2019 RI Prelims Paper 1 Soln

© Raffles Institution [Turn over 2019 Preliminary Examinations H2 Physics Paper 1 Solutions 1 C 33 344volume (12) 7200 cm33 r  2 C Let t be the time from release of ball to mo ment of impact with floor of elevator, s be the distance travelled by elevator during time t and u be the initial speed of both ball and elevator at point of release. 2 2 22 22 1 (3.5)2 12 (9.81)2 11(3.5) 2 (9.81)22 112 (9.81) (3.5)22 0.796 s su t t su t t ut t ut t tt t         3 D Constant force (resultant) implies constant acceleration 22 2 2 02 2 vu a s va d va d    Since p = mv, 2p ma d 4 C  cos20 cos50 cos50 0.68404cos20 sin20 sin50 0.68404 sin20 sin50 5.0 9.81 49 N AB A B ABC BB B TT T T TTT m g TT m g Tm g             Since the force triangle is isosceles, TB = 5.0 g = 49 N 5 C difference in vertical height, h = x sin  Phelium = P + gh = P + xg sin  6 C     1000 110110350 9.81 2700 W3.0 60 3.0 60 RFhmghP tt      7 D Upon release, the mass will fall vertically downwards, stretching the spring. The mass will undergo oscillations losing GPE and gaining KE and EPE before coming to rest due to air resistance. Hence at any point during oscillation, the loss in gravitational potential energy = gain in elastic potential energy + gain in kinetic energy + energy dissipated. However, at its maximum displacement below its equilibrium position, kinetic energy is zero. Hence only option D is correct. 20  70  70  OR TA TB 5.0 g

2 © Raffles Institution 8 B 1150 2 0.26 m s23 0 6 0vr            9 B    2 22 700 0.5985000 9.81 c mv F vr Wm gr g   10 A 22 '6 4 0 0 0.94'6 6 0 0 gr gr         Hence, g’ is 6% less than g. Alternatively, one can use 2002 2 0.0625 6400 gr gr           . So the difference is about 6%. 11 D The rotation of the Earth results in the acceleration of free fall being smaller than the gravitational field strength. 12 A Total number of moles of gas is constant. Since PV = nRT, we take the initial number of moles in the smaller (A) and larger bulb (B) to be n and 8 n respectively. Total number of moles = 9n. At new equilibrium, the pressure will be the same for both. ' ' (80 273.15) (8) (9 ') (9 ') (10 273.15) fA fB PV n RT n R P V nn R T nn R      Dividing: '(353.15) 8 (9 ')(283.15) 9 ' 8 353.15 ' 283.15 9 8 353.15 1' 283.15 '0 . 8 1 9 4 0.8194 0.18 Vn Vn n nn n n n nn nn n n         13 C pp c a l LL (150 25 ) (0.250)(2130) (21.7)( ) (0.400)(130)(125 ) (532.5) (21.7)( ) 52(125 ) 10.7 C mc C mc                        14 D At t = 0.4 s, a = 0 and v = vmax = xo = -12 0.157 m soxT  

3 © Raffles Institution [Turn over 15 B The vertical displacement is given by cosyA t  . At 1 2,ty A