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