DHS 2021 JC1 Promos Physics (Solutions)

DUNMAN HIGH SCHOOL 2021 PHYSICS H2 (YEAR 5) 1 Answers to Promotional Exam Section A 1 2 3 4 5 6 7 8 9 10 C B C A C B D B A D 11 12 13 14 15 16 17 18 19 20 D A C D D A B A C B 1 C mass of 20c coin: 3.85 g or about 4×10−3 kg (Order of magnitude 10−3 kg) thickness of a sheet of paper: 0.1 mm or 1×10−4 m (Order of magnitude 10−4 m) an average apple has a mass of 100 g or weight of 1 N (Order of magnitude 100 N) temperature of a person’s body: 36.5 oC or 310 K or 3×102 K (Order of magnitude 102 K) 2 B pqf Cm k= ( ) 1 2 1 2 1 2 kg s comparing the exponent: i: of s: n terms of base un o its s 12 f kg: k 0 g qp qq p q p q −− − =−  = = +  =− =− = 3 C X = P – R ⇒ X = P + (−R) ⇒ vector addition of P and –R, i.e. the beginning of vector −R is placed at the end of vector P; the vector sum X is drawn as the vector from the beginning of P to the end point of –R. 4 A ( )( ) ( )( ) o 21 2 2o1 2 After releasing from rest, the mass slides down the inclined surface with constant acceleration sin30 . Using , we have 0 0.80 sin30 0.80 1.6 m (to 2 s.f.) ag s ut at sg = =+ = + = 5 C ( ) ( ) ( ) 22 22 2 Using 2 for uniform acceleration along t he vertical direction, and taking upward direction as positive, we have 0 sin 2 sin 2 v u as v g H vH g   =+ = + − =

DUNMAN HIGH SCHOOL 2021 PHYSICS H2 (YEAR 5) 2 6 B force = rate of change of momentum = gradient of momentum-time graph. Since the resultant force is increasing, the graph has increasing gradient with time. 7 D action-reaction forces act on different objects and are of the same type of force. 8 B at equilibrium W = U 32 = (0.012)(0.3)  (9.81)  = 910 kg m-3 9 A When tension = 400 N, length = 37 mm and extension = 2 mm. strain energy = 1 2Fx = 1 2(400)(2 x 10-3) = 0.40 J 10 D This equation is not valid as there is an acceleration. 11 D efficiency = useful power output/ power input = (260/320) x 100 = 81% 12 A At highest point, T is lowest (=0) and speed is minimum. T + mg = mv2/r so g = v2/r or v = √rg = √(0.25(9.81) = 1.6 m s-1 13 C When released from P, the direction of motion is towards planet A, in the direction of lower potential. By conservation of energy, loss in potential energy = gain in kinetic energy m = 1 2 mv2 v = √2() = √2(62.3 – 13.4) x 106 = 9.89 x 103 m s-1

DUNMAN HIGH SCHOOL 2021 PHYSICS H2 (YEAR 5) 3 14 D total energy = −GMm/2r, potential energy = −GMm/r, kinetic energy = GMm/2r As r increases, total energy and potential energy increases while kinetic energy decreases 15 D Isothermal processes do not change U as temperature is constant, U1 = 0. Thus, Q = −W1 However, an adiabatic compression will cause temperature to increase as state of gas moves to a higher isotherm, so U will increase. Note that U2 = W2 since Q = 0 We find that |W1| < |W2| (from area under graph), so U1 + U2 = U2 > Q , so the answer is D. 16 A The rate of heat loss of the hot liquid is constant during cooling before solidification and during the freezing During cooling: 𝑃𝑡 = 𝑚𝑐∆𝜃 𝑃(1) = 𝑚𝑐(4.0) During freezing: 𝑃𝑡 = 𝑚𝑙𝑓 𝑃(25) = 𝑚𝑙𝑓 𝑃(1) 𝑃(25) = 𝑚𝑐(4.0) 𝑚𝑙𝑓 𝑐 𝑙𝑓 = 1 25 × 1 4.0 = 0.01 17 B Since product of pV for T2 is four times that of T1, complied with the fact that pV = nRT, T2 is four times that of T1. Since 1 2 𝑚〈𝑐2〉 = 3 2 𝑘𝑇, 𝑐𝑟𝑚𝑠 ∝ √𝑇 𝑐2 𝑐1 = √𝑇2 √𝑇1 = 2 1

DUNMAN HIGH SCHOOL 2021 PHYSICS H2 (YEAR 5) 4 18 A Same current flows through both lamps in series. Hence, P  R. 1 3 24 24 6 V4 X X X Y Y Y X X X P R V P R V V V V == = − == 19 C Consider a fixed point, 4Q will pass through the point in one period. So, current = total charge time = 4Q 1/ f = 4Qf 20 B 𝐸𝑏𝑎𝑡𝑡 = 𝑉𝐼𝑡 18.0 = 𝜀(0.025)(80) 𝜀 = 9.0 V 𝐸𝑟 = 𝐼2𝑟𝑡 18.0 − 11.0 − 4.0 = 0.0252𝑟(80) 𝑟 = 60 𝛺

DUNMAN HIGH SCHOOL 2021 PHYSICS H2 (YEAR 5) 5 Section B 1 (a) (b) (i) 1. ( )( ) 2 QQ Q 22 at t at= [A1] 2. ( )( ) 2 Q same distance for each third of journey: 0.52 QQ at at t t t=   = [A1] (ii) (c) Since distance travelled by both trains is the same, [C1] = 0.962 [A1] Correct shape [A1] Correct time values [A1] P P Q Q Q P (2 )( ) (1.5 )( ) 4 3 t at t at t t = = Q P 2.5total time taken by train Q for the jour ney 2.5 4 0.962total time taken by train P for the jour ney 3 3 3 t t = = =  Correct shape [A1] Correct time values [A1] [C1] [C1]

DUNMAN HIGH SCHOOL 2021 PHYSICS H2 (YEAR 5) 6 Marker’s comments: Most have drawn part (a) graph correctly. Part (b) (i) expressions were generally written correctly by majority of candidates, though, for part (b) (ii), some did not draw the second third of the journey having half the time interval of the first third. Part (c) proved difficult for some candidates. Either they calculated an incorrect total time taken by each train, or did not calculate correctly the ratio tQ/tP. 2 (a) (i) gain in kinetic energy = loss in gravitational potential energy 1 2 mv2 = mgh [C1] 1 2 v2 = 9.81 (4.5) v = 9.4 m s-1 [A1] Alternative method: Using equation of motion for constant acceleration, v2 = u2 + 2as [C1] = 0 + 2 (9.81)(4.50) v = 9.40 m s-1 [A1] (ii) p = mv = (250) (9.40) = 2350 kg m s-1 [A1] (iii) Using conservation of momentum, 2350 = 2250V [C1] V = 1.04 m s-1 [A1] (b) (i) loss in k.e. = 1 2 mv2 = 1 2 (2250)(1.04)2 [C1] = 1220 J [A1] Marker’s comments: loss in kinetic energy does not equal to gain in potential energy due to the presence of frictional forces. In this case, there is also loss in gravitational potential energy besides the loss in kinetic energy. (ii) work done by frictional force = loss in kinetic energy + loss in potential energy F(0.25) = 1220 + 2250 (9.81)(0.25) [C1] F = 2.7 x 104 N [A1] Marker’s comments: many students forget to include the loss in gravitational potential energy. Alternative method: Using equation of motion for constant acceleration, v2 = u2 + 2as 0 = (1.04)2 + 2a(0.25) a = −2.2 m s-2 [C1] Using Newton 2nd law: W – F = –ma F = W + ma = 2250 (9.81 + 2.2) = 2.7 x 104 N [A1]

DUNMAN HIGH SCHOOL 2021 PHYSICS H2 (YEAR 5) 7 (c) As pile gets deeper in the ground, friction from the ground will increase. [B1] The loss in kinetic energy of the hammer and pile is the same. [B1] Marker’s comments: the loss in kinetic energy is the same as the hammer is raised to the same height each time. 3 (a) It is a single point whe