TJC 2022 Prelim P2 Guide

H2 Physics Prelim 2022 Solutions 1 (a) Thrust = resistive force since velocity is constant. Power = Fv = Dv D x 15 = 90000 D = 90000/15 = 6.0 kN C1 A1 (b) F = Thrust Power = Fv F – D = ma P/v – D = ma 120000/15 – 6000 = 800a a = 2.5 m s-2 M1 A1 (c) As the velocity increases with time, the resistive force also increases . The driving force decreases as the speed of the boat increases since driving power stay constant, the net force on the boat decreases and hence acceleration decreases with time. Acceleration drops to zero when the resistive force is equal to the driving force on the boat. B1 B1 B1 B1 2 (a) (i) Gravitational force provides centripetal force. Same gravitational force acting on star and planet. ( ) ( ) 2 27 28 27 2 98 8 2 2.0 10 6.0 10 2. m1 5.2 10 1.68 1031 1 0 10 1 306.0 0 G p p s s ps s p s F m r m r rr r r r  =   =  = =  =  == C1 A1 (a) (ii) ( ) ( ) ( ) 2 2 2 2 11 27 2 8 29 6 2 6.67 10 2.0 10 21.68 10 5.2 10 1.16 10 s GC sp ss sp ss FF Gm m mrd Gm m mr Td T T − = = =   =  = C1 A1

(a) (iii) Planet moves in front of star periodically. A1 (b) (i) Between starting height 6.40 x 106 m & ending height 7.225 x 106 m 6 6 -1 () area under g-r graph 6.74 10 to 7.61 10 J kg dg dr g dr =− = = =    2300 Um = =  = 1.55 x 1010 to 1.75 x 1010 J OR 2 MgG r= and MG r =−  gr =− ( ) 2300( )f i i i f fU m m g r g r   =  = − = − 662300( 9.75 6.4 10 7.625 7.225 10 )= −   +    1.60 x 1010 J C1 M1 M1 C1 M1 M1 (b) (ii) ( ) 10 11 66 24 111.6 10 6.67 10 2300 6.40 10 7.225 10 5.85 10 kg E s E f f s E E i i GM m GM m x U rr M U M U −  =  − = − =− − −   =      C1 A1 (b) (iii) Advantages to low polar orbit: • High(er) resolution imaging/clearer images • Image more of the planet as the Earth spins underneath the satellite Geostationary: • Remain at the same position in the sky so satellite dishes can keep locked on to signal/no steerable dishes needed/ send & receive signals all the times/continuous/uninterrupted/stable signals • Higher orbit means greater coverage of the signals A1 A1 3 (a) angular speed of grating, ω = 2 π / T = 2 π / 3.0 = 2.1 rad s -1 A1 (b) The peaks represent the positions of constructive interference/maxima. B1 B1

The effect of diffraction through the slits of diffraction grating causes interference fringes of higher order to have lower intensity than the zeroth order maxima. (Peak C corresponds to the zeroth order of the diffraction pattern whereas peaks B and D the first order and Peaks A and E the second order.) (c) (i) Time interval = 3.7 x 0.1 = 0.37 s θ = ωt = 2.1 x 0.37 = 0.78 rad (ecf) C1 A1 (c)(ii) Using the grating equation n λ = d sin θ, d = 1 x 10-3 / 550 sin θ2 = 2 x 5.5 x 10 5 x λ λ = 640 nm (ecf) C1 A1 (c)(iii) 1 Peak E is preferred as the angle θ is larger so the percentage uncertainty for calculating the wavelength is smaller. C1 (c)(iii)2 From n λ = d sin θ n < d/ = 1 x 10-3 / 550 x 640 x 10-9 = 2.8 Hence, the highest order observed is 2nd order maxima. Total number of orders observed = 2n +1 = 5 C1 C1 A0 (d) Diagram must show that since the slit separation remains the same, the fringe separation remains the same. Peaks are less intense (poorer contrast) and less sharp (broader and less defined). M1

4 (a) (i) 2.25/0.0108 = 208 Ω C1 (a)(ii) & (b) (i) (a)(ii) correct area shaded (b)(i) Line R should be a straight line passing through (0,0) and (1.60 V, 10.0 mA) B1 M1 (b) (ii) Draw a horizontal line on Fig. 4.1 such that the pd across 160  and LED1 adds up to 3.0 V. I = 6.2 mA B1 A1 (iii) With two LEDs, effective resistance of the circuit is lowered, leading to higher current from battery. With higher current, the potential difference across the 160Ω resistor is larger, leaving less potential difference across the LEDs. OR With two LEDs in parallel, effective resistance between the LEDs is lowered. By Potential Divider Principle, the potential difference across each LED is also lowered. From the graph, it can be seen the current decreases with voltage, so the power (P=VI) dissipated in each LED is lower, so brightness is dimmer. B1 B1 B1 I/ mA V/ V (a)(ii) (b)(i) R (b)(ii)

(c) B1 5 a DERQ and CFSP [B1] bi As current flows through the slice in the presence of the magnetic field, it will experience a magnetic force to the left, by Fleming Left Hand rule. Hence a potential difference is generated across DERQ and CFSP. When the current is undeflected at equilibrium, force on charge due to magnetic field equals force due to electric field Bqv = Eq E = B v Electric field across faces DERQ and CFSP is constant E = VH / d , where d is the distance between faces DERQ and CFSP VH = Bvd [B1] [B1] [A1] bii since I = nAqv where A is the cross-sectional area through which current flows, A= dt , and t is the distance between faces CDEF and PQRS. () ( )( ) () H Iv n dt q I BIV Bd n dt q ntq = == [M1] [C1] [A0] biii For copper, n is very large hence VH is very small, making it difficult to measure. [B1] 3.0 V 160 Ω S1 S2 LED1 LED2 160 Ω

6 ai The magnetic flux density at a point is defined as the force acting per unit current per unit length on a current carrying conductor when the conductor is placed at right angles to the magnetic field. M1 B1 aii As AB moves from P towards Q, magnetic flux linkage over the area ABCD enclosed by the frame increases resulting in an induced e.m.f. generated in the frame by Faraday’s law. Since the frame is a conductor, induced current flows. By Lenz’s Law, the induced current flows in an anticlockwise direction. This results in a magnetic force that acts on AB towards the left which oppose motion. Hence the frame slows down. Alternative: As AB moves from P towards Q, magnetic flux linkage over the area ABCD enclosed by the frame increases resulting in an induced e.m.f. generated in the frame by faraday’s law. Frame is a closed circuit so induced current will flow in the frame. By conservation of energy, KE of the frame is converted to heat energy. Energy is loss as P= I2R. Hence frame slows down B1 B1 B1 B1 B1 B1 aiii As AB enters the field, the increase in magnetic flux (linkage) Φ = 𝐵𝐴 = 𝐵(𝑤𝑥) where x is the distance AB has moved past P. Hence, the magnitude of induced emf is given by 𝐸 = 𝑑Φ 𝑑𝑡 = 𝐵𝑤 𝑑𝑥 𝑑𝑡 = 𝐵𝑤𝑣 Induced current that flows in the frame, 𝐼 = 𝐵𝑤𝑣 𝑅 Magnetic force which acts on AB is in opposite direction to motion is the braking force which slows the frame. Braking force is thus 𝐹 = 𝐵𝐼𝑤 = 𝐵2𝑤2𝑣 𝑅 B1 B1 B1 B1 A0 bii Distance 2L = Area under v – t graph = ½ (26 + 16 x 2 + 10 x 2 + 6 x 2 + 3.5)(5) = 234 m PQ = L = 117 m, Accuracy: L = 105 to 129 m C1 A1 biii Braking force only acts on the train when there is a changing magnetic flux through the frame. There is no braking force on the train when the whole frame is within the magnetic field. Increasing the length PQ does not change the exit speed. M1 A1

7(a)(i) ( ) 2 14 17 5.0 10 2.6 10 1.3 10 W PI P ISS P =  = =   = C1 A1 (a)(ii) 1.3 × 1017 W A1 (a)(iii) ( ) 2 17 23 2 4 1.3 10 4 6400 10 253 W m PPI S r − ==  =  = M1 M1 A0 (b)(i) x-axis marked with an arrow labelled M at 900-1100 nm B1 (b)(ii) ( ) max 6 1000 2900 2.9 10 nm Tk k kK = = = C1 A1 (b)(iii) max T = 2.90 x 106 nm K Sun: max = 2.90 x 106 / 5800 = 500 nm