2021 RI Prelims H2 Phy Paper 3 Soln

© Raffles Institution [Turn over 2021 Raffles Institution Preliminary Examinations – H2 Physics Paper 3 – Suggested Solutions 1 (a) vut a  Sub into the 2nd equation    22 22 1 22 2 2 vu vuvusu v aa as v u vu a s      B1 Students are required to simplify the equation and not leave it in its raw form (eg   1 22 vu vuvusu v aa    and expect to get the full mark. (b) Displacement required 0.233.0 2.1 1.015 m2  Minimum speed of release required means that velocity at 1.015 m is zero.   22 2 1 2 0 2 9.81 1.015 4.46 m s vu a s u u     B1 M1 A1 OR, using conservation of energy Loss in kinetic energy = Gain in GPE     2 2 2 1 1 02 1 2 2 9.81 1.015 4.46 m s yf i yf i y y y mu mgh mgh ug h hg s u u        The displacement of the ball has to be taken with reference to a fixed point on the ball. Students are required to either calculate the di splacement of the ball from the centre of the ball as it is released (2.1 m from the ground) to the centre of the ball as it enters the net (3.0 m from the ground + radius of the ball); or from the base of the ball (2.1 m from the ground minus the radius of the ball) to the hoop (3.0 m from the ground). A few students assumed height to be displacement. (c) (i) Horizontally  1 cos 6.7 1.3 cos50 6.7 8.02 m s1.3cos50 xxsu t u t u u       M1

2 © Raffles Institution (ii) Vertically       22 2 11 sin50 9.81 1.322 18.0sin50 1.3 9.81 1.32 0.3226 m yy ysu t a tu t          height = 2.5  0.3226 = 2.18 m M1 A1 A1 Students should realise that displacement is a vector that takes into account the direction of velocity. Thus, there is no need to calculate the displacement to the max height and then the displacement again from the max height before adding everything together to find the final displacement. Again, some students forgot to include the negative sign for acceleration (assuming up as positive) and ended up with 8.65 m (this is approximately 3 storeys high) and yet did not think that this number was plausible. Some students also assumed that the ball travelled at constant speed. If this is so, the ball will not return to the ground. 2 (a) Gravitational potential at infinity is zero. Since gravitational force is attractive in nat ure, to bring a mass from infinity to a point in the gravitational field, the direction of the external force is opposite to the direction of displacement of the mass. This results in negative work done by the external force B1 B1 It is important that students include a statement that the work done by the external force is negative, rather than ending the explanation as external force is opposite in direction to the displacement of the test mass from infinity to the point in the gravitational field. A common mistake among students is to write “T he work done by the external force is in opposite direction to the displacement” when the entities that are in opposite directions should be force and displacement rather than work and displacement. (b) (i)

3 © Raffles Institution [Turn over   6 7 2 acceleration gradient 10 30 10 2.00 0.00 10 1.00 m s       M1 A1 Values ranging from 0.97 to 1.03 are accepted. Negative value for acceleration is also accepted. Some students dangerously linked the formulae 2 GMg r and /GM r  (many ignored the minus sign completely which is obviously wrong) using gr  . Some even thought that this is the same formula as GPE mgh . They can’t be more wrong. The above relations are only correct for point masses (or any other spherical mass distributions) and don’t work in general cases, eg for g and  between two masses. Many students also did not draw the tangent line accurately. A simple way to judge if the tangent is good is to see if the overlapping parts of the tangent and the curve on the left and right hand sides of the point of interest are balanced. Here are some examples: (ii) Since acceleration is assumed to be constant, 22 2vu a s 22 7 7 11 22 1 . 01 . 0 01 . 01 0 1.00 10 J mv mu mas    M1 A1 Students are reminded that they should show their formula to illustrate that they know the correct physics. A student who wrote   71.0 9.81 1.0 10 could be applying the wrong physics GPE m mgr commented earlier. Hence, only students who have illustrated that they understood the physics correctly are given full credit in this part. less overlap more overlap same overlap: good tangent.

4 © Raffles Institution (iii) B1 B1 Despite the question stating clearly that candidates are use vertical arrows, many candidates used slanted arrows, showing that they do not know which part of the graph (the vertical axis!) indicates the change in gravitational potential. For those who understood the question, most of them drew the arrows accurately. A small number drew the arrows too short or too long and were penalised. 3 (a) The thermodynamic temperature of an ideal gas is proportional to the mean kinetic energy of the molecules of the gas. A1 “mean translational kinetic energy of an ideal gas molecule 3 2Ek T ” is provided in the Formulae list. Hence, it is important to stat e that the temperature is the “thermodynamic temperature”. Statements such as “An increase in the temperature will cause an increase in the mean kinetic energy of the molecules” are not acceptable as it does not mean that it is a linear relationship. The increase can be exponential or polynomial or any other functions. Internal energy is not acceptable as this ques tion is about the energy of the molecules. Internal energy is a macro term that applies to a system or gas. (b) (i)  2 22 3 23 1 13 22 10 . 0 3 0 3 1.38 10 27 273.1522 6.02 10 499 m s mc k T c c            M1 A1 A common mistake is to regard the mass of a molecule as 0.030 kg. Another common mistake is to regard the number of molecules to be 6.02  1023 or number of moles to be 1 when the gas is at a pressure of 1.0  105 and volume of 0.029 m3.

5 © Raffles Institution [Turn over (ii) PV = nRT 1.0  105  0.029 = n  8.31  (27 + 273.15) n = 1.16 M1 B1 (iii) The root-mean-square speed of the molecules increases with temperature. Since the volume of the oven remains the same, there is an increase in the number of collisions per unit time of the molecules and the walls of the oven, causing an increase in pressure. Or For each collision between a molecule and the wall of the oven, there is a greater change in momentum of the molecule. This leads to a greater force exerted on the wall and an increase in pressure. B1 B1 Students should note that the number of marks for this question is only 2 marks and focus their explanations on what happens to the molecules when the temperature is increased. Explanations invoking Newton’s Second and Third Laws can be included if the number of marks for this question is much more. (iv) PV = nRT 1.0  105  0.029 = n2  8.31  (220 + 273.15) n2 = 0.7076 mass of air = (1.163  0.7076)  0.030 = 1.37  102 kg M1 A1 4 (a) (i) , (ii) & (iii) Many students ignored the ‘standing wave’ mentioned in the question, and treated it as a progressive wave. Note that, since we are plotting the actual displacements of the particles, we are plotting t