(ACJC) 2024 J2 H2 Physics Prelim P3 Solutions

Anglo-Chinese Junior College 2024 H2 Preliminary Exam Paper 3 Guide H2 (9749) Physics JC2 2024 Page 1 of 9 Qn Suggested Answer 1 (a)(i) The linear momentum of a body is defined as the product of its mass and its velocity and it is in the direction of the velocity. (a)(ii) The total final momentum of a system after a collision is equal to total initial momentum of the system before the collision provided no net external force acts on the system. (b)(i) Applying the principle of conservation of momentum, 0 2.0 1.0 2.0 (1) A A B B AB BA m v m v vv vv += =− =− Applying law of conservation of energy, 2211 12 (2)22 A A B Bm v m v+= Substitute equation (1) into equation (2), 22 22 1 11(2.0) (1.0)(2.0 ) 1222 2.0 12 2.0 m s AA AA A vv vv v − += += = 14.0 m sBv −= (b)(ii) A and B are moving towards each other. At any instant of time, the speed of B is twice that of A, distance covered by B is twice that of A when they collide. xA + xB = 0.90 xA + 2xA = 0.90 xA = 0.30 m 2 (a) An ideal gas obeys the equation pV = nRT where p is the gas pressure, V is the volume of the gas, n is the number of moles of gas, R is the universal gas constant, and T is the thermodynamic temperature of the gas. There are no intermolecular forces between the gas molecules, therefore an ideal gas has no potential energy. (b)(i) 54 3 23 4.8 10 2.3 10 100 (65 273.15) 1.38 10 PV Nkt PVN kT − = =   = +   242.37 10 (3 s.f.)= (b)(ii) 23 24 3 2 3 (1.38 10 )(65 273.15)(2.37 10 )2 E NkT − = =  +  41.66 10 J (3 s.f.)=

Anglo-Chinese Junior College 2024 H2 Preliminary Exam Paper 3 Guide H2 (9749) Physics JC2 2024 Page 2 of 9 (b)(iii) 2 2 65 75 13 22 Since and are constant, (65 273.15) (75 273.15) rms rms rms rms mc kT mk cT c c =  += + 0.986 (3 s.f.)= (c)(i) The product of pressure and volume of the gas at B and C are the same. (c)(ii) process work done on the gas /  104 J heat supplied to the gas /  104 J increase in internal energy /  104 J A → B −1.92 4.80 2.88 B → C 3.05 −3.05 0 C → A 0 −2.88 −2.88 54 3 4 () 4.8 10 (6.3 2.3) 10 100 1.92 10 J ABWD p V=−   − =− =−  3 (a) As the tube is partially submerged, there is upthrust. The weight and upthrust are equal in magnitude and acts in opposite directions. Hence, the resultant force is zero. (b) At equilibrium, weight of the tube and sand upthrust Mg Ahg = = When further displaced and taking downwards as positive, ( )Mg A h x g Ma− + = Ahg Ahg Axg Ma Axg Ma Agax M    − − = −= =−