Broadrick Prelim Ans

4E Physics Prelim 2024 Marking Scheme Paper 1 (40 Marks) 1 2 3 4 5 6 7 8 9 10 D D C B B D B C D B 11 12 13 14 15 16 17 18 19 20 D C A B B C A C B A 21 22 23 24 25 26 27 28 29 30 A A B C A B C D A C 31 32 33 34 35 36 37 38 39 40 D C D C B A C B D B Paper 2 Section A (70 Marks) Qn Solution Marks 1 (ai) Scalar quantities have magnitude only while vector quantities have both magnitude and direction. [1] (aii) mass, distance, speed, etc. [1] (b) Vector Diagram Scale clearly indicated Correct parallelogram / tip-to-tail diagram drawn Tension in X = 3.1 N (+/- a range) [1] [1] [1]

2 Qn Solution Marks 2 (a) In the first 2.0 s, the ball has increasing velocity. [1] (b) Terminal velocity = gradient of graph after 2.0 s 50 30 26.7 /3.125 2.375 m s   (+/- a range) [1] [1] (c) Initially the only force acting on the ball is gravitational force pulling it downwards. As the ball falls, gravity accelerates it and its speed increases. As speed increases, air resistance opposing the motion of the ball increases. When the air resistance opposing the motion of the ball is equal to the gravitational force acting on the ball, resultant force is zero. When resultant force is zero, there is no more acceleration and the ball moves at constant speed known as terminal velocity. [1] [1] [1] (d) Graph [1]

3 Qn Solution Marks 3 (a) Taking moments about the pivot, sum of clockwise moments = sum of anticlockwise moments 90 500 30 500 30 16790 F F N      [1] [1] (b) If the radius of wheel Y is increased, the perpendicular distance between the pivot and the force increases. Since Moment = F x d, when the perpendicular distance increases, the force required decreases. [1] [1] (c) 60 100 60 100 10 500100 60 100 10100 1.2500 P t F d d d m             [1] [1] 4 (a) 4 5 4 5 9 4 10 6 10 9 4 10 606 10 A B A B A A F F A A F F N            [1] [1] (b) pressure of air = pressure in liquid column + atm pressure 5 5 5 5 913600(10)( ) 1.0 106 10 9 1.0 106 10 0.368 36.813600(10) B atm B B atm B F gh PA Fgh PA h h m cm                [1] [1] (ci) The difference in mercury level remains unchanged. [1] (cii) The difference in glycerine levels is larger than the difference in mercury levels. [1]

4 Qn Solution Marks 5 (ai) 480 (2 60) 0.025 480 (2 60) 2300000 /0.025 v v v Pt ml l l J kg        [1] [1] (aii) In reality, there is energy transferred to the internal store of the surroundings. Since Latent heat = m x lV, and the mass of water vaporised is less than in an ideal experiment, the latent heat calculated will be larger than the true value. [1] [1] (bi) During boiling