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Solutions to Mock Physics Promo Exam Section A C D D B D C C D A C B B B C C B B A B B C C B A Section B 1(a) The principle of conservation of momentum states that the total momentum of a system remains constant provided there is no resultant force acting on the system. [2] (b)(i) (ii) 1. In an elastic collision, the total kinetic energy of the system is conserved. [1] 2. In a head-on collision, the subsequent motion of the particles takes place along the same straight line. [1] 1. velocity of separation = v 2. Let vA = final speed of neutron vB = final speed of deuteron [1] By conservation of momentum, v v A 2v B (1) velocity of approach velocity of separation v 0 v B v A (2) [4] (2) 2 (1) v 3v A 1 vA v 3 2. (a) 45 [1] (b) Using s = ut + 1/2at2 20 = (u cos)t t = 20 / (u cos) Using v = u + gt 0 = u sin - gt1 t1 = (u sin )/ g [1] 2t1 = t 2(u sin 45) / g = 20 / (u cos 45) u = 14.0 m s-1 [1] (c) (d) [1] [1] [1] [1] (e) 3. Horizontal: t2 = 10 / (14.0 cos 45) t2 = 1.01 s Vertical: Using s = ut + 1/2at2 s = (14 sin 45)(1.01) + (1/2)(-9.81) (1.01)2 s = 4.99 m > 2.40 m Pass is not intercepted. [1] [1] (a) The specific latent heat of fusion of a substance is defined as the quantity of heat required per unit mass to change the substance from the solid phase to the liquid phase without a change in temperature. [2] (b) The contact surface between the ice and the heater is larger when crushed ice is used. [1] This maximizes the heat transferred from heater to ice. Less heat is lost to the surrounding. [1] (c) Since the temperature of ice is lower than the room temperature, heat is absorbed from the surroundings and the ice melts. [2] (c) (i) (ii) (iii) 4. [1] Heat supplied by heater =Pxt = 15 x (20 x 60) = 1.8104 J Mass of ice melted by absorbing heat from the heater = m2 – m1 = 8.4 – 3.0 = 5.4 g E = m Lf or P x t = (m2 – m1)Lf 4 1.810 = 0.0054 Lf Lf = 3.33 x 106 J kg-1 (a) correct v and a in diagram (b) FSP = (c) Centripetal force = gravitational force GMm r2 mr2 [1] [1] [1] [1] [1] [2] [1] = GMm r2 [1] (d) GM r3 [1] Since = 2/T => r3 T2 rJ3 T J2 = rE3 [1] T E2 3 rJ = 5. = (1.50 x1011 ) 3 (11 .9) 2 (1) 2 rJ = 7.82 1011 m [1] (a) a = - 2 x [1] (b) (i) [1] A = 4.0 cm k= 0.20 x9.81 = 39.24 N m-1 0.05 B =w = ( (ii) (iii) 2 ) T [1] k 39.24 14.0 rad s-1 M 0.20 vmax = w x0 T [1] [1] = 14.0 x 0.04 = 0.56 m s-1 [1] 2 2 0.449 s w 14.0 [1] No. of oscillations made in 25 s = 25 55.7 0.449 [1] Kinetic Energy / J (iv) 0.0314 -0.04 Shape : Labelling & axis : 0.04 displacement / m 1] [1] Section C 1(a) (i)horizontal component = 20 cos 45o = 14 ms-1 vertical component = 20 sin 45o = 14 ms-1 [2] [1] (ii) change in velocity can be found by vector triangle or by resolution of vectors. 45o v v-u 900 -u Correct triangle [1] Magnitude of change in velocity v-u = (202 +202) = 28 ms-1 Direction is perpendicular from the wall [1] [1] 1(b) (i) The conditions for static equilibrium : The resultant force acting on the body is zero [1] and the resultant torque about any axis on the body is zero [1]. (i) The three forces in equilibrium should meet at a point since the resultant is zero. So R acts along the direction of the spine. [1] 800 N Use cos 30o = (½ R) / 800 30 o R = 1390 N 120o 800 N R Method + substitution = 2 marks Answer = 1 mark (ii) 1. Work out moments about the pivot. The resultant torque about pivot should be zero. [Concept= 1 mark] 0.6 L cos 60o x W + 0.85L cos 60o x 50 = 0.6L sin 30o x F Each term in the equation should be given credit [ Substitution = 2 mark, one for clockwise, one for anticlockwise moments] F = 870 N [1] 2. The force R will not act along the spine[1]. The resultant force along the ydirection has increased although the resultant force along the horizontal direction remains constant. So angle of R from horizontal will be increased [1] [correct Answer with appropriate explanation = 2 marks) 2 (a) (i) (ii) (iii) (b) (i) (ii) (c) (d) PV = nRT (1.50 x 105)(1.30 x 10-3) = n x 8.31 x 300 n = 0.0782 [1] [1] PV = nRT (1.00 x 105)(1.80 x 10-4) = n x 8.31 x 300 n = 0.00722 [1] ni + np = nf 0.0782 + 0.00722 = (Pf x 1.30 x 10-3) / (8.31 x 300) Pf = 1.64 x 105 Pa [1] [1] Work done on the air = area under graph = ½ (1.50 + 1.25)105 x (0.13 x 10-4) + ½ (1.25 + 1.11)105 x (0.10 x 10-4) + ½ (1.11 + 1.00)105 x (0.10 x 10-4) = 4.02 J (accept 3.81 to 4.22 J) ∆U = q + w = 6.1 + 4.02 = 10.1 J (accept 9.9 to 10.3 J) Increase in temperature. [1] [1] [1] [1] [1] [1] Sources must be coherent [1] Sources must have roughly the same amplitude [1] Separation between the sources is several wavelengths [1] Sources must be either unpolarised or have the same plane of polarisation [1] D (i) using x [1] a 6.0 10 7 2.0 a = 2.4 10-4 m [1] 5.0 10 3 (ii) Change 1 From x λD a As a increases, fringe separation increases. [1] [1] Change 2 The bright fringes are less bright; intensity reaching the screen decreases. [1] The narrowed slits allow a decreased amount of light to pass through for interference. [1]