(DHS) 2024 Prelim Phy P3 Ans final

1 2024 DHS H2 Physics Prelim Paper 3 Suggested Solutions Section A 1 (a) Resultant force on it must be zero in any direction. B1 Resultant torque on it must be zero about any axis (of rotation). B1 MC: Quite badly done. Missing key words in conditions stated. Marks penalized if there is no mention of about any axis (for torque) or in any direction (for force) in answers. (b) (i) MC: The question specifically mention to draw in Fig 1.1, but some drew a separate diagram instead. BOD awarded if T2 is replaced by weight of the 5.0 kg, Force by wall is not a normal force since it is not 900. Many students drew the force by the wall wrongly, they either assumed its horizontal, or along the beam. (ii) Let x be the distance of c.g of beam from the hinge. Taking moments about hinge, sum of clockwise moments = sum of anticlockwise moments 120 (5 sin 60o) = 5g (5 sin 70o) + 20g (x sin 70o) C1 x = 1.57 = 1.6 m A1 MC: Poorly done, because many could not obtain the correct clockwise or anticlockwise moment about the hinge. Some made mistakes when determining the necessary angle needed for calculations. (iii) Vertical summation of forces: FY + 120 (sin 40o) = 5g + 20g FY = 168.1 N Horizontal summation of forces: weight of beam force by wall on beam 40o tension, T2 tension, T β Any 2 correct pairs of forces B1 Other 2 correct pairs of forces B1 With correct direction and label C1

2 FX = 120 (cos 40o) = 91.93 N F =√FX 2+FY 2 = √91.932+168.12 = 192 N A1 tan  = 91.93 168.1  = 29o A1 Force is at an angle of 41o with the beam A1 MC: Many failed to resolve horizontally and vertically. A few did not calculate the numerical angle of the force by wall, while many did not express the angle correctly with the beam. 2(a) In uniform circular motion, the speed of the metal ball is constant, but its velocity is constantly changing direction. Since acceleration is the rate of change of velocity, the metal ball experiences an acceleration. B1 B1 OR In uniform circular motion, the speed of the metal ball is constant, but there is a net non-zero force acting on the metal ball according to Newton’s 1st law because it is moving in a circular path. From Newton’s 2nd law for constant mass, there must be an acceleration in the same direction as the net force. B1 B1

3 CKW MC • Most candidates demonstrated a lack of understanding regarding the significance of uniform circular motion. Paraphrasing the question, candidates should explain why a metal ball moving at a constant speed in circular motion is still considered to be accelerating. • Some candidates mentioned the presence of a centripetal force without explaining why the centripetal force exists • Some candidates explained why the metal ball experiences a centripetal acceleration and not an acceleration. • Some candidates quoted Newton’s 2nd law in general or in mathematical form, without explaining why acceleration cannot be zero. • For the A-level syllabus, because of the general case of variable mass systems, a force can be defined as the rate of change in momentum, and it is the change in momentum that leads to the emergence of a force. 2(b)(i) Horizontal component of normal contact force provides the centripetal force: N cos = mr2 --- (1) The weight is balanced by the vertical component of the normal contact force: N sin = mg --- (2) (2) / (1) : tan  = g r 2 B1 C1 A0 weight of ball mg Normal contact force N

4 CKW MC • Many candidates did not provide proper statements for the “show” question. • Quite a number of candidates recognized that tan  is a ratio of mg and mr2 mathematically. However, they do not go on to explain how tan  comes about from Newton’s 2nd law, in the horizontal direction, and Newton’s 1st law, in the vertical direction. 2(b)(ii)  = 2f = 2(3) = 6 rad s−1 Since v = r, tan  = g r 2 = 9.81 0.10(6) 2  = 15° C1 A1 CKW MC This part was generally well done, except for careless mistakes in calculations, e.g. forgetting to square, or conversion of units. 2(c) A1 CKW MC • 22 1tan For same and , g grr = →  . The graphs shows that as  increases r decreases. • Some candidates did not realise there were 2 asymptotes. • Many candidates did not label the origin. 3 (a) (i) When the gas molecules that are in continuous random motion collides with the inner wall of the bubble and rebounds, there is a change in velocity and hence a change in momentum. B1 By Newton’s Second Law, the bubble walls will exert a force on the gas molecules. By Newton’s Third Law, the gas molecules will exert an equal an opposite force on the inner walls of the bubble. B1 The force per unit area exerted by the gas molecules on the inner walls r r 0 0

5 of the bubble gives rise to the pressure of the bubble. B1 MC: Generally, the question was not very well done despite being rather lenient in the marking. Many responses were essentially a regurgitation of the answers that was used in Junior High/O -level without bringing in concepts learnt at the A -levels. Students need to be aware that the level and depth of their response must evolve in proportion to the level of the exam that they are sitting for. Some common misconceptions/issues include: • Gas molecules moving around and colliding with each other which causes a change in momentum (this is a violation of ideal gas assumption and also the wrong reason for the change in momentum of gas particles resulting in pressure) • Pressure was due to the collision of the gas molecules with water molecules (Pressure of gas is due to collisions of gas molecules with the inner walls of its container – appropriate terminologies should be used in explanations) • No explicit links made to Newton’s 3 rd law in their explanations. Many jumped straight to stating that since the gas molecules experienced a change in momentum, the gas molecules exert a force on the inner walls of the bubble. • There is a change in momentum of the inner walls of the bubble when the gas molecules collide (Note that while this is technically not a misconception, since mwall >> mparticle, by using conservation of momentum, vwall ≈ 0 both before and after collision . - Therefore, it is more meaningful to centre the discussion based on the gas molecules.) (ii) Since temperature constant: 1 1 2 2 3 1 1 2 5 2 5 2 53 2 53 4 ( )( )3 4(2.4 10 ) (0.015) 2.4 10 14(1000)(9.81)3 3.3050 10 m 3.3 10 m pV p V p r p h g V V V   − − =  =−   =  −  = = MC: Poorly done. Common (more eye-catching) issues include: • Erroneously assuming Patm = 101325 Pa or 1.01 × 105 Pa. Values were given in the question to calculate the pressure of the atmosphere in this context. (Very common mistake) • Wrong formula used for the volume of sphere. A number of candidates used 𝜋r2 instead. • Using diameter instead of radius to compute initial volume. C1 A1

6 (b) (i) The First Law of Thermodynamics states that the increase in internal energy of a system B1 is the sum of the heat supplied to the system and the work done on the system. B1 MC: Very commonly tested definition, which has very specific keywords that have been clearly outlined by the syllabus outcomes. Hence close to zero variance in answers will be accepted. (except minor things like swapping the order of Q and W in the definition) Common mistakes: • Change