CJC 2018 A level H2 Physics Answers

1 H2 Physics 9749 – 2018 A Level Exam Paper 1 1 Essential Question(s):  How to calculate percentage uncertainty of a quantity from other quantities? o Need an equation linking these quantities. Then apply the relevant rules to calculate the percentage uncertainty (product-quotient or addition-subtraction rule). o Note that we are needed to determine the percentage uncertainty of g. We therefore need to make g the subject of the equation. Make g the subject of the equation: T=2πඨL g T2 4π2 = L g g=L4π2 T2 ∆g g × 100% = ൬∆L L + 2 ∆T T ൰ × 100% %g = %L + 2%T %T = %g − %L 2 %T= %g − 0.05 6.25 × 100% 2 %T= 2% − 0.05 6.25 × 100% 2 = 0.6% (1 s.f.) Answer: A Examiner’s comments: Most candidates evaluated the percentage uncertainty in L correctly and also realised that there was a need to halve in order to deduce the percentage uncertainty in T from the percentage uncertainty in T2. The difficulty that candidates had was in realising that the percentage uncertainty in L needed to be subtracted from, rather than added to, the percentage uncertainty in g. 2 Essential Question(s):  What is the difference between distance and displacement, and how can we deduce these quantities from the graphs? Integrating v with respect to t gives change in displacement. Unlike displacement, distance is a scalar, hence the magnitude of the integral is considered to find distance. Change in displacement for P is positive, while for Q it is zero. Hence displacements are different.  either option A or B. For P, the change in displacement is the same as the distance travelled, since the object is travelling on a straight line, in the same direction for the entire journey.

2 For Q, the distance travelled is shorter as the total area bounded by the graph and the time axis is smaller compared to the total area for P. Therefore, both displacement and distance are different for P and Q. Answer: A Examiner’s Comments: Most candidates found this question challenging. The difficulty they experienced was in comparing the two graphs in terms of distance moved. Most candidates correctly deduced that the displacements were different. Only a minority of candidates correctly compared the total magnitudes of the areas of the triangles defined by the graph lines and the time axes to deduce that object P moves through a greater distance than object Q. 3 Essential Question(s):  As momentum is a vector, how do we interpret the directions mathematically for vectors that are in opposite directions?  How to calculate impulse? impulse = ∆p = pf- pi = mvf- mvi = m(vf- vi) = m[vf-(-18 vi)] Taking direction of final velocity as positive. = ൫80×10-3൯[23-(-18 )] = 3.3 Ns Answer: B 4 Essential Question(s):  Is total momentum conserved? What is the condition for total momentum to be conserved?  What kind of collision is this? (Elastic? Inelastic? Perfectly inelastic?)  As momentum is a vector, how do we interpret the vector directions mathematically for vectors that are in opposite directions? Not net external force acts on the system of two trolleys, hence total momentum remains constant throughout and equal to the total initial momentum. Stick together  perfectly inelastic collision Take direction to the right as positive. Total initial momentum = ݌ହ.଴ ௞௚ +݌ଶ.଴ ௞௚ = 5.0 × 4.0 + (−2.0 × 3.0) = 14 Ns By conservation of momentum, Total momentum at any instant = Total initial momentum = 14 Ns  either Option C or D

3 Total final momentum = (5.0 + 2.0) V = 14 Ns V = 2.0 m s-1 (to the right) Answer: C 5 Keyword(s):  Long-handed broom  non uniform  weight not at geometric centre  “balances”….”horizontally”  broom is in equilibrium Essential Question(s):  What are the conditions for the object to be in equilibrium, before and after mass is hung?  What forces act on the broom before and after the mass is hung?  Draw 2 FBDs: before and after mass is hung. Before mass is hung, broom is in equilibrium which means that Weight acts through the pivot O initially  CG of the broom is thus 1.05 m from end X After mass is hung, and the broom is moved 0.27 m to the right, the CG is also shifted 0.27 m to the right. Pivot remains at the top of the chair. Applying the principle of moments, taking moments about the pivot, Sum of anticlockwise moments = Sum of clockwise moments 200 x 9.81 x [1.05 – (0.27+0.10)] = Mass of broom x 9.81 x 0.27 Mass of broom = 504 g = 500 g (2sf) Answer: A 6 Keyword(s):  “extra weight” – on top of what the bridge was already supporting Essential Question(s):  What contributes to the total weight supported by the bridge?  water on the bridge and the barge  How does the weight supported by the bridge change when the barge is on it? When barge enters the bridge, there is water displaced which in this case (according to principle of floatation) equals to the weight of the boat. Water displaced flowed away to both sides of the bridge. Hence, the water on the bridge is reduced by an amount of weight exactly equal to weight of the barge. Thus there is no change to the loading on the bridge. Watch: https://www.youtube.com/watch?v=sFdpyxWcKwo Answer: A 7 Keyword(s):  Constant speed  no net force on car (resistive force equals to driving force of car) Essential Question(s): 200 x 9.81 Mass of broom x 9.81 0.27 m 0.10 m 1.05 – (0.27+0.10) m 1.05 m

4  How is power related to force? [Poutput = Fdrivingv]  How is force related to speed? [resistive force is proportional the square of its speed but since resistive force equals to driving force, driving force is also proportional to square of its speed.]  How is speed varied? Speed has double from 20 m s-1 to 40 m s-1 Poutput = Fdrivingv = kv2.v = kv3 When speed is doubled, power increases by a factor of 23 = 8 Therefore, the new power required (output) is 8 x 23 = 184 kW Answer: C 8 Essential Question(s):  What determine GPE?  What determines EPE? GPE = mgh  S is at a lower height, hence GPE is lower at point S.  either option B or D. EPE, U = qV, where q = -e  U = -eV V is the electric potential at a point Vs is lower electric potential compared to VR Therefore, the electric potential energy at S is higher (less negative / more positive) compared to electric potential at R (more negative / less positive). Answer: B 9 Essential Question(s):  For a disc rotating at uniform circular motion, how does angular velocity vary for points on the disc? (all points on the same disc vary with same angular velocity)  How does angular displacement vary with angular velocity? (ω = ௗఏ ௗ௧) After the same time duration of revolution, since dt is the same for both points and ω is the same for both, their angular displacement will be the same as well. Answer: D 10 Essential Question(s):  What is required for circular motion? (centripetal force)  What are the forces acing on the stone at the highest point? Which forces provide for the centripetal force required? Resultant force acting towards the centre = T + W This resultant force provides for the centripetal force required. T + W = centripetal force required T + W = Mrω2 T = Mrω2 - W Answer: A

5 11 Essential Question(s):  What happens if the spacecraft fires its rockets as it follows the path?  Without the spacecraft firing its rocket, what is currently affecting the spacecraft motion as it moves? The context of the question is on “as it follows the path”. This means that the spacecraft is already moving initially. As the spacecraft is in the vicinity of the Earth, gravitational force (FG) is currently acting on the spacecraft. By Newton’s 3rd law, as the spacecraft exerts a force on the rocket as it fires the rocket, there will be an e