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