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 temperat