2021 RI Prelims H2 Phy Paper 2 Soln

© Raffles Institution [Turn over 2021 Raffles Institution Preliminary Examinations – H2 Physics Paper 2 – Suggested Solutions 1 (a) (i) When bodies in a system interact, the total momentum of the system remains constant, provided no net external force acts on it. B1 B1 It is important to include the word “net”. There can be external forces acting on the system. As long as the “net” external forces is zero, conservation of momentum can be applied. Note that conservation of momentum is not only confined to collision problems. Hence no credit is given for simply st ating “Total momentum before and after collision is constant”. Also, there is no need to confine the discussion to just “two bodies”. (ii) Relative speed of approach = relative speed of separation u  (u) = vB  vA 2u = vB  vA …………. (1) Conservation of momentum mA u  mB u = mA vA + mB vB …………. (2) mA(1) + (2): (2 mA u ) + (mA u  mB u ) = (mA vB  mA vA ) + (mA vA + mB vB) 3 mA u  mB u = mA vB + mB vB 3 AB B AB mmvu mm   B1 B1 B1 Some candidates worked backwards from the result and quote some expressions. No credit is given if there are no bas is and no explanation given for these expressions. Candidates who tried to use conservation of kinetic energy will end up with very messy expressions and most likely do not obtain the result needed. For elastic collisions, using the relative speed relationship is neater. The sign convention is given in the question (taking right to be positive). Candidates using the wrong convention may still end up with the required answer, but no credit is given for such case, as candidates failed to follow the instructions given. (b) (i) 1 3 3(0.62) (0.059) (4.4)0.62 0.059 11.7 m s AB AB mmvutennis mm      B1 Some candidates confused the collision between the basketball and tennis ball with that of the collision with the ground. (ii) The impact force is much greater than the weight of the tennis ball / basketball. (Hence, gravitational forc e, which is an external force, is neglected when applying conservation of momentum.) B1 The scenario in part (b) (which is a vertical motion) is an extension to the result in part (a) (which is a horizontal motion). Hence the force of gravity (which is an external force) may made the application of conservation of momentum invalid. The

2 © Raffles Institution reason why it may still be applied is that the impact force is much larger than the gravitational force (weight) of the objects. This is due to the short time duration of impact between the two objects. Repeating what the question has already pr ovided will not earn the credit, e.g. stating that the “speed of the basketball before and after impact with the ground is the same” will not earn any credit. Simply stating that there are no external fo rces is also not correct, as gravitational force is an external force. Stating that “gravitational force is negligible” is not good enough. It should be “negligible compared to the impact force”. “Friction” is also not a consideration as the objects are not moving sideway between themselves. As the objects are “released” from a height, they should fall vertically (and not at an angle). Hence stating “head-on” or “collide vertically” will not get any credit as these are trivial observations in this context. Note that the conservation of momentum is obtained from Newton’s 2 nd and 3rd Law. There is no consideration of energy invo lved here. Hence candidates stating that the assumption of no loss of energy do not get the credit. Other trivial suggestions like mass not changed, there is no wind, etc, do not get the credit. (iii) Basketball is much greater in mass as compared to the ball, hence mB is negligible compared to mA. Using (a)(ii), 1 3 3 3(4.4) 13.2 m s A B A mvu m u      M1 A1 Candidates should make effort to explain their workings. 2 (a) The moment of a force about a point is defined as the product of the force and the perpendicular distance of the line of action of the force from the point. B1 The definitions in the lecture notes should be memorised properly. They are already as concise as they can be without losing any significant meaning. Also do try to read through the sentence after writing as there were many cases of “distance from the point to the pivot” appearing in the definitions.

3 © Raffles Institution [Turn over (b) Since the airplane is travelling at constant speed, it is in equilibrium Vertically:   561.5 10 9.81 1.47 10 NLW m g     Horizontally: 68.0 10 NTD  clockwise torque = anti-clockwise torque       56 0.75 8.0 10 1.47 10 0.75 1.38 m TyW y y      C1 M1 A1 Many students did not specify their pivot and seemed to either be calculating torque (instead of moment as stated in their soluti ons). These two quantities are inherently the same but would be good to specify what the calculation is correctly. (c) As the airplane accelerates forward, the box will tend to move at its original velocity. Hence, the box will tend to move backwards relative to the airplane floor. Friction on the box by the floor therefor e acts forward (to oppose this relative motion). M1 A1 Simply stating N1L followed by a “hence” does not make an explanation valid. Many students correctly stated N1L but did not give any context as to what was happening to the box, and then followed up by using N2L to explain the direction of friction. This is not answering the question. It is necessary to state the movement of the box with reference to the plane. Many students incorrectly stated that there is no resultant force on the box as the box is stationary (wrt to the plane) or that there is a backward resultant force acting on the box as the plane accelerates. (d) Applying principle of moments, taking pivot at the C.G.,    55 5 2.0 10 10 2.0 10 20 30cos30 2.31 10 N P P     M1 A1 Many students incorrectly resolved the distance/perpendicular component of P. 3 (a) In one second, the volume of air that is swept by the blades is   2 53 1 50 20 1.57 10 m VA v     M1 A1 Most students calculated the volume flow ra te correctly. A few candidates do not know the formula for the area of a circle. (b) 5 5 In 1 s, mass 1.20 1.57 10 1.88 10 kg V    B1 This part presents little challenge to most students.

4 © Raffles Institution (c) (i)   22 52 2 7 1Rate of loss KE 2 1 1.88 10 20 152 1.65 10 W dm uvdt        M1 A1 A number of students wrongly wrote the loss in KE as  21 2 mu v when it is supposed to be  221 2 mu v  . Some students use the “” sign excessively or carelessly omitting them for convenience and were penalised. It will be helpful to remember that: “Gain” = “Final  Initial” “Loss” = “Initial  Final” This will save you from introducing the “” sign haphazardly. (ii)   5 5 Force 1.88 10 20 15 9.42 10 N dm uvdt      M1 A1 Quite a number of students applied the familiar “ dmFv dt ” here wrongly, with v = 20 or 15. A significant number of students misunderstood the deceleration of the wind to be 5 m s2 (calculated using uva t  ). The actual situation experienced by the wind when it reaches the windmill is very di fferent. The deceleration from 20 m s 1 to 15 m s 1 happens over a time much shorter than 1 second, while the mass of wind undergoing the deceleration at any instant is much less than the value obtained in (b). In other words, while F = ma is valid, the m is not 1.88 105 and the a is not 5. Students who