TMJC_2025_H2 Physics_P2_Soln

1 Tampines Meridian Junior College 2025 JC2 Preliminary Examination H2 Physics Tampines Meridian Junior College 2025 JC2 H2 Physics Preliminary Examination Paper 2 Suggested Solution 1 (a) -1 -1 -2 -1 2 [ ][ ][] [] m s (C) [M1]N m s (A s) kg m s A kg s [A1] ve F ve F   = = = = = −1 if use C and/or N −1 if poor presentation (b) (i) Zero error or wrongly calibrated scale [B1] (ii) Reading scale from different angles [B1] Must explain if write parallax error. (c) (i) −==  = 3 3.00 [B1]4.9 10 612 X VR I (ii)  =+  =+  =   =   0.03 0.1 612 3.00 4.9 20 [M1] 610 20 [A1] X X X X XX R V RV R R RR I I (d) When plotting a graph, a best fit line shows the average trend of the data points, thus reducing random error present. Those readings which are too far out from the graph are not taken into account so the results will be more reliable. [B1] OR By using multiple sets of values and plotting them on a graph, any consistent deviation from the expected trend can be observed. If all points lie away from the expected line or curve in a similar manner, this may indicate the presence of a systematic error. OR When several data points are collected and plotted, it becomes easier to identify anomalous readings (readings that deviate significantly from the

2 Tampines Meridian Junior College 2025 JC2 Preliminary Examination H2 Physics general pattern or trend). These readings are not taken into account so the mean value of resistance of X will be closer to the actual value.. 2 (a) At top of trajectory, vertical component of velocity is zero and horizontal component of velocity is constant, hence it will have minimum KE [B1] or At top of trajectory, gravitational potential energy of object is greatest hence kinetic energy is the lowest since only 2 forms of energies are involved. [B1] (b) Min KE is when object is at the highest point with only horizontal velocity. ½ m vx2 = 12.5 J vx = ux = = -112.5 10 ms [M1]0.5 0.25 = = -1 cos40 10 [M1] 13 ms [A0] ou u (c) (i) =+ − = + − =− = 2 2 1 2 taking upwards as positive, 1100 13sin40 ( 9.81) [C1 ]2 3.74 s (rej) or 5.45 s [A1] yy o s u t at tt tt (ii) = = = 10(5.45) 55 m [A1]xxs u t allow ecf (iii)  − − − =+ = + − =− = + = = =  1 2 2 1 1 taking upwards as positive, 13sin40 ( 9.81)(5.45) 45.1 m s [C1] (45.1) (10) 46 m s [A1] 45.1tan ( ) 77 (clockwise) below horizontal [ A1]10 yy y v u at v v

3 Tampines Meridian Junior College 2025 JC2 Preliminary Examination H2 Physics 3 (a) It is due to the difference in pressure between the bottom surface and top surface. [B1] (b)(i)   B1oo W mg Vg = = (b)(ii)   B