2017 RI H2 Physics P1 solutions

2017 Year 6 H2 Physics Prelim Paper 1 Mark Scheme © Raffles Institution 9749/01 1 A 2 D Option A: The mass of a typical car is about 1500 kg. The volum e of a typical car is about 3.0m×0.7m×1.5m (lower part) + 2.0m×0.7m×1.5m (upper part)  5 m3. So, the density of a car is about 300 kg m3. Option B: Typical A4 paper is labelled as 80 gsm (or grams per square metre). So an A4 paper of dimension 210 mm  297 mm has a mass of 80gm –20.210m0.297m = 5 g. Or, a pack of 500 sheets of A4 paper weighs about 2 kg. Hence, each sheet is about 4 g. Option C: A safe lifting speed is about 1 m s 1. So, the useful power is about mgv  103×10×1 = 104 W. Option D: Volume of a class room is about 10×10×3=300 m 3. Room temperature is about 27 degrees = 300 K. One atmospheric pressure is about 10 5 Pascal. Hence, n = pV/RT = (105×300)/(8.31×288)  105 moles. 3 A Only in region A does the speed change, i.e. there is acceleration. In regions B, C and D, the velocities are constant, i.e. there is no acceleration. 4 D 13.5sin20 10sin5.0 2.0686 m syu   Taking upward displacement as positive, 2 2 1 2 1300 2.0686 (9.81)2 8.03 s , 7.61 s (N.A.) yy ysu t a t tt tt      5 B 11Area under - graph (1 3)(2 ) (2)( ) 322 30 (2.0)(3 ) 5.0 at v x x x pm v x x          6 D By applying conservation of momentum and conservation of kineti c energy to an elastic collision, one would arrive at the conclusion that the relative speed of approach is equal to the relative speed of separation. Refer to notes for derivation. 7 A With A intact, the system would be in equilibrium without F. With B removed, and the system requires F to be in equilibrium. Hence, the moment due to F (and the resulting friction) should be equal to that due to the weight of B. With F in place, translational equilibrium in the horizontal directio n requires a friction f equal in magnitude to F but acting in the opposite direction to F. F and f form a couple. Weight of section removed (B) = 48  36 = 12 N. Method 1: Taking centre of mass of cylinder A as pivot Moment due to the weight of B = 12 × r/3 = 4r Moment due to the couple of F and f = F × 2r Equating the two moments, F = 2 N

© Raffles Institution 9749/01 2 Method 2: Taking point of contact between cylinder and ground as pivot Friction f does not contribute a moment, since it passes through the point of contact. Moment due to the weight of B = 4r Moment due to F = F × 2r It yields the same answer as the previous method. 8 C The scale will measure the following forces: i. weight of beaker of water (Z) ii. weight of object at bottom of beaker (X) iii. force exerted by suspended object on water (Y) Hence, the total force = X+Y+Z 9 A 22 12 122 1 12 2 11 2 2 2 22 1 1 , , 22m 2m mm m4 m ppKK p p Kp Kp   10 C At the same distance d from the sphere, the electrostatic force F experienced by both particles is the same