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