2022 RI Prelims H2 Phy Paper 2 Soln

Raffles Institution Year 5-6 Physics Department 1 2022 H2 Physics Preliminary Examination Solution Paper 2 1 (a) (i) particle A particle B 9mu 0 3T 2T –3mu T t p 4.2mu 1.8mu

Raffles Institution Year 5-6 Physics Department 2 (ii) As there are no resultant external forces acting in the horizontal direction on the system of particles A and B, total momentum of the system is conserved in the horizontal direction by the principle of conservation of momentum. Since the momentum of particle A decreases by 7.2mu (36 small squares) after the collision, the momentum of particle B should increase by 7.2mu (36 small squares) so that the total momentum remains constant at all times. OR Since the momentum before the collision is 6.0 mu, the total momentum of the system remains constant at 6.0mu at all times. (b) Solution 1 relative speed of approach  34ABuuu u u   relative speed of separation 4.2 0.2 4BAvv u uu   Since the relative speed of approach and the relative speed of separation are equal, the collision is elastic. Solution 2 kinetic energy,  2 2 211 22 2 k mv pEm v mm  before collision:    2222 ,, 2 , 93 922 2 92 A before B before k before AB mu muppEm u mm m m   after collision:    2222 ,, 2 , 1.8 4.2 9222 9 2 A after B after k after AB mu muppEm u mm m m   Since the kinetic energy of the system before and after the collision remains the same, the collision is elastic. (c) By Newton’s second law, resultant force R dpF dt , which is given by the gradient of the graph in Fig. 1.1. During collision, from tT to 2tT , the gradient of the graph for particle A is 7.2mu T . The gradient of the graph for particle B is 7.2mu T . Hence the horizontal resultant force on parti cle A, which is the force on A by B has magnitude 7.2mu T and the horizontal resultant force on particle B, which is the force on B by A also has the same magnitude of 7.2mu T . Since the gradients have opposite signs, this imply that the two forces are opposite in direction. This is consistent with Newton’s third law as the force on A by B and the force on B by A are equal in magnitude and opposite in direction.

Raffles Institution Year 5-6 Physics Department 3 Comments (a) (i) Generally well done. The most careless mistake made was to read the vertical axis values wrongly and either draw 4.2 mu at the wrong position (e.g. small squares above 4.0mu) or draw it at the right position, but quote the value of 4.1mu instead. Please note that each small square is equivalent to 0.2mu on the vertical axis. (ii) Weak responses failed to include the condition for momentum conservation, which is that there must be no resultant external forces on the system. Weak candidates tend to forget either the entire condition, or leave out the word “resultant”, which is key – in this case, there could be external forces on the system (e.g. weight, normal contact force) but since these forces cancel out, the resultant external force is therefore zero. Henc e, total momentum of the system is conserved. Other weak responses did not include sufficient details about how the graphs were drawn by including relevant values (e.g. 6.0mu, increase/decrease by 7.2mu). (b) Generally well done. Most mistakes were careless mistakes, which mostly involved: 1. Using the wrong mass / speed values (e.g. stating particle B has mass 3m instead of m, or it has initial speed u instead of 3u) 2. Confusing momentum for velocity (e.g. st ating particle A’s initial kinetic energy is    21 992 mm u instead of  21 92 mu ) 3. Using the kinetic energy formula 2 2 p m incorrectly (e.g. stating particle A’s initial kinetic energy is  2 9 2 mu m instead of   2 9 29 mu m ) Some candidates were also unable to correctly state the reasoning for why it is an elastic collision, especially with equating the relative speed of approach with the relative speed of separation. (c) Many candidates lost the first mark because of an improper definition of what dp dt or the gradient of the momentum-time graph meant – it refers to the resultant force on the particle, not just the force. This was marked particularly strictly as it was a specific weak point identified amongst our candidates in the 2021 A-Level Physics Examinations. Since the question asked for how the graphs demonstrated Newton’s Third Law, candidates are expected to go into sufficient detail about what aspects of the graph demonstrate the Law. For example, including the expression 7.2mu T for the gradients of the graphs during the collision is expected, in order to compare their equal magnitudes. Similarly, when referring to the forces being opposite in direction, candidates should specifically pinpoint that it is the opposit e signs of the gradient of the graph that demonstrate this fact. Weak responses atte mpted at the above, but no credit was given to responses that did not show enough detail.

Raffles Institution Year 5-6 Physics Department 4 2 (a) Since the wheel is in rotational equilibrium,      5.0 25 N0.20 Tr T r (b) Since the box is in equilibrium, horizontally, (not required to answer this part) 0 sin20 xx x TR RT   vertically, 0 cos20 yy y RTm g RT m g     By the principle of moments, taking moments about point M,  2 yx y hRd Th T                    1 2 0.80 cos20 sin20cos20 2 0.80 sin20 0.5cos20cos20 0.80 25 sin20 0.5cos2025cos20 2.0 9.81 0.3766 0.377 m xy y hdT TR TTTm g T Tm g motor ground box rope 0.80 m 0.80 m T mg Rx Ry M

Raffles Institution Year 5-6 Physics Department 5 (c) Solution 1 As  increases, the increased clockwise mo ment about point M due to the tension exceeds the maximum anticlockwise moment possible due to the normal contact force when the normal contact force is at the corner ( 0.40 md  ) of the box. The box will rotate clockwise about the bottom right corner. Solution 2 As  increases, d increases to balance the increasing moment due to the tension and if 0.40 md  , this implies that the contact force needs to act outside the box for the box to be in equilibrium. Since it is not possible for the contact force to act outside the box, there is a maximum value for  . Solution 3 As  increases, the horizontal component of the tension in the rope increases. If the horizontal component exceeds the maximum fr iction on the box by the ground, the box will slide. Comments (a) Most candidates are able to arrive at the correct answer. Some thought that it is equal to the weight, while some others incorrectly included a cos 20° here. (b) Poorly attempted. Many candidates used weight to calculate moments, but since M is the mid-point of the base of the uniform square box, the weight of the box does not cause any moment about point M. Many candidates used either one the horiz ontal or the vertical component of T to form the equation using Principle of Moments, neglected the fact that both components of T cause clockwise moments about M. Some candidates tried to find the perpendicular distance from M to the rope, but only few managed to correctly solve using this method due to incorrect calculations of this distance. (c) Decently attempted. Weak responses mainly did not explain wh y equilibrium is not maintained beyond the angle (some even merely paraphrased the question), or were not specific about the direction of moments in their explanation. Some candidates also mentioned that the r ope will break beyond the maximum angle, but that is not answering the question about equilibrium.

Raffles Institution Year 5-6 Physics Dep