RVHS_H2_Physics_P1_Soln

River Valley High School Pg 1 of 6 H2 Physics 9749 JC2 Preliminary Examinations 2022 Solutions to J2 Preliminary Examination Paper 1 1 A 11 C 21 D 2 D 12 A 22 C 3 B 13 B 23 B 4 D 14 D 24 A 5 A 15 C 25 A 6 C 16 C 26 A 7 B 17 D 27 C 8 C 18 B 28 B 9 B 19 A 29 C 10 D 20 B 30 D

River Valley High School Pg 2 of 6 H2 Physics 9749 JC2 Preliminary Examinations 2022 Qn Key guide 1. A Considering units of the equation, we have s2 = kg m2 units of K , so units of K = kg m2 s-2 = (kg m s-2) m = N m 2. D Since air resistance is negligible, the gradient of both graphs will be the same. From the v-t graph, it can be seen that at any time, the difference in the area under graph for X and Y will be increasing hence the separation will be increasing. 3. B Note that the carriages and the engine have a common acceleration, say a. Let the mass of each carriage be m. Consider both carriages, Newton’s 2nd law gives 𝑇 = (2𝑚)𝑎, so 𝑎 = 𝑇 2𝑚 Consider the back carriage. Tension pulling the back carriage is then 𝑚𝑎 = 𝑚 ( 𝑇 2𝑚) = 𝑇 2 4. D During acceleration, 𝑊 − 𝑚𝑔 = 𝑚𝑎, so 𝑊 = 𝑚(9.81 + 2.0) During deceleration, 𝑚𝑔 − 𝑊′ = 𝑚𝑎, so 𝑊′ = 𝑚(9.81 − 2.0) So 𝑊′ 𝑊 = 9.81−2.0 9.81+2.0 = 0.661 5. A Since the same load is applied, for X, if extension is L, then extension for Y will be 2L For Z, the extension will be the greatest as the top two springs will stretch by the same amount of L while the lower spring will have twice the extension due to the weight being supported by only one spring 6. C Work done = 120 cos 37o x 5.0 = 479 = 480 W 7 B Initially, when no force is applied to the piston, force on piston by air in container = force on piston by air in atmosphere = pressure x area = ( )( ) 33100 10 3.5 10 − = 350 N ( )( )( ) 3 3 3100 10 3.5 10 80 10 T −−   = ( )( )( ) 33new pressure 3.5 10 160 10 0.5T −− new pressure = 25 kPa new force on piston by air in container = pressure x area = ( )( ) 3325 10 3.5 10 − = 87.5 N v t x y

River Valley High School Pg 3 of 6 H2 Physics 9749 JC2 Preliminary Examinations 2022 For piston to remain stationary, Force on piston by air in container + F = Force on piston by air in atmosphere 87.5 + F = 350 F = 260 N (262.5) 8 C Let X be the gas compressed isothermally (no change in temp). There is heat exchange between the gas and the surroundings. Let Y be the gas compressed and is isothermally isolated from surroundings. There is no heat exchange between gas and the surroundings. For X: Since temperature does not change due to heat exchange between gas and the surroundings, p1V = p2(0.5V), p2 = 2 p1 For Y: Using pV = nRT, when V decreases, pressure increases and leads to larger speed of the molecules. This leads to higher temperature of gas. A: No heat is given to both gases during compression. B: Internal energy of Y is higher due to higher temperature of gas