TMJC 2023 H2 Physics P2 MS

1 Tampines Meridian Junior College 2023 JC2 Preliminary Examination H2 Physics Tampines Meridian Junior College 2023 JC2 H2 Physics Preliminary Examination Paper 2 Suggested Solution 1 (a) 2 2 1 1 2 11.0 (0.80) ( 9.81)(0.80) [M1- correct and v alues]2 2.674 2.67 m s [A0] yy yy y S u t at u Sa u − = + −= +− = = (b) ( ) ( ) 1 tan 10 2.67 [M1]tan 10 15.1 m s [A1] y x x u u u − = ° = ° = (c) (15.1)(0.80) 12.1 m [C1- for calculating best value] 5.0 0.08+ 0.15 [C1: correct substitution]100 0.80 0.15( ) 1.8 m 12 2 m [A1] xx x xx xx xx xx S ut S Su t Sut SS SS = = = ∆∆ ∆= + = = ∆= = ±∆ = ± (d) (i) Air resistance acts in the same direction / downward as the weight of ski jumper. [B1] Hence, the vertical acceleration of the ski jumper is greater than that of free fall or net force is downwards. [B1] Air resistance decreases with decreasing speed. Hence magnitude of vertical acceleration decreases. [B1] At highest point, vertical acceleration is g or free fall (as there is no resistance). [B1] Maximum 3 marks for any of points stated above. (ii) This position increases her surface area to generate upwards lift force. The more the lift, the more time she stays on the flight and thus, the further she will travel. [B1] Or: This position decreases her exposed surface area horizontally (or more streamlined/ reduce the air resistance). Hence, she can travel more horizontally.

2 Tampines Meridian Junior College 2023 JC2 Preliminary Examination H2 Physics Or: This position will lower her CG such that there is a counter moment to the moment created by air resistance. 2 (a) (i) The resultant force in any direction is zero. The resultant torque about any point/axis is zero. [B1] [B1] (ii) [B1] arrow from base of spine and correctly going through the point of intersections of all the forces. (iii) Taking the base of the spine as the pivot, sin12 (0.40) sin70 (0.65) sin70 (0.65) sin12 (0.40) 7.3 QP Q P °= ° °= ° = [M1] [A1] (b) (i) Vertical direction: Fnet = 0 as the suitcase did not move in the vertical direction. 30sin20 11 (9.81) 97.6 N yT NW N N += °+ = = [M1] [A1] (ii) The resultant force acting on the body is proportional to the rate of change of its momentum, and the change takes place in the direction of the resultant force. [B1] (iii) Horizontal direction: Fnet = ma as the suitcase accelerated horizontally. -2 30cos20 8.5 11 1.79 m s xT f ma a a −= °− = = [M1] [A1] base of spine Q P

3 Tampines Meridian Junior College 2023 JC2 Preliminary Examination H2 Physics 3 (a) Rate of change of angular displacement [B1] (b) (i) By conservation of energy, Loss in Ep = Gain in Ek ( )( ) 2 2 1 1 0 [C1]2 19.81 4.5 5.5 [M1]2 14 m s [A0] mgh mv m mv v − = − += = Do not accept use of kinematics equation. (ii) 1 14 4.5 3.1 rad s [B1] v rω − = = = (iii) (Tension – weight) provides for the centripetal force. [B1] ( )( ) ( ) 2 21465 9.81 65 [C1] 4.5 3469 3470 N [A1] netF ma vT mg m r T T = −= −=   = = (iv) Parabolic path with initial horizontal velocity [B1] 4 (a) (i) V [M1] potential at A= 0-3.2V=-3.2V[A1] 0040 80 3 2..outV = ×= (ii) [C1] R [A1] 80 32 48 48 1200040 .. . . . x x VV=−= = = Ω (b) (i) [C1] V [C1] = 2.16 V [A1] 100 80 44 44100 80 80 44 44 8044 44 120 . . . .. out out out total R R R ×= = Ω+ = × = ×+

4 Tampines Meridian Junior College 2023 JC2 Preliminary Examination H2 Physics (ii) When temperature gets hotter, resistance across the thermistor decreases [M1] and hence effective/combined resistance across AB decreases. By applying potential divider rule, potential difference across BC increases [M1] which will turn on the fan. Thus, fan should be connected across BC [A1] 5 (a) (i) The magnetic flux density of a magnetic field is defined as the force per unit current per unit length (of conductor) acting on a straight current-carrying conductor placed at right angles to the magnetic field [B1]. (ii) Current in clockwise direction through wire [A1] (iii) T [A1]74 504 10 20 4189 10030 oB nIµπ −−= =× × ×= × ... (iv) (4.189 )(1.6 10 )(10) [C1] =6.7 10 N [A1] 4 19 22 10 F Bqv −− − = = ×× × (v) The magnetic force is acting perpendicular to the direction of travel/ velocity of the electron [M1] and this will only affect direction of velocity and not its magnitude [A1] Alternative answer: The magnetic force is acting perpendicular to travel/ velocity of the electron [M1] and therefore no (net) work done [A1] on the electron, resulting in no change in kinetic energy (b) (i) No net force acting on electron, Magnetic force= Electric force [C1] qE= Bqv E= Bv = 4.19 x 10-3 Vm-1 [A1] (ii) Downwards vertical arrow labelled E [B1] 6 (a) α : protons decrease by 2, neutron decrease by 2 [B1] β : protons increase by 1, neutron decrease by 1 [B1] γ: no change to number of protons and neutrons [B1]

5 Tampines Meridian Junior College 2023 JC2 Preliminary Examination H2 Physics (b) γ has no charge, so does not experience magnetic force [B1]. So its path should be undeflected [B1] α and β hav e opposite charges, so should experience magnetic forces in opposite directions [B1]. So their paths should curve in opposite directions [B1] (c) (i) Mass is converted [B1] into kinetic energies of nickel-60 nucleus and β particle (also energy of neutrino) as well as electromagnetic energy (photon energy) of the γ radiation [B1] (ii) Energy released = (mass loss) c2 = (59.9338 – 59.9308 – 5.4348 x 10-4) (1.66 x 10-27) (3.00 x 108)2 [C1] = 3.67 x 10-13 J [C1] = 2.29 MeV [A1] 7 (a) (i) Sound cannot be polarised [B1] as it is longitudinal [B1]. (ii) The particles are more tightly packed in solid steel [B1] than in air. Hence it is faster for energy to be transferred between particles. (iii) I is inversely proportional to square of r. 2 1 rΙ ∝ [B1] iv.1 ( ) ( ) 12 -1 6 10lg 120 1.0 10 1.0 W m [B1] 20lg 120 20 10 20 Pa [B1] X X X X p p Ι Ι − − = × = = × = iv.2 Estimated area of an eardrum = 1 cm2 = 10-4 m2 [B1] (accepted area between 2 x 10-5 m2 to 2 x 10-4 m2; based on circular area between diameter of 0.5 cm to 1.5 cm) iv.3 ( ) 4 3 20 10 [C1] 2.0 10 N [A1] F PA − − = = = × (b) (i) To avoid causing noise disturbances to people living on land. [B1] OR As no human lives on the ocean, no one will be affected by the noi se disturbance. [B1] (ii) Any of the following (or other sensible suggestions): [B1] Minimise the drag force at high speed as air is thinner at high altitude.

6 Tampines Meridian Junior College 2023 JC2 Preliminary Examination H2 Physics Minimal obstacle / traffic at high altitude so that it can travel at high speed without obstructions. Minimise intensity of noise at sea level. (ii) Speed = (1.4)(295) = 413 m s-1 [B1] (c) (i) Resonance [B1] (ii) Wavelength = 2(0.91 – 0.54) [C1] = 0.74 m [A1] (iii) Speed of sound = fλ = (480)(0.74) [C1] = 355 m s-1 [A1] (iv) -1 22 273331 [M1]273 344 m s [A1] v += =