TJC H1 PHY P2 Solutions

Solution to 2016 H1P2 1 (a) Rate of change of velocity A1 (b)(i) x = 72 t A1 (b)(ii) y =½ (9.81) t 2 = 4.91 t 2 A1 (c) 100 = 4.91 t 2 t = 4.5 s M1 A1 (d) v = [(72 x 4.5) – 125] / [4.5] v= 44.2 m s -1 (160 km h -1 ) M1 A1 2 (a) Product of force and perpendicular distance of the line of force from fixed point/axis B1 (b)(i) Label normal contact force N 1 at wall; normal contact force N 2 at the ground and friction at the ground correctly Label weight of man and weight of ladder correctly B1 B1 (b)(ii) As system is in equilibrium, N2 = (72 + 40) x 9.81 = 1100 N M1 A1 (b)(iii) cos = 6.0/12.0   = 60 (or other methods) Let L be the length of ladder, Taking moments about D, N1sin60 x L = (72 x 9.81 x cos 60) (3L/4) + (40 x 9.81 cos 60) (L/2) N1 = 419 N M1 A1 3 (a) Electrical resistance of a conductor is defined as the ratio of the potential difference across it to the current flowing through it. B1 (b)(i) When S1 is the only switch closed, lamps A and C are in series giving a resistance of 30.0 Ω. Thus, resistance of a lamp is 15 Ω. Or when all three switches are closed, lamps A and C are in parallel with lamps B and D giving a resistance of 15 Ω. Thus, resistance of a lamp is 15 Ω. B1 (ii) faulty lamp: lamp E nature of fault: lamp is short-circuited B1 B1 (iii) Short-circuited lamp could cause excessive current to flow in the circuit that could cause damage to the power supply / other lamps / blow fuse in power supply. B1 (c)(vi) Using V = I R R = V / I = 12.0 / 0.40 = 30.0  B1 (c)(v) The lamp filament in (v)1 has a higher resistance than (iii) because it is hotter when operating at normal brightness. B1

2 4(a) Magnetic flux density is defined as the force per unit length acting on a straight conductor with unit current placed perpendicular to the magnetic field. B2 4(b)(i) B1 – Correct direction B1 – Three concentric circles with increasing spacing 4(b)(ii) B1 – Uniform field near poles of magnetic B1 – Stronger field on top, weaker field below 4(b)(iii) Towards the bottom of the page. B1 4(b)(iv) Magnetic force = (0.50)(1.5)(0.10) = 0.075 N B1

3 5 (a)(i) resistance of LDR = 0.20 kΩ B1 (ii) (A logarithmic scale compresses the scale so that the widely differing values can be shown easily on one graph. B1 (iii) On the logarithmic scale in Fig. 7.1, the graph is similar to lg R against lg I. It is a straight line with gr adient = -1 and y -intercept = lg 10000, the relation is true. B2 (b) B2 (c)(i) 𝐼 ≥ 50.0𝑊𝑚−2 ⇒ 𝑅𝐿𝐷𝑅 ≤ 0.2𝑘Ω When RLDR = 0.2 k 𝑉𝑅 = 𝑅 𝑅+𝑅𝐿𝐷𝑅 × 12 = 9 R = 0.60 k B1 M1 A1 (ii) It can be used as a burglar alarm. B1 (iii) Below 1 W m -2, the LDR’s resistance changes too rapidly with intensity such that any small fluctuation in intensity will cause a trigger. B2

4 6 (a) (b) (i) (ii) (iii) (i) (ii) (iii) (iv) Axes drawn with correct labelling [including u and v] straight line 1.acceleration 2.displacement area = displacement s = ½ (u+v) t --eqn(1) gradient a = v-u/t, so v = u+at substitute into eqn (1) to get s =ut+1/2 at2 a = 1.6/2.0 = 0.80 ms-2 1. s = ½ at2 = ½ x .80 x 2.02 = 1.6 m 2. By Newton’s 2nd law, T - mg =ma T = mg + ma = 35x9.81 + 35x0.80 =371 or 370 N 1. t = (32 – 1.6)/1.6 = 19 s total time = 19 + 2 = 21 s 2 Tension is Less At constant speed T = mg only [or equiv.] 1. Using force resolution: T sin 300 = total mass  g = 539 N Correctly resolving vertically at A TAB = 1078 N 2. AC undergoes Compression Force by support at point A should be horizontally outwards in order to balance T cos 30. B1 B1 A1 A1 M1 M1 A1 M1 A1 M2 A1 M1 A1 A1 C1 M1 A1 M1 A1

5 7 (a) When two or more waves of the same kind exist simultaneously at a point in space, the resultant displacement of the waves at any point is the vector sum of the displacement due to each wave acting independently. B2 (b) sketch has amplitude = 3.0 ± 0.1 cm Correct displacement values at previous peaks to produce correct shape M1 A1 (c) (i) Sources whose phase difference is constant. B1 (ii) The waves from the transmitters superpose/interfere The path difference between the 2 waves varies as the satellite moves. Hence, the amplitude of the superposed signal varies. (OR path difference change from being an integral multiple of the wavelength for maxima to being half integral multiple of the wavelength for minima as satellite moves.) B1 B1 (iii) 1. 3 fringe separations in 7.7 km  x = 7.7 3 = 2.57 𝑘𝑚 2. ∆𝑥 =  𝐷 𝑎  𝐷 = 2.57×160 1.2 = 342 km C1 A1 M1 C1 A1 (d) (i) 1. Reflected waves and incident waves superposed to form a standing wave. When detector D passes through nodes, it will detect minimum intensity ( minima ) but when it passes through antinodes, it will detect maximum intensity ( maxima ). B1 B1 2. Measure distance between 2 nodes ( minima ) or 2 antinodes ( maxima ) Distance between 2 nodes ( antinodes ) =  2 B1 B1 3. 𝑓 = 𝑣  where v = speed of microwaves = 3.00 x 108 m s-1 B1 B1 (ii) Place a polarizer between T and D, and rotate Signal will drop to zero if microwaves are plane-polarized B1 B1

6 8 (a) (i) Photoe