2020 RI Prelims H2 Phy Paper 3 Soln

© Raffles Institution 9749/03   2020 Raffles Institution Preliminary Examinations – H2 Physics Paper 3 – Suggested Solutions Section A 1 (a) (i) The direction of motion after collision will be along the same line of motion before collision. OR The direction of velocities after collision will be along the line joining their centres of gravity. (ii) Their relative speed of separation is equal to their relative speed of approach. (b) Take direction to the right as positive. By conservation of linear momentum,     12 1 2 12 12 22 4.0 2 2.0 2 0 2 ----- (1) mu m u mv mv m m mv mv mv mv     Since collision is elastic,    12 2 1 21 21 4.0 2.0 6.0 ----- (2) uu v v vv vv        Subst. (2) into (1),   11 1 1 1 02 6 . 0 01 2 3 12 4.0 m s3 mv m v mv v       Velocity of A is –4.0 m s–1 i.e. body A moves to the left with speed 4.0 m s–1. From (2),  1 2 6.0 4.0 2.0 m sv   Velocity of B is 2.0 m s–1 i.e. Body B moves to the right with speed 2.0 m s–1. (c) In the collision in Fig. 1.3, the styrofoam increases the duration of impact as it is compressible. Thus, for the same change in momentum, the impact forces experienced by the balls are smaller compared to the collision in Fig. 1.2. Hence the external gravitational forces (weight) on the balls are comparable in magnitude to the impact forces and they contribute to the change in the momentum of the system such that the momentum of the system is no longer conserved. Comments (a) (i) Did not accept “parallel” to initial line of motion, “horizontal axis” etc. (ii) Did not accept “kinetic energy is conserved”, “12 21uu v v ” as question asks to “describe subsequent motion” (i.e. qualitativ e description of speed/ velocity is required). (b) Some are still unaware that 11 2 2 11 2 2mu mu mv mv  and 12 21uu v v are vector equations, so they forget to substitute numbers with negative signs for the velocities which are in the opposite direction. Note that u1 and u2 in the question are speeds.

2    © Raffles Institution 9749/03 (c) Poorly done. This part reveals common misconception about the principle of C.O.M. Many students reasoned that since the speeds after collision have reduced (as styrofoam has absorbed some energy), the total momentum is not conserved. This is far from the truth. Think of this example: suppose the col lision in (b) results in a completely inelastic collision, then (4.0) 2 ( 2.0) 30mm m v v   , final v < u1 or u2 but total momentum is still conserved. Basically, students need to understand one has to use a vector sum approach when adding individual momenta. Some students mixed up C.O.M. with cons ervation of kinetic energy and erroneously concluded that the collision being inelastic (loss of k.e.), violates C.O.M. Another common issue is students did not recognize the force exerted by the styrofoam is an internal force (not an external force) , acting within the new system of Fig. 1.3. Students should have seen questions where a spring is attached to one colliding body and the spring force is regarded as an inte rnal force in the system, else ideal gas molecules bouncing off each other due to contact forces (also internal forces) in a closed system. What constitutes as external force is the gravitational force acting on bodies A and B. For the system in Fig. 1.2, gravitational force could be treated as negligible simply because the duration of impact was very short (hence collision force is much bigger than gravitational force/ or impulse due to gravity is negligible during the short instant). This is not so for Fig. 1.3. Very few students explained correctly.     2 (a) (i) Gravitational potential is taken to be zero at infinity. Since gravitational forces are attractive , the work done by an external force in moving a mass from infinity to the point is negative as the displacement and external force are in opposite directions. (ii) The work done per unit mass by an exter nal force in bringing a small test mass from infinity to a point 70.98 10 km (OR point Q) from the center of Star X is 12 13.0 10 J kg  . (iii) 1Since gravitational potential GM rr   12 12 3.0 10 1.0 10 QR QR r r          12 77 12 3.0 10 0.98 10 2.94 10 km1.0 10 Rr      (iv) work done by the external force   12 15 1200 3.0 1.0 10 2.4 10 J QRm       

3    © Raffles Institution 9749/03 (b) (i) Gravitational force provides the centripetal force for the stars to orbit about the common centre of mass of the system.     22 122 12() YZ YZ GM M Mx Mxxx Since  for both stars is the same, 28 1 30 2 1.45 10 0.005532.62 10 Z Y xM xM      *If the ratio of masses is quoted directly to calculate, it must be stated that C is also the centre of mass. (ii) If they do not have the same angular velocity, the gravitational force between them on either star will not be towards C for them to orbit around C. OR The two stars must always be on opposite sides of C so that the gravitational force between them is always pointing towards C, providing the centripetal force. OR If they do not have the same angular velocity, the centre of mass at point C will not be stationary. Point C should be stationary as there is no net force acting on the system. Comments (a) (i) In answering qualitative questions, students are reminded to do so concisely, after having fully understood the requirements of the question. Students are to use appropriate and accurate keywords/phrases/terms and structure their sentences in some logical sequence as well as avoid beating around the bush. For this part, many students defined what gr avitational potential is. This is not required. Students were penalised due to one or more of the following mistakes/oversights/ambiguities:  Stated that external force acts in a direction opposite to gravitational force, when it should have been “opposite to displacement” as how work is defined i.e. the product of force and displacement (parallel to the force). In addition, phrases like “direction of motion” or any other non-quantifiable terms used in place of displacement are not acceptable.  Stated to the effect that work done by an external force is opposite to displacement. Work is a scalar quantity and does not have an associated direction. So the above-mentioned statement is incorrect.  Did not state that gravitational potential at infinity is zero. Some erroneously mentioned that gravitational potential at infinity is negative. In addition, quite a number of students did not make it clear that the (negative) work is actually done by an external force. (ii) Common mistakes in this part are not stating “per unit mass” in “work done per unit mass” and also not explicitly stating that the work per unit mass is done by an external force. (iii) Generally well done other than mistakes invo lved in the transfer of the numerical answer in terms of kilometres which some students mistook to be in metres. Several students made computational error when they calculated the factor GM from information given of point Q, ending up with the wrong order of magnitude.

4    © Raffles Institution 9749/03 (iv) Students who made mistakes here either didn’t understand that external force and displacement act in opposite directions hence work done by external force should be a negative value or they got the order of subtraction wrong i.e. initial gravitational potential energy at R  final gravitational potential energy at Q, when it should have been the other way round. (b) (i) A very common mistake here is that students equated the gravitational force on star Y to the gravitatio