HACH A-Level Physics



M1.?????????(a)????? (i)?????nucleon number is the number of protons and neutrons OR mass numberproton number is the number of protons OR atomic number 1(ii)?????14 – 6 = 8 1(iii)?????specific charge = 6 × 1.6 × 10–19/(14 × 1.66 × 10–27 )specific charge = 4.1 × 107 (C kg–1) 3(b)?????(i)??????isotopes are variations of an element that have sameproton/atomic number but different nucleon number OR different number of neutrons 2(ii)?????4.8 × 107 = 6 × 1.6 × 10–19 /(A × 1.66 × 10–27)A = 6 × 1.6 × 10–19/(4.8 × 107 × 1.66 × 10–27)A = 12 Number of neutrons = 12-6 3[10]?M2.(a)???? (i)????? Number of complete waves passing a point in one second / number of complete waves produced by a source in one second / number of complete vibrations (oscillations) per second / number of compressions passing a fixed point per second1(ii)???? 180° phase difference corresponds to ? λUse of v = fλ with correct powers of 100.33 (m)3(b)???? (i)????? Do not have the same frequencydo not have a constant phase difference2(ii)???? Waves meet antiphaseUndergo superpositionResulting in destructive interference3(iii)???? T = 100 msUse of T = 1 / f or beat frequency (?f) = 10 Hz500 (Hz) (allow 510 –their beat frequency)3(c)???? (i)????? Only box ticked: Quality1(ii)???? Add regular alternating voltages togetherWith appropriate amplitudesWhere frequencies of voltages match the harmonics of sound / where frequencies are multiples of 440 HzAllow 2 for sampling sound (at twice max frequency ) B1Convert to binary ( and replay through D to A converter). B13[16]M3.(a)???? The molecules (continually) move about in random motion?Collisions of molecules with each other and with the walls are elastic?Time in contact is small compared with time between collisions?The molecules move in straight lines between collisions?ANY TWOAllow reference to ‘particles interact according to Newtonian mechanics’2(b) ????Ideas of pressure = F / A and F = rate of change of momentum?Mean KE / rms speed / mean speed of air molecules increases?More collisions with the inside surface of the football each second?Allow reference to ‘Greater change in momentum for each collision’3(c) ????Radius = 690 mm / 6.28) = 110 mm or T = 290 K ? seenvolume of air = 5.55 × 10-3 m3?n × 29(g) = 11.4 (g)? n = 0.392 molUse of pV = nRT = ?p = 1.70 × 105 Pa ?Conclusion: Appropriate comparison of their value for p with the requirement of the rule, ie whether their pressure above 1 × 105 Pa falls within the required band?Allow ecf for their n V and T?6[11]M4.B[1]M5.(a)???? (i)????? 230 × √2 = 325 (V) ?(2 × 325 =) 650 to 651 V ?allow doubling their incorrect peak voltage (162.6 × 2) by use of 2 as an attempt to find peak-to-peak for 1 mark but not just 2 × 2302(ii)???? (use of P = V2/R)P = 2302/12 ?P = 4.4 × 103 (W)? cao2 sig. figs. Incorrect answer must be supported by working ?Allow their incorrect answer (a)(i)2 ÷ 12Or 3252 ÷ 12 as a use of for 1 markAlternativeFor first markI = and P=VI allowing their incorrect answer(a)(i) or 325 as sub for V for 1 markAnswers 8.8 kW (325V) and 35 kW (650V)3(b)???? (i)????? there is a pd / voltage across the cable ?pd / voltage across cooker is 230 V minus this pd / voltage ?2nd mark depends on 1st mark in allThe current is lower due to the resistance of cable / The current is lower as circuit resistance increases ?pd across oven is lower since V=I × Resistance of element ?orResistance of the cable is in series with element ?Voltage splits (in ratio ) across these resistances ?2(ii)???? resistance of cable = 2 × 3.15 × 0.0150 = 0.0945 ?Allow power 10 error here ?Or???=228 V ? caoAllow their incorrect Rcable correctly substituted for 2nd marking3(iii)???? 230 ? their (b) (ii) or 19 (A) quoted for current or equivalent seen in equation (230 / 12.0945) ?(P =) 34.2 to 42.3(W) ? correct workingecf as P = (230- (b)(ii))2 / their Rcable2(iv)???? minimise power loss / maximise efficiency of oven / ensure element gets as hot as possible ?avoid overheating / fires ?not just to carry a large current / larger pd across elementEither order2[14]M6.C[1]M7.D[1]M8.C[1]M9.C[1]M10.B[1]M11.A[1]M12.B[1]M13.C[1]M14.(a)???? (i)??????X must have a negative charge?to conserve charge?second mark dependent on first i.e. conserve charge alone scores nothingcan gain second mark by showing balanced equation2(ii)?????X must be a baryon?to conserve baryon number?here two marks are independent i.e. conserve baryon number alone scores 1 markcan gain second mark by showing balanced equation2(iii)????K-: s OR strange anti-up?????K+: u OR up anti-strange?K0: d OR s OR down anti-strange OR strange anti?down?in each case the symbols or words can be in either ordermust be a bar over anti ? quarkcan be upper case letters e.g. U3(iv)????(strangeness on LHS is -1)strangeness on RHS without X is +2 / strangeness of X is -3 ?thus sssORstrangeness on RHS without X is +2 / strangeness of X is -1?thus sdd??correct strangeness without X on RHS is minimum working needed for first marknext two marks awarded for correct quark structure3[10]M15.(a)???? Total mass of spacecraft = 3050 kgChange in PE = 1.9 × 1011(J)2 sfcondone errors in powers of 10 and incorrect mass for payloadAllow if some sensible working4(b)???? Chemical combustion of propellant / fuel or gases produced at high pressureGas is expelled / expands through nozzleChange in momentum of gases escapingequal and opposite change in momentum of the spacecraftThrust = rate of change of change in momentumMax 3N3 in terms of forces worth 13(c)???? 0.031(4) (m s-2)1(d)???? Use of rocket equationv = 1200 ln 996 (m s–1)Condone 1000 (m s–1)3(e)???? (i)????? Use of correct mass 108 kg0.0198 NAllow incorrect powers of 10 and mass3(ii)???? Use of v = Correct substitution v = 0.86 (m s-1)Recognisable mass – condone incorrect power of 103(iii)???? Impulse = 25 N × 4.8 = 120 N s(120 = 108 v so) Velocity = 1.1 m s-1Clear conclusionie explanation/comparison of calculated velocity with escape velocity from (e)(ii)May use F = ma approach 3[20]?E1.????????? This proved to be one of the most accessible questions on the paper with many students securing full marks. The majority of explanations of nucleon number, proton number and isotope were clear although a minority did confuse number of neutrons with number of protons in there definition of isotope. The calculations and deductions pertaining to the nucleus and one of its isotopes were in the main well set out and in many cases this helped generate correct answers. A minority, as has been the case in previous sessions, did include the mass of electrons in the specific charge calculation even though the question clearly refers to a nucleus.??E2.(a)???? (i)????? Acceptable definitions were given by a good majority of the students. Those who failed to produce a satisfactory response usually omitted reference to time.(ii)???? Most gained credit for the use of v = fλ . The common errors were ignoring the k in kHz and not calculating λ/2.(b)???? (i)????? This question was a ‘twist’ on a commonly asked question that requires students to explain what is meant by waves being coherent. This question required students to identify that the tuning forks had different frequencies and would not have a constant phase difference when they arrive at a point so would not be coherent. This proved to be too challenging for many students.(ii)???? This was poorly done and fewer than half the students were able to give at least one acceptable point worthy of credit and there were relatively few who gained full credit. One can only speculate that students have difficulty understanding interference that occurs due to changes in phase difference that take place at a point with time as is the case in this instance.(iii)???? A high proportion of the students gained credit for use of f = 1/T and many of these arrived at the correct beat frequency. Many did no more than this and relatively few of these went on to calculate the correct frequency of the fork that emitted the lower frequency.(c)???? (i)????? Almost three quarters of the students selected the correct response to this question.(ii)???? Relatively few appreciated the meaning of synthesis of sound ie the process of adding together sinusoidal waves of appropriate frequencies and amplitude to produce a required sound. Students were given compensatory credit for explaining the process of sampling a sound and storing it digitally.E4.With 87% of students choosing the correct answer, few students had any difficulty with this question. The remaining answers were fairly evenly spread between the three distractors.E5.Most students were able to determine the peak voltage of 325 V in part (a)(i). Unfortunately a considerable proportion of students neglected to double this value to determine the peak-to-peak voltage.In part (b)(ii), students were told that the rms voltage across the heating element was less than 230 V and asked to explain why this was the case. Many weaker responses restated information that had been given without adding any new detail. Others would state that there was a voltage drop across the connecting cable without developing this point to explain why this meant that the rms voltage would be lower across the element. Better responses were able to produce a two stage argument to support their explanation.The calculation in (b)(ii) was challenging for most students. The working seen was often unclear with students carrying forward non-relevant data from (a)(i) and (a)(ii). Grade A students generally produced work that was well laid out and direct in approach. Some students used incorrect methods for determining the current such as dividing 230 V by 12 ohms or dividing 4400 W by 230 V, neither of these approaches took account of the new resistance in the circuit and how this would impact on the current value.Similarly, part (b)(iii) caused problems for all but the most able students. Many students took the answer to part (b)(ii) and substituted this into with no clear appreciation of the task. This perhaps indicated a mechanical approach to calculations by some students in which preceding parts are blended to get the final answer without really understanding the nuances of the task.E6.This question required students to work out the wavelength of the sound wave, and then calculate the phase difference of two parts a certain distance apart. 45% of students correctly identified the correct answer. Approximately 35% thought A was correct, using π, rather than 2π, as the phase difference for two points a whole wavelength apart, perhaps.E7.57% of students correctly identified D as the appropriate answer. The other students were split almost evenly between the distractors, with A being slightly more popular.E8.Surprisingly the most popular distractor here was B, perhaps highlighting students’ confusion with mass and weight. 70% were able to identify the correct answer.E9.The correct answer was given by 43% of students. Unsurprisingly perhaps, A was the most popular distractor.E10.54% of students were able to perform this relatively straightforward calculation. Surprisingly D was the least popular distractor, suggesting students had more problems with correctly using seconds rather than minutes than dealing with the m in mA.E11.Students familiar with the characteristic for a fixed resistance were probably led to answer B without reading the question. This proved to be the most popular answer despite it being incorrect. Approximately 20% were sufficiently careful with their reading, or sufficiently familiar with the practical, to give the correct answer, A.E12.This question, on angular speed, had also appeared in an earlier examination. The facility in 2016 was 62%, an improvement over the 56% facility last time. Over 20% of the students chose distractor A (75 rad s-1) by confusing the diameter of the wheel with the radius.E13.This question tested students’ interpretation of a velocity-time graph for simple harmonic motion. The question asked for the incorrect statement to be selected; this generally causes problems for students who read the questions too superficially. However on this occasion 70% of them gave the correct answer. Maybe the 13% of students who selected distractor D, which is clearly a correct statement, had forgotten the wording of the question by the time they reached it.E14.The first three parts of this question were well done and students demonstrated that they were familiar with the relevant conservation laws and could also quote the quark structures of the kaons. A minority did lose marks in (a)(i) however, because they did not state the charge of particle X and merely stated that it was charged.(a)(iv) was much more discriminating and weaker students did not really appreciate that conservation laws were needed to make the deduction and only the strongest students identified that the strangeness of X was either ?3 or ?1 depending on the quark structure they used for the K0.E15.(a)???? Most students gave the answer to 3 significant figures, although 2 sf was what was required. The correct mass (3050 kg) was chosen by those who used the correct formula, but some students used no mass in calculating the potential only.(b)???? Many stated that the propellant/fuel was ejected through the nozzle. The statements about the momentum of the exhaust gases were often confused. The most popular way of deriving thrust was by attempting to use Newton’s 3rd law but the statements were often incomplete.(c)???? The simple use of F = ma was easily achieved by most students.(d)???? Although some students attempted to use conservation of momentum, most realised that the rocket equations was needed. There is the same confusion over the meaning of the symbols vf and mf. Some used mf = 1720 kg instead of 1330 kg, and others, after correctly calculating vf = 996 m s?1, went on to subtract this from the exhaust gas speed, thus sacrificing a mark.(e)???? (i)????? Most students chose the correct formula, but many forgot to square the radius, and others chose the wrong mass. The original mass of the spacecraft (3050 kg) was the most popular erroneous value, although even the mass of the Earth was seen occasionally.(ii)???? Nearly everyone started with the correct formula but two common errors ensued. Some forgot to take the square root and others did not convert 2.0 km to meters. Also some gave the answer to 1 sf (0.9 m s?1) thus losing a mark.(iii)???? Those who calculated that the velocity change of the probe was 1.1 m s?1 followed with the right conclusion. Some students used the wrong mass but could still gain the third mark with a correct comparison. ................
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