2013 JJC H2 Physics P3 solutons

Page 1 of 11 JURONG JUNIOR COLLEGE JC2 Preliminary Exam 2013 H2 Physics Paper 3 solutions Qn Suggested solution Remarks 1(a)(i) & (a)(ii) Applying vy = uy + at and substituting vy = 0 at maximum vertical height, 0 = (20.0 sin 60) + (- 9.81)t t = 1.77 s or t = 17.3 9.81 = 1.77 s (a)(i) [1] v H is horizontal and equal to 10.0 m s -1 (a)(ii) [1] vV is sloping and straight [1] v-intercept = 17.3 m s -1 [1] t-intercept between 1.7 and 1.8 s or gradient = 9.81 m s -2 (b) Object reached maximum vertical height h at t = 1.77 s. maximum vertical height h = area under v-t graph from t = 0 s to t = 1.77 s (= area of blue triangle below) = ½ (1.77)(17.3) = 15.3 m Comments: Many candidates did not use data from the graph (as requested by the question [1] equating area under graph to h [1] ans Note: Max [1] for candidates who did not consider area under graph in arriving at answer vV vH

Page 2 of 11 JURONG JUNIOR COLLEGE JC2 Preliminary Exam 2013 H2 Physics Paper 3 solutions Qn Suggested solution Remarks (2a) Gravitational Potential Energy is the energy possessed due to the relative position of 2 masses whereas Electric Potential Energy is the energy possessed due to the relative position of 2 charges. Comments: Most candidates, who try to state the definition of gravitational and electric potential energy, define them wrongly as “work done per unit mass in moving...” or “work done per unit positive charge in moving…” 1 1 (bi) dm/dt = (dV / dt) = 1000 x 1.4 = 1400 kg s-1 (Shown) Comments: Most candidates did not explain properly the product of 1000 and 1.4. 1 (ii) Rate of change in GPE = (dm/dt)gh = 1400 (9.81)(750) = -10.3 MJ s -1 (Loss) Comments: Most candidates forgot that the change of GPE was a loss and did not put a negative sign. 1 for eqn 1-sub 1 – end answer with negative sign 3(a)(i) U is the increase in internal energy of system q is the thermal energy/heat supplied to system w is the work done on system Comment :Many candidates wrote U as change in internal energy. [1] with all underlined points (a)(ii) Solid which expands on melting Solid which contracts on melting U + (3) +/- (4) q + (2) +/- (2) w - (1) + (1) (1) Expansion implies work done by system and w is negative by definition. Vice versa for contraction. (2) Thermal energy has to be supplied to the system for melting to occur. (3) Expansion implies intermolecular separation larger than the equilibrium intermolecular separation (where potential energy is minimum), resulting in greater potential energy. (4) Apply 1 st law of thermodynamics. Contraction implies intermolecular separation smaller than the equilibrium intermolecular separation (where potential energy is minimum), resulting in greater potential energy. For (3) and (4), random kinetic energy remains constant since melting occurs at constant temperature. Hence internal energy increases. [1] for every two correct signs [3] for all correct

Page 3 of 11 JURONG JUNIOR COLLEGE JC2 Preliminary Exam 2013 H2 Physics Paper 3 solutions Qn Suggested solution Remarks (b)(i) At constant temperature, the random kinetic energy of the gas molecules remains unchanged. The internal energy of the gas which is the sum of the random kinetic energy of its molecules remains unchanged. Comment : Quite a number of candidates did not mention that internal energy of an ideal only depends on the random K.E. Many of they also did not equate internal energy to temperature directly. [2] for correct reasoning and conclusion The part in bold must be mentioned in either (b)(i) or (b)(ii), else max [3] of possible 4 marks (b)(ii) The work done on (compressing) the gas increases the random kinetic energy of its molecules. The internal energy of the gas increases. Comment : Many candidates thought that temperature is unchanged since there is no heat exchange. [2] for correct reasoning and conclusion 4a Newton’s law of gravitation states that the gravitational force between 2 point masses is directly proportional to the product of their masses and inversely proportional to the square of their separation, R. i.e. 2 GMmF R 2/M GMgF m R Where g is defined as force per unit mass and m is a test mass. 1 for Newton’s Law 1 for definition of g as force per unit mass 1 for mentioning m as a test mass Comments: Many students neglected to mention “point masses” while quoting Newton’s law of gravitation. The majority also did not distinguish between M and m (i.e. mention that m is a test mass or small mass). bi 11 30 12 -1 24 2 (6.67 x 10 )(5.2 x 10 ) 1.2 x 10 N kg(1.7 x 10 ) GMg R    1 – sub 1 – answer Comments: Generally well done except for a handful of students who forgot to square the denominator while substituting and hence got an incorrect answer. bii The neutron star is assumed to be a point mass. 1 Comments: Generally well done. Other answers accepted include: star’s radius is assumed to be constant, star is a perfect sphere, star has uniform density etc. biii 22 24 7 - 2 22 (1.7 x 10 ) = 1.52 x 10 m s0.21aR R T           1- Sub 1 - answer

Page 4 of 11 JURONG JUNIOR COLLEGE JC2 Preliminary Exam 2013 H2 Physics Paper 3 solutions Qn Suggested solution Remarks Comments: Generally well done except for a handful of students who forgot the equation for centripetal acceleration. biv On the surface of the star, the gravitational field strength is much greater ( approximately 105 times ) than the centripetal acceleration of the particle. Hence the gravitational force on the particle is sufficient to provide the centripetal force to maintain the particle in circular orbit on the surface of the star. This is why a particle will not leave the surface of the star. 1 – both points must be mentioned. 1 – correct conclusion only with correct explanation Comments: Most students were able to mention the fact that the centripetal acceleration was smaller than the gravitational field strength and hence particles will not leave the surface of the star. However, very few mentioned the crucial detail that because of this, the gravitational force is sufficient to provide the centripetal force to keep the particles in orbit (hence preventing them from flying off). Quite a few misread the question to mean that the particles will leave the surface of the star and thought they were being asked if this was due to the high speed of rotation of the star or not. 5 (ai) Resistivity is the proportionality constant relating the resistance of a circuit component to its length and cross-sectional area. It is a property of the material and is dependent on temperature. Comment : Most candidates did not realize that resistivity is a constant of proportionality … (as stated above), but tried to state that resistivity is proportional to the cross-sectional area, and inversely proportional to the length. Students should realize that resistivity, being a property of a material, is independent of the dimensions of the sample/resistor. Many gave incorrect answers such as “resistivity is the resistance per unit length”, or “measure of the ability to conduct electrical current”, which is “conductivity” not “resistivity”. Students should also differentiate the word “material” from “sample”, “resistor” or “conductor”. 1 mark 1 mark (ii)    6 9 1.50 10 1.2 6362.83 10 lR A         Ans: 1 mark (bi) p.d. across 1.2 m of nichrome wire 636 18 10.1636 500  V p.d. across 0.050 m of nichrome wire 0.050 10.08 0.4211.2 V Comment : A common error is to calculate : Vc = 0.050 18 0.751.2 VV . Another error is to use the potential ratio method, but with an incorrect denominator. Vc = 26.5 18 0.906500 26.5 VV  .