2017 RI H2 Physics P2 solutions

2017 Year 6 H2 Physics Prelim Paper 2 Mark Scheme 1 (a) Acceleration is defined as the rate of change of velocity with respect to time. [B1] (b) (i)  1 sin 9.0 ( 9.81 sin26 )(0.70) 5.99 6.0 m s (2 s.f.) vu g t         [M1] [A0] (ii) By the principle of conservation of energy, 22 22 22 gain in GPE loss in KE 1 ()2 2 (9.0) (6.0) 2(9.81) 2.29 2.3 m ( 2.s.f) mgh m u v uvh g      [M1] [A1] (c) (i) y y y 1 tan5.0 6.0 cos26 (tan5.0 )(6.0cos26 ) 0.472 0.47 m s v v v v         x [M1] [A0] (ii) 22 yy y 22 yy 22 2 2 (0.47) (6.0 sin26 ) 2( 9.81) 0.341 0.34 m y vu a s vus a       [M1] [A1] (ii) The momentum of the sphere is not conserved during the collisi on because there is a net force acting on it due to the contact force by t he ceiling and the gravitational force on it by the Earth. [B1] (d) Linear displacement up the slope (must use ruler). Shape of 2 disjointed parabolas (peak must be obvious). [B1] [B1] y/m 0 2.3 x/m 2.64 5 Vx Vy V

2 © Raffles Institution 9749/02 2 (a) Vector quantities have magnitude and direction, whereas scalar quantities only have magnitude. [B1] (b) (i) Drawn to scale, where vi = initial velocity vf = final velocity v = the change in velocity = vf – vi Directions and labels of all vectors are correct. Length of v is correct (accept 4.5 – 4.8 cm) [B1] [B1] (ii) Using cosine rule, Using scale drawing, 22 if i f 22 1 1 1 1 2 ave 2 cos 4.6 cm 50.0 65.0 2 50.0 65.0cos45 46 km h 46.1 km h 12.8 m s 12.8 m s 12.8 0.427 m s30 vv v v v v va t                     [M1] [A1] (c) (i) 12 i 1f 2 25.0 30.050.0 65.060 60 53.33 53.3 km (3 s.f.)     ss v tv t [A1] (ii) 1i 1 1i 1 1i 1 i1 1i 1 i 1 i 1 i1 2f 2 f 2 f 2 12 i 1 ,where 25.0 0.5( ) ( ) 0.5 50.0 60 60 0.208 0.416 0.625 km 30.0 0.5() () () 0 . 5 6 5 . 0 60 60 0.250 0.542 0.792 km ()                         sv t sv tsv t vtsv t v t v tvt sv t v t v t ss v t f2() 0.625 0.792 1.417 1 km (1 s.f.)   vt OR [M1] [M1] [A1] OR N v vi vf v vi

3 © Raffles Institution 9749/02 max min max min 25.5 30.550.5 65.5 54.758 km60 60 24.5 29.549.5 64.5 51.925 km60 60 54.758 51.925 1.417 1 km22        s s sss [M1] [M1] [A1] (iii) Total distance 53 1 km (both values have the same precision) [B1] 3 (a) The principle of conservation of momentum states that the total momentum of a system remains constant, provided no net external force acts on it. [B1] [B1] (b) (i) Taking the velocity to the right as positive. (Statement must be written.) By the principle of conservation of momentum, XX YY X Y 1 () (2.2)(6.0) (1.0)( 2.0) (2.2 1.0) 3.5 m s mu mu m m v v v        Since we have taken velocity to the right as positive, the blocks move right. Note: Accept – 3.5 m s–1, if taken velocity to the left as positive. [M1] [B1] (ii) ave t () (1.0)(3.5 ( 2.0)) 0.35 15.7 16 N (2 s.f.) YY pF mv u t       [M1] [A1] (iii) 1. Weight W and upthrust U creates a net moment on the system which causes the system of blocks to rotate anticlockwise and oscillate. The blocks eventually reach equilibrium when there is no net force on the system where upthrust and weight are equal. [B1] [B1] X Y X Y system oscillates till it reaches equilibrium B G W U X Y U W

4 © Raffles Institution 9749/02 2 . By N2L, ww x y upthrust weight of blocks () (0.10 0.10 )(1000) (2.2 1.0) 0.32 m       Vg m m g h h [M1] [A1] 4 (a) The work done by a force is defined as the product of the force and the displacement in the direction of the force. [B1] ( b ) Consider an object of mass m being raised upwards from height h1 to height h2 by an external force near the Earth’s surface. The external force required to lift the object at a constant sp eed is equal to its weight mg. Increase in gravitational potential energy Ep = work done by external force = force  displacement in the direction of force = mg (h2 – h1) = mg Δh (Although it is not necessary for the speed to be constant, it is important that the speeds at the starting and ending point are the same so that th ere is no change in kinetic energy.) [B1] [M1] [A0] (c) (i) By the principle of conservation of energy, the decrease in th e gravitational potential energy of the blocks goes into the increase in kineti c energy of the blocks and the work done against friction. Note: All energy gains and losses must be described to be awarded the mark. [B1] (ii) By the principle of conservation of energy, QP loss in GPE of system loss in GPE of block Q gain in GPE of block P ( sin30 ) (3.0)(9.81)(0.45) (1.0)(9.81)(0.45 sin30 ) 11.04 11 J (2 s.f.) mg h mgh        [M1] [A0] (iii) By the principle of conservation of energy, loss in GPE gain in KE work done against friction gain in KE loss in GPE work done against friction 11.04 (6.3)(0.45) 8.20 8.2 J (2 s.f.)     [M1] [A1] h1 h2 F mg

5 © Raffles Institution 9749/02 (iv) By the principle of conservation of energy, loss in KE loss in GPE gain in EPE work done against friction 2 2 18.20 (3.0)(9.81) (1.0)(9.81)( sin30 ) (800) (6.3)2 400 18.225 8.20 0 0.168 m (reject 0.122 m) ee e e ee e        [M1] [A1] OR 2 2 Alternatively, considering the whole motion: loss in GPE gain in EPE work done against friction gain in EPE loss in GPE work done against friction 15 () ()22 15(800) (9.81)(0.45 ) (6.3)(022 ke g h e f h e ee        2 .45 ) 0 400 18.225 8.20125 0.168 m (reject 0.122 m) e ee e     Allow e.c.f. of KE from part (iii). [M1] [A1] (iv) A larger incline results in a smaller loss in the gravitationa l potential energy of the system. Hence, the maximum compression of the spring decreases. [B1] 5 (a) Internal energy is determined by the state of the system and it can be expresses as the sum of a random distribution of kinetic and potential energ ies associated with the molecules of the system. [B1] (b) pV = NKT (1.05  105)  (2.9  104) = N  (1.38  1023)  (303) N = 7.282  1021 = 7.28  1021 [M1] [A1] ( c )   23 21 21 3 2 3 1.38 10 3032 6.272 10 6.27 10 J KEk T          [A1] ( d ) ( i )    21 23 3 2 3 7.282 1 0 1.38 10 357 3032 8.14 J UN k T         Allow e.c.f. from part (b). [M1] [A1] (ii) Since gas A undergoes an adiabatic compression, Q = 0. From the 1st law of thermodynamics, U = Q + W 8.14 = 0 + W W = 8.14 J [A1]

6 © Raffles Institution 9749/02 (iii) For gas A, constantPV T  . 454 52.1 101.05 10 2.9 10 1.71 10 Pa303 357 A A p p      For gas B, PV = constant. (1.05  105)  (2.9  104) = pB  (2.1  104)  pB = 1.45  105 Pa 6 (a) (i) Equation of graph is V = E – Ir, where gradient = –r and vertical-intercept = E. 0.000 1.050Gradient 0.700 2.000 0.500 0.700 r     E  1.050  0.700×0.500  1.400 V or read from graph (both 3 d.p.) Note: Deduct [1] for wrong d.p. in this part, not the cover page. [M1] [A1] [A1] (ii) The maximum resistance of R is 2.10  or is not infinite, and hence the current cannot be reduced below 0.500 A. [B1] (b) (i) R = r [B1] (ii) Since R = r, V = E/2 = 1.400/2 = 0.700