2021 JPJC Prelim H2 Physics P2 Solutions

2021/JPJC/Prelim/9749/02 [Turn over Answers to 2021 JC2 Preliminary Examination Paper 2 (H2 Physics) Suggested Solutions: No. Solution Remarks 1(a)   2 sin 9.81 sin30 4.905 4.9 m s (shown) ag       [1] correct working shown 1(b)    22 2 1 sin30 0.50 1.0 msin30 2 2 4.905 1.0 3.13 m s h d d v u as v v        [1] for correct d [1] correct equation and substitution [1] correct answer 1(c)       2 2 1 2 12.0 3.13sin30 9.81 2 0.50 s horizontal distance 3.13cos30 0.50 1.35 m      s ut at tt t [1] for correct equation and substitution [1] for correct t [1] for correct distance 1(d)(i) In the system of object and trolley, there are no external forces acting on them in the horizontal direction. The only horizontal forces are contact forces (action-reaction pair) acting on each other when object hits the trolley. Hence, total momentum in t he horizontal direction remains constant (or is conserved). [1] for explanation [1] for conclusion 1(d)(ii)         1 1 Applying conservation of momentum, rightward as +ve, 0.35 3.13cos30 1.2 4.0 1.2 0.35 2.48 m s final speed 2.48 m s Direction is to the left f f v v          [1] for correct equation and substitution [1] for correct answer [1] for correct direction 2(a) From the given expression 21 3pc  , we have 21 3 Nmpc V ----- (1) (where m is the mass of 1 molecule) Using the ideal gas equation pV NkT , we have NkTp V ------ (2)

2 2021/JPJC/Prelim/9749/02 No. Solution Remarks Equating (1) and (2), we have 21 3 Nm NkTcVV  21 3 m c kT 213 22m c kT Since k is the Boltzmann constant, the average kinetic energy 21 2 mc of a molecule of mass m, is directly proportional to the thermodynamics temperature T. (shown) [1] for equating (1) and (2) [1] for correct derivation of expression [1] for the correct conclusion 2(b)(i) Total pressure,     total atm 35 5 5 15 1.03 10 9.81 1.01 10 2.526 10 2.53 10 Pa p h g p        [1] for correct method and numerical substitution [1] for correct answer 2(b)(ii) Using pV nRT ,    2 2 1 1water tank 1 21 2 7 3 5 5 53 2.32 10 9.4 10 2.526 10 8.633 10 8.63 10 cm p V pV pVV p         [1] for correct expression and numerical substitution [1] for correct answer 2(c) Pressure at depth of 35 m,     2 atm 35 5 35 1.03 10 9.81 1.01 10 4.547 10 Pa p h g p      Volume of air at depth of 35 m and temperature of 19 °C, 2 2 1 1 21 35m 15m 12 21 21 5 5 5 53 2.526 10 19 273.15 8.633 10 24 273.154.547 10 4.715 10 cm p V pV TT pTVV pT                      [1] for correct method and numerical substitution [1] for correct volume

3 2021/JPJC/Prelim/9749/02 [Turn over No. Solu