IJC H1 PHY P1 Solution

Innova JC 2016 Prelim H1 Physics Paper 1 Q Ans Q Ans Q Ans 1 A 11 D 21 B 2 B 12 D 22 C 3 B 13 C 23 A 4 A 14 B 24 C 5 A 15 B 25 D 6 D 16 B 26 C 7 B 17 B 27 D 8 D 18 A 28 C 9 D 19 A 29 C 10 B 20 D 30 D Solution to 2016 H1 Physics Paper 1 1 The dimensions of a typical household washing machine are 450mm×450 mm×700mm. So its volume is 1.4×108 mm3 ; its mass is around 70 – 80 kg ; power consumption 500 W The spin rate is around 20 cycles per sec. [Switched options in original question] A 2 21 2W kx ∆𝑊 𝑊 = ∆𝑘 𝑘 + 2 ∆𝑥 𝑥 ∆𝑊 𝑊 = 2 100 + 2 0.002 0.050 = 0.10 (10%) B 3 𝑅 = 𝑉 𝐼 = (𝑃 𝐼 ) (1 𝐼) = 𝑃 𝐼2 Power = Energy  time = Force × displacement  time SI base unit for power = (kg m s-2)(m) / s = kg m2 s-3 SI base unit for electrical resistance = kg m2 s-3 / A2 = kg m2 s-3 A-2 B 4 At the maximum height, the acceleration is 9.81 m s-2 not zero. A 5 The slope of the s-t graph gives the value of the instantaneous velocity. A 6 The force of P on Q and the force of Q on P are an action-reaction pair. Hence the two forces are equal in magnitude but opposite in direction. The F-t graph of Q on P is the mirror image of P on Q. Alternatively, D

  initial,P initial,Q after,P after,Q after,Q initial,Q after,P initial,P QP Total momentum before Total momentum afte r Area under F-t graph for Q Area under F-t graph for P P P P P P P P P PP          7 Newton’s 2nd law defines the net force acting on an object as being directly proportional to the rate of change of momentum of the object, where the constant of proportionality is 1. Hence, from this definition, options A and C (rate of change of energy), and option D (Rate of change of force) can be eliminated. water exerted by wall on water jet exerted by water jet on wall exerted by w all on water jet change in momentum of the water jet per unit time. The magnitude of the force exerted by the water je pF t FF    t on the wall is equal to the force exerted by the wall on the water jet. Thus the magnitude of the force exerted by the water jet on the wall is numerically equal to the change in momentum of the water jet per unit time. B 8 For a car travelling at constant speed, the net force acting on the car is zero. Consider the forces acting parallel to the slope, Sum of forces upward and parallel to slope = sum of forces downward and parallel to slope engine force = component of weight parallel to slope + resistive force engine force = mg sin  + F D 9 Increase in force on arrow = Increased in total horizontal forces on arrow 2 (120 cos 55) – 2 (100 cos 65) = 53 N D 10 The change in velocity, - - (- ) R Final velocity Initial Velocity QP QP    B

11 Weight is a vertically downward force. Hence Option A and C are not correct. The forces should form a closed triangle as there is equilibrium. Hence, only D is correct. D 12 Taking pivot at where W is acting, as L is moved outwards, the clockwise moment due to L is increased. To maintain rotational equilibrium, the anti-clockwise moment due to R has to increase. Hence, the perpendicular distance has to increase and R moves to the right. D 13 P = Fengine v Since car is travelling at constant speed, P = Fresistive v Fresistive  v2 Fresistive = kv2 (800) = k (20)2 k = 2 (Fresistive) = (2) (40)2 Fresistive = 3200 N P = Fresistive v P = (3200) (40) P = 1.28 x 105 W C

14 Initial drop, ΔGPE = ΔKE m(9.81)(0.80 – 0.08) = 0.75 m = 0.106 kg Rebound: ΔKE = ΔGPE = 0.106(9.81)(0.45 – 0.08) = 0.39 J B 15 Gas is expanding, hence work is done by the gas. Work done by gas = p ΔV = 3.0 x 104 x 25 x 10-4 x 5 x 10-3 = 0.375 J Work done on gas = - Work done by gas = - 0.375 J B 16 ∆𝜃 2𝜋 = ∆𝑥 𝜆  𝜋/3 2𝜋 = 0.40 𝜆 Since phase difference of 3  = 0.40 m, the wavelength = 0.40 × 6 = 2.40 m Thus, speed =   = 200 x 2.40 = 480 ms-1 B 17 From definition : Intensity = Power / Area Power = Intensity x Area …….(1) Intensity  Amplitude2 ………..(2) Based on (1) and (2), we have Power  Amplitude2 x Area Since power = energy / time,  E  A2 S …………… (3) E' E = (A' A) 2 (S' S) E' E = ( 2A A ) 2 ( 1 2S S ) = 4 × 1 2 = 2 Rate of energy transfer is increased by 2 times B 18 Period, 40052 11 ..fT  s From the graph, 8 intervals is equivalent to 1 period. Hence 8 intervals is equivalent to 0.40 s and 1 interval is equivalent to 0.05s. Since the wave is travelling to the left, it takes 1 interval for the Q to be at the zero displacement position of the wave. Hence the shortest time elapsed is 0.05 s. A 19 The sound generated by the vibrating string is a progressive longitudinal wave. [Switched options in original question] A 20 At the 1 st order bright fringe, the waves meet at in phase with the phase difference of 2𝜋. From the graph, 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑥 𝑜𝑓 1𝑠𝑡 𝑜𝑟𝑑𝑒𝑟 𝑏𝑟𝑖𝑔ℎ𝑡 𝑓𝑟𝑖𝑛𝑔𝑒 𝑓𝑟𝑜𝑚 𝐶 = 2.4 𝑚𝑚 (phase diff 2𝜋) 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝐶𝑃, ∆𝑥 = 4.2 𝑚𝑚 (phase diff ∆𝜃) Using ratio, ∆𝜃 2𝜋 = ∆𝑥 𝑥 𝑃ℎ𝑎𝑠𝑒 𝑎𝑛𝑔𝑙𝑒 𝑎𝑡 𝑄 = 4.2 2.4 × 2𝜋 = 7 2 𝜋 Equivalent phase difference = ( 7 2 𝜋 − 2𝜋) = 3 2 𝜋 (𝑝𝑟𝑖𝑛𝑐𝑖𝑝𝑙𝑒 𝑣𝑎𝑙𝑢𝑒) D

21 240 V, 60 W lamp 10 V, 2.5 W lamp Resistance     22 240 96060 VR P     22 10 402.5 VR P Theoretical working current    60 0.25240 PIA V   2.5 0.2510 PIA V Actual current flow in the circuit is 250 V  (960  + 40 ) = 0.25 A Since the actual current equals to the theoretical current needed by each lamp, both lamps will work normally. B 22 The current in both external and internal resistors will be the same as both resistors are in series.  2 2 power in external resistor (2 ) 2power in internal resistor IR IR C 23 A. This is the definition of Emf. Hence this option is correct. B. The amount of electrical energy converted to other forms of energy in the external circuit is defined as the terminal p.d. The terminal p.d. is only equal to the emf if (1) there is no current in the circuit or (2) there is no internal resistance. C. Depends how many components there are. Even in the case of one component, it is likely that the terminal p.d. is lower than the emf. D. See reasoning in option B. A 24 Using  lR A       lR wt lt wR t = (2.0 × 10–5)(30 x 102)  (1.2 103)(40 x 103) = 1.25 x 107 m C 25 Total resistance, 66 k = R + (1/R + 1/R)1 + (1/R + 1/R + 1/R)1 66 k = 11R/6 R = 36 k D 26 At A, both the magnetic fields of X and Y are directed upwards. Hence the field would be non-zero. At D, both the magnetic fields of X and Y are directed downwards. Hence the field would also be non-zero. At B, both the magnetic fields of X and Y are directed opposite to each other but do C

not cancel out each other as the field due to X (which is closer) will be stronger than the field due to Y. Hence the field would also be non-zero At option C, both the magnetic fields due to X and Y are equal in magnitude but opposite in direction. Thus the net field will be 0. 27 Since the current balance experiences a turning moment about the pivot, we can infer that the magnetic force is either into or out of the page, and the force is acting on the segment which is 0.093 m long. B 3 32 7.4 10 7.4 10 Moment 0.23 ( . . ) 0.23 ( 0.093) 0.23 9.6098 9.61 6 10 A 3. F BL              I I I I D 28 As light intensity increases, the rate of incidence of photons increases. As rate of incidence of photons is proportional to the rate of emission of photoelectrons, the photocurrent increases proportionally as well. C 29 K.E. of mass, 𝑬 = 𝟏 𝟐 𝒎𝒗𝟐 = 𝟏 𝟐 (𝒎𝒗)𝟐 𝒎 = 𝒑𝟐 𝟐𝒎 where p is the momentum of the mass ⟹ 𝒑𝟐 = 𝟐𝒎𝑬 ⟹ 𝒑 = √𝟐𝒎𝑬 From de Broglie equation 𝝀 = 𝒉 𝒑 = 𝒉 √𝟐𝒎𝑬 C 30 D E3 - E1