CJC 2015 A level H2 Physics Answers

1 H2 Physics 9646 – 2015 A Level Exam Paper 1 1 Keywords:  Base quantity  SI units Answer: A 2 Keywords:  Average speed Essential Questions:  How do you determine the average speed of the vehicle?  How do you determine the total distance travelled from a speed-time graph? Answer: D 3 Keywords:  Thrown Horizontally  Air resistance is negligible Essential Questions:  What are the forces acted on the stone after it leaves the cliff?  Which component of the velocity is affected? Use kinematics equations to determine the x and y components of the final velocity just before it hits the surface of the water. Answer: C 4 Keywords:  Propelling force is constant  Water runs out of the tanker Essential Questions:  What happens to the mass of the water tanker?  What are the changes in the variation of the resistive forces and ultimately, the net force? Answer: A 5 Q: What physics relationships hold for a head-on elastic collision?  (1): Total momentum is conserved throughout the motion AND  (2): Total KE before the collision equals to total KE after the collision.  (3): A consequence of solving (1) and (2) simultaneously is that the Relative velocity of approach between the two bodies before the collision equals to their Relative velocity of separation after their collision..\ Take direction to the right as positive. Applying (1), we get: mu1 + m(-u2) = mv1 + mv2  u1 – u2 = v1 + v2  Tallies with option B Applying (2), we get: ½ m u1 2 + ½ m u2 2 = ½ m v1 2 + ½ m v2 2  u1 2 + u2 2 = v1 2 + v2 2  Tallies with option C Applying (3), we get: u1 – (-u2) = v2 – v1  u1 + u2 = v1 – v2

2  No such option. Answers: B and C Note: the proposed answer by Cambridge was C, but later found to be an error as both B and C are valid. 6 Keywords:  Diagram showing a section of canal  Extra weight Essential  What is the meaning of extra weight ?  If there is extra weight, where does it comes from?  Before the barge cross the bridge, how much weight does the bridge supports?  When the barge is at the bridge, what are the forces acting on the barge? Proposed Solution: When the barge moves onto the bridge (aqueduct), it is displaces an amount of water and the weight of the displaced water is equal to the weight of the barge. The displaced water moves off the aqueduct as the boat moves onto it. Thus there is no extra weight supported by the bridge. Answer: A 7 Keywords:  Held below the surface of water  Released and floats to the surface  Fully submerged  Floating Essential questions 1. How forces are balanced when the ice is floating? 2. How can weight be expressed in terms of density and volume? 3. What expressions can be written when fully submerged and floating? Proposed Solution: Let the volume of ice be Vice = A(Surface area) x H (thickness) , ρice = density of ice, ρw = density of water, d = height of ice submerged in water, Vf = volume of water displaced when ice floats and Wice = weight of ice Ice held below surface of water  Fully submerged, Upthrust UH = ρw Vice g = ρw (AH) g ----- (1) Ice floats  Upthrust Ud = ρw Vf g = ρw (Ad) g ----- (2)  Upthrust = Weight of ice  ρw (Ad) g = Wice ρw (Ad) g = ρice (AH) g ⟹ ܪ ݀= ߩ௪ ߩ௜௖௘ Therefore from (1) and (2) ,

3 (1) (2) ⇒ ܷு ܷௗ =ܪ ݀= ߩ௪ ߩ௜௖௘ = 1.0 0.9 = 1.1 Answer: D 8 Keywords:  Constant speed  Total Frictional proportional to the square of speed Essential Questions:  What is the net force at force speed?  What is the engine force at constant speed?  How is power related to speed at constant speed? Answer: C 9 Key info: - Gas - Pressure and volume graph (Note the units) - NET work done BY gas in ONE CYCLE Essential questions: 1) How to determine ‘work done’ from a p-V graph?  In general, where p is not necessarily constant, W = ∫ p dV  Thus work done W is determined from the area under p-V graph 2) How to find NET work done?  Sum up the work done at every stage of the cycle. Following the sign convention for W, include ‘+’ sign for work done ON gas and ‘-‘ sign for work done BY gas.  For a cycle, net work done = area enclosed within the ‘loop’ 3) When is there work done BY gas, and, when is there work done ON gas? In this scenario, is there NET work done BY gas or ON gas?  There is positive work done BY gas when the gas is expanding (increasing in volume); there is positive work done ON gas when the gas is being compressed (decreasing in volume).  In this thermodynamic cycle, the area under the graph during the expansion stage is greater than that during the compression stage, hence there is positive net work done BY gas…anyway, all the options given are positive. Answer: A 10 Key info: - horizontal circle - forces on the string (Always take note of whose free-body diagram the question is asking for) Essential questions: 1) How do forces arise?  Forces arise due to interaction between 2 bodies; can be contact or non-contact force 2) What other body/bodies are interacting with the string?  The vertical pole that the upper end of the string is tied to, and, the ball on the lower end of the string. String is massless, so no gravitational force acts on it.  But all four options A to D gives two forces…what is the difference between the options?  the directions of the forces!  A string can only experience tensile forces, and tensile forces act long its length! Answer: C

4 11 Key info: - Minute hand  1 revolution in 60 minutes (this is the period) - 3.00 m long  radius of circular motion - Magnitude of angular velocity Essential questions: 1) How is ω related to the information given (i.e. period T and radius r)?  ω = 2π/T, where T = 60 mins  r is not needed Answer: B 12 Key info: - meteorite… initially zero velocity relative to planet - …. falls from a large distance to the planet… - assume meteorite is initially at infinity - no atmosphere  no air resistance - gravitational potential - r from centre of planet - Which expression…maximum kinetic energy…as it hits the surface Essential questions: 1) How to link KE to potential?  Both are related to energy. Apply conservation of energy.  Loss in GPE = Gain in KE  GPEi – GPEf = KEf - KEi  Based on key info in the question, assume GPEi = 0 and KEi = 0  Thus, 0 – m multiplied by potential at surface = max KE – 0 Answer: A 13 Essential Questions: 1. What does it mean by “which force = mass x acceleration equation”? F = ma, it is the simplified version of Newton’s second law. F in this equation refers to the resultant force on the satellite according to the context of the question. 2. What is the resultant force on the satellite? The centripetal force (as according to the question, the satellite rotates in a circle, i.e. circular motion) 3. Where is centripetal force pointing at? (Why is this important? Because then you will be able to know which force should be larger). It is pointing towards the Sun as the satellite rotates around the Sun. Hence the resultant force on satellite = Centripetal force = m(Rω2) = Gravitational force by Sun on Satellite – Gravitational force by Earth on Satellite Answer: D 14 Essential Question: 1. What feature do we see on the graph? A straight line with negative gradient passing through the origin. This graph represents the defining equation for Simple Harmonic Motion: a = -ω2 x, where ω is a constant. Hence the graph is a against x. Answer: A 15 Essential Questions: 1. What does the change of GPE and EPE compares with the value at the centre mean? It means that we take the centre point as reference point for both its GPE and EPE, i.e. set them as zero. 2. How do these energies relate to one another? Total energy = KE + EPE + GPE = constant.

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