DHS 2019 JC1 MYE Physics (Solutions)

DUNMAN HIGH SCHOOL 2019 PHYSICS H2 (YEAR 5) 1 Answers to Mid-Year E xam MCQ Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 A B C C D D C D A D Q11 Q12 Q13 Q14 Q15 C A B B B Structure Questions 1 (a) 23 46 () 3 Lrv r gπη π ρ ρ= − Units of left hand side: (kg m−1 s−1) (m) (m s−1)2 = (kg m2 s−3) [M1] Units of right hand side: (m3) (kg m−3) (m s−2) = (kg m s−2) [M1] Since the units for left and right hand sides are not equal, the equation is not correct. [A1] (Comments: Students need to be more mindful of the presentation of their answers. They should not include physical quantities in equations that consist of units as well. In addition, students should write a statement to conclude whether the equation is physically correct.) (b) (i) Irregularities in diameter of sphere. [A1] (Comments: Students need to explain clearly the type of random error involved, instead of merely stating a generic and insignificant error.) (ii) Measure the diameter along different orientations of the sphere and take average. [A1] (Comments: The improvement stated should be linked to the random error stated in (b)(i) and it should be explained clearly.) (iii) 22F Br v ρ= 22 FB rvρ= 3 22 8 10 (0.03) (1000)(0.24)B −×= = 0.15432 [C1] ρ ρ ∆∆ ∆∆ ∆=+ ++ 2( ) 2( )BF r v BF r v ∆ = + ++0.1 0.1 0.01 12( ) 2( )0.15432 8.0 3.0 1.00 24 B [M1]

DUNMAN HIGH SCHOOL 2019 PHYSICS H2 (YEAR 5) 2 ∆= × =0.15432 0.1725 0.0266B = ±0.15 0.03B [A1] (Comments: Many s tudents were careless with the conversion of units. The alternative correct method to solving this part of question is to calculate the maximum and minimum B and subsequently find∆B .) (iv) The measurement velocity gives rise to the greatest uncertainty in the value for B [A1] because ∆  2 v v = 0.08 is the largest. [M1] (Comments: Many students did not get any credit, as they were not able to state a correct reason for choosing velocity as the answer. A large number of students were confused with the terms ‘uncertainty’ and ‘fractional uncertainty’.) 2 (a) (i) horizontal component = 63 cos 14o = 61.1 m s−1 vertical component = 63 sin 14o = 15.2 m s−1 [A1] (Comments: Some answers in (i) had their horizontal and vertical components mixed up.) (ii) horizontal displacement = (63 cos 14o) (4.9) = 300 m [A1] (iii) vertical displacement = (63 sin 14o)(4.9) – (0.5)