2024 TJC H2 Prelim Paper 1 Solutions

2024 H2 Physics Paper 1 Solutions 2024 H2 Prelim Paper 1 Solutions: 1 2 3 4 5 6 7 8 9 10 A B B C A A D A D B 11 12 13 14 15 16 17 18 19 20 C C B A A A C A A C 21 22 23 24 25 26 27 28 29 30 D C A C D B D C D B 1 A Units of ∆𝑃 = 𝑘𝑔𝑚−1𝑠−2, units of 𝜌 = 𝑘𝑔𝑚−3 Units of ∆𝑃 𝜌 = 𝑚2𝑠−2, hence n =0.5 2 B Accurate → How close they are to the true value Precise → How close the measured readings are from one another Readings d/ mm Ave Value Largest – smallest value 1.49 1.46 1.52 1.50 1.4925 0.06 1.48 1.58 1.51 1.40 1.4925 0.18 (not precise) 1.35 1.37 1.42 1.42 1.39 0.07 1.32 1.37 1.41 1.50 1.4 0.18 3 B Vrelative to ground = 22350 100− = 335 km/h 4 C 5 A Change in momentum = area under F-t graph Area under F-t graph from 0 to 10 seconds = 0.5 x (8+4) x 12 + [(-0.5) x (2) x (12)] = 60 N s Change in momentum = final momentum – initial momentum 60 = mv – 0 v = 200 m s−1 Vplane = 100 km/h Vplane = 350 km/h Vrelative to ground

2024 H2 Physics Paper 1 Solutions 6 A 2 221000 2 0 10 0 50 0 314 N net 2 2 2 Force by wall on water= F mass per unit time v m ρ(Vol)Δv Δv since m ρ(Vol) Δt Δt ρ r v(0 - v) = - ρ r v . ) ( . ) . −   =   = = = =  = − (  =− By Newton’s 3rd Law, Force by water on wall = Force by wall on water = 0.314 N 7 D On each cuboid ; Volume of water displaced is the same, therefore U is constant for all cuboids of equal volume. T + U = mg T +U =Vg (where  is density of material/cubpid) T = Vg – U Since cuboid W has largest density, T on string is the largest. 8 A Extra energy stored = area of trapezium = ( )( )1212 2 1 xx TT −+ 9 D Since P and Q have the same angular velocity, the linear speed (v = rω) of P is greater than Q as r1>r2. Also, centripetal force (F = mrω2) is greater for P than Q as r1>r2. 10 B At P: N + mg = mv2/r N = mv2/r – mg rgv = when N  0 Since GPE at P and Q is the same, KE at P = KE at Q. At Q:mg – N = mv2/2r N = mg - m(rg)/2r N = mg/2 F x X1 T2 X2 T1

2024 H2 Physics Paper 1 Solutions 11 D By conservation of energy, KEi + GPEi = Kef + GPEf    + − = + −       20½ 7 GMm GMmmvRR = 12 7 GMv R 12 C The antenna will have the same speed as the space -craft and is still bound in orbit. There is still gravitational force acting on it which causes it to move in circular motion. 13 B 21 3=  PV Nm c V and m remain unchanged  Nc P  2 = constant  ( ) N cN. PP  = 2 5 3 3 0 10 4 2  N = 2.0 105 m2 s2 14 A This is an SHM question and the Ep graph is A, lowest at the equilibrium and maximum at the amplitudes (opposite from the KE-r graph). 15 A ma = mg – R Object will remain in contact when R is greater than 0. Object will lose contact when R = 0, ma = mg For SHM, a = ω2xo = (2πf)2xo = g xo = g (2πf)2 = 9.81 (2π(2.0))2 = 9.81/(2𝜋(3.0))2 = 0.028 m 16 A I kA2= Resultant amplitude when antiphase = A – 0.6 A = 0.4 A Resultant intensity 220 4 0 16 0 16( . ) . .= = =RI k A kA I

2024 H2 Physics Paper 1 Solutions 17 C Using oII 2= cos  For diagram 1, 2 = 8 cos2    = 60o For diagram 2, when rotated another 90o, I 28 90 60=+ cos ( ) = 6 W m-2 18 A 20 75=tan   = 14.9o Using dn =sin where d N 1= , o 310 14 9 2300 − =sin .    = 429 nm 19 A E field point from high V to low V  E field point towards right Since electron is -vely charged, electric force on electron points towards X or away from Y. Change in EPE from X to Y = EPEY - EPEX = -eVY – (- eVX )= - 70 eV + 100 eV = +30 eV 20 C LLR AA = ( ) 2 '' ' '5 4.0 2 ' 10 R L A R L A R d d d d d R = = = 21 D In the driver circuit, the p.d. across XY should be minimized, so the resistance of the NTC thermistor needs to be maximised, according to potential divider rule. Hence, the temperature should be low. The terminal p.d. of the test circuit should be maximised, so the resistance of the LDR needs to be maximised. Hence, the environment should be dark. 22 C ss pp VN VN=  sV =  =130 230 122500 V Peak power = ()oV R == 2 22 12 486 W

2024 H2 Physics Paper 1 Solutions 23 A Taking moments about pivot, ( )( ) 3 cos30 0.12 sin302 150 0.40 60 10 cos30 0.12 sin30 2 0.029 T B LFL B B −  =    =   = 24 C Force causing deflection acts upwards hence magnetic force is upwards. Using Fleming’s left hand rule, current is opposite in direction to motion of particle so it must be negatively charged. If direction of magnetic force is upwards, electric force must be downwards so electric field must be upwards 25 D Maximum flux linkage occurs when the B field lines are perpendicular to coil, i.e. either  3 anticlockwise. When flux linkage is maximum , induced e.mf. will be minimum ( zero ) and vice-versa. 26 B Step up transformer  no. of turns in secondary coil Ns > no. of turns in primary coil Np Since magnetic flux  in both coil is equal as it is an ideal transformer, Ns  > Np  Ideal transformer  pps p s s VNI I V N== < 1 27 D 28 C E = ½ m v2 = p2/ 2m p = (2mE)1/2 (m of alpha particle = 4u) = (8uE)1/2  = h / p = h / (8uE)1/2 = h / 2(2uE)1/2 29 D  will not pass through paper  is stopped by a few cm of aluminium foil only  can pass through thick steel 30 B ( )( ) 2 268 8 6.0 10 3.00 10 60 60 1.5 10 W E mcP tt − ==  =  =

2024 H2 Physics Paper 1 Solutions