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)(9.81)(4.9)2 [M1] = – 43 m [A1] (Comments: Some students substitute values into the equation of motion with no regard to the different direction between initial vertical velocity and the acceleration of free fall) (iv) tan θ = (43/300) [M1] θ = 8.2o [A1] (Comments: This is not the angle that the velocity at B makes with the horizontal) (v) Distance is the actual path travelled. [B1] Displacement is the straight line between A and B or minimum distance between A and B. [B1] (b) new path lower highest point, shorter range, asymmetrical, (larger angle at which ball hits ground) (Comments: Deduct one mark for each mistake)

DUNMAN HIGH SCHOOL 2019 PHYSICS H2 (YEAR 5) 3 3 (a) The rate of change of momentum of a body is proportional to the resultant force acting on the body and occurs i n the direction of the resultant force . (Comments: For definitions, underlined keywords must be seen before marks are awarded.) (b) 5 forces correctly labeled [B1] W B > T [B1] WA = N (if seen) (Comments: A lot of candidates lose marks because they: (1) did not use ruler, (2) did not draw using the correct magnitude, (3) did not spell the forces in full. When the questions ask specifically to draw forces, the above criteria are applied by Cambridge’s markers too.) spring 0.50 m frictionless pulley A Normal Contact force, N Tension, T Tension, T Weight, WA B Weight, WB [B1] [B1]

DUNMAN HIGH SCHOOL 2019 PHYSICS H2 (YEAR 5) 4 (c) Block A: T = mA a ––– (1) Block B: WB – T = mB a ––– (2) (1) + (2): WB = (mA + mB) a [B1] 0.50 × 9.81 = (0.30 + 0.50) a [B1] a = 6.13 m s–2 [A0] (Comments: For “show” questions, all steps need to be clearly developed. All substitutions need to be shown, e.g. must show (0.3 + 0.5) rather than 0.8 directly. Candidates who write WB = (mA + mB)a directly without showing the two equations (1) and (2) above, are, by right, should not deserve the full credits. Due to leniency of marking, benefit of doubt is given to them. In future examinations, it is advisable to either formed two equations first as shown above, or write explicitly: “Applying F = ma to whole system, then write WB = (mA + mB)a. ) (d) From (1): T = 0.30 × 6.131 = 1.84 N [A1] (Comments: Common mistakes: Some candidates thought that a = 9.81 ms -2, ignoring the answer in the previous part.) (e) Using v2 = u2 + 2as, v2 = 0 + 2 (6.131) (0.50) [C1] v = 2.48 m s–1 [A1] (Comments: Common mistakes: Some candidates thought that a = 9.81 ms -2, ignoring the answer in the previous part.) (f) Using v = u + at, 2.476 = 0 + 6.131 t [C1] t = 0.404 s [A1] (Comments: Common mistakes: Some candidates wrote time = distance/velocity , treating this as a con stant velocity question, ignoring the fact that there is acceleration.)

DUNMAN HIGH SCHOOL 2019 PHYSICS H2 (YEAR 5) 5 (g) Resultant force = 0 N, acceleration is zero. Block B: WB – T = 0, Block A: T – Fspring = 0 0.50 (9.81) – 360 x = 0 [M1] x = 0.0136 m [A1] (Comments: Common mistakes: A large number of candidates use tension = 1.84 N to try to solve this part, without realizing that the value of tension is different after the spring is compressed.) (h) By conservation of energy, mBgh = 1 2 k xmax2 (0.50)(9.81)(0.50+xmax)= 1 2 (360) xmax2 180 xmax2 – 4.905 xmax – 2.4525 = 0 xmax = 0.131 m Alternatively, mBgh + 1 2 (mA+mB)v2= 1 2 k xmax2 0.50(9.81)(xmax)+ 1 2 (0.80)(2.48)2 = 1 2 (360) xmax2 (Comments: Good to see that a good number of candidates try to use conservation of energy to solve. However, they are not aware that object B continues to lose gravitational potential energy as the spring is being compressed.) 4 (a) (i) 1. mass = density × volume C1 = 920 × 6.4 × 10 4 × (28 + d) A1 2. either 920 × 6.4 × 104 × (28 + d) or 1030 × 6.4 × 104 × d A1 (Comments: The majority obtained the correct expressions. A few incorrectly quoted mass as ratio of density and volume. Some used the incorrect value of density in the required expression.) (ii) 920 × 6.4 × 104 × (28 + d) = 1030 × 6.4 × 104 × d C1