2018 MJC P3 Soln

2018 MJC H2 Physics Prelim Exam Paper 3 Suggested solutions 1 (a) (i) Newton’s first law of motion states that a body continues at rest or at constant / uniform velocity unless acted on by a resultant (external) force. [A1] Comments: 1. many students wrote uniform motion or constant speed instead of uniform velocity (and this is penalised) 2. many students did not indicate “resultant” (ii) resultant force (in any direction) is zero [B1] resultant moment / torque (about any axis) is zero [B1] Comments: 1. very well done (b) (i) Using principle of moment and taking moment about the bottom of ladder: [B1] clockwise moment = anticlockwise moment N L sin 60 = W 2 L cos 60 N L sin 60 = 80 2 L cos 60 [B1] N = 23 N [A0] Comments: 1. poorly done, a number of students either left blank or tried to balance the forces (which they would not be able to get the answer) 2. many students did not indicate using principle of moment (or sum of clockwise moment = sum of anticlockwise moment) nor indicate which point to take moment about (ii) Resolve vertically: Y = W = 80 N Resolve horizontally X = N = 23 N [C1 for both equations] force R = 2 2X Y = 2 223 80 = 83 N [A1] angle = tan–1 Y X = tan–1 80 23 = 74 Direction: 74 clockwise above horizontal [A1] R

Meridian Junior College H2 Physics Paper 3 JC2 Preliminary Examinations 2018 2 OR by vector triangle [C1] get R [A1] get angle [A1] Comments: 1. generally able to get answer although in some cases the direction is not written clearly (a diagram showing the direction of R would help) (iii) 1. (Due to the person’s weight) there is now greater downward force on the ladder, and so (to maintain equilibrium) the floor exerts a larger upward vertical force on the ladder. [A1] Comments: 1. some students used Newton’s third law or action and reaction 2. (Due to the person’s weight) there is now a greater anticlockwise moment about the ladder bottom, and so (to maintain equilibrium) the wall exerts a greater clockwise moment and hence greater horizontal force. [A1] Comments: 1. many students thought the horizontal force remained unchanged because no horizontal component of force is exerted by the person 2 (a) 10.5 9.7uncertainty in 0.4 K 2 Accept: 10.2 9.7 10.5average 10.133 K 3 uncertainty in 10.13 9.7 0.433 0.4 K Comments: 1. Not well done. Many students stated 0.1 K, 0.2 K, 0.5 K etc, making no reference to the data given. Some students presented the uncertainty to 2 or 3 s.f.

Meridian Junior College H2 Physics Paper 3 JC2 Preliminary Examinations 2018 3 (b) -1 -1 4.125 11.8 200.0 [C1]0.309 10.133 3110 J kg K [A1] Q IVtc m m Comments: Common mistakes include: 1. applying wrong equations. It was common to see students equating Q = It (amount of charge = current × time) with Q (heat) = mc . Students must be aware of what the symbols in the equations represent. 2. not converting 309 g to kg 3. not calculating the mean temperature of all 3 sets of data (c) 1 1 0.3 0.002 0.5 3 0.4 [M1]11.8 4.125 200.0 309 10.1 0.077721 3110 0.077721 241.7 200 (1 s.f.) [M1] 3100 200 J kg K [A1] c V t m c V t m c c I I Alternative: m m m max min min 1 1 4.127 12.1 200.5 0.306 9.7 3373 3373 3110 [M1] 263 300 (1. s.f.) [M1] 3100 300 J kg K [A1] ax ax axV tc m c c I Comments: 1. Not well done. Students who used the equation Q = It get zero credit for this part too.

Meridian Junior College H2 Physics Paper 3 JC2 Preliminary Examinations 2018 4 2. Vast majority of the cohort did not present the absolute uncertainty to 1 s.f., and/or did not present c to the same c. 3. Some students did not make c the subject before applying the equation for fractional uncertainty. (d) No heat loss to surrounding. or All electrical energy is converted to heat. [B1] Comments: 1. Very well done. 2. No credit was given for students who stated that I and V are constant values, since these are values stated in the question. 3 (a) Newton’s second law of motion states that the rate of change of momentum is proportional to the net / resultant (external) force (acting on it) and the change (of momentum) takes place in the direction of the (net) force [A1] Comments: 1. a number of students mentioned acceleration instead of momentum (b) (i) 221 1 480020 12 12 [C1]2 2 800 672 m [A1] s ut at Comments: 1. very well done (ii) 1 480020 12 [C1] 800 92 m s [A1] v u at Comments: 1. very well done (iii) 1. 6 work 4800 672 [M0] 3.23 10 J [A1] Fs

Meridian Junior College H2 Physics Paper 3 JC2 Preliminary Examinations 2018 5 Comments: 1. very well done 2. 2 22 2 6 work gain in KE 1 1 1 1= 800 92 800 20 [M0] 2 2 2 2 =3.23 10 J [A1] mv mu Comments: 1. a number of students did not minus the initial KE (iv) impulse = change in momentum = mv – mu = (800)(92) – (800)(20) [C1] = 5.76 104 N s [A1] OR use impulse = Ft = (4800)(12) = 5.76 104 N s Comments: 1. very well done 4 (a) curve from +15 m s–1 steepest at first then gentler and gentler [B1] gradient at v = 0 should be same as that of original line. [B1 provided correct graph shape] areas under graph above and below x-axis are similar (when the curve hits the straight line). [B1 provided correct graph shape] Comments: 1. poorly done, most students could not get the last two marks velocity / m s–1 time / s Fig. 4.1

Meridian Junior College H2 Physics Paper 3 JC2 Preliminary Examinations 2018 6 (b) greater resultant (downward opposing) force, so lesser height [A1] Comments: 1. many students wrongly stated there is less net force; many thought that there is a net upward force when object is moving up and air resistance reduced the net upward force 5 (a) No resultant (external) force (acts on the system) so (by principle of conservation of momentum) the total momentum is conserved. [A1] Comments: 1. a number of students say not conserved because collision is inelastic (b) conservation of momentum: (0.800)(9.2) + 0 = (3.200) V V = 2.3 m s–1 [C1] initial kinetic energy = 221 1 0.800 9.2 33.856 J 2 2 mv final kinetic energy = 221 1 3.200 2.3 8.464 J 2 2 mv % loss in kinetic energy = 33.856 8.464 100% 75% 33.856 [A1] Comments: 1. some students equate final speed and KE as zero (c) conservation of energy: 2 2 A&B 1 1 2 2 m v kx 2 21 13.2 2.3 2500 [M1]2 2 0.082 m [A1] x x Comments: 1. a number of students only use the mass of ball B

Meridian Junior College H2 Physics Paper 3 JC2 Preliminary Examinations 2018 7 6 (a) Straight lines in uniform radial pattern centred on charge [B1] Arrows pointing inwards [B1] Comments: 1. Many students did not label the field lines. BOD was given 2. This part was very well done. Mistakes include: uneven separation of E field lines or wrong direction of arrows. (b) Electric field points inwards, because E points from points of higher potential to lower potential. OR potential is negative suggests that the sphere is negatively charged and hence electric field points inwards. [B1] Electric field is numerically equal to the potential gradient and is stronger where the potential gradient is stronger (closer to the charged sphere). The stronger E field is shown by the closer spacing between the E field lines. [B1] Comments: 1. This part was very poorly done. Most students stated that E is proportional to V, making no reference to potential gradient. 2. It was also common to see answers that made no reference to the drawing in (a). (c) (i) If V is inversely proportional to x, then Vx = constant [B1] Multiplying any 2 values [B1] to conclude that Vx is constant: x / m V / V Vx / V m 0.19 1.50 × 105 28.5 × 103 0.