2013 JJC H2 Physics P1 solutions

Page 1 of 7 JURONG JUNIOR COLLEGE PHYSICS DEPARMENT JC2 Prelim Exam 2013 H2 Physics Paper 1 solutions Qn Ans Suggested solution (Font Arial, 11 pt) 1 C         2 12 sin kg m s kg A sA m FB I L FFB I L B IL         2 C The changes in horizontal and vertical components of the ball’s velocity means that the force acting on the ball by the horizontal surface has components in the x and +y directions. 3 D Taking downwards and to the right as positive, ux = 5.0 cos 30 and uy = 5.0 sin 30 Applying v2 = u2 +2as to the vertical component of motion just before the ball hit the floor, v2 = (5.0 sin 30)2 + 2(9.81)(2.5) = 55.3 Speed just before ball hit floor = = 8.61 m s-1 4 A 1F7 5 2 5Fm a a 2 m s m (10 15)      For the 10kg mass, Resultant force=10(2)=20 N 5 D if 12 12 2 1 11 1 pp m(u) m( v ) M(v ).................(1) u v v ......................(2) v u v Subst.(2) (1) m(u) m( v ) M(u v ). (Mv Byconservationof momentum, Relativespeedof approach = Relativespeedof separation into         m) u(m M) 6 C By definition.

Page 2 of 7 JURONG JUNIOR COLLEGE PHYSICS DEPARMENT JC2 Prelim Exam 2013 H2 Physics Paper 1 solutions Qn Ans Suggested solution (Font Arial, 11 pt) 7 D The 3 forces acting on the rod must must all act through the same point. Since there is an upward component provided by F, there should be a upward component provided by the unknown force. Thus the force on the rod by the wall should be pointing in the option A direction. By Newton’s 3rd Law, the force acting on the wall by the rod should be equal in magnitude and opposite in direction, therefore the ans is option D. 8 C Since the car is moving at constant speed, forward force = frictional force WD by friction = Fd = 3800 x 300 = 1140 kJ 9 C KE = ½ mv2 = ½ m(u2 + 2as) = mas since u = 0 This means that KE is directly proportional to displacement (linearly related) Consider when x=0 (At ground level), KE will be maximum Consider when x=H (At top of tower), KE will be zero Hence graph of KE vs x is best represented by C 10 D Consider the forces acting on the body of 1 kg at top T top + mg = mv2/r Consider the forces acting on the body of 1 kg at the bottom T bottom – mg = mv2 /r Note: v is the same when body is at top and bottom. Hence T top + mg = Tbottom – mg Difference in tension = 2mg = 2(1)(10) = 20 N 11 B T sin = mv2/r -----------(1) T cos = mg ------------(2) (1)/(2) tan  = v2/rg (1)2+(2)2 T2 = m2g2+ m2v4 / r2

Page 3 of 7 JURONG JUNIOR COLLEGE PHYSICS DEPARMENT JC2 Prelim Exam 2013 H2 Physics Paper 1 solutions Qn Ans Suggested solution (Font Arial, 11 pt) 12 C A is wrong as the geostationary satellite must be at a fixed distance above the Earth equator. B is wrong as linear speed is proportional to the distance away from the centre of the earth (v=r) hence the speed of the satellite can never be the same as the speed on the equator. D is wrong as the earth is rotating from west to east hence the satellite must follow the same direction. 13 C Gravitational potential at X = -8 kJ kg-1. Hence gravitational potential at Y is -4 kJ kg-1 as gravitational potential is inversely proportional to r. Hence change in GPE = m x change in gravitational potential = (2)(final gravitational potential – initial gravitational potential) = (2)(-4 – (-8)) = + 8kJ 14 B Across the broken filament, the p.d is equal to the voltage source at the mains so filament 5 is broken. Since no current flows through the circuit, there is no p.d across the lamps in which the filaments are not broken. 15 B Current in 3.0  resistor:   22 2 12 3 4 2PI R I I I      A Voltage across 3.0  resistor:     23 6VI R V  V Voltage across internal resistance r = 10 – 6 = 4 V Current through internal resistance r = 4 A    44 1 . 0VI R r r    16 B Does not obey Ohm’s law as it is not a straight line passing through origin. When V > 1.8 V, its resistance is not constant. 17 D When resistance of LDR is 200 , p.d across LDR 750 40 32200 750  V When resistance of LDR is 2000 , p.d across LDR 750 40 112000 750  V 18 B Solid Melting point/ C Specific heat capacity/ J kg-1 K-1 ∆T from 20 °C to melting pt Energy required for ∆T/ kJ A 80 1200 60 72 m B 100 800 80 64 m C 150 600 130 78 m D 300 250 280 70 m Hence solid B will melt first.

Page 4 of 7 JURONG JUNIOR COLLEGE PHYSICS DEPARMENT JC2 Prelim Exam 2013 H2 Physics Paper 1 solutions Qn Ans Suggested solution (Font Arial, 11 pt) 19 D A: Density decreases because the same mass of gas occupies a larger volume. B: Larger volume implies smaller frequency of collision of gas molecules with walls of syringe. C: Apply first law of Thermodynamics. Q is assumed zero because piston is drawn outwards quickly. W is negative as work is done by gas. Hence U is negative, implying a decrease in the temperature of gas and root-mean-square speed of the atoms. 20 C Car suspension system is an example of critical damping. 21 B The P.E against displacement graph will follow U shape graph as P.E= 221 2 mx 22 C Only transverse waves can be polarized, but not longitudinal waves. (Some sunglasses have polarized lens, i.e. the lens are actually a form of polarizing filters and block out certain orientations of light.) Option A: Reason for this is due to the diffraction of sound waves but not light as the waves encounter the obstacle (the corner). Option B: Reason being light travels faster than sound. Option D: Reason being light may be considered as consisting of photons each with energy hf. 23 B Based on the waveforms at time zero and using the Principle of Superposition, the resulting stationary wave pattern will start off at position Y. The resultant of the the subsequent paths of P and Q will give you wave X. So the order will be Y X Y Z. Note: In every cycle, the stationary wave pattern will become perfectly flat (position Y) twice. 24 B For zero intensity to occur at the light sensor, the polaroids’ axes must be perpendicular to each other (for eg. with Polaroid A’s polarizing axis perfectly vertical and Polaroid B’s polarizing axis perfectly horizontal, as shown in diagram below). By placing a third polaroid C with its polarizing axis at an angle, in between Polaroids A and B, a small amount of light that has passed through Polaroid A will also be able to pass through Polaroid C. Likewise, a small component of the light that is able to pass through Polaroid C will also be able to pass through Polaroid B. Polaroid C

Page 5 of 7 JURONG JUNIOR COLLEGE PHYSICS DEPARMENT JC2 Prelim Exam 2013 H2 Physics Paper 1 solutions Qn Ans Suggested solution (Font Arial, 11 pt) 25 D An antinode would be formed at the open end of the tube while a node would be formed at the closed end of the tube. Hence, 17 cm = distance between node and adjacent antinode = /4   = 68 cm 26 D Electric field strength, E  2 1 x , Electric potential, V  1 x 27 A Electric field lines are represented as directed lines from high to low potentials. The negative charge on the top plate will induce positive charge on the surface of the bottom plate by repelling electrons to the earth. 28 A     m()TTB q . / 22        A is right as Bqv mv Bqv mv B is wrong as the speed of the electron remains constant once it is inside the magnetic field C is wrong as r mv Bq D is wrong as F BqV 29 C torque=(NBIL)x=40 0.010 0.0050 0.0080 0.0160 2.6 10 7 Nm     30 C Recall that magnetic flux density, B = Φ/A Hence, the smallest cross-sectional area will give the largest variation in magnetic flux density for the same amount of magnetic flux (magnetic flux in concentrated within the soft- iron ring). Points to note: At any time the magnetic