RI 9646 2013

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 1 2013 A-Level H2 Physics Suggested Solutions Paper 1 1 A The scattering of data points around the best fit line suggests that the results are imprecise. Using 2212 t2s ut at sg= + ⇒= and value of gradient, g = 9.804 m s−2 (close to 9.81 m s−2) and is accurate. 2 C Normal A4 paper is 80 grams per square meter (gsm). Area of A4 paper is 21.0 cm by 29.7 cm. (Use ruler to measure exam question paper). Mass of a single sheet = 80 × 0.210 × 0.297 = 5 g 3 A 22 22 2 22 3 to 4 11 22 1112 ( 2) (1)22 8.00 m s 11 (4) (3) 28 m22 s ut at s at aa a saa − = + ⇒= = − = =−= 4 D Velocity of stone passing edge of cliff on its way down is 10 m s−1 downwards 2211 (10)(1.2) (9.81 )(1.2) 19.1 m22s ut at= += + = 5 C For elastic collision, relative speed of approach = relative speed of separation, u1 − u2 = v2 − v1. Only option C is correct. 6 B 2 m Ax Avtt p mvF Avtt ρ ρ ρ ∆∆ = =∆∆ ∆∆= = =∆∆ 7 A pressure due to fluid (600)(9.81)(0.200) 1177 Paghρ= = = 8 D For equilibrium, the third force must form a closed vector triangle with the other two forces. Hence, it must act in the Z direction. 9 C 1001400 9.81 1.6 109872 W0.20 20 FvP = = × ×× = 10 A The sum of the three types of energies must be 120 J. Option B is wrong because kinetic energy at bottom must be 0. Option C is wrong because the total energy in middle is not conserved. Option D is wrong because kinetic en ergy throughout the motion cannot be constantly at 0. 11 B There are only two forces acting on the ball, the weight of the ball and the tension in the thread. The resultant of these two forces provides the required centripetal force for the ball to move in a circle.

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 2 12 D 2 13 7 2 72 14 2 1 5.0 102 2.447 10 (2.447 10 ) 9.98 10 m s0.6 c mv v va r − − = × = × ×= = = × 13 A 2 2 2 64when 2 , 16 2 64when 4 , 4 4 GMg r rg rg = = = = = = = 14 A Centripetal force is provided by gravitational force 2 22 At P, 11At Q, 99(3 ) Q GMa r GM GMaa rr = = = = 15 D GPE is linear as it is proportional to height. KE is an inverted parabola with a value of zero at the equilibrium position. EPE is a curve with increasing values as extension increases (displacement downwards). 16 C 1 max 0 2 (75)(0.01) 4.71 m svx ωπ −= = = 17 C ' 5 and ' 2 2 0.45 0.4 0.6 final escape PVn RT TTPP VPnn RT n nnn = →→ = =  = −= 18 C 11 22 12 1 2 12 12 Pm h Pm h PP m m PP mm = + = + −= − −= − l l ll l 19 B 2 0 max 0 kx vx ω= I= Thus maximum speed of air molecules increases. Speed of wave is independent of intensity and is dependent on temperature and mass of a molecule in the medium.

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 3 20 A The crests are locations of nodes of the stationary wave, which have no amplitude and has no effect on the powder. 2 0.2 m 330 3300 Hz0.1 vf f λ