(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    − − = −= =−  (c) Since , A, g and M are constants, acceleration is directly proportional to displacement. The negative sign shows that acceleration is in the opposite direction to displacement.

Anglo-Chinese Junior College 2024 H2 Preliminary Exam Paper 3 Guide H2 (9749) Physics JC2 2024 Page 3 of 9 (d) 2 Ag M  = ( )( )( ) ( ) 2 34 23 2 4 1.2 10 5.5 10 9.81 4 120 10 Ag TM Agf M f     − − = =  =  1.2 Hz (2 s.f.)= (e)(i) Correct shape. (e)(ii) Spiral inwards clockwise starting from positive maximum displacement (downwards positive). 4 (a)(i) A polarised wave is a wave whose vibrations are restricted to one plane. (a)(ii) 2 10 Apply Malus' Law, (cos40 )=II 22 20 (cos40 ) (cos80 ) 0.018 (2 s.f.) =   = II I (b)(i)1. Displacement of wave laterally (left and right) from the central axis of the tube represents vibrations of air molecules vertically (up and down) along the axis of the tube. The vibrations are parallel to the direction of propagation of the incident wave down the tube. v x

Anglo-Chinese Junior College 2024 H2 Preliminary Exam Paper 3 Guide H2 (9749) Physics JC2 2024 Page 4 of 9 (b)(i)2. Correct stationary wave pattern. (b)(i)3. Correct number of nodes and location of nodes. (b)(ii) 1 35.0 cm2 70.0 cm   = = 340 0.700 486 Hz (3 s.f.) f = = (c)(i) 9 5 sin 1 sin90 (540 10 ) 5.00 10 dn n  − = =   3.70 3 (round down to nearest whole number) n= = Total number of maxima (3 2) 1 7 =  + = (c)(ii) 9 5 sin 1 sin 2(542 10 ) 5.00 10 32.820 dn    − = =  = 9 5 sin 1 sin 2(539 10 ) 5.00 10 32.616 dn    − = =  = Angular separation is 0.204° Since angular separation is less than 0.30°, the two wavelengths cannot be distinguished at the second order. 5 (a) 6 4 Electric force 20000.78 10 5.0 3.12 10 N Vq d − − = = = Direction of electric force is rightwards towards Q. N N ½ λ ¼ λ

Anglo-Chinese Junior College 2024 H2 Preliminary Exam Paper 3 Guide H2 (9749) Physics JC2 2024 Page 5 of 9 (b)(i) Forces must be labeled clearly. (b)(ii) The weight of particle X has the same order of magnitude as the electric force. (c) 4 -2 4 Horizontally, 3.12 105.0, 0, 6.25 m s 4.9 10 9.81 x x xs u a − − = = = =  22 -1 Using 2 , 7.91 m s xx xx x v u a s v =+ = 2 -1 1Using , 2 1.26 s Vertically, Using 23 (9.81)(1.26) 35.4 m s xx Yy Y s a t t v u a t v = = =+ =+ = 22 -1 Magnitude of resultant velocity 7.91 35.4 36.3 m s (3 s.f.) =+ = (d) Since electric force and weight are of constant magnitudes and directions, the resultant force would also be of constant magnitude and direction. As the velocity has a component along the resultant force and a component perpendicular to it, the path taken by the particle would be parabolic. (e) V = 2000 V 6 3 Change in EPE ( 0.78 10 )(1000 ( 1000)) 1.56 10 J qV − − = = −  − − =−  6 (a) Force per unit length acting on a straight, current-carrying conductor, carrying unit current and placed at right angles to an external magnetic field. u electric force weight resultant force

Anglo-Chinese Junior College 2024 H2 Preliminary Exam Paper 3 Guide H2 (9749) Physics JC2 2024 Page 6 of 9 (b)(i) 74 10 80 2 0.80B   − =  7 19 3 sin90 4 10 80 (1.6 10 )(1.0 10 )2 0.80 F Bqv   − − =  =    213.2 10 N−= (b)(ii) Force drawn perpendicularly to the tangent of the path at Q. (b)(iii) At P, the ion will experience a magnetic force to the right by FLHR. This will make the ion move in a circle. As ion moves nearer the cable, magnetic flux density due to the cable increases ( B α 1/r), increasing the magnetic force (F = Bqv). Thus, the radius of circular path becomes smaller. Eventually there will be a point where the ion is moving parallel to the cable. The magnetic force will then cause the ion to turn back, with the radius of the circular path becoming larger as the ion moves further away from the cable. (c)(i)1. The magnetic field of the magnet is not uniform and the magnet’s velocity