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)   B1ofU V g = (c) In order to sink,          M1 A1 o o o f of WU V g V g (d) As the (downward) velocity of the object increases, the (upward) drag force (or fluid resistance) is increasing. [B1] Hence the downward acceleration (or net force) decreases. [B1] (also accept: hence the velocity increases at a decreasing rate) When the (sum of) drag force and upthrust is equal to the weight [B1] The net force is zero and acceleration is zero, resulting in terminal velocity.

4 Tampines Meridian Junior College 2025 JC2 Preliminary Examination H2 Physics 4 (a) Consider the top most position. When the angular speed is just enough for person to stay in contact with the wall , person is about to lose contact with the wall. As such, normal contact force = 0 at top most position Component of weight parallel to floor of cylinder provides for the centripetal force [B1] ( ) ( ) 2 2 1 sin = mr 9.81 sin50 8.0 [M1] 0.97 rad s [A0] mg    − = = (b) At the bottom most position, The resultant of the normal contact force from wall and component of weight parallel to floor provides for the centripetal force. ( )( )( ) ( )( ) 2 2 sin = mr [C1] 65 8.0 0.97 + 65 9.81 sin50 980 N [A1] wall wall wall N mg N N − = = (c) ( )( )=    = Gain in gravitational potential energy 65 9.81 2 8.0 sin50 [C1] 7800 J [A1] (d) Average power work done= time gain in gravitational potential energy= time 7800= [C1]12 2 0.97 2400 W [A1]   = Allow ecf from (c)

5 Tampines Meridian Junior College 2025 JC2 Preliminary Examination H2 Physics 5 (a) Simple harmonic motion is defined as the motion of a particle about a fixed point such that its acceleration is proportional to its displacement from the fixed point and is directed towards the point (or opposite in direction to the displacement). [B1] (b) 2 2 2 2 -------- (1) 1 -------- (2) [C1 for both (1) & (2)]2 sub (1) into (2) 2 2 0.72 [M1]2(0.86) 0.30 kg [A0] K K K p mv E mv pE m pm E = = = = = = (c) (i)  = = = rod pendulum 1 2 [M1]0.55 9.81 0.82 m [A1] TT L L (ii) (2 ) 0.50 (2 )1.82 1.73 rad [B1] t T  = = = (d) (i) Damping due to viscous forces [B1] (ii) ( ) 22 total 2 2 2 22 2 22 4 4 1 2 12 2 12 2 12(0.30) (0.016 0.020 ) [M1]2 1.05 2 1.93 10 J decrease in 1.93 10 J fi o o f i o o E m x mx T E E m x x T E     − − = =  − = −  =−   =−  = [A1]

6 Tampines Meridian Junior College 2025 JC2 Preliminary Examination H2 Physics 6 (a) From graph, resistance = 1.85  (b) Effective resistance of the 4 components: 1 terminal 11 3.111 [C1 for working]2.5 3.6 1.85 4.5 Hence, terminal p.d.: 3.111 6.0 [M1]3.111 0.70 4.9 V [A0] effR V − = + =  ++ = + = Alternative: ( )( ) = = = + =− =− = terminal 6.0 1.5744 A3.111 0.70 6.0 1.5744 0.70 [M1] 4.9 V total V R V E r I I (c) = + = + = terminal 1.85 4.5 4.9 [M1]1.85 4.5 0.771 A [A1] VI (d) ( )( ) ( ) = = = == + thermistor thermistor 0.771 1.85 1.43 V [B1] 1.85 4.9 1.43 V [B1]1.85 4.5 TVR OR V I (d) For the galvanometer to read zero, the potentials at C and D must be the same (i.e. no p.d. between C and D). ( ) ( ) thermistor thermistor For p.d. across CD to be zero, 2.5 4.9 2.008 V [C1 for working to show ]2.5 3.6 1.43 1.0 0.71 m [A1]2.008 AC AB AC AC AC AB AB AC AC AB AB VV V V V VL VL VLL V = = = = + = =  = = OR

7 Tampines Meridian Junior College 2025 JC2 Preliminary Examination H2 Physics ( ) ( ) thermistor thermistor For p.d. across CD to be zero, 4.9 1.43 [C1 for working to show ]2.5 3.6 1.776 1.776 1.0 0.71 m [A1]2.5 AC AC AC AC AC VV R VV R L =  == + = == 7 (a) (i) Number of protons = 57, number of neutrons = (141 – 57) = 84 [B1] (ii)   ( ) ( )( )   − − − − = − + − = = = =   = = = 27 27 2 227 8 10 Mass defect, (141 57)(1.009) 57(1.007) 140.9 11 [M1] 1.244 1.66 10 2.06504 10 kg Binding energy 2.06504 10 3.00 10 [M1] 1.858536 10 J [M1] 1161.585 MeV 1162 m mc MeV [A0] (iii) 1807B.E. per nucleon of Pu-239 239 7.56 MeV 1162B.E. per nucleon of La-141 141 8.24 MeV [M1: for both energies] = = = = Since plutonium-239 nucleus has a lower binding energy per nucleon, it is less stable against fission [B1] and is more likely to undergo nuclear fission. [A1] (b) (i) 141 141 0 57 58 1La Ce −→+ Electron or beta particle [B1] (ii) There is emission of a neutrino (or anti-neutrino, or third particle) along with an electron. [M1] The energy/momentum is shared between