VJC 2022 Prelim P2 Guide

2022 VJC Prelim H2 P2 Suggested solution 1(a) His explanation is not correct as even though the force that the truck acts on him is equal and opposite to the force that he acts on the truck, the 2 forces act on different bodies [APPLICATION] so they do not cancel out. [CONCLUSION] As Adam pushes the truck, the truck exerts a friction on the surface of the roadside [be specific: friction at where?] . By Newton’s 3 rd law [CONCEPT], the surface of the roadside will exert an equal and opposite friction back on the truck. [APPLICATION] The truck does not move because the friction that the surface of the roadside exerted on the truck and the force exerted on the truck by Adam are acting on the same object and are equal and opposite, therefore there is no net force on the truck, [APPLICATION] and hence by Newton’s 1st law [CONCEPT], it doesn’t move [CONCLUSION]. (b)(i) The principle of conservation of momentum states that total momentum of the system before the collision is equal to the total momentum of the system after the collision, provided no net external force acts on the system. (ii) 1. Total initial momentum of the system is 0 since nobody is moving. By conservation of momentum [CONCEPT], Taking upwards as positive, Total initial momentum of system = total final momentum of system 0 = MballoonVballoon + mmanvman 0 = (320)Vballoon + 80(2.5) Vballoon = -0.625 m s-1 Since upwards is taken to be positive, -0.625 m s -1 would indicate that the balloon is moving in the negative direction, which is downwards. 2. Speed of the balloon is zero [because initial total momentum = 0]. 2(a) (i) (a)(ii) Total upward forces = total downward forces [CONCEPT] T + mg = U kx + mg = U L = 𝑉𝜌𝑔−𝑚𝑔 𝑘 = 2 3( 5.0 650)(1000)(9.81)−(5.0)(9.81) 160 where x = L L = 7.86 x 10-3 m (b)(i) Taking pivot at A, total clockwise moment = total anti-clockwise moment, [CONCEPT] ( 𝐴𝐶 2 cos 65°)(1200) + (𝐴𝐶 cos 65°)(2000) = 𝑇(𝐴𝐵) ( 𝐴𝐶 2 cos 65°)(1200) + (𝐴𝐶 cos 65°)(2000) = 𝑇( 3 4 𝐴𝐶) T = 1.47 x 103 N Upthrust Weight Tension

(b)(ii) Since tension exerted on the boom is at an angle, there will be a horizontal component. [APPLICATION] For the boom to be in equilibrium, there must be a horizontal force to counter this force. [CONCEPT: FORCES IN EQUILIBRIUM] Therefore, the force at the hinge is not vertical but at an angle to provide a horizontal component to counter the horizontal component of the tension. [CONCLUSION] 3(a) Using pV = nRT, [CONCEPT] Volume Vf = nRT p = Total mass Molar mass RT p = ( ) 5 8.31 273 100350 18 1.0 10 +   = 0.60271  0.60 m3 (b) Work done, W = pV [CONCEPT] = p(Vf - Vi) = p(Vf - m  ) = 1.0 x 105 (0.6027 - 0.350 1000 ) = 60235  6.0 x 104 J (Note: Volume of a gas at atmospheric pressure is so much larger than its volume as a liquid, so it’s OK not to include Vi in the calculation.) (c) U = Q – Wby [CONCEPT] = mLv – Wby = (0.350 x 2.26 x 106) – 60235 = 730770  7.3 x 105 J (d) The increase in internal energy takes the form of an increase in potential energy due to intermolecular forces of attraction. 4(a)(i) Since g and r are constant, so a is proportional to x. Negative sign shows that a and x are in opposite direction. [APPLICATION] Hence the ball undergoes simple harmonic motion [CONCLUSION] (ii) 2 = g r and  = 2 T [CONCEPT] 2 = 9.81 0.28 = 35 T = 1.06 s  = 0.53 s

(b) Sketch: time period constant (or increases very slightly) drawn lines always ‘inside’ given loops, up to given time duration successive decrease in peak height 5(a)(i) Faraday’s law states that the emf induced in a conductor is proportional to the rate of change of magnetic flux linkage. (ii) The steel string near the permanent magnet gets magnetised (and produces its own magnetic field). [APPLICATION] When the string vibrates, the magnetic flux density at the location of the coil changes. [APPLICATION] There is a changing magnetic flux linkage with the coil [APPLICATION] so, according to Faraday’s law [CONCEPT], an emf is induced in the coil [CONCLUSION]. (iii) Nylon string cannot be magnetised. (b) The nodes of the stationary wave are located at the clamps, giving the first harmonic, so the length of 64 cm is equal to half a wavelength of the wave. /2 = 64 cm (need to explain properly and not just write the equation) Frequency of the wave, 300 234.4 Hz2(0.640) vf = = = [SHOW: GIVE MORE DP]  230 Hz (c) maximum vibrational velocity of the string, max 0 0 2 1 (2 ) [CONCEPT] (2 )(230)(1.50 10 ) 21.68 m s   − − == = = v x f x Maximum induced emf, max max 32 [CONCEPT] (4.50 10 )(2.00 10 )(21.68)  −− = =   BLv = 1.95  10−3 V or 2.0  10−3 V 6(a)(i) As the half-life of X (in years) is very long, it means that the decay constant is very, very small. Thus the fraction of nuclei that would decay during the time of measurement is very low, and the number of nuclei N in the sample remains almost constant. [APPLICATION] Since activity, A = N [CONCEPT], then the activity will remain almost constant. [CONCLUSION] (ii) 1. Decay constant of Y, 1/2 ln 2 0.693 1.5 60 60 Y t ==  = 1.283 x 10-4 s-1 = 1.3 x 10-4 s-1

2. Equilibrium is reached when the rate of production of Y (from the decay of X) is equal to its rate of decay of Y. Hence, the number of isotope Y in the sample will stabilise at a constant value. 3. Thus, the activity of Y is about 1.1 x 107 Bq. Amount of Y, 7 4 1.1 10 1.283 10 Y Y Y AN  − = = =  8.6 x 1010 atoms b(i) By t oN N e −= : [CONCEPT] 56 tN Ne −= Take ln on both sides, ln 5 ln 6 t=− 9 11 ln 6 ln 5 3.683 104.95 10t − −= =  yrs = 3.7 x 109 years (ii) Decay of Th-232 will give rise to a radioactive series where there will be a number of radioactive daughter products before ending up as the stable Pb -208. It is assume d that these intermediate radioactive daughter products have very short half-lives (much shorter than that of Th -232) so the number of intermediate daughter products are insignificant compared to Th-232 and Pb-208. (iii) If the assumption is not valid, there would have been more Th-232 in the beginning which have decayed and is still in the form of the intermediate daughter products, therefore the fraction of undecayed Th-232 is less than 5 6 . This means that the rock would have been older, as a longer time would have been elapsed, thus answer for (b)(i) will be an under-estimate. 7(a) Potential difference used to accelerate the ions, V = 1060 − (−225) = 1285 V The gain in kinetic energy of the ion = its loss of electric potential energy 21 2 mu q V= , where u is the speed of the ion. [CONCEPT] 19 41 25 2 2(1.60 10 )(1285) 4.323 10 ms2.2 10 qVu m − − − = = =   414.3 10 ms −= (b)(i) The passage says it takes 4 days (or more) to eject 1 kg of xenon. maximum mass of xenon ejected per second, 1 kg 4 days m t  = 611 2.894 10 kg s4 24 60 60 m t −− = =     = 2.9  10−6 kg s−1

(ii) Maximum force exerted on the ejected xenon, ( ) 64 [CONCEPT] 2.9 10 (4.3 10 ) 0.1247 N − =  = =   = pF t m ut By Newton’s 3rd law, the maximum thrust = F = 0.12 N (c) Assume that, at the start, the mass of Deep Space 1 (MDS) and the mass of the xenon (mXe) are at rest. So their total initial momentum pinitial = 0 Applying the principle of conservation of momentum, [CONCEPT] their total final momentum pfinal = pinitial = 0 So the momentum of DS must be equal and opposite to that of Xe, DS DS Xe XeM v m v= 4 3 (74)(4.3 10 ) 740 kg(4.3 10 ) Xe Xe DS DS mvM v = = =  (d) The positive ions wi