2022 RI Prelims H2 Phy Paper 1 Soln

Raffles Institution Year 5-6 Physics Department 1 2022 H2 Physics Preliminary Examination Solution Paper 1 Qn Ans Solution 1 D 4.5 335.61 15.1 20 (1 s.f.)100v    1(340 20) m svv    v is rounded off to the nearest tens place. 2 A 4 4 4 4 2 34 24 unit of unit of unit of unit of unit of kg m s m kg s Ks m K P AeT Wt AeT Fd tAeT mad tAeT              3 C Take direction to the right and upwards as positive. 3.4cos30 5.0cos60 xx x x su t st u     13.4cos305.0sin60 9.81 7.2240 m s5.0cos60yyyvua t          1 tan , where is the angle that the final velocity makes with the horizontal 7.2240tan 70.9115.0cos60 y x v v        70.911 30 40.911 41       OR     22 12 2 11 2 5.0sin60 2 9.81 3.4sin30 7.2183 m s or 7.2183 m s (reject) yy y y y vu a s v         1 tan , where is the angle that the final velocity makes with the horizontal 7.2183tan 70.8975.0cos60 y x v v        70.897 30 40.897 41      

Raffles Institution Year 5-6 Physics Department 2 4 A Based on the given graph, the direction upwards is positive. At time t, the ball is at its maximum height, and its displacement is +S1 (or +S2). At time 2t, the ball is back at its initial point and is accelerating as it moves downwards (negative velocity). Hence its displacement is zero (S1 = S2) and the gradient of the displacement-time graph is negative. At time 3t, the ball hits the ground since its velocity is –2u and acceleration is constant at –g. Its displacement is –S3 from the initial point. From 2t to 3t, since the ball is accelerating as it moves downwards, its speed increases. Hence the gradient of the displacement-t ime graph should be negative with increasing magnitude. 5 C When P starts to slide: Consider block P: PPfm a Consider block Q: QQfm a PP QQma ma Since PQmm , PQaa  Friction on Block P causes it to decelerate as it moves to the left. Friction on Block Q causes it to accelerate as it moves to the left. 6 A from the time the column is dropped to when it just reaches the surface of the soil, increase in K.E. = decrease in G.P.E 21 02 2 Mv MgH vg H   from the time the column enters the soil to the time when it comes to a stop, by Newton’s second law and taking direction downwards as positive, R dpF dt 2 where is the average resistive force 0 2 21 fi pMg f f t ppMg f t MvMg f t Mg HMv HfM g M g M gtt g t          P Q NP mPg f f NQ mQg direction of motion

Raffles Institution Year 5-6 Physics Department 3 7 B When a quarter of the cube is removed, the C.G. changes to the new geometrical centre and the weight decreases. Upthrust remains unchanged initially. Hence UW and there is a resultant force upwards. Since the upthrust and the weight are now not acting along the same line of action, there will be a resultant clockwise moment. 8 B  221 2 kx yEm v v where xxvu and yyvug t   221 2 kx yEm u u g t  21() 2 pyE mgh mg u t gt  From the above equations, the variation wi th time of the energies is a quadratic relationship. By the conservation of energy, the increase in G.P.E. is equal to the decrease in K.E. as the ball moves upwards to its highest point a nd the decrease in G.P.E. is equal to the increase in K.E. as the ball moves downwards back to the ground. At 0t  and when ball is back to the ground,  221 2 kx yEm u u and 0pE  At maximum height, 0yv  , yut g 21 2 kxEm u , 2 21 22 y py uEm g h m g m u g     9 D since 0vu m vFm am u tt      2 22 2 21K.E. K.E.22 2 mv Ft tmv mm m 22 2 2 2 K.E. 0gain in kinetic energy of body A gain in kinetic energy of body B K.E. 0 22 8 A B AB AB AB BA tt mm tm tm tm tm             removed liquid W U

Raffles Institution Year 5-6 Physics Department 4 10 D As the sphere moves from highest to lowest point, increase in K.E.  decrease in G.P.E.  2211 222mv mu mg L 22 122 2vg L u  When the sphere is at the lowest point,        2 2 2 2 122 2 2 0.40 2 9.81 0.50 0.5 2.5 0.40 9.810.50 24.62 25 N mvTm g L mvTm g L mg L u mgL        11 B Since the total energy 2 T GMmE R of the satellite decreases due to work done against drag forces, its orbital radius R decreases. As R decreases, kinetic energy 2 k GMmE R increases, hence orbital speed increases. Since 2 22 3 2 2GMm mRm R T R TR      The orbital period T decreases as R decreases. 12 A Gravitational force provides the centripetal force. 2 2 GMm v m RR  3 22 4 43 3 GRGMvG R v RRR            22 22 1 1.25 4 1 (7.90)1.25 16 1.7665 1.77 km s MM M M M E EE MM M vR R v R R v        

Raffles Institution Year 5-6 Physics Department 5 13 D       1.5 8.31 273.15 25 20 3.71000 50222 Pa 50 kPa R pV nRT nRTp V mRT MV        14 D 2 2 1 3 33 Nmpc V pVk Tc Nm m      23 1 27 3 1.38 10 273.15 1003 719.04 720 m s 18 1.66 10 rms kTc m         15 D Using Qm c  heat gained by metal = heat lost by liquid      20 3 2.5 100 20 750 7.5 8.5 750 20 90.59 91 C ff ff f f mc m c            16 A 22 0 1 2Em x  2 22 2 011 1 22 3 9 p xEm xm E     17 B At 4T , the mass is at the equilibrium position where 25 cmd  . At 2T , the mass is at the highest point of its oscillation. Hence the amplitude of the oscillations is 15 cm. 18 C After passing through the first polariser, intensity is halved, i.e. 220 W m . Subsequently, the intensity after passing through the 2nd and 3rd polarisers is  22 2 211 0022 cos cos 90 cos sinfII I      0 2cos sin fI I  0 2 22 . 5 1sin2 2 2 40 2 fI I   22.5 

Raffles Institution Year 5-6 Physics Department 6 19 B Table shows sin n d   order violet (400 nm) red (700 nm) 1 400 nm d 700 nm d 2 800 nm d 1400 nm d 3 1200 nm d 2100 nm d Since 1200 nm 1400 nm dd  , 3rd order spectrum overlaps with the 2nd order spectrum. 20 B The constant electric force on the electron is pointing vertically downwards, in the direction of increasing potential.  22VV VEV E x xxd d         where x is the vertical distance work done by external force,    fiWqV e VV   From X to Y: This is along an equipotential line. Since there is no change in potential, the work done is zero. From Y to Z: decrease in potential energy    22 2 cos30 sin60YZ Z Y VV VW e V V e x er er dd d           From Z to X: increase in potential energy    22 2 cos30 sin60ZX X Z VV VW e V V e x er er dd d          21 B increase in K.E. = decrease in E.P.E. 21 02 mv q V 2 2 qV qEdvv E d mm    for the same charge to mass ratio 22 2 11 1 2 2( 2) 2 vE d vE d Edvv Ed v   

Raffles Institution Year 5-6 Physics Department 7 22 C 12 12 12 111 T T T VVV RR R RR R    II I Since the resistors are in parallel, the p.d. across each resistor is the same. 23 C XXVR I The graph for X is a straight line with positive gradient and it passes through the origin. YXVE R I The graph for Y is a straight line with negative gradient and positive vertical intercept, E. The current through the 2 resistors is always the same at any time. The total p.d. across the 2 resistors is always E. Hence as the reading of Y decreases, the reading of X increases. 24 D For a long straight wire, 0 2B d   I At Q,         earth due to P due to R 7 5 5 5 55 1 41 0 1 . 0 2.0 10 2 20 . 3 0 1.8667 10 T (northwards) 1.8667 10 2.