RI 2020 Y5 H2 Physics TP Soln

© Raffles Institution [Turn over 2020 Year 5 Term 3 Timed Practice H2 Physics Solutions 1 D Specific heat capacity of water is about 4 200 J kg−1 K−1 and it takes about 5 min for the water in a kettle to boil P × t = m C ∆θ (O-level syllabus) P × 300 = 1 × 4200 × (100 − 30) P = 980 = 1000 W 2 C Average of results of method P = 9.76 m s−2 Average of results of method Q = 9.56 m s−2 Range of results of method P = 10.12 − 9.51 = 0.61 m s−2 Range of results of method Q = 9.61 − 9.51 = 0.10 m s−2 3 A ( ) ( ) 2 tot 22 22 tot tot let be the time taken to travel 1 2 10.50 2 1 2 2 time taken for remaining 0.5 2 1 0.41 ts s at s at s at at at tt s tt = = ⇒= = = = −= 4 A gradient of the velocity-time graph is acceleration 5 B Assuming that A moves up the slope − −= → −= → AA BB T mg f ma mg T ma s in 3 0 [1] [2] [1] + [2]: ( ) ( )( ) − −= + − −=  =−= B A ABmg mg f m m a gg f fg sin30 13.0 4.0 7 .0 1.02 7 2.81 N 6 B ( ) ( ) ( ) 1 12 21 12 2 1 1.0 3.5 7.0 1.0 1.0 7.0 1.6 0.70 m s mu m u mv mv − += + += + =− v v 7 C 4TF m a ma= = 11 43xF T m a T ma ma ma−= ⇒= − = 22 xTm a Tm a= ⇒= 1 2 3 1 T T =

2 © Raffles Institution 8 B 22 7313 1.3 10=−=A Taking moments about corner of platform: 731(4.3) (9.0)( ) 10 5.7 N = = F F 9 D Pressure in a water column is dependent only on the depth of liquid + atmospheric pressure. 10 B By COE, loss in gain in KPEE= Since they are launched with the same kinetic energy, both stones will gain the same amount of potential energy when they reach their highest point. 12 11 22 2 2 Gain in Gain in 2 1 2 PPEE m gh m gh mgh mgh hh = = = = 11 D ( )θ= + = × +× =  0sin 240 1500 9.81sin3.2 240 17 18.04 kW = 18 kW P mg v 12 (a) (i) unit of t = s 2 mmmunit of s 2 m m m s m Lwh A gd − ×× ×× = = ×× × B1 B1 (ii) 1. There may be wrong coefficients. 2. There may be missing or additional terms. B1 B1 (b) (i) Micrometer screw gauge / travelling microscope / digital vernier caliper B1 (ii) ( ) 23 6 0.32 10 1.60 4 0.500 0.23 1.12 10 m ρ − − π× × = ×× =×Ω M1 A1 (iii) 2 dVL dV L ∆ ∆∆∆∆= +++ I I ρ ρ 6 0.01 0.02 0.1 0.042 0.32 1.60 50.0 0.231.119 10 − ∆ = +++×  ρ ∆ρ = 2.81 × 10−7 = 3 × 10−7 Ω m M1 A1 (iv) ρ = (1.1 ± 0.3) × 10−6 Ω m B1 platform A 1.7 cm 6.0 cm F 1.3 cm 4.3 cm

3 © Raffles Institution [Turn over (v) Any 1: Use an ammeter/voltmeter with greater precision / resolution / least count Use a wire of thicker diameter / longer length Increase potential difference/current B1 Markers’ Comments (a) (i) Students should present their workings clearly. (ii) Many students have tried to state what is wrong with the derivation of the equation. Only those explanations with the correct physics concepts will be accepted. (b) (iii) The most common mistake made is not rounding off the actual uncertainty to one significant figure. Many students did not use the standard form for their answer, ending up with many zeros in their answer and making careless mistakes. (v) A common mistake is to suggest taking repeated readings and finding the average of the readings. This method minimises the effects of random error but does not reduce the percentage uncertainty. 13 (a) Any 1: Speed is a scalar while velocity is a vector. Speed has magnitude and no direction while velocity has both magnitude and direction. Velocity is speed in a specific direction. Velocity is rate of change of displacement while speed is the rate of change of distance. B1 (b) (i) Parabolic / parabola B1 (ii) 2 2 1 2 150 (7.0sin30 ) ( 9.81)2 3.5694 or 2.8558 (rejected) 3.57 s s ut at tt tt t = + − = °+ − = = − = M1 (iii) 7.0cos30 (7.0cos30 ) 3.57 21.6 m x x u x ut = = = ×=   C1 A1 (iv) 1 1 22 1 7.0cos30 6.062 m s 7.0sin30 ( 9.81)(3.5694) 31.52 m s 32.1 m s tan 5.200 79.1 below the horizontal x y xy y x v u at v v vvv v vθ θ − − − = + = = = +− = − = += = = =    M1 A1 M1 A1

4 © Raffles Institution (c) Markers’ Comments (b) (ii) A few candidates did not define sign convention, and this led to the formation of an incorrect quadratic equation. (iv) Some candidates used unconventional methods to describe the direction of the final velocity. A well labelled diagram of the final velocity (in the resolved directions) with angle θ labelled clearly would suffice. Workings for the determination of θ should be clearly shown. (c) Some candidates did not score full credit due to inconsistencies in the range of the parabolic path they drew. The difference in horizontal distance between the 2 paths should be increasing. 14 (a) The rate of change of momentum of a body is proportional to the resultant force acting on it and occurs in the direction of the force. B1 B1 (b) (i) ( )( )1500 80 0 30 47400 N sp mv∆= ∆= + − = − Magnitude = 47400 N M1 A1 (ii) ( )( )47400 1500 80 80 0.375 s aveF ma p mat t t = ∆ =∆ = +∆ ∆= M1 A1 (iii) The wall exerts an external force on the system. Linear momentum is not conserved. M1 A1 (iv) ( )( ) = = = aveF ma 3.0 80 240 N or ( )( )∆= =∆ = ave pF t 3.0 30 0.375 240 N M1 A1 (v) The harness provides an extra backward force on the helmet/head. M1 This prevents the head from moving forward while the body is being held/restrained by the seatbelt, thus preventing injury to the neck. OR The force on the neck due to the head will be lower. B1 A 50 m 30° x 7.0 m s−1 B 1 mark for peak (lower and to the left) 1 mark for shorter range

5 © Raffles Institution [Turn over Markers’ Comments (a) This portion was quite well answered. The common mistakes were that the definition was shortened by omitting words here and there which changed the meaning of the definition entirely. Missing terms include “net/resultant force” and “rate of change” to name a few. Another mistake was that students rearranged the definition to say that the resultant force was proportional to the rate of change of momentum. This implies that the resultant force is brought about by a change in momentum (or a change in the state of an object’s motion) and not the other way around. (b) (i) Many students used “change” too casually. A change in a quantity, by convention in Physics, is used to denote a difference between the final and the initial state, in this case final initialpp p∆= − . Since the object comes to a rest, the change in momentum should have been negative. Benefit of doubt was given since the question asked for the magnitude. Other students who correctly calculated the change in momentum sometimes forgot to state the magnitude in the answer line and gave a negative number as the magnitude of a vector which makes no sense. (iii) There is no need to state the principle of conservation of linear momentum (COLM). Comparing the total momentum of the car and the dummy before and after the collision is not an application of the principle of conservation of linear momentum. Stating that the final momentum of the car and the dummy reduces to zero is not a valid application of COLM. Many students wrongly included the wall as part of the system as well and wrongly concluded that the total momentum of the car, dummy and wall was conserved. (iv) A common mistake was to use the change in momentum of the car and the dummy (answer in (b)(i)) multiplied by the time duration to get the average force. This would give the average horizontal force exerted on the entire system of the car and the dummy not the dummy’s head. Some students also calculated the vertical force exerted by the neck on the head ins