Exampro a-level Physics (7407/7408)
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3.4.1.8-Conservation-of-energy
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[10] M9. (a) ½ Fx or ½ kx2 C1 29.4 mJ A1 2 (b) (i) amplitude clearly marked on diagram - must touch lines or be an accurately drawn equivalent distance B1 1 (ii) idea of interchange of p.e. and k.e. B1 appropriate use of elastic p.e. at start of cycle and of gravitational p.e. highest point + some k.e. in between. B1 2 [5] M10.(a) kinetic energy = mgh (1) = 0.37 J (1) (b) υ = (1) = 2.22 ms–1 (1) (c) Fc = 2.9 N [or 3.0 N if g = 10 used] (1) (d) T = Fc + W = 4.4 N (1) [6] M11.(a) (i) F = W sin θ = W × (1) correct angle (1) = 46 N (1) [2 out of 3 if no working shown] (ii) P (= Fυ ) = 410 W (1) (4) (b) (i) kinetic energy of bicycle + rider ⇒ gravitational potential energy (1) (ii) initial Ek = gain in gravitational Ep = 2.8 × 103 (J) (1) h = 4.1 m (1) distance = 62 m (1) alternative: (F = ma) a = = 0.657 (m s–2) (1) v2 = u2 = 2as (1) s = = 62 m (1) (max 3) [7] M12.(a) (i) and (ii) (iii) A + B = constant (1) loss in potential energy = gain in kinetic energy for A and B [or potential energy at P = kinetic energy at Q for A and B] (1) reason for C being below B e.g. transfer to heat [or work done against friction] (1) (6) (b) (i) clear reference to energy υC (1) = 31(.3) m s–1 (1) (ii) = 3.9(2) × 103 N towards centre of circle (1) (iii) gain in gravitational potential energy ( = mgh sin θ) = 620 × 9.8 × 60 × sin 20° (1) = 1.25 × 105 J (iv) 620 × 9.8 × 50 = (F × 60) (1) + 1.25 × 105 (1) F = 3000 N (1) alternative (iv) calculation of acceleration = (–)8.0 m s–2 (1) use of F + mg sin θ = ma (1) F = 3000 N (1) (max 9) [15] M13.(a) loss of potential energy = m × 9.81 × 6.0 (1) gain in kinetic energy = loss of potential energy (1) ½ mv2 = 58.9 m gives v = 10.8 (m s–1) (≈ 11m s–1) 3 (b) loses potential energy (as it moves to B) (1) gains kinetic energy (as it moves to B) (1) regains some potential energy at the expense of kinetic energy as it moves from B to C (1) some energy lost as heat (due to friction) (1) 4 The Quality of Written Communication marks are awarded for the quality of answers to this question. [7] M14.(a) (i) Ep = mgΔh (1) = 5.8 × 10-2 × 9.8(1) × 1.5 = 0.85 J ✓ (ii) 0.85 J (1) (allow C.E. for value of Ep from (i)) (iii) (use of Ek = ½mv2 gives) 0.85 = 0.5 × 5.8 × 10-2 × v2 (1) (allow C.E. for answer from (ii)) (v2 = 29.3) v = 5.4 m s-1 (1) (iv) (use of p = mv gives) p = 5.8 × 10-2 × 5.4 (1) (allow C.E. for value of v from (iii)) = 0.31 N s (1) 7 (b) (allow C.E. for value of p from (iv)) = 31 N (1) [or a = = 540 (m s-2) (1) F = 5.8 × 10-2 × 540 = 31 N (1)] 2 (c) egg effectively stopped in a longer distance (1) hence greater time and therefore less force on egg (1) [or takes longer to stop hence force is smaller as [or acceleration reduced as it takes longer to stop thus force will be smaller] [or some energy is absorbed by container less absorbed by egg] 2 [11] M15.(a) (i) mass per sec ( = density × vol per sec) = 1000 × 1.4 (1) = 1400 kg (s–1) (ii) loss of Ep per sec = 1400 × 9.8 × 750 (1) = 1.0 × 107 J (s–1) (1) (1.03 × 107 J s–1) (allow C.E. for value of mass per sec from (i)) (iii) efficiency = 0.2 (1) (allow C.E. for value (ii)) 6 (b) (i) (use of P = IV gives) Irms = (1) = 80 A (1) (ii) power output = (0.95 × power input) = 0.95 × 2.0 (MW) = 1.9 (MW) (1) = 6.9 A (1) [or I for 100% efficiency = 7.3 (A) (1) I for 95% efficiency = 95% of 7.3 = 6.9 A] 4 [10] M16. (a) (use of T = 2π gives) T = 2π (1) = 1.8 s (1) 2 (b) mgh = ½ mv2 (1) v = (1) (= 0.63 m s‑1) vmax = 2πfA = (1) A = (1) (= 0.18m) [or by Pythagoras A2 + 7802 = 8002 gives A = (1) ( = 180 mm) (or equivalent solution by trigonometry (1) (1)) vmax = 2πfA or = (1) = (1) (= 0.63 m s‑1) 4 (c) tension given by F, where F – mg = (1) F = 25 × 10‑3 = 0.26 N (1) 2 [8] M17. (a) (i) area = 120 × 106 (m2) (1) mass = 120 × 106 × 10 × 1100 = 1.3 × 1012 kg (1) (ii) (use of Ep = mgh gives) ΔEp = 1.3 × 1012 × 9.8 × 5 = 6.4 × 1013 J (1) (allow C.E. for incorrect value of mass from (i)) (iii) power (from sea water) = [or correct use of P = Fv] = 3000 (MW) (1) (allow C.E. for incorrect value of ΔEp from (ii)) power output = 3000 × 0.4 (1) = 120 MW (1) (allow C.E. for incorrect value of power) [7] M18. (a) (i) (use of Ep = mgh gives) Ep = 70 × 9.81 × 150 (1) = 1.0(3) × 105 J (1) (ii) (use of Ek = ½mv2 gives) Ek = ½ × 70 × 452 (1) = 7.1 × 104 J (1) (7.09 × 104 J) 4 (b) (i) work done (= 1.03 × 105 – 7.09 × 104) = 3.2(1) × 104 J (1) (allow C.E. for values of Ep and Ek from (a) (ii) (use of work done = Fs gives) 3.21 × 104 = F × 150 (1) (allow C.E. for value of work done from (i) F = 210 N (1) (213 N) 3 [7] M19. (a) component (parallel to ramp) = 7.2 × 103 × sin 30 (1) (= 3.6 × 103 N) 1 (b) mass = = 734 (kg) (1) a = = 4.9(1) m s–2 (1) 2 (c) (use of v2 = u2 + 2as gives) 0 = 182 – (2 × 4.9 × s) (1) s = 33(.1) m (1) (allow C.E. for value of a from (b)) 2 (d) frictional forces are acting (1) increasing resultant force [or opposing motion] (1) hence higher deceleration [or car stops quicker] (1) energy is lost as thermal energy/heat (1) Max 2 [7] M20. (a) potential energy to kinetic energy (1) mention of thermal energy and friction (1) 2 (b) (use of ½ mv2 = mgh gives) ½ vh2 = 9.81 × 1.5 (1) vh = 5.4(2)ms–1 (1) (assumption) energy converted to thermal energy is negligible (1) 3 (c) component of weight down the slope causes acceleration (1) this component decreases as skateboard moves further down the slope (1) air resistance/friction increases (with speed) (1) 2 (d) (i) distance (= 0.42 × 5.4) = 2.3m (1) (2.27m) (allow C.E. for value of vh from (b)) (ii) vv = 9.8 × 0.42 (1) = 4.1(l) m s–1 (1) (iii) v2 = 4.12 + 5.42 (1) v = 6.8 m s–1 (1) (6.78 m s–1) (allow C.E. for value of vh from (b)) 5 [12] M21. (a) use of E = mgh (1) 2680 J (1) 2 (b) use of v2 = 2 as (1) 9.1 m s-1 (1) 2 (c) increases the time taken for the athlete to come to rest/reduced deceleration force = mass × acceleration/mass × change in velocity/time (1) or momentum argument or energy argument (1) 2 [6] M22. correct boxes ticked gravitational potential energy (1) B1 potential energy and kinetic energy (1) B1 elastic potential energy ( in ‘rope’) (1) (condone gravitational PE in addition) B1 [3] M23. (a) (i) 65 × 9.8 × 35 seen and evaluated to 22295 or 2231 or 22300J B1 (ii) correct substitution of 65 kg and either 11000J or 18 m s–1 in ke formula seen B1 18.4 (18.397) (m s–1) to at least 3 sf B1 (iii) distance = energy loss/force or work done/force or numerical equivalent C1 64-64.3 using Ep = 20kJ or 79 – 81(m) using 22.3 kJ A1 (iv) friction B1 air resistance B1 further detail eg friction at ski-ice surface or caused by need to move air when passing through it B1 8 (b) (i) time = Δv/a or numerical equivalent C1 6.4(3) – 6.6(6.57) (s) A1 (ii) use of appropriate kinematic equation C1 (57.8 – 60.4) 58 m or 60 (m) to 2 sf A1 4 [12] M24. (a) (i) (EK = ½ mv2 =) 0.5 × 68 × 162 (1) = 8700 or 8704(J) (1) (ii) (ΔEP = mgΔh =) 68 × 9.8(1) × 12 (1) = 8000 or 8005 (J) (1) (iii) any three from gain of kinetic energy > loss of potential energy (1) (because) cyclist does work (1) energy is wasted (on the cyclist and cycle) due to air resistance or friction or transferred to thermal/heat (1) KE = GPE + W – energy ‘loss’ (1) (owtte) energy wasted (= 8000 + 2400 - 8700) = 1700(J) (1) 7 (b) (i) (u = 16 m s–1, s = 160 m, v = 0, rearranging s = ½ (u + v) t gives) 160 = ½ × 16 × t or t = or correct alternative (gets 2 marks) (1) = 20s (1) Download 0.7 Mb. Do'stlaringiz bilan baham: 
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