2021 Prelim H2Phy P2 Ans

1 2021 DHS H2 Physics Prelim Paper 2 Suggested Solutions 1 (a) time B2 (electric) current allow amount of substance allow luminous intensity any two of the above quantities, 1 mark each (b) (i) = 0.562 s A1 percentage uncertainty = (2% + 8%) / 2 (= 5%) C1 or fractional uncertainty = (0.02+0.08) / 2 (= 0.05) ΔT = 0.562 x 0.05 = 0.028 s C1 T = (0.56 ± 0.03) s A1 (ii) 1. total energy = max kinetic energy = ½ (200 x 10-3)(2 /T)2(5.00 x10-2)2 M1 = A1 2. total energy = = 3.15 x 10-2 J A1

2 2 (a) sum / total momentum of bodies is constant B1 or sum / total momentum of bodies before = sum / total momentum of bodies after for an isolated / closed system / no (resultant) external force B1 (b) (i) 4.0 vA = 6.0 vB C1 EK = ½mv2 C1 ratio = M1 = 1.5 A0 (ii) 0.48 = EK of A + EK of B = EK of A + (EK of A / 1.5) = 5/3 × EK of A C1 EK of A = 0.29 (0.288) J A1 (iii) curve starts from origin and has decreasing gradient M1 final gradient of graph line is zero A1 3 (a) (i) EK = ½mv2 = 0.5 × 0.40 × 0.302 = 1.8 × 10–2 J A1 (ii) (change in) kinetic energy = work done on spring / (change in) elastic potential energy C1 1.8 × 10–2 = ½× F × 0.080 C1 FMAX = 0.45 N A1 (iii) a = F / m = 0.45 / 0.40 = 1.1 m s–2 A1 (iv) 1. constant velocity / resultant force is zero, so in equilibrium B1 2. decelerating / resultant force is not zero, so not in equilibrium B1 (b) curved line from the origin M1 with decreasing gradient A1

3 4 (a) two sine waves in antiphase (one dotted, one solid line) B1 4 antinodes and 5 nodes shown B1 (b) a second sine wave of same wavelength on same axis B1 separated by a quarter of the wavelength B1 (c) plane parallel wavefronts before the opening B1 circular wavefronts showing diffraction after the opening with same wavelength and B1 greater than the gap width B1 5 (a) M1 M1 A0 (b) (i) From (a)   C1 A1 (ii) C1 A1

4 (iii) C1 A1 (c) Over-estimate. 0.20 V is actually the p.d. across the internal resistance cell r as well as that of the ammeter. B1 B1 6 (a) (i) to increase flux linkage (with secondary coil) B1 (ii) e.m.f. (induced only) when flux (in core/coil) is changing B1 constant / direct voltage gives constant flux / field B1 (b) (i) NS / NP = VS / VP C1 NS = (52 / 150) × 1200 = 416 turns A1 (ii) 0 ms or 7.5 ms or 15.0 ms or 22.5 ms A1 (c) (i) either mean power = V2/ 2R and V = 52 (V) C1 R = 522/ (2×1.2) = 1100 (1127) Ω A1 or mean power = V2/ R and V = 52 /√ 2 (= 36.8 V) C1 R = 36.82/1.2 = 1100 Ω A1 (ii) sinusoidal shape with troughs at zero power B1 only 3 ‘cycles’ B1 each ‘cycle’ is 2.4 W high and zero power at correct times B1 7 (a) In the magnetic field, the magnetic force acting on the ion

5 provides the centripetal force for the ion to move in uniform circular motion. Thus Since , which is independent of r. Assume that the time taken by the ion to cross the gaps is negligible compared to the time taken by the ion to travel in the magnetic field. B1 B1 B1 (b)(i) C1 A1 (b)(ii) In order for the nucleus to accelerate when it crosses the gap, freq. of the alternating voltage = orbital freq. of the nucleus B1 A1 (b)(iii) The work done by the magnetic force on the ion is zero since the magnetic force is always perpendicular to the velocity of the ion. Each time the ion crosses the gap, it gains a kinetic energy of qV due to the work done on it by the electric field given by Fd = qEd = q(V/d)d where F is the electric force acting on the ion, d is the separation between the gaps and E is the strength of the uniform electric field between the gaps. In one revolution, the ion will cross the gap two times. Thus, the total gain in its kinetic energy is 2 qV = 4eV, since the charge of the helium nucleus is 2e. A1 A1

6 8 (a) A longitudinal wave is one in which the oscillation of the molecules of the wave is along the direction of transfer of energy of the wave. A1 (b) A = v2 = (2700)(3100)2 = 2.59 x 1010 A1 Units of A = (kg m-3)(m s-1)2 = kg m-1 s-2 = Pa A1 (c) A1 (d) A1 (e) The waves should be weaker after traveling longer distances, B1 hence direct waves should show larger amplitude than reflected waves. B1 (f) (i) SD8, t = 0.40 s SD8 = (3.1)(0.40) = 1.24 km A1 (ii) SXD8, t = 0.60 s SXD8 = (3.1)(0.60) = 1.86 km A1 Route 2 Route 1 Route 2

7 (g) Assume SX = XD8 Using Pythagoras Theorem, depth d = √( XD8 )2−( OD8 )2 = √ ( SXD8 2 ) 2 −( SD8 2 ) 2 M1 = = 0.70 km A1 (h) (i) Graph is lower, B1 Any peak value at a time factor about (3.1/2.4) = 1.3 times later B1 S X D8O d

8 (ii) Top graph is same, B1 Lower graph is lower and asymmetric as shown B1 (iii) Any one of the following:  An extra layer of rock halfway down that can cause partial reflection  Double reflection before reaching detector  Some refraction takes place at intermediate level (as a result of density changes) B1 - THE END -