2013_IJC_H2_Physics_P2 Solutions

2013 H2 Physics Prelim 2 Exam – Paper 2 Solutions  1  (a)  (i)  VH = 13 cos 50o = 8.36 m s1. (Horizontal line around 8.5)     [C1]         (ii)  VV = 13 sin 50o = 9.96 m s1. (straight line passing through coordinates (0, 9.96)  & (1.02, 0)   Using v = u + a t,  0 = 9.96 + (9.81) t,  t = 1.02 s  [C1 for negative gradient straight line graph], [C1 for acceptable coordinates  points on graph]      (b)   Maximum height is the area under the straight line graph,     s = ½ (9.96)(1.02) = 5.06 m       [A1]  Accept answers between 4.96 and 5.16 m      (c)   The value in (b) will be smaller [A1] due to a smaller value of gravitational  potential energy as there will be work done against air resistance. [M1]    Or, the resultant downwards acceleration will be higher, stone will not be able  to travel as high.             2  (a)   gravitational force provides the centripetal force [B1] GMm / r2 = m r ω2 (must be in terms of ω) [B1] T2 = (42 / GM) r3 = and (42 / GM) is a constant [B1]       (b)  (i)  (9.39 × 106)3 / (7.65)2 = (1.99 × 107)3 / T2 [C1] T = 23.6 hours [A1]      (ii)  Almost ‘geostationary’ or satellite would take a long time to cross the sky or the moon has the same angular speed as the rotation of the planet. [B1]      (c)  (i)  1.  12 2 12() GM M RR [A1]          2. M1R12. [A1]        (ii)  M1 has a slower speed due to its smaller orbital radius and hence has a smaller acceleration. [M1] By Newton’s second law, for the same gravitational force between M1 and M2, [M1] M1 must be of a bigger mass. [A0] Alternatively, since both masses experience the same gravitational force (N3L)  same centripetal force  since R1 is smaller than R2, M1 has to be larger than M2.             3  (a)   Any increase in the internal energy of a system is the sum [B1] of the heat supplied to the system and the work done on the system. [B1] The internal energy of a system depends only on its state. (bonus mark)     (b)  (i)  11 1 11 2 p VT p VT 11 11 12 12 p Tp Vor orp Tp V or calculate n = 0.053 and substitutes in pV = nRT

555 to 580 K (567 K) depending on data used from the graph    (ii)  Attempt to find area enclosed. Number of squares = 80  6 small squares (3 to 3.4 large squares) Energy per square = 0.50 J Work done ON the gas = 40 J ( 3 J)      (iii)  increase in internal energy = 0 J [B1] Net work done on the gas = + 40 J By first law of thermodynamics,Q net = -wnet = -40 J [A1]            4  (a)  (i)       (ii)       (iii)        (b)   Speed of the 8M mass = squareroot [ ½ M (5)2 / (8M/2) ] = 1.776 m s-1 [M1] Since this speed is less than the min speed required in (a)(iii), [M1] the 8M mass will not reach point C. [A1]              5  (a)  (i)  When the magnet is approaching the coil, there is an increase in magnetic flux threading through the coil. When the magnet is leaving the coil, there is a decrease in the magnetic flux threading through the coil. [