EJC Physics 2023 J2 H1 MYE PAPER 2 MS

©EJC 2021 9749/J1H2MYE/2021 EUNOIA JUNIOR COLLEGE JC2 MIDYEAR EXAMINATIONS 2023 8867 PHYSICS MARK SCHEME Paper 2 – Structured Qns Answer Marks 1ai (vernier/ digital) callipers Examiner’s comment: Common mistakes: micrometre screw gauge B1 1aii 0.0004percentage uncertainty = 100%0.0420 0.95% × = Examiner’s comment: Common mistakes: A lot of students did not leave the answer in the correct number of significant figures 2 s.f. For THIS MYE ONLY, they are awarded full mark despite wrong s.f. due to leniency in marking. B1 1bi kg m−3 = kg × mn/m −3 = n−1 n=−2 Examiner’s comment: Common mistakes: Cambridge does NOT recognise [ ] as a short cut symbol for “unit of”. Please write “unit of” in the future examinations where applicable. “Show” questions must show ALL steps. Do NOT skip steps. M1 A1 1bii 2M rL M rL ρ ρ ∆∆ ∆∆=++ 0.001 0.0004 0.0001percentage uncertainty = 2 100%1.072 0.0420 0.1242 2.1%  + +× = C1 C1 A1 1biii 2 3 3 3 1.072 0.0420 2337 (kg m )2.094 0.1242 0.021 2337 49 (kg m ) (2340 50) kg m ρ ρ ρ − − − − ×= =× ∆= × = = ± C1 C1 A1

2 ©EJC 2023 8867/J2H1MYE/2023 Qns Answer Marks 2ai By conservation of linear momentum, 5 0 5 1 5 Y X xY X XY p p mv m v mv mv m mm +=+ = + =+ ( ) ( ) ( ) 2 2 2 2 total kinetic energy of X and Y after collision total kinetic energy of X and Y before collision 1 2 1 5 2 5 1 5 0.20 XY X mm v m v v v += = = C1 C1 A1 2aii ratio = 1 B1 2b horizontal line from (0 ms, 0 squares) ending at (20 ms, 0 squares) and horizontal line from (40 ms, 4.0 squares) ending at (60 ms, 4.0 squares) straight line from (20 ms, 0 squares) ending at (40 ms, 4.0 squares [= 4.0 cm vertically]) Examiner’s comment: Note: From t = 0 to t = 20 ms, the line needs to be visible too despite on the axis. B1 B1

3 ©EJC 2023 8867/J2H1MYE/2023 Qns Answer Marks 3a Rate of change of momentum is proportional to the resultant force Takes place in the direction of the resultant force B1 B1 3b Apply N2L to each of the masses. For 8kg mass, F=ma 150 +8(9.81)sin30° − T = 8a [1] For 4kg mass, F=ma T = 4a [2] Solving the equations, T = 63 N a = 15.8 m s −2 Accept if students solving a using the whole system for this case. M1 M1 A2

4 ©EJC 2023 8867/J2H1MYE/2023 Qns Answer Marks 4a Let force in rod be 𝐹𝐹. Resultant of force in rod and component of weight towards the centre provides centripetal force Acceptable: “Component of resultant force towards centre” Not acceptable: “resultant force provides centripetal force” Assuming 𝐹𝐹 points towards the centre of the circle, 𝑚𝑚𝑚𝑚 𝑐𝑐𝑐𝑐𝑐𝑐 50° + 𝐹𝐹 = 𝑚𝑚 𝑣𝑣2 𝑅𝑅 (1.5)(9.81) cos 50° + 𝐹𝐹 = (1.5) �9.52 2.0 � 𝐹𝐹 = 58 N Since 𝐹𝐹centripetal > 𝑚𝑚𝑚𝑚 𝑐𝑐𝑐𝑐𝑐𝑐 50° , 𝐹𝐹 points towards the centre of the circle. Rod is in tension.