EJC 2021 J1 H2 Promo P2 MS

©EJC 2021 9749/J1H2MYE/2021 EUNOIA JUNIOR COLLEGE JC1 Promotional Examination 2021 9749 PHYSICS MARK SCHEME Paper 2 Structured Questions Qns Answer Marks 1(a)(i) straight line from (0, 5) to (4, − 34) or (4, − 34.2) B1 1(a)(ii) flat portion from (0, 5) to (2, 5) straight portion from (2, 5) to (4, − 15) or (4, − 14.6) B1 Comments: Many candidates failed to indicate values along the axis despite circling or underlining “quantitatively” in the question paper. Quite a number of curves were seen despite the scenario being a free fall with negligible air resistance (and therefore the g radient to the v-t graph should be constantly g = 9.81 m s-2). A small number of candidates failed to follow the instructions to (i) end their graph at t = 4 s and (ii) label their graphs. Revision pointers: Kinematics H201.1 and H201.2 t / s v / m s-1 0 5 − 34 − 15 2.0 4.0 A B 0.5 2.5

2 ©EJC 2021 9749/J1H2MYE/2021 Qns Answer Marks 1(b) Method 1 area under v-t graph gives displacement (for difference in displacement between the 2 stones, look at red + green area) = 59 m C1 correct area identified A1 Method 2                         2 1 2 2 2 steady ascend freefall 2 22 12 54 5 2 5 2 0. 1 2 1 9.81 42 5 4 5 1 2 1 9.81 22 11 9.81 4 9.81 222 58.9 58 25 m 2 3 m .5 m 8 s ut u at s at t ut ss                              accept 59 m C1 58.5 C1 0.38 or taking sum of the above 2 numbers A1 Comments: Most did not realise that Method 1 involving the area of 2 trapeziums will be far easier. Of those who were unsuccessful with using the equations, many neglected the period of constant speed in B. Many candidates were able to solve for the correct distance but failed to draw the proper v-t graphs earlier. This is a sign of incomplete understanding and these students should revise. t / s v / m s-1 0 5 − 34 − 15 2.0 4.0

3 ©EJC 2021 9749/J1H2MYE/2021 Qns Answer Marks 1(c) line starts from (0, 5) decreases at decreasing rate ends at t = 4 s, above (4, − 34.2) gradient at v = 0 demonstrably same as (a)(i) [e.g. via dotted tangent line] B1 B1 Comments: Revision pointers: H202 Kinematics Notes pg 21 - 23. t / s v / m s-1 0 5 − 34 − 15 2.0 4.0

4 ©EJC 2021 9749/J1H2MYE/2021 Qns Answer Marks 2(a)(i) total linear momentum of isolated system of interacting bodies before and after collision remains constant if no net external force acts on system B1 Comments: Candidates need to be precise with definitions. Many failed to mention “isolated”, “system”, and “if no net external force acts on system”. Some candidates unnecessarily limited their definitions to the case of two interacting bodies, which is not accepted as the principle holds even for multiple bodies so long as the conditions are satisfied. Revision pointers: Dynamics Lecture H203.3 2(a)(ii) flat line p = 39 kN s B1 Comments: Generally well-done. question. However, a significant number of students mistakenly sketched a horizontal line in-between the graphs for A and B. 2(b)(i) rate of change of momentum of a body is directly proportional to the resultant force acting on it and in the direction of the resultant force B1 Comments Similar to (a)(i), definitions should be precise. Students need to define rate of change of linear momentum in terms of the resultant force (the order matters) and to also include the direction of this rate of change of linear momentum. Revision pointers: Dynamics Lect ure H203.1 2(b)(ii)     33 on A 17 12 10 12 17 10 3330 N or 1 5 1 5 pF .. t          M1 A1 Comments Careless mistakes include ignoring the fact that the vertical axis is measure in kN s (and not N s) or dividing by 6 seconds, instead of duration during which the lorries’ momenta changed. Other mistakes include assuming that the vertical axis is the velocity measurement of the lorries, instead of its linear momentum. Revision pointers: Dynamics Lecture H203.1

5 ©EJC 2021 9749/J1H2MYE/2021 Qns Answer Marks 2(c)     22 223 5 3 Total initial KE of system 22 17 10 22 10 2 1500 2 3000 1.77 10 J ii AB pp mm                       22 2233 5 Total final KE of system 22 12 10 27 10 2 1500 2 3000 1. 7 10 0J ff AB pp mm                   final total kinetic energy of system not same as initial total kinetic energy, inelastic M1 A1 Comments Marks are not awarded for c omparing relative speed of approach with relative speed of separation , as the question specifically requires the candidates to compare energies. Other mistakes include misidentifying the linear momentum as velocities or mistaking the vertical axis as N s instead of kN s. The workings and the values calculated must be correct in order to achieve the M1 mark here. The description needs to specifically mention kinetic energy that was reduced. There were descriptions which incorrectly stated total energy was reduced; this would have violated the principle of conservation of energy. Revision pointers: Dynamics Lecture H203.4

6 ©EJC 2021 9749/J1H2MYE/2021 Qns Answer Marks 3(a) no net force in any direction no net torque about any point B1 B1 Comments Generally well done. Common mistakes included leaving out either one of the two conditions for equilibrium, not specifying the direction in the case of net force, not specifying pivot for net torque. A worrying number of candidates mistakenly stated the principle of moments as the condition for rotational equilibrium instead. In addition, stating the summation of forces being zero in the vertical and horizontal direction is incomplete, as this definition would only hold along 2D and doesn’t hold for 3D (in and out of plane of paper). Revision pointers: Forces Lecture H204.3 3(b) B1 B1 Comments Generally well done. The common mistake was indicating the directions wrongly, especially for fw. Revision pointers: Forces Lecture H204.3 NG NW W fW fG

7 ©EJC 2021 9749/J1H2MYE/2021 Qns Answer Marks 3(c) Method 1: WGvertical equilibr u im f W N Let ladder by length L by Principle of moments about point of contact with floor, sum of clockwise moments = sum of anticlockwise moments       WW W G W G WW cos s tan in cos t 2 2 2 an 2 2 L W N W W f W N W W f Nf N L N L               B1 B1 Method 2: GWhorizontal equil ibrium fN Let ladder by length L by Principle of moments about point of contact with wall, sum of anticlockwise moments = sum of clockwise moments     G W G G G W G G 2 cos s 2 ta t in co 2 n a t s an 2 n L N W W W L f L N W Nf NN N                B1 B1 Method 3: Let ladder by length L by Principle of moments about point of contact between floor and wall, sum of anticlockwise moments = sum of clockwise moments     W WG WG G ta s cos sin 2 co2 2 n tan L N W L N L N N WN N W              B1 B1

8 ©EJC 2021 9749/J1H2MYE/2021 Qns Answer Marks Method 4: WGvertical equilibr u im f W N GWhorizontal equil ibrium fN Let ladder by length L by Principle of moments about centre of gravity of ladder, sum of anticlockwise moments = sum of clockwise moments  W G G WG G W WG G tan tan t 2 sin si an tan ta s n co cos2 2 2 2 n GW W W L ff N W N N L L Lf N f N N N N N N W                             