RI 2015 A-Level H2 Phys Solution

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 driving force: 31.15 10 Nf = × because the net force on the car is zero. When the speed increases to 40 m s−1 (speed is doubled), because the frictional force is proportional to the square of the speed, the fric tional force increases by a factor of 4. By N2L, since the velocity is again constant (net force is 0), the driving force is equal to this new frictional force 33 new 4 4 1.15 10 4.6 10 NFf = = × ×=× The new power output is then 3new power output 4.6 10 40 184 kW=× ×= 9 A For the processes with no volume change, no work is done either on or by the gas. For the process with increasing volume, work is done by the gas, whereas for the process with decreasing volume, work is done on the gas (equal to the negative of the work done by the gas). Hence net work done by the gas work done by t he gas work done on the gas 600(5 3) 400(5 3) 400 J = − = −− −= where the work done in each process is given by work done pV= ∆ 10 C The force the string acting on the sphere is along the string away from the sphere. By N3L, the force on the string by the sphere is in the opposite direction. This condition alone rules out options A, B and D. The force due to the pole on the string must be equal in magnitude and opposite in direction to the force due to the sphere on the string. This is because the net force on the string must be zero. Note that the string is massless, hence it does not require a nonzero net force (the centripetal force) in order to undergo circular motion.

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 3 11 B The minute hand takes 1 hour (3600 seconds) to complete one revolution. The angular velocity is then 312 1.75 10 rad s3600 πω −−= = × Note that the angular velocity does not depend on the length of the minute hand (the radius of the circle). 12 A The meteorite initially has zero potential energy (since it is at a large distance from the planet, where ϕ is zero). The final potential energy of the meteorite is m(ϕ at R), which is negative, since ( ϕ at R) is negative. The corresponding positive value is given by m(magnitude of ϕ at R). By conservation of energy, Gain in KE Loss in GPE (magnitude of at R) 0 (magnitude of at R) m m ϕ ϕ = = − = Note that the mass of the planet, M, should not appear in the expression, since it is built into the potential function ϕ. 13 D The net force acting on the satellite is equal to the difference between the gravitational attraction due to the Sun ( S 2 GM m R , the larger force), and that due to the Earth ( E 2 GM m r , the smaller force). The net force acts as the centripetal force, which points towards the Sun (assumed to be stationary at the centre of the circle). The radius of the circular orbit of the satellite is R, hence 2S E 22centripetal force GM m GM m mRRr ω=−= × 14 A For simple harmonic motion, one has acceleration = − ω2×displacement which fits the graph (a line with negative gradient passing through the origin). 15 A The KE at the centre position (the equilibrium position) is the maximum KE, which is given by 22 0 11Max KE 2.5 1.2 1.8 J22mv= = ×× = which rules out options C and D. Between options A and B, the columns for GPE are identical. The decrease in GPE from top to bottom is given by 3.7 ( 3.7) 7.4 J−− = . By conservation of energy, elastic potential energy should increase by the same amount (because KE is zero for both top and bottom). Option A satisfies this condition, while option B does not.

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 4 16 A The first law of thermodynamics is UQW∆=+ where ∆U is the increase in internal energy of the system, Q is the heat absorbed by the system, W is the work done on the system. Translating the options into equations using the above symbols, one has A: Q UW= ∆− B: Q UW= ∆+ C: WQ U= −∆ D: W UQ= ∆+ Only option A is consistent with the original first law equation. 17 C Method 1: ideal gas equation Before the tap is open: For flask X, according to the deal gas equation, XX XX X X , pVp V n RT n RT= = Similarly, for flask Y, YY YY Y Y , pVp V n RT n RT= = Thus, one has XX YY XY pV pVnn RT ++= (1) After the tap is open: By conservation of mass and the ideal gas equation, one has XY pVnn RT+= (2) Combining Equations (1) and (2), one has XX YYpV p V p V= + . Method 2: internal energy For an ideal gas of fixed mass, the internal energy is proportional to temperature: internal energy ∝ nRT = pV. (For monoatomic gas, the constant of proportionality is 3/2.) Thus, by conservation of energy, internal energy in X + internal energy in Y = total final internal energy XX YYpV p V p V= + 18 C The pulse reaches point O after 1 s. At point O, the wave is reflected, after which it undergoes a phase shift of π (meaning that the downward displacement becomes upward and vice versa), and travels to the left.