (a) Inelastic collision2.4 Momentum (b) (1200)(30 ) + (1000)( – 20) =(a) Inelastic collision = (1200 + 1000)v v = 7.27 m s-1 .(b) (c) Inertia (d) The larger the mass and velocity, the higher thec) mv = 2 x 100 = 2 x 1 = 2 kg ms-1 = = . momentum. Momentum = mass x velocity . (d) FIZIKMOZAC 2010
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Truck with higher momentum Question 17 is difficult to stop or higher (a) The position between the two impact if accident occurs balls are equal. The distance between the two2.6 Impulsive force balls increase.Question 14 (b) Gravitational force (a) Change of momentum (c) (i) gravitational acceleration, g (b) (i) Force on surface B is larger (ii) 10 m/s2 than surface A (iii) Mass does not affect g (ii) Time of impact on surface (d) (i) Velocity decreases A is longer than on surface (ii) moves against gravitational B force. (iii) Constant/unchanged (iv) The shorter the time of Question 18 impact, the larger the (a) Gravitational force force //force Inversely (b) Surface area of the feather is proportional to the time larger of impact (c) Velocity increases constantly / (v) Sponge/mattress constant acceleration (c) Higher mass high velocity, Final velocity is constant. high impact / high momentum / Final velocity for water droplet high impulsive force is higher than final velocity for feather.Question 15 (d) Final velocity is inversely(a)(i) gravitational potential energy proportional to surface area (ii) kinetic energy (e) Graph(b)(i) mgh = ½ mv2 v2 = 2gh = 2 x 10 x 2.5 v = 7.07 m s-1(ii)(c)(i) soft/spongy Lengthen the time of impact Reduce the impulsive force(ii) use parachute To reduce velocity/momentumQuestion 16 (a) Impulsive force (b) High impulsive force/short impact time (c) 0.05 x 5 = 0.25 kg m/s (d) Use sponge, mattress / soft material 2.9 Equilibrium of forces2.8 Gravitational Force FIZIKMOZAC 2010
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Question 19 (ii) R = mg – ma / R < mg (a) Diagram Question 22 (a) Force is anything that can move a stationary object // stop a moving object // change direction / shape / speed of an object. (b) Label force (b) F = ma a = 5 / 2 = 2.5 m s-2 (c) 5 N but in opposite directionQuestion 19(spm 08) (c) (i) Fx = 120 cos 30⁰ = 104.4 N(a)(i) magnitude: equal (ii) FY = 120 sin 30⁰ = 60 N Direction: opposite (ii) zero Question 23 (iii) equilibrium of forces (a) Force is anything that can(b)(i) Acceleration move a stationary object // (ii) there is resultant force, force is stop a moving object // change directly proportional to direction / shape / speed of an acceleration object.Question 20 (b) Diagram (a) 650 = 300 + P P = 350 N (b) The bicycle moves with constant velocity. The resultant force equals zero. (c) Velocity increases / accelerate (d) Inertia (e) The time of impact is short. (c) Fx = 1500 x cos 60⁰ = 750 N High impulsive force. (d) (i) Fx : to make car move forward // overcomeQuestion 21 frictional force (a) 500 N (b) 1. At rest (ii) Fy : to lift the car off the 2. moving down or up with ground // to move the car constant velocity upwards// to overcome (c) F = R – mg the weight of the car (d) There is a resultant force acts upward. R = mg + ma(e)(i) decreases Question 24 FIZIKMOZAC 2010
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(a) 500 N (ii) energy lost to heat/frictional (b) (i) increased force (ii) decreased (iii) Unchanged Question 28 (c) Resultant force = 0 (a) Gravitational potential energy (d) R = (46)(10) = 460 N (b) (i) E = mgh = (30)(10)(2.5) F = mg - R = ma = 750 J 50a = 500 – 460 (ii) E = ½ mv2 = ½ (30)(4)(4) a = 0.8 ms-2 = 240 J (c) (i) Energy consumes in (b)(i) isQuestion 25 larger than in (b)(ii) (a) Attraction force (ii) energy lost to heat / (b) (i) same magnitude, opposite frictional force direction (ii) equal in magnitude, Question 29 opposite direction (a) (i) Diagram 29.2 is further than (c) (i) zero in diagram 29.1 (ii) equal (ii) decreases (d) The net force is zero, at rest (b) Streamline The net force is zero, moves (c) W = 70 x 10 = 700 J with constant velocity. (d) (i) kinetic energy to (e) Equilibrium of forces gravitational potential energy to kinetic energy2.10 Energy (ii) sound / heatQuestion 26 (a) To gain maximum kinetic Question 30 energy before he begins to (a) (i) Trolley Q has less mass jump. (ii) Newton’s second law Kinetic energy increases with (b) velocity. (b) To gain elastic potential energy from the pole. Change to gravitational potential energy. (c) mgh = 60 x 10 x 52 = 3 120 J (d) 10 m s-2 (e) To lengthen the time of impact between the athlete and the (c) Velocity of the trolley mattress. decreases So reduce the impulsive force (d) Trolley Q moves faster along frictionless slopeQuestion 27 (e)(i) mgh = 0.5 x 10 x 0.7 (a) Work = force x displacement = 3.5 J (b) (i) 220 x 0.5 = 110 J (ii) 20 x 10 x 0.5 = 100 J (ii) Total energy at the top of (c) (i) Work done in 4(b)(i) is larger the slope: FIZIKMOZAC 2010
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Kinetic energy + gravitational potential energy = 3.5 + 3.0 = 6.5 J2.11 ElasticityQuestion 31 (a) (i) Newton, N (ii) e is directly proportional to W (iii) Hooke’s Law (b) (i) 12 N (c) Gradient = 12 = 240 N/m 0.05 (d) E = ½ Fx = ½ x 10 x 0.04 = 0.2 J (e)Question 32 (a) (i) Hooke’s Law (b) (i) elastic potential energy (ii) E = ½ x 60 x 0.08 = 2.4 J (c) (i) 80 N (ii) Spring Q The elastic limit for spring Q is 100 N which more than 80 N. FIZIKMOZAC 2010
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