DHS 2022 JC1 Promos Physics (Paper 2 Solutions)

DUNMAN HIGH SCHOOL 2022 H2 PHYSICS (YEAR 5) 2022 DHS H2 Physics Promo Exam Mark scheme Section B Structured Questions 1(a)(i) Allow anything between 20 – 20000 Hz B1 1(a)(ii) Allow anything between 10 – 400 nm B1 1(b) A vector is a physical quantity that has both magnitude and direction. B1 1(c)(i) Arrow labelled R in a direction ≈ 10o – 15o north of west. B1 1(c)(ii) 2 2 2 o 1 28 95 (2 28 95 cos115 ) 110 m s v v − = + −    = C1 A1 2(a) 45o A1 2(b) Take up and right as positive. Horizontally, o( cos 45 ) 4.00 (1)ut = −−−−−−− Vertically, oo o ( sin45 ) ( sin45 ) (9.81)( ) (2 sin45 ) / 9.81 (2) u u t tu − = − = −−−−−−− Sub (2) into (1) and solving, 2 -1 39.24 6.26 m s u u = = C1 C1 A1 Comment: It was evident that most students did not practise solving kinematics problems. R

DUNMAN HIGH SCHOOL 2022 H2 PHYSICS (YEAR 5) 2(c) 22 o2 Using ' 2 ', 0 (6.26sin45 ) 2(9.81)( ) 0.999 m v u as s s =+ =− = A1 2(d) asymmetrical shape, smaller horizontal range, and smaller maximum height. B1 Comment: Symmetrical shapes were drawn in most scripts. 3(a) ( )mV = ( ) ( ) ( ) 341.0 10 1.5 10 5.0 1.6 1.2 (kg)−=      = C1 A1 3(b)(i) 1.2 5.0 6.0 N s p m v =  =  = A1 3(b)(ii) 6.0 1.6 3.8 N F = = A1 3(c) By Newton’s third law, the force exerted by the water on the wall is equal in magnitude (but opposite in direction) to the force exerted on the water by the wall, (so) 3.8 N. B1 3(d) 443.8 1.5 10 2.5 10 Pap F A −= =  =  A1 3(e) momentum change of water is equal and opposite to momentum change of the wall OR total change in momentum of water and wall is zero B1 Comment: The statement should refer to the specific interaction between the water and wall, instead of just stating the principal of conservation of momentum. 4(a) Net force in any direction is equal to zero Net torque about any axis of rotation is equal to zero B1 B1

DUNMAN HIGH SCHOOL 2022 H2 PHYSICS (YEAR 5) Comment: Very few students stated the correct conditions. Resultant force and resultant torque were commonly written instead of net force and net torque. Zero net force and zero net torque must be clearly stated. 4(b)(i)1. cos = 6.0/12.0 →  = 60 (or other methods) Let L be the length of ladder, Taking moments about axis through lower end of ladder, N1 x L sin 60o = (72 x 9.81) (3L/4) cos 60o + (40 x 9.81)(L/2) cos 60o N1 = 419 N C1 A1 Comment: Majority of students lacked the skills in resolutions of vectors, in this case, forces in the x and y direction. Wrong moments of forces were taken and equated. This part was either left totally blank or badly done. 4(b)(i)2. Vertically, N2 = (72.0 + 40.0) x 9.81 = 1100 N A1 Comment: Wrong moments of forces were taken and equated, instead of equating the forces vertically. 4(b)(ii) magnitude of friction by ground on ladder = N1 = 419 N Hence magnitude of total reaction force by ground on ladder = 221100 419 1180 N+= A1 Comment: Most often, the understanding of resultant of forces was lacking among the students in this part. 4(b)(iii) 1o1100tan ( ) 69.1419 − = R drawn correctly at an angle of 69.1o clockwise from the horizontal. A1 Comment: Only a few students could draw the correct direction of R. 5(a) Consider a stationary object of mass m which moves a horizontal displacement s under the action of a constant net force F. B1 F v s u = 0

DUNMAN HIGH SCHOOL 2022 H2 PHYSICS (YEAR 5) Since the force is constant, the object moves with a constant acceleration a given by Newton’s second law of motion: F = ma When the object undergoes a displacement s, its final velocity v can be found from the following equation of motion: v2 = u2 + 2as or a vs 2 2 = since u = 0 The work done on the object W = F.s =       a vma 2 2 = 2 2 1 mv = kinetic energy Ek of the object. [Answers with only equations and substitutions, without physics explanations or context will score 1 mark max] Note: - Accept: Consider an object starting from rest and travelling with constant acceleration for a distance s. - 2nd B1 mark is only awarded if the working is logically and mathematically consistent. Students must explain in some form of another how 𝑢 is eliminated from 𝑣2 = 𝑢2 + 2𝑎𝑠 if no context was given. - Did not accept answers that states/suggests that 𝑎 = 0. (i.e. contradictory statements such as “box is travelling at constant velocity under a force F” or workings that use 𝑠 = 𝑣𝑡) B1 A0 5(b)(i) change in GPE = (330)(1.1 4.0) 960 JYXmgh mgh− = − =− A1 5(b)(ii) work done against resistive force = 960 – 540 = 420 J distance moved = 420 / 52 = 8.1 m C1 A1 5(b)(iii) Air resistance increases with speed / child may change body shape along the slide hence air resistance changes. Accept also: B1

DUNMAN HIGH SCHOOL 2022 H2 PHYSICS (YEAR 5) - The normal force on the child changes due to the curvature of the slide, hence friction force is not constant. - Slide may be unevenly coated with water and hence friction force varies at different points of the slide Do not accept: - If type of force (friction or air resistance) is not identified in the answer. - Some parts of the slope are smoother than others, hence friction force is different. (Students should specify the cause of the difference in smoothness) Common misconceptions: - Friction increases with increasing speed - Friction increases when surface area of contact between the child and slide increases - Air resistance is directly proportional to speed/velocity - Air resistance/friction changes with acceleration 6(a)(i) − == = oo 2 sin40 9.81sin40 6.31 m s ag A1 Comment: Students who got wrong here were unable to resolve vectors. 6(a)(ii) ( )( ) − == = 1 speed of the ball at the bottom of the s lope, 2 2 6.31 0.56 2.66 m s v as A1 6(b)(i) ( )( ) − −  ==  = 232 2 72 10 1.5 magnitude of centripetal force, 12 10 1.4 N mvF r C1 A1 Comment: Mistakes were made in the conversion of units and forgot to square the speed in the calculation. 6(b)(ii) at the top of the loop, centripetal force acting on the ball is provided by the sum of the force due to the track and the weight of the ball. force due to the track acting on the ball = centripetal force – weight of the ball = 1.4 – (72 x 10-3) (9.81) = 0.69 N direction of force due to the track acting on the ball: vertically downwards. A1 B1 Comment: Direction of force was not properly stated clearly. 7(a) direction of force on a (small test) mass OR B1

DUNMAN HIGH SCHOOL 2022 H2 PHYSICS (YEAR 5) path in which a (small test) mass will move Accept: Line of action of force on a (small test) mass. Do not accept if students wrote: - Direction of force on an object OR path in which an object will move. (Students must define “line of gravitational force”