Tiêu đề 2023 GCE A level H2 Physics 9749 Ans.pdf ✅

1 2023 H2 Physics (9749) Solutions Paper 1 Easy: 1, 2, 7, 8, 18, 27 and 28. Students are good in questions requiring mathematical skills. Challenging: 5,10, 19, 22, 24 and 25. Students find conceptual questions more challenging. 5 (Unable to find C.G of a complex shape) 10 (Memorise the proof of equation without being clear of assumption made) 19 (Confusing path of proton in a E-field and in B-field) 22 (Unfamiliar circuit not seen in past years, unsure of how to set up a potentiometer circuit with correct orientation of the polarity of the emfs) 24 (Careless in analysing the direction of the magnetic field) 25 (Unable to carefully analyse the path of a charged particle given different orientation of E and B-fields)
2 1 Ans C A, B and D are constant errors that causes all results to differ from the true value by a fixed amount by the same amount thus systematic. 2 Ans D Displacement of car between t = 0 and t – t = area under v-t graph between t = 0 and t – r = area of trapezium = 1⁄2 ® (p+ q) 3* Ans A E = p2 /2m where p is the momentum when KE is E P = (2mE)1/2 Since F = rate of change of momentum = p/t Ft = p = (2mE)1/2 Ft  E 1/2 4 Ans C For elastic collision, relative speed of approach = relative speed of seperation ➔ : uA – uB = vB – vA u1 – (-u2) = v2 – v1 (Taking into account the sign convention) u1 + u2 = v2 – v1 5* Ans B (challenging) Let B has 4 units of mass and A has 8 units of mass, The equivalent total moments of A and B about their CG from P as the entire sheet with 12 units of mass with its C.G at distance d from P Therefore, 8 (6) + 4(2) = 12 (d) d = 4.7 cm 6 Ans B KEi/t = 1⁄2 mv2 /t = 1⁄2 (9.7)(4)2 = 77.6 W
3 KEf/t = 1⁄2 mv2 /t = 1⁄2 (9.7)(1.5)2 = 10.9 W Power input to generator = 77.6 – 10.9 = 66.7 W Power output from generator, P = 60/100 (66.7) = 40 W P = IV When V = 12 V I = 3.3 A When V = 24 V I = 1.7 A 7 Ans C 1 year ➔ 2 1⁄2 year ➔  8 Ans B F = gm F/ m = g 9 Ans C (repeated qn) For gravitational field, path along the field lines and in circular motion (perpendicular to the field lines) are possible without spacecraft firing its rocket 10* Ans A (challenging) When only 1 D motion is considered in this simple model, the derivation is p = mNv2 /V where v is the velocity perpendicular to the shaded wall. Comments: This question proved challenging. The most common incorrect answer was option C, indicating a recalled version of the proof rather than an application of the proof to the specific case in the question, where all N molecules are travelling perpendicular to the shaded wall. 11 Ans B U  T (in Kelvin) 0 oC → 273 K → U 273 oC → 546 K → 2(273 K) → 2U 12 Ans C Q = mc  = Pt/mc Since radient of P is steeper than than Q for the part when temp increases