2020 9749 H2 Physics MS EJC Suggested

©EJC 2021 9749/H2 PHYSICS 2020 PHYSICS SUGGESTED MARK SCHEME Maximum Mark: 190 9749 October/November 2020 Paper 1 Multiple Choice Question Key Question Key Question Key 1 B 6 C 11 B 2 D 7 B 12 A 3 D 8 A 13 A 4 A 9 D 14 B 5 C 10 B 15 B 16 A 21 A 26 D 17 C 22 B 27 C 18 C 23 A 28 D 19 D 24 B 29 A 20 C 25 B 30 D Notes: Candidates found Questions 8, 16, 19, 22 and 24 more challenging. Question 8 distractor B – did not account for two ropes distractor C – confused the air densities distractor D – did not account for the weight of air inside the balloon Question 11 Takes 1 hour (3600 s) for a minute hand to complete one revolution. Question 14 Note that line MN is NOT an isothermal change. Using PV = nRT, temperature is decreasing. Question 16 Note that the energy of the oscillations reduced by E / 4 rather than to E / 4. Question 19 Single slit equation accounts for only one side of the central maximum. The width of the central maximum is double of this. Question 24 Note that the question asks for potential difference, not potential.

2 ©EJC 2021 9749/H2 PHYSICS 2020 Paper 2 Structured Questions Notes: Candidates need to take care with definitions and make sure that all detail is included, and correct scientific language is used, especially in definitions. Stronger candidates used the word ‘per’ to indicate the division of two quantities. In ‘show that’ questions, stronger candidates made each step in the logic of the proof or calculation clear in order to show how to achieve the formula or answer. Stronger responses were often short and pertinent to the question asked. Qns Marks 1(a) the cyclist exerts a forward driving force which equals in magnitude and opposite to the frictional force. B1 (weight of the cyclist is equal in magnitude and opposite in direction to the normal contact force from the horizontal ground) By Newton's 1st law, the cyclist continues at constant velocity since there is no resultant force B1 1(b) ( ) ( )( )( ) 2 D D 1 1 2 2 2 22 11 4109 m s08 8 12 03 2 c Av F cA .. v .. F ρ ρ − = = = = M1 Method 1 ( ) ( ) ( )( ) ( ) ( )( )( ) ( ) max min 1 2 2 24 2 201 2 08 7 11 03 08 9 13 03 4 1 4 2 s.f. v . .. . . . vv . ∆=  =  = − − M1 M1 Method 2 ( ) ( ) ( ) D D 11 11 22 22 1 2 00 1 01 00 2 2 22 1 2 0 32 1 2 00 1 01 00 2 11 41092 22 1 2 0 32 1 4 2 s. 0 88 . 0 f 88 cvF A vFc A . ..vv .. .. . . . . . .. ρ ρ ∆∆∆ ∆∆=+ ++ ∆= + + + = + ++  = M1 M1 111 1 m sv −±= A1 Notes: one s.f. for the final uncertainty in v

3 ©EJC 2021 9749/H2 PHYSICS 2020 Qns Marks 1(c)(i) Work done = force x displacement in the direction of force Power = work done/time Power = (force ×displacement)/time Power = force × velocity B1 B1 1(c)(ii) ( )( )( )( ) 2 D 3 D 3 1 2 1 2 1 12 03 2 1 142 W 0 88 250 c Av cA P Fv .. v ..