2021 RI Promo Sect BC Soln

Raffles Institution Year 5-6 Physics Department   1   2021 Year 5 H2 Physics Promotional Examination Solution Section B 1 (a) (i)    upthrust 0.040 1410 9.818910 0.062097 0.062 N liquidUV g    (ii) At terminal velocity, the resultant force on the sphere is zero.   1 0.040 9.81 0.062 0.26 0.040 9.81 0.062 0.26 1.2708 1.27 m s T T T mg U v v v       (iii)  0 0.040 1.2708 0 4.5 0.011296 0.0113 N T average mvpF tt    (b) *Curved, increasing gradient (zero gradient at 0 s) from 0 s to 2.0 s, with label *Straight line, constant gradient (smooth transition at 2.0 s) from 2.0 s to 4.5 s, with label (c) When moving at terminal velocity, there is a constant drag force acting upwards on the sphere by the liquid. By Newton’s Third Law , the sphere exerts an additional reaction force downwards on the liquid. This increases the normal contact force bet ween the container of liquid and the mass balance, as compared to X when the sphere is instantaneously at rest and experiencing a downwards acceleration (upthrust less than weight). Hence, Y is larger than X. 4.5 s / m t / s 0 1.0 2.0 3.0 4.0 5.0

Raffles Institution Year 5-6 Physics Department   2   Comments (a) (i) Please note the following for all show questions: 1. All working and numerical substituti on of values have to be written down explicitly. 2. Computation is required and please wr ite down the more precise value prior rounding off to the value given in the question. (ii) A significant handful of students made the following mistakes: 1. Inaccurate free body diagram of the s phere i.e. missing forces such as weight, upthrust or drag force. 2. Careless mistakes in computation. (iii) Partial credit is given for the following: 1. Change in velocity, v is computed as initial finalvv or there is lack of clarity in the substitution forv . 2. Ne gative sign was included when asked for magnitude. (b) Many students did not label the graphs explicitly despite values provided on the horizontal axis. Students tend to forget that the initial gradient for the s–t graph should be zero. A handful of students were confused with the time that the sphere takes to reach terminal velocity. It should be 2.0 s and not 2.5 s. A significant handful of students drew the v–t graph instead of the s– t graph. No credit will be given. (c) This question was very poorly done. Common misconceptions include that the reading of the mass balance is given by the  ‘normal contact force exerted by the sphere on the mass balance’  ‘net force acting on the sphere’  ‘moving sphere exerting a force’  ‘resultant/net force acting on the mass balance’ The mass balance reading is due to the norma l contact force between the container of liquid and the mass balance. There is no cont act force that exists between the sphere and the mass balance since they are not in direct contact. Since the reading on the mass balance is due to the normal contact force between the container of liquid and the mass balance, it would be most appropriate (and conv enient) to select the container of liquid as the system for your free body diagram analysis. Students need to be more mindful of the terms th at they use. There is a stark difference between a force and a net/resultant force. The reading of the mass balance is due to a physical force that is acting on the mass balance. This force is not the net force (which according to Newton’s Second Law gives rise to an acceleration). At terminal velocity, the net force on the sphere is zero but there are reaction forces acting on the container of liquid by the sphere due to the upthrust and drag force that the container of liquid acts on the sphere according to Newton’s Third Law. Students have to be more careful with their free body diagram analysis as it is impor tant to account for all the physical forces acting on the selected system of interest. A significant number of students were not aware that immediately after the string was cut, the sphere was not in equilibrium. i.e. the sphere experiences a downward acceleration since wei ght is greater than upthrust, which causes it to sink.

Raffles Institution Year 5-6 Physics Department   3   2 (a) (i) Since the lower portion of the string is horizontal, the string forms a right angle triangle with the two fixed points on the rod. Let r be the length of the lower portion of the string.     22 2 22 2 2 48 16 64 16 16 3 0 Lr L r Lr L L r r LLr       since 0, 3 0 3 LL r rL      (ii) Let  be the angle that the upper portion of the string makes with the rod.  44cos 55 L L Consider the vertical forces on the ring,  cosTm g 5 cos 4 mgTm g  (iii) Consider the horizontal forces on the ring,   2sinTT m r 2 138 8 52 555 3 4 3 TgTT m gmr mr m L L         2 3 g L (iv)   236 3 gvr L g L L (b) When the angular speed is increased, the centripetal force, provided by the tensions in the two portions of the string, required to keep the ring in circular motion increases. The increase in tension results in a greater vertical component of the tension in the upper portion of the string. Hence there will be a resultant vertical force acting upwards on the ring, causing it to accelerate and rise. Comments (a) (i) Many students did not read the question carefully. The total length of the string (horizontal part + slope part) is 8L. (ii) Some students referenced the angle  without indicating it in the diagram. (iv) The final answer should be simplified, e.g.,  1L L should be L .

Raffles Institution Year 5-6 Physics Department   4   (b) Many students claimed that the horizonal component of T provides for centripetal force. They failed to realise that there is also tens ion in the horizontal part of the string. Hence the increase in centripetal force required is only due to an increase in tension, rather than a change in the angle  .   Some students argued that the angle    increases because  cosTm g and T is increased. However, if the angle does indeed change, then the lower string is no longer horizontal and hence needs to be factored in the equation. 3 (a) The graph shows that the acceleration of the ball is a constant negative value when it is in mid-air with a sharp spike in acceleration when the ball hits the ground. This means that the magnitude and direction of the acceleration of the ball does not vary with displacement when the ball is in mid-air. There is only a momentary change in magnitude and the direction of the acceleration when the ball hits the ground. For simple harmonic motion, the magnitude an d direction of the object’s acceleration changes such that its magnitude is directly proportional to its displacement and its direction is always opposite to that of displacement. Hence the ball is not undergoing simple harmonic motion. (b) (i)  Total energy of system at higest point 0 since potential energy is zero at lowest (eqm) point 0.0040 J PKEE mg h mg     OR  ,max Total energy of system when 0 since potential energy is zero at lowest (eqm) point 0.0040 J PKEE mg h mg    (ii)  22 2 ,max max 0 22 0 2 0 2 0 2 11Total energy 22 10.0040 2 10.0040 2 0.0080 0.15 0.0080 2.8125 2.81 m KEm v m x mg m x gmg m x L xL          

Raffles Institution Year 5-6 Physics Department   5   (iii) When the pendulum is raised vertically by 0.30 cm, its total energy (maximum EK) would be ¾ of the total energy when it is raised by 0.40 cm. Method 1: Hence the EK graph will shift downwards such that the maximum EK shifts downwards by 1 large square.