2026 EJC H2 Physics Prelims P3 MS

Qns Answer Marks 1(a) Moment of a force is the product of the force and its perpendicular distance to the pivot. Examiner’s Comments Generally well done by candidates. B1 1(b)(i) 6.8cos35 6.8 8.3 mcos35 extension 8.3 7.0 1.3 m y y x = ==  =− = Examiner’s Comments Generally well done by candidates. M1 A0 1(b)(ii) W: weight of beam (or gravitational force by earth on beam) N: normal contact/reaction force by bricks on beam F: force by spring on beam (or spring force / tension / elastic force) (EITHER) R: contact/reaction force by wall/hinge on beam (OR) Rvertical: vertical component of contact/reaction force by wall/hinge on beam Rhorizontal: horizontal component of contact/reaction force by wall/hinge on beam B1 W N R F

Qns Answer Marks (Alternatively): normal contact/reaction force by wall/hinge on beam Correct position and direction of all forces Length of arrow of W longer than N Examiner’s Comments This question was poorly attempted by candidates. Please take note that you must write in full (or include a legend that indicates what each symbol represents in your diagram) when the question asks for labelled diagram. You must indicate the type of force (e.g. contact force) and by which object on which object. Many candidates did not include the contact force by wall on beam in your diagram. For candidates who drew the arrow for R, many responses are incomplete. For N as shown in the diagram: Many candidates wrote weight of bricks which is incorrect. Weight of bricks acts on brick, hence is not the force to be drawn for free body diagram of beam. Also, please take note that forces are vectors. So when the question asks for labelled diagram, please be mindful about the length of the arrows drawn for each force. The length of arrows represents the magnitude of the forces. In this question, the length of arrow for W must be longer than the length of arrow for N. B1 1(b)(iii) For rotational equilibrium, taking moment about the pivot O, sum of clockwise moment = sum of anticlockwise moment ( ) ( ) ( )( ) ( )( )( ) ( )( )( ) ( )( ) 1 21 0 70 . N 1.8 3 4 sin35 6.8 120 9.81 1.8 190 9.81 3.4 sin35 6.8 2168 N 2200 N or 220 or 1692 N m . 2 17 1. 3 0 13 N W F F F FF FFkk xx − + =  + =  = == = = = = == 1 11 1669 N m 1700 N m (2 s.f.) 1700 N m (2 s.f.) − −−== Accept both 2 s.f. and 3 s.f. Note: accept 1670 N m-1, 1690 N m-1 and 1700 N m-1. M1 A1

Qns Answer Marks Examiner’s Comments This question was poorly attempted by candidates. Some candidates used the masses of bricks and beam for calculation instead of their respective weights. Some candidates did not include the perpendicular distance from the line of action of the force to the pivot O. Some candidates used ( )( )cos35 6.8F  for the anticlockwise moment contributed by the spring force which is incorrect. 1(b)(iv) ( )( ) 2 2 1 2 1 1700 2 1437 1400 J (2 s.f.) 1.3 PE kx= = = = Note: Can also use 211 22 orPP FE Fx E k==  Accept both 2 s.f. and 3 s.f. Allow ECF. Note: accept 1400 J, 1410 J, 1430 J and 1440 J Examiner’s Comments Generally well done by candidates. C1 A1 1(b)(v) As the pile of bricks is moved towards the pivot O, the clockwise moment about the pivot (due to the bricks) decreases (as the perpendicular distance from pivot to line of action of the force decreases). (The total clockwise moment decreases). A lower anti-clockwise moment is needed to maintain equilibrium. Therefore, the force exerted by the string (on the beam) decreases. Examiner’s Comments Generally well done by candidates. A few candidates mentioned that clockwise moment increases which is incorrect. A few candidates mentioned that the anticlockwise moment is contributed by the bricks and clockwise moment is contributed by the tension in the string which is incorrect. B1 B1

Qns Answer Marks 2(a) g and r are constants so a is proportional to x the negative sign shows that a and x are in opposite directions/ a is always directed towards the equilibrium Marker’s Comments This question was generally well done. Some students simply quoted defining features of SHM without relating the specific evidence from the equation to the relevant part of SHM. Others quoted both pieces of evidence without specific which evidence maps to which part of SHM. B1 B1 2(b) 2 2 2 1 since and so 9 81 0 28 5 9 rad s 5 92 gxa a x r g r . . . ( . )     − =− =− = = = Marker’s Comments This question was generally well done. Some students thought that angular frequency was f M1 A1 2(c)(i) Negative sine graph max a = 2xo = 5.92 (0.05) = 1.7 m s-2 Marker’s Comments This question was generally well done. B1 B1

Qns Answer Marks (c)(ii) since this is a light damping oscillation, period remain the same as before and remains constant over time Amplitude consistently decreases and first peak must be less than 5.0cm Marker’s Comments Some students wrongly assumed that the first amplitude would remain at 5.0 cm. Others thought that the decrease in amplitude would only happen after each period, e.g. drawing the same decreased amplitude for the first and second humps, then the same further decreased amplitude for the third and fourth humps. Some students considered the period to be slightly larger than the undamped oscillation. While this is correct, most of these candidates failed to ensure that this increased period still needs to remain constant over time. Since this can be difficult to draw, students are therefore advised to consider the same period as that of undamped oscillation. B1 B1 (c)(iii) 2 2 2 2 2 2 2 2 Loss in energy 11 22 11 0.090 5.9 0 050 0.090 5.9 0 025 M122 0 294 0 29 J A1 oi ofm x m x .. .. =− =    −    == Marker’s Comments The classic mathematical mistake committed for this question includes that of thinking that (x22 – x12) is the same as (x2 – x1)2. Other issues included the omission of square for angular frequency  and failure to convert units to that of SI units. M1 A1 x/ cm T 5.0 t/ s 0 −5.0

Qns Answer Marks Since the value of angular frequency  is taken from part 2(b), full ecf credit was awarded. However, some students had left part 2(b) blank and so were unable to gain any credit for this part. In such cases, students should provide a reasonable value for the earlier part (e.g. not “0” or “1”) to continue with this part. Qns Answer Marks 3(a)(i) water at 5.0 C to water at 0 C water water 220 4.18 (5. 0 0) 4598 J 4600 J (2 s.f.) Q m c T  = =   − = = Examiner’s Comments Generally well done by candidates. Some candidates did not take note of the units such as kg and 11J g C−−  which results in incorrect calculation. B1 3(a)(ii) water at 0 C to ice at 0 C water f 220 334 73480 J 73500 J (3 s.f .) Q m L  = = = = Note: accept 73000 J and 73500 J. Examiner’s Comments Generally well done by candidates. Some candidates did not take note of the units such as kg and 1J g− which results in incorrect calculation. Some candidates calculate the heat released for the water from 5 C water to 0 C ice which is incorrect. B1 3(b) ( ) ice at -8.0 C to ice at 0 C ice ice 500 2.09 0 8.0 8360 J (3 s.f. ) Q m c T = =   − −  =