DHS 2021 JC1 MYE Physics (Solutions)

DUNMAN HIGH SCHOOL 2021 PHYSICS H2 (YEAR 5) 1 Answers to Mid-Y ear Exam Section A 1 2 3 4 5 6 7 8 9 10 B B B C B C A A B C 11 12 13 14 15 D D D C A 1 B Airbus A330: 870 km h−1 Boeing 737-300: 780 km h−1 2 B Let the top mark be and lower mark be . Let the time the ball passes the top mark be and when it passes the lower mark, . 270Hence the speed of the ball is 2.00 a b ba b ab t t bavv tt ttv ba v ba −= = − ∆ +∆∆ ∆ +∆= + − 2 mm 0.04 s 2 0.04 270 mm 2.00 s 270 2.00 a batt = += +− 3 B The gradient of the displacement-time graph gives the velocity. Car M is accelerating as it has an increasing velocity (since gradient increases). Car N is not accelerating as it has a constant velocity (since gradient is constant). Statement is incorrect. 4 C 2 2 1 2 1 0.5 For 1st half, 0.5 0.5 --- (1) For entire journey, 0.5 --- (2) Solve (1) and (2) 0.71 s ut at L at L aT tT = + = = = 5 B By Newton’s 3 rd Law of Motion, action and reaction forces are equal in magnitude and opposite in direction and they act on different bodies.

DUNMAN HIGH SCHOOL 2021 PHYSICS H2 (YEAR 5) 2 6 C 7 A 8 A Since all cuboids are fully submerged, volume of displaced water is the volume of cuboid which is the same for all cuboids. Since upthrust is the volume displaced water x density of water x g, upthrust is the same on all cuboids. 9 B Gain in GPE = 400 x 9.81 x 1200 = 4.7 MJ Power output = Work/time = 4.7 x 10 6 / 120 = 39kW Power input = 39/0.8 = 49kW From Newton’s 2nd Law of Motion, T – mg = ma (1) Mg sinθ – T = Ma (2) Adding the equations gives Mg sinθ – mg = (m + M)a Since the right hand side is positive, then Mg sinθ – mg > 0 or sinθ > m/M a a T Mg sinθ T mg extension e spring length/ cm 3 0 The corresponding extension of spring is shown on graph. From graph, spring constant k = F/e = 5/3 N cm-1 When force = 7 N, F = ke 7 = (5/3) e so extension e = 4.2 cm

DUNMAN HIGH SCHOOL 2021 PHYSICS H2 (YEAR 5) 3 10 C Since body is moving up the track with constant velocity, there is no change in its K.E. power supplied = rate of increase in G.P.E. + rate at which work is done against the resistive forces ( ) ( )( ) ( )70 9.81 20sin30 120 20 9267 9300 W mg h Fd tt ∆= + = °+ = = 11 D ( ) 22 1 1 2 2 2 minimum tension occurs at the top: 5.00.30 12.5 N0.60 9.6 N maximum tension occurs at the bottom: 12.5 15.4 N vTWm r T vTWm r T += = = ⇒= −= = ⇒= 12 D mE 22 Smm SE Sm 2 S force on the Earth due to the Moon force on the Earth due to the Sun MMG rrM MM MrG r = =   13 D = = ⇒= = = ⇒ 2 2 2 2 2 at the point a distance fr 2 om Earth's centre for points outside the Earth of mass , ----- (1) at the surface of Earth of radius , 10 ----- (2) (1) 5 (2) : 1 2 M GMg r GM x R GM R Rx x Rx =7.0(9.81)(20 sin 30o) +12(20) = 930 W

DUNMAN HIGH SCHOOL 2021 PHYSICS H2 (YEAR 5) 4 14 C ( ) ( ) ( ) ( ) ( ) 11X Y YX X YX Y 1 11 YX 1 Y X YX gravitational potential, is inversely proportional to 8 kJ kg 4 kJ kg2 4 kJ kg 8 kJ kg 4 kJ kg 1 kg 4 kJ kg 4 kJ GM r r r Lrr rL UUm φ φ φ φ φφ φφ φφ −− − −− − = − ⇒ ⇒ = ⇒= = − = − − =− −− = + −= − = + = + 15 A 2 2 1 33 22 MmG mrr M GMG rr ω ωω = ⇒ = ⇒= 

DUNMAN HIGH SCHOOL 2021 PHYSICS H2 (YEAR 5) 5 Section B 1 (a) (i) check for zero error (on micrometer)/zero the micrometer [B1] (ii) take readings along the length of the wire/at different points and take average value [B1] (iii) take readings spirally/around the wire at the same point and take average value [B1] (b) unit of D = [units of k] [units of d] units of k = 1 [A1] unit of Q = [units of C] [units of ρ] [units of g ] [units of (D − kd)3/2] [M1] kg s−1 = [units of C] (kg m−3) (m s−2)1/2 (m)3/2 units of C = m [A1] Comments: Mass flow rate is the mass of a substance which passes per unit of time, Its unit is kg s−1 in SI units. However, many quoted its unit to be m s−1. Many also failed to see that terms D and kd must have the same units as they are combined using subtraction. (c) Determining the direction of Δv using Δv = v f – vi [M1] Since Δv is 90° to the surface of the ramp, Δv is 70° with respect to the horizontal [A1] ( ) ( )( ) 2 22 o 2 2 o f i fi cosine rule: 2 cos 40 5.0 5.0 2 5.0 5.0 cos 40v v v vv∆ =+− = + − [M1] 13.4 m sv −∆= [A1] Comments: Quite a number drew the vector diagram for Δv incorrectly as vf +vi, instead of vf + (–vi). Graphical method to determine Δv is acceptable. 20o vf horizontal ramp −vi ∆v 20o 40o

DUNMAN HIGH SCHOOL 2021 PHYSICS H2 (YEAR 5) 6 2 (a) The project i le undergoes a trajectory symmetrical about the centre (or parabolic trajectory), with vertical displacement sy = 0. [B1] Taking upwards as positive and considering vertical motion, ( ) 2 2 1 2 10 sin 2 2 sin (shown) y yo o oo o s u t at u t gt ut g θ θ = + = − = [C1] Alternative method: Since the projectile undergoes a parabolic trajectory, Taking upwards as positive and considering vertical motion, Vertical component of initial velocity, sinyuu θ= [C1] Vertical component of final velocity, sinyvu θ= − sin sin 2 sin (shown) yy o o o v u at u u gt ut g θθ θ = + −= − = [C1] Alternative method: Taking upwards as positive and considering vertical motion of the projectile, At the highest point of its trajectory, vertical component of velocity, 0yv = 0 sin sin where is the time taken for it to reach the peak of its trajectory yyv u at u gt utt g θ θ = + = − = [C1] Since the projectile undergoes a parabolic trajectory, its total flight time is twice the time taken to reach the peak of its trajectory. 2 sin2 (shown)o utt g θ= = [C1] Comments: Most students can prove it using one of the 3 methods mentioned.

DUNMAN HIGH SCHOOL 2021 PHYSICS H2 (YEAR 5) 7 (b) horizontal displacement of projectile = horizontal displacement of cart + 45 m or ( )cos 45u t vt θ = + [C1] 1 45cos cos 45 2 sin 9.8135cos 23 45 2 35 sin23 16.08 16.1 m s vu t gu u θ θ θ − = − = −  = °−  ×° = = Comments: Some students will forget to minus 45 from displacement, or add 45 instead. (c) Comments: Some students left numerical values as 35 cos 23 / sin23. It is preferred to calculate them out. Some mixed up vx and vy, others did not read question and did not include any values. For all sketches it is recommended to include values if they are known. (d) Launch the projectile with a larger speed or lauch the projectile at a larger angle