TMJC 2023 H2 Physics P1 MS

1 Tampines Meridian Junior College 2023 JC2 Preliminary Examination H2 Physics Tampines Meridian Junior College 2023 JC2 H2 Physics Preliminary Examination Paper 1 Suggested Solution 1 A 11 A 21 B 2 C 12 C 22 D 3 C 13 C 23 A 4 B 14 A 24 A 5 B 15 D 25 C 6 D 16 C 26 B 7 D 17 A 27 B 8 B 18 A 28 D 9 D 19 A 29 A 10 C 20 B 30 C 1. Ans: A 2 21 [][] [ ][ ] kg m s[ ] k g s AA m F BIL FB IL B − −− = = = = 2. Ans: C 6 7 10 13 42.195 km 42195 m 4.2195 10 cm 4.2195 10 mm 4.2195 10 m 4.2195 10 nm µ = = × = × = × = ×

2 Tampines Meridian Junior College 2023 JC2 Preliminary Examination H2 Physics 3. Ans: C ( ) ( ) 2 2 2 2 2 At time : 1 2 1 2 At time 2 : 1(2 ) 2 2 11 24 422 t s ut at x gt t s ut g t g t gt x = + = = + = = =  Hence distance fall in 2nd time interval t = 4x – x = 3x 4. Ans: B () W N ma N W ma W −= = −< By Newton’s third law, the magnitude of the force on boy by floor is the same as the magnitude of the force on the floor by the boy. weight of boy, W force on boy by floor, N

3 Tampines Meridian Junior College 2023 JC2 Preliminary Examination H2 Physics 5. Ans: B At equilibrium, CW ACW PQ PQ Tx Tx TT = = = PQ PP QQ P X QY TT WUWU W gV W gVρρ = −=− −= − Since XYρρ > , PQUU> since both has the same volume. Therefore, PQWW> , PQmm> . TQ TP x x P Q WP TP UP WQ TQ UQ

4 Tampines Meridian Junior College 2023 JC2 Preliminary Examination H2 Physics 6. Ans: D Taking right as positive 2 22 222 2 222 Using conservation of momentum: 0 ------ (1) U sing conservation of KE: 1 11 02 22 ------ (2) (1 ) (2) : () 0 P PQ P PQ P PQ P PQ PQ P Q mu mv mv uvv mu mv mv uvv vv v v += + = + += + = + − + −+= 20 0 and PQ P QPP P vv v v uv u = ⇒= ⇒=−= Hence, P becomes stationary and Q moves to the right with the same speed as P. When Q collides with R, the same occurs, Q becomes stationary while R moves to the right with the same speed. 7. Ans: D Force exerted by the motor, F = 40 – 25 = 15 N Output power of motor, P = F v , where v is speed of rotation (at circumference) of wheel. Hence P = 15 [ 2π(0.080) x 10 / 1] = 75 W 8. Ans: B Tension of cord provides for the centripetal force. ( ) ( )( ) 2 2 2 0.80 3.242 0.72 0.27 m mvT r mvkx r x x = = = = Unstretched length = 0.72 – 0.27 = 0.45 m

5 Tampines Meridian Junior College 2023 JC2 Preliminary Examination H2 Physics 9. Ans: D 22 22 Net forces acting on satellite, S E S E Fnet ma Mm MmG G ma Rr M MaG G Rr = −= = − 10. Ans: C Escape speed, 2 14 2 0.2281 1 9 m mE E Em v GMv r Mv r v Mr v Mr = ∝ = × = ×== 11. Ans: A 221 2 PE mx ω= 12. Ans: C From Graph 1, 0.45 0.15 0.30 mλ =−= From Graph 2, 31.6 1.2 0.4 ms 0.4 10 sT −=−= =× 1 3 0.30 750 m s0.4 10vf T λλ − −= = = = × 13. Ans: C Average wavelength of light = 550 nm. Using Rayleigh criterion, limiting angle 9 4 3 550 10 1.964 10 rad2.8 10 − − − ×= = ×× Using small angle approximation, taking the distance between pixels as the “arc length”, 4 4 3.0 10 1.5 m1.964 10 sr sr θ θ − − = ×= = =× Note if wavelength of light is not given: Using wavelength as 400 nm, distance = 2.1 m Using wavelength as 700 nm, distance = 1.2 m Hence answer must lie between 1.2 m to 2.1 m.

6 Tampines Meridian Junior College 2023 JC2 Preliminary Examination H2 Physics 14. Ans: A For the maxima to coincide, the angle of diffraction are the same for both the red and blue light. Since sindn θλ= , and θ is the same, 411 3 685 5 red red blue blue red blue blue red nn n n λλ λ λ = = = = Hence 3rd order of red will coincide with 5th order of blue. 15. Ans: D The gas molecules are very far apart hence no intermolecular forces and the potential energy between molecules is zero. 16. Ans: C 2 rms 2 rms 2 rms 2 rms 2 rms 2 rms rms rms 13 ()22 () () 1 (thermal equilibrium)() () () 2.5 1.6 He He Ne Ne He Ne HeNe He Ne Ne He m c kT mc T mc mc cm mc cm cm = ∝ = = = = = 17. Ans: A From Z to X, ZZ X X ZX ZX pV T pV pV TT UT UU ∝ < < ∝ <

7 Tampines Meridian Junior College 2023 JC2 Preliminary Examination H2 Physics 18. Ans: A ( ) 4 1 (since Q is constant) 0.10 180 0.20 90 kV o surfaceP surface P P QV r V r rV Vr V πε= ∝ = = ×− =− 19. Ans: A Explanation: Initially effective resistance is zero. As resistance of variable resistor is increased, the effective resistance increases until it is R/2. By potential divider rule, terminal p.d will increase. 20. Ans: B Explanation: using potential divider, potential at top left corner will be 6 V while potential at bottom right corner will be 4 V. Thus potential difference as read by voltmeter will be 2 V. 21. Ans: B Explanation: By using RHGR, wires at the top left and bottom right corner of square will produce a B-field that cancels off one another since they have same current and at a same distance apart from center. Using RHGR, wire with 2I will produce a B field diagonally to top left of square. 22. Ans: D Explanation: Recall like currents attract, unlike current repel. Both X and Z will cause a leftward force on Y. Adding up, total magnitude will be 3 F 23. Ans: A Explanation: The soft iron core increases the flux linkage through coil Q. Hence the rate of change of flux linkage is larger, resulting in a larger induced e.m.f. in coil Q. However, the frequency of the signal remains unchanged as the frequency of the alternating voltage in P remains unchanged. 24. Ans: A ( )( )( )0.54 0.10 2.2 0.12 VBLvε = = = With reference to an electron in the rod, using Fleming’s LHR, the force on the electron is downwards. Hence end B will be at a lower potential.

8 Tampines Meridian Junior College 2023 JC2 Preliminary Examination H2 Physics 25. Ans: C s sp p NVV N= Power 2 222 2 s p p sps P N VN NVV R R NR  = = = 26. Ans: B Consider a one-period time frame, 1. Square the graph. 22 46 2116V = = 2. Determine the mean over one period. ( ) ( ) 3 2 3 2116 0.25 10 5291.0 10 areaV period − − × = = =× 3. Take square root. 2 . .. 529 23 Vr msVV = = = 27. Ans: B max max 0 hf E E hf hf φ= + = − The x-intercept is the threshold frequency. A large work function will result in a larger threshold frequency. The gradient of the graph is Planck’s constant and hence should not change. 28. Ans: D ( )( ) 31 6 26 -1 0.29.11 10 5.8 10 100 1.0568 10 kg m s p mv∆∆ − − = = ×× = × 34 26 8 6.63 10 1.0568 10 6.3 10 m px h Min x ∆∆ ∆ − − − ≥ ×= × = ×

9 Tampines Meridian Junior College 2023 JC2 Preliminary Examination H2 Physics 29. Ans: A A range of (kinetic) energies indicates a range of speeds for the β particles. Since beta particles are emitted with a range of speeds, the products of a beta decay process cannot just consist of the daughter nuclide (product nuclide) and the beta particle as this would imply definite speeds for both products, in order for linear momentum to be conserved. Option B is wrong. There is no such observation. Neutrino is chargeless. The total charge of the decay products is equal to the charge of the parent nuclide. Option C is wrong. It is a true observation, but the loss in mass during β decay is due to conversion to energy released, and not the existence of the neutrino. Option D is wrong. There is no such observation. Neutrino is chargeless so has no ionising power, and therefore cannot be observed in a cloud chamber. 30. Ans: C Since the count rate drops significantly when there is a piece of paper, the radiation is stopped by the paper. Therefore it must be α radiation, which has low penetrative power. The background count rate is 24 min-1. Hence initial count rate due to source is 532 – 24 = 508 min-1 After two half-lives, count rate due to source =