DHS 2020 JC1 Promos Physics (Solutions)

DUNMAN HIGH SCHOOL 2020 PHYSICS H2 (YEAR 5) 1 Answers to Promo Exam MCQ Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 D B D D C D D C B A Q11 Q12 Q13 Q14 Q15 D A B B B Structured Questions 1(a) 2 of the following: -temperature, kelvin -current, ampere -amount of substance, mole -luminous intensity, candela 1 mark for each correct quantity and unit, correctly spelled (no caps). [2] MC: It is surprising that more than half the cohort cannot remember the SI base quantities and units. Some students didn’t write in words, instead using only symbols hence no credit. Lastly, for units, the spelling should be all small letters, i.e. kelvin instead of Kelvin. This point wasn’t penalized for this exam, but please remember next time. (b) units of E: kg m2 s-2 , units of ρ: kg m-3 , units of g: m s-2, units of A: m2, units of L: m C1 units of C: kg m2 s-2 / kg2 m-6 m2 s-4 m2 m3 C1 = kg-1 m s2 A1 [3] MC: Some students left answers in J or N, which are not base units. (c) radius = (10.0 ± 0.1) mm Volume V = π r2 h = π (10)2 (1.5) = 471.2 mm3 --- (M1) ΔV / V = 2Δr / r + Δh / h ΔV = (2Δr / r + Δh / h) V = [2(0.1/10) + (0.1/1.5)](471.2) = (0.02 + 0.067) (471.2) = 40 mm3 (1 s.f) --- (M1) Therefore V = (470 ± 40) mm3 --- (A1) [3] MC: Many students left answer as (471 ± 40) mm3. This is a misconception. Note that the uncertainty is in the “tens” decimal position, hence the quantity should be also rounded to the same decimal position i.e. “tens” position.

DUNMAN HIGH SCHOOL 2020 PHYSICS H2 (YEAR 5) 2 2 (a) Use of mgh = ½ m v2 C1 h = 14.7 or 15 m A1 (b) (i) vY = uy + ayt = 0 + (9.81) (1.6) C1 = 15.7 or 16 m s-1 A1 (ii) Straight line of positive gradient B1 S tarting at 0 m s-1 and ends at 16 m s-1 at 1.6 s. B1 ( iii) resultant v2 = 15.72 + 172 C1 v = 23 or 23.1 m s-1 A1 MC : Majority of candidates did well in this question. A few used the slope angle to calculate the vertical component of velocity at C which was incorrect. Q3 MC For similar questions, refer to - 2019 DHS Promo Structured Question 2 - 2020 DHS Dynamics Lecture Example 13 - 2015 A-level H1 Q7 - 3(a) 0.76 + 0.48 = 1.24 m s-1 A1 (b) 1.24 m s-1 A1 MC For e.c.f, ans should be identical to (a)

DUNMAN HIGH SCHOOL 2020 PHYSICS H2 (YEAR 5) 3 (c) Before collision, After collision, By conservation of linear momentum, 5.2(0.48) 2.3(0.76) 5.2 2.3 0.748XYVV− =+= --- (1) Using law of restitution, 1.24 YXVV= − --- (2) Solving, -10.281 m sXV = − -10.959 m sYV = C1 C1 A1 MC Momentum is a vector; The conservation of kinetic energy equation i.e. ( ) ( ) ( ) ( ) 22 2 2111 15.2 2.3 5.2 (0.48) 2.3 (0.76)222 2 XYVV += + ( ) ( ) 225.2 2.3 2.52656XYVV += --- (3) is acceptable for C1 However, it is more difficult to solve equations (1) and (3) compared to equations (1) and (2). The other possible answer from equations (1) and (3) is -10.48 m sXV = & -10.76 m sYV = − , which is physically impossible, because it is as if X and Y never collide, and has to be rejected. VY Vx

DUNMAN HIGH SCHOOL 2020 PHYSICS H2 (YEAR 5) 4 (d) Change in momentum of X, Xp∆ = -1(5.2)( 0.281) (5.2)(0.48) 3.96 kg m s− − ≈− M1 A1 MC Candidates should use physical quantities related to X for calculation. Change of a physical quantity = final physical quantity – initial physical quantity. (e) -13.96 kg m s A1 MC Y XY YX Yp Ft Ft p∆ = ∆= − ∆= − ∆ 4(a) Boat has greater volume compared to solid block (more empty space within). Hence, it displaces more water and can achieve sufficient upthrust to support its weight. OR Combined density of reshaped clay and enclosed air is less than density of water B1 B1 MC Common misconception(s): - justification using surface area - state cause of upthrust - failure to relate to context of question and state in general why a solid object floats on a liquid using principle of floatation i.e. weight of object = upthrust; candidates are expected to compare the situations of the clay block and clay boat - confusion between clay block and clay boat e.g. stating volume of water displaced is more than volume of clay boat, and not clay block. 4(b)(i) T = W = mg = 0.200 9.81( ) = 1.96 N A1 MC - 4.s.f acceptable due to ambiguity of 200 g.

DUNMAN HIGH SCHOOL 2020 PHYSICS H2 (YEAR 5) 5 4(b)(ii) V = m ρ = 0.200 8000 = 2.5 × 10−5 m3 A1 4(b)(iii) T + U = W T = W − U T = 1.96 − 2.5 × 10−5 ( ) 800( ) 9.81( ) = 1.76 N A1 4(c) B1: Constant gradient up till h = x B1: no increase thereafter MC Candidates should relate upthrust to volume of fluid displaced; once the object is fully submerged, the volume of fluid displaced is equal to the volume of the object. 5(a) A1 MC - = +TF FR - Power required in kW = / 1000TFv (b) FT = 220 + ½(0.995) (1.29)(32)2 = 877 N C1 A1 Upthrust / N h / m

DUNMAN HIGH SCHOOL 2020 PHYSICS H2 (YEAR 5) 6 (c)(i) Power supplied by engine = 102 x 0.746 = 76.1 kW A1 (c)(ii) Power required < power supplied by engine due to energy loss (Loss could be due to other energy uses in the car such as radiator, headlights, air-conditioner, radio or other internal frictional forces in the car along the transmission line from engine to the wheels. /Thermal energy produced by engine.) B1 MC Difference in power required in Fig. 5.1 and power supplied by engine in (c)(i) for v = 45 m s-1 is 76.1 - 67.7 = 8.4 kW which is ~11% of power supplied of engine and thus it is significant. In Fig. 5.1, power required already take into account total resistance i.e. air resistance F + friction with road R “… the car is travelling at maximum speed on a level road.” 6 (a) (i) ω = x π 62.33 10 2 C1 = 2.69 x 10-6 rad s-1 A1 (ii) a = r ω2 = (3.82 x 108) (2.69 x 10-6) C1 = 2.76 x 10-3 m s-2 A1 (iii) F = ma = (7.35 x 1022)(2.76 x 10-3) = 2.03 x 1020 N A1 MC : Majority of candidates did well in this part. A few used the formula for gravitational force of attraction to calculate (a)(iii). No credit given as there was no data on the mass of Earth in the question.

DUNMAN HIGH SCHOOL 2020 PHYSICS H2 (YEAR 5) 7 (b) Acceleration is directed towards the centre [B1] and thus it always perpendicular to velocity [B1], hence it does not change the magnitude of the velocity (i.e. speed). MC : A few candidates mentioned “acceleration is perpendicular to the direction of speed” which was wrong as speed was not a vector. Maj ority of candidates obtained 1 out of 2 marks. (c)(i) Potential at infinity is taken to be zero. [B1] Due to the attractive nature of the gravitational force, work done by an external force to bring any mass from infinity to that point is always negative. [B1] Hence the potential at any point must always be negative. MC :