ADSS Prelim P2 Ans

Admiralty Secondary School Marking Scheme & Marker’s Report 4E Pure Physics (6091/2) PRELIMINARY EXAMINATION 2024 PAPER 2 SECTION A [70 marks] Qn. Description Mark Marker’s Report 1a t = 5 s to t = 50 s: increasing acceleration t = 50 s to t = 100 s: constant acceleration of 20 m/s2 1 1 1 MP1 & MP2 were well answered, however many students missed out MP3 of the calculation of the acceleration. Few students still confused between acceleration and velocity, citing constant velocity and acceleration. b upward force – weight = ma upward force – (1.6 x 106 x 10) = (1.6 x 106) x 20 upward force = 4.8 x 107 N 1 1 Not well done. Many students did not take into account the weight and equated the upward force to ma.

2a Acceptable scale (min 1cm: 20N) Resultant force = 88 N Direction = 90 o anticlockwise from horizontal Correct drawing 1 1 1 1 Most students were able to get 2 marks out of the 4 marks, with many drawing the wrong direction of the resultant force. Students also did not read the question carefully, with the direction of the resultant force often measured from the tension given and not with the horizontal. b 88 N The tyre is in equilibrium, thus resultant force is 0 N. Hence, the weight must be equal and opposite in direction to the resultant force of both tension due to the wires. 1 1 Well done. Ecf was given to students, hence students were able to at least obtain MP1. However, MP2 was rarely achieved, with many missing out that the resultant force is zero. 3a It states when a body is in equilibrium, the sum of clockwise moment about a pivot is the equal to the sum of anticlockwise moment about the same pivot. 1 Few students did not mention about the body being “in equilibrium” and missed out moments must be taken from the same pivot. 3b moment 1 220 100 0 24 Nm . . F d     1 1 Well done. c By principle of moments, sum of clockwise moments sum of anticlockwise moments 0 24 0 10 3 N m(10)=3 m=0.3 . . W W kg    1 1 Could not be done as 10 cm was not shown in the diagram. d Widen the base area of the bin/construct the base using a denser/heavier material/ add weight to the base 1 Well done.

4a 3 504000 10100 800 kg/m P h gρ ρ ρ      1 1 Not well done. Many students substituted P to be 104000, the pressure of the gas, instead of taking the difference between Patm and Pgas. bi 104000 0 05 5200 N Work done 85200100 416 J . F P A F d           1 1 Mostly well done. However some students cited the wrong formula of P = FA. bii When the cylinder is heated, the gas particles in cylinder gain kinetic energy and collide with the walls of cylinder with greater force and more frequently. This causes the piston to move outwards and volume of gas increases, the number of particles per unit volume decreases. Particles collide with the walls of cylinder less frequently. Hence, pressure (force per unit area) exerted on the walls remains the same. 1 1 1 Not well done. Students were not able to obtain the full marks of the question, often missing out MP2. 5ai Ek = ½ mv 2 = ½ x 500 x 5.5 x 5.5 = 7.56 x 103 J 1 Well done. aii Ep = mgh = 500 x 10 x 6 = 3.00 x 10 4 J Work done against friction = 3.00 x 104 - 7.56 x 103 = 2.24 x 10 4 J 1 1 Not well done. Many students used the formula of W = Fs and used the weight log as F and did not realise that s has to be in the same direction of the force used. b Friction causes the log to slow down from Y to Z. All the energy from the kinetic store of the log is transferred to the internal store of the log mechanically to do work against friction. 1 1 Few students still lack the terms in describing the change in energy stores.

6ai 600 60 569 931 560 016 39344 J/g =39300 J/g ( . . ) v v v Q ml l l      1 1 Many students did not take the difference in masses. aii Overestimation. Energy from the internal store of the immersion heater could also have been transferred to the internal store of the surroundings and the beaker by heating. 1 1 Generally well done but a few students could recognise that there is energy transferred to the internal store of the surroundings but cited it as an underestimation. b θ      36000 39344 28 5 70 0 864 g ( ) . ( ) . vQ ml C m m 2 1 Not well done. Students are still not familiar with the calculation of heat capacity type of questions.

7ai Rarefaction 1 Mostly well done. aii Half amplitude Double period 1 1 MP1 was mostly obtained. However, few students could not recognise that half the frequency meant period is doubled. b Distance travelled by sound = 2(900) = 1800 v = d/t = 1800 / (1.2) = 1500 m s-1 1 1 Generally well done. 8a the waste gases are warmer and less dense than surrounding air so they rise and carry the smoke with them 1 Mostly well done. Few students did not recognise that this is a convection question and attempted to answer using electrostatic charging concepts which were unsuccessful. b The smoke particles will be repelled by the negative metal grid as like charges repel and attracted to the positive collecting plate as unlike charges attract. Therefore, leaving the smoke particles remain attracted to the metal grid. 1 1 Many students only obtained MP1 but many missed out “remain” as a keyword in MP2. c 1 Many students drew arcs for the electric field instead of straight lines.

9ai Radioactivity is a random process by which an unstable atomic nucleus loses it energy by emission of electromagnetic radiation or particle(s). 1 Not well done. Students do not know the definition of radioactivity. aii There is background radiation in the environment like radon gas in the air. 1 Many students missed out the example of background radiation in their explanation. b As Uranium-234 and Thorium-230 decays, the number of protons and neutrons both decreases by 2, thus implying that an alpha particle has been emitted. 1 Generally well done but few students saw it as a “gain in protons and neutrons”. c 536 – 44 = 492. After 68 hours which is equivalent to two half- lives, the reading on the detector is 492 × ቀ1 2ቁ 2 = 123 counts per second. 123 + 44 = 167 1 1 A few students did not calculate two half-lives instead they only calculated for one half-life. A few students also did not account for the background radiation in the final answer.

Qn. Description Mark Marker’s Report 10ai When pd is negative, the current across the diode remains the same / is independent of the p.d. When pd is positive, the current across the diode increases at an increasing rate. 1 1 Many students did not take into account the value of current for negative pd. aii1 when p.d. = 1.00 V, Ibatt = Idiode + Itungsten Ibatt = 0.60 + 0.50 = 1.10 A 1 1 Generally well done. aii2 R = V / I = 0.80 V / 0.10 A = 8 Ω 1 1 Not well done. Students could not see that for a series circuit, the current is the same and should look for the corresponding pd when current is the same. aiii The current becomes too large for the diode and causes the diode to fuse/melt, breaking the circuit. 1 Generally well done but a few students missed out keywords like current being too large and cited the voltage being too large instead. .bi Efficiency is the ratio (percentage) of the useful energy output is to energy input. 1 Students used “power” instead of energy and missed out certain keywords. bii energy outputefficiencyenergy input 2 1energy output 240 0 5 1100 2 52 J . . .