(DHS) 2024 Prelim Phy P1 Ans final

1 2024 DHS H2 Physics Prelim Paper 1 Suggested Solutions Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 A A B D B A C B C D Q11 Q12 Q13 Q14 Q15 Q16 Q17 Q18 Q19 Q20 C C C D C D A C A C Q21 Q22 Q23 Q24 Q25 Q26 Q27 Q28 Q29 Q30 A D A B C D D B B A 1 A By dimensional analysis, Units of √𝑔𝜆 = √m2 s−2 = m s−1 2 A s2 = s – s1 By Pythagoras theorem, s2 = √1402 + 1502 = 205 cm tan  = 150 140 or  = 47o Hence  = −17o anti-clockwise with respect to positive x-axis 3 B a = (v – u)/t = (0 -15)/1.2 = -12.5 m s-2 v2 = u2 + 2as 02 = 152 + 2(-12.5)s s = 9.0 m Closest distance = 9.0 + 15(0.10) = 10.5 m x 1200 s1 =150 cm s =140 cm −s1 =150 cm 30o s2 

2 4 D For a (perfectly) elastic collision between two bodies, the relative speed of approach is equal to the relative speed of separation. relative speed of P approaching Q Option relative speed of Q separating from P equal ? ( ) ( ) − −− = 1 1.30 0.50 1.80 m s A ( ) ( ) 10.58 0.22 0.36 m s −−=  B ( ) ( ) 10.40 0.40 0.00 m s −−=  C ( ) ( ) 10.99 0.19 1.18 m s −− − =  D ( ) ( ) 11.30 0.50 1.80 m s −− − =  5 B By Newton’s Third Law of Motion, the magnitude of the force exerted on m1 by m2 is equal to the magnitude of the force exerted on m2 by m1. 6 A Air resistance increases proportionately with speed. Resultant force = Weight – Air resistance 7 C Work done by a constant force on an object is defined as the product of the force and its displacement in the direction of the force. 8 B (0.5 x power output/power input) = 0.80 (0.5 x 200 000 x 250/power input) = 0.80 Power input = 31.3 MW 9 C Period, 𝑇 = 60 𝑥 60 s 𝜔 = 2𝜋 𝑇 = 2𝜋 60 × 60 = 1.75 × 10−3 rad s−1 10 D At the Earth’s surface: 2 GMmW R= At orbit: 22 1 49 49( 6 ) GMm GMm WF R R R = = = + 

3 11 C For geostationary satellites, all satellites orbit the Earth at a fixed distance from the Earth, r, and the orbital period T is the same as Earth’s rotation of 24 hours. Hence, all geostationary satellites also have the same angular velocity by 2 T  = (options A and D are eliminated) For a satellite, the gravitational force FG provides the centripetal force. Therefore, 2 2 2 2 GMm mrr GM rgr   = == Therefore, the centripetal acceleration which is dependent on the radius and orbital period is the same for geostationary satellites. (Option B is eliminated) Even though the speed of the satellites are the same, as given by the equation vr = , the mass of the satellites may be different. So by 21 2KE mv= , the kinetic energy of the satellites may be different. (Option C is the answer) 12 C Since the mixture is in thermal equilibrium, gases X and Y have the same temperature. Since avgT KE , the average kinetic energy of gases X and Y are the same. Therefore, the average kinetic energy of molecules of X is 6.0 × 10-21 J. 13 C Let Q be the heat supplied to the liquid per minute, Considering the heat capacity of the liquid: ( )4Q mc T mc=  = Considering the latent heat of the liquid: (40)Q mL = Therefore, 114 K40 160 Q c m QL m −  ==  

4 14 D The time required for the object to move from P to Q , half a complete oscillation, is half a period, 2 T . Thus, A and B are eliminated since the options are more than 2 T . Consider the time taken t required for the object in Simple Harmonic Motion (SHM) with amplitude A to move from P to X, thus 2cos2 A At T =  ➔ 6 Tt = and the required answer is given by −=2 6 3 T T T 15 C Object Q experiences more damping. Hence amplitude of Q is smaller. The graphs do not pass through the origin, so B is wrong, leaving C as correct answer. 16 D 17 A Using 2 x  = , 1 82  = ➔ 4  = compression rarefaction compression P Q R S T U V

5 18 C Constructive interference occurs at I, II while destructive interference occurs at III, IV. Since sources are in phase, path difference =  and 2 at II and I respectively, path difference = 1.5 and 2.5 at III and IV respectively. 19 A An electric field strength upwards means that the lower plate is at a higher potential than the upper plate (Options C and D are eliminated) ( ) 36 4000(4 10 ) 16 22 V lower upper upper upper VE d V V Ed V V − = −= − − =  = =− 20 C For X, RX = L A = l 𝑙2 For Y, RY = L A = 2l 2𝑙2 = RX 21 A I = nAve v = I nAe = 𝐼 ned2 4 = 4I ned2 Since I, n and e are constants, as x increases, d decreases and v increases. The graph is non-linear as v is not linearly related to d. I II III IV

6 22 D R = V I for resistor, resistance = 10  (constant) for filament lamp, resistance = 5.0  at 1.0 V At about 1.2 V, the graphs for diode and filament lamp intersects and they both have the same I, V values and hence same resistance. 23 A Before the negative ions of mass m and charge q enter the metal container, its velocity v is found from the following: qV = 1 2mv2 or v = √2qV m For the ions to travel undeflected, electric force qE = magnetic force Bqv q ( V d) = Bq (√2qV m V Bd = √2qV m q m = V Bd222 24 B Direction of magnetic field produced by wire X at point P is as shown. Using Fleming left hand rule to determine force on wire Y , where the vertical component of Bx is upwards in the plane of the paper. filament lamp diode resistor Bx

7 25 C rate of heat dissipation P = Irms 2 R = Io2 2 R for half-rectified wave, Irms = I𝑜 2 so rate of heat dissipation Pnew = Irms 2 R = Io2 4 R = P 2 26 D From Fleming left hand rule, electrons at Q will experience a force towards O. Hence Q is at higher potential than P or an induced current will flow from Q to P. 27 D Using the de Broglie’s relation, the magnitude of the momentum of a photon of wavelength 𝜆 is hp = . From Newton’s 2nd law, the force exerted by 1 photon upon reflection off the mirror is 2dp p hF dt t  = = = For n photons: 2nhF = 28 B 19 19 19 3.5 10 1.6 10 (0.25) 3.1 10 J photon sE eV    −− − =+  = +  = 29 B From max hc KE =+ : As the intensity of light affects the number of photons incident on the metal rather than the energy carried by each photon, hc  , the maximum KE of an electron is independent of the intensity. 30 A The diagram shows that most of the alpha particles passed through the gold foil undeflected or deflected by small angles. This indicates that the size of the nucleus relative to the size of the atom is small, as most alpha particles interacting with the gold atoms were not close enough to the nucleus to experience significant electrical repulson and hence, were only deflected by small angles or remain undeflected. ~ THE END ~ P Q R rotation direction