RI 9646 2015

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 1 2015 A-Levels H2 Physics Suggested Solutions Paper 1 1 A The amount of substance is a base quantity, and the mole (mol) is its SI unit. The other three quantities (charge, energy and force) are not base quantities. 2 D total distance travelled = area under the curve 1 1118 30 40 18 (18 24) 50 24 30 2400 m2 22=× × + × +× + × +× × = average speed = total distance travelled / total time 12400 16 m s150 −= = 3 C The horizontal component of the velocity is 40 m s−1, which is constant in time. The vertical component of the velocity after 3.0 seconds = u + gt = 0 + 9.81×3.0 = 29.43 ≅ 30 m s−1 2 2 22 1 fin horizontal vertical 30 40 50 m svv v −= + = += 4 A The acceleration, given by (net force/mass), increases, since the net force (equal to the propelling force) is constant, whereas mass decreases due to the leakage. Note that, since the water is leaked vertically down, it does not exert a horizontal force on the tanker. 5 C Since the collision is elastic, the initial kinetic energy and the final kinetic energy are equal: 2222 12 12 2 222 12 12 initial KE final KE 11 11 22 22mu mu mv mv uu vv = +=+ +=+ where, in the last equation, mass m is cancelled because both spheres have the same mass. NOTE If one considers the conservation of momentum instead, option B is also correct. Note that u1, u2, v1, v2 are spe eds, not velocities. In order to apply conservation of momentum, one has to use velocities instead. Taking rightwards as positive, the corresponding velocities are u 1, −u2, v1, v2. Applying the principle of conservation of momentum, one has 1 2 12 12 12 ()mu m u mv mv uu vv +−= + −=+ which is option B. 6 A By Archimede’s principle, upthrust on the barge = weight of water displaced, and by the principle of floatation, upthrust on the barge = weight of the barge. Combining the above two principles, one has weight of the water displaced = weight of the barge Thus, as long as the water level remains the same everywhere, the bridge supports the same weight, regardless of the position of the barge.

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 2 7 D When the block of ice is fully submerged, the volume of water displaced is equal to the volume of the ice, hence water iceupthrust gVρ= where we have applied the fact that the upthrust is equal to the weight of the liquid displaced. When the ice is floating, the upthrust is equal to the weight of the ice, by the principle of floatation: ice iceupthrust gVρ= Hence the ratio between the two upthrust is equal to the ratio between the density of water and that of ice: water ice ice ice upthrust when fully submerged 1.0 1.1upthrust when floating 0.9 gV gV ρ ρ= = ≈ 8 C When the speed is 20 m s−1, the driving force F is solved using 3 3 power output 23 10 20 1.15 10 N Fv F F = ×= × = × The frictional force f at this speed is equal to the dr