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 wall