TOPIC 6. Further mechanics. 6.1 Further momentum. Answers to Student Book 2 questions Energy in collisions. 6.1.

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1 TOPIC 6 Further mechanics 6.1 Further momentum Energy in collisions 1 (a) J ev (c) 5.46 MeV 2 (a) 3.90 m s 1 E = 6.6 J kg, so likely to be an alpha particle More collisions 1 (a) N s 3.24 s 2 (a) Cue ball velocity is virtually the same as the black, but on a downward 45 trajectory. It is highly likely to enter the bottom corner pocket. Kinetic energy is conserved, so it is an elastic collision. 3 v imp = m s 1. This is faster than the speed of light. 4 Students own answers. For example: The table top could be marked with a grid of squares (or graph paper) for improved position determination. The size of the ball bearings could be reduced, so that their position is more accurately determined; and also, so that the trajectories change more on collision, reducing the percentage errors in trajectory direction determination. 6.1 Answers to Exam-style questions 1 A 2 C 3 B 4 Considers momentum (1) Calculates momentum of Xenon or spacecraft (1) Calculates a second momentum Or calculates speed of spacecraft (1) A statement that the prediction is correct Or a statement that the increase is (about) 8 m s 1 (1) (only award this mark if based on correct calculations ) (Calculation to find the speed of the Xenon or either mass scores max 3) Momentum of Xenon = 0.13 kg m s 1 = 3900 kg m s 1 Momentum of spacecraft = 486 kg 8 m s 1 = 3888 kg m s 1 Or Momentum of Xenon = 0.13 kg m s 1 = 3900 kg m s 1 Momentum of spacecraft = 486 kg v v = 3900 kg m s 1 / 486 kg = 8.02 m s 1 5 (a) Sum of momenta before (collision) = sum of momenta after (collision) Or the total momentum before (a collision) = the total momentum after (a collision) Or total momentum remains constant Pearson Education Ltd Copying permitted for purchasing institution only. This material is not copyright free. 1

2 TOPIC 6 Further mechanics Or the momentum of a system remains constant (1) Providing no external / unbalanced / resultant force acts Or in a closed system (1) (i) Use of equation(s) of motion sufficient to get answer (1) Initial speed = 1.1 (m s 1 ) (1) s = (u + v)t/ m = (u + 0) 1.3 s / 2 u = 1.06 m s 1 (ii) Constant acceleration/deceleration (accept constant force) (1) (iii) Use of momentum = mv ecf v from (i) (1) Calculates momentum after collision using correct mass (1) Speed of pellet = 117 or 124 or 129 (m s 1 ) (1) Momentum after = ( ) g 1.06 m s 1 = 104 g m s 1 Momentum before = momentum after Speed of pellet = 104 g m s 1 / 0.84 g = 124 m s 1 (c) (i)* (QWC Work must be clear and organised in a logical manner using technical wording where appropriate) Mention of momentum (1) (d) Pellet (bounces back so) has negative momentum / velocity Or momentum after = momentum of car momentum of pellet (1) Pellet undergoes a bigger momentum / velocity change Or mass of car is less (1) (ii) Reference to greater horizontal momentum / force (1) [The question says that the calculations are correct, the question is about the assumptions made. Do not credit a statement that the GPE is correct. MP1 is for the assumption that the KE after firing is the same as the max GPE. Do not credit energy loss due to air resistance or sound] E k E grav of pendulum correct Or KE after collision is correct (1) E k in collision not conserved Or not an elastic collision Or inelastic collision (do not credit just KE is lost ) (1) Some energy becomes heat (1) E k (of pellet before collision) is greater than 0.16 J (1) 6 (a) Conversion of MeV to J (1) Use of E k = ½ mv 2 (1) Max velocity = (m s 1 ) (1) v = Mev kg J velocity = m s 1 (i) Correct momentum of any particle seen, e.g. Nux (must contain u) (1) (ii) Correct equation from conservation of momentum (allow even if u not shown) (1) Rearrange for z (dependent on second mark) (1) Nux = 14uy + Nuz Nz = Nx 14y Kinetic energy is conserved (iii) See ½ Nux 2 Or ½ Nuz 2 Or ½ 14uy 2 (1) Pearson Education Ltd Copying permitted for purchasing institution only. This material is not copyright free. 2

3 TOPIC 6 Further mechanics Clear statement that E k nitrogen atom = E k neutron before E k neutron after Or E k nitrogen atom = E k lost by neutron (1) (c) (i) Use of equation, N in the denominator must be included, given with y = Or y = (1) (ii) In equation given use of: N + 1 with y = Or N + 14 with y = (1) In equation given use of: N + 1 with y = And N + 14 with y = (1) For hydrogen 2Nx = (N + 1) For nitrogen 2Nx = (N + 14) Equating gives (N + 14) = (N + 1) (so N = 1.06) Collision might not be elastic Or Speed (of particles) approaches speed of light (so mass increases) (1) 6.2 Circular motion Circular motion basics 1 (a) rad (= 0.63 rad = π 5 rad) rad s 1 3 (a) (i) 3.5 rad s 1 (ii) 4.7 rad s 1 (iii) 8.2 rad s m s 2 4 (a) rad s m s 1 (c) m s % Centripetal force N ( = 1.4 rad s 1 ) 2 (a) 2.54 N (c) W = 736 N, so centripetal force is much smaller. Centripetal force is provided by weight, so reaction is less than weight by the amount of the centripetal force (in order that there is a resultant to provide centripetal acceleration). At South Pole, zero centripetal force, so reaction is 2.54 N larger than at equator. Pearson Education Ltd Copying permitted for purchasing institution only. This material is not copyright free. 3

4 TOPIC 6 Further mechanics 6.2 Answers to Exam-style questions 1 C 2 B 3 B 4 (a) Use of F=mv/t or F = ma (1) Answer = N (1) E.g. F = / 3.5 Arrow down labelled mg/w (1) Arrow up labelled, e.g. R / reaction / force from seat (1) Equal length vertical arrows from a clear single point / centre of mass and bottom (1) (c) 4mg mg Or 3mg (1) (m)v 2 /r seen (1) Answer = 110 (m) (1) E.g. 3mg = mv 2 /r r = (57) 2 /3g (d) Use of KE/PE conservation (1) Answer = 23 (m s 1 ) (1) E.g. ½ m(57) 2 = ½ mv 2 + mg 139 v 2 = ½ (57) (e) Using (m)g only (1) Answer r = 54 m [allow ecf] (1) E.g. mg = mv 2 /r r = (23) 2 / (a) Conversion from per minute to per second (1) Conversion from revolutions to radians (1) 20 revolutions = 20 2π/60 (= 2.1 rad s 1 ) Use of rω 2 (1) Answer in range 6 13 m (1) m s 2 (correctly corresponding to radius estimate) (1) 6 (a) Use of v = 2πr/t Or v = rω and T = 2π/ω (1) t = s [24.6 minutes] (1) t = 2πr/v t = (2π 61 m) / 0.26 m s 1 t = 1473 s Use of F = mv 2 /r (1) F = 11 N (1) F = kg (0.26 m s 1 ) 2 / 61 m F = 10.7 N Pearson Education Ltd Copying permitted for purchasing institution only. This material is not copyright free. 4

5 TOPIC 6 Further mechanics (c) (i) Three arrows all pointing to the centre of the circle (accept freehand and lines of varying length) (1) (ii)* (QWC Work must be clear and organised in a logical manner using technical wording where appropriate) Maximum at C / bottom and minimum at A / top (1) At C contact / reaction force (R) greater than weight (1) (accept R W = mv 2 /r or R = W + mv 2 /r) At A contact / reaction force is less than the weight (1) (accept W R = mv 2 /r or R = W mv 2 /r) Any statement that centripetal force / acceleration is provided by weight / reaction (1) Or centripetal force is the resultant force This is a QWC question so a bald statement of the equations can score the marks but to get full marks there must be clear explanation in words. 7 This is as per the experiment on page 27: Description of apparatus and activity (1) Measure time for ten revolutions and divide by ten (1) to find time period (1) Repeat for various different hanging masses which will give F = mg (1) Control variables: radius of revolution (confirmed with marker on string), mass of rotating mass (1) Plot a graph of F (y-axis) against v 2 (x-axis) will verify the equation if it shows direct proportionality (1) Pearson Education Ltd Copying permitted for purchasing institution only. This material is not copyright free. 5

6 TOPIC 7 Electric and magnetic fields 7.1 Electric fields Electric fields N N C m s 2 4 (a) V m m s 1 (c) (d) Parallel plates with field lines with arrows from + to /ground; parallel equipotentials; some correct labelling of potential difference across field, similar to fig C. Proton has same magnitude charge as electron but is more massive; reduced acceleration; charge is of opposite sign, so acceleration is in opposite direction Millikan s oil drop experiment 1 (a) Negative Equal sized vertical arrows up and down from the drop, labelled weight or mg (down) and electrostatic or coulomb force (up). (c) E = V m 1 [Or N C 1 ] (d) mg = qe: use of equation (1) (e) 3 Charge = C (1) Radial electric fields 1 (a) Radial field emanating from circle representing dome (drawn clearly as non-point source); field lines from surface of dome drawn so they would come from centre of circle; arrows on field lines away from sphere; no field inside dome, similar to fig A. Equipotentials concentric with dome surface; increasing separation of equipotentials; surface voltage is V, so equipotential labelling should decrease from that following 1/r relationship. (c) V m 1 (d) 4500 V 2 It would feel no force as the electric field at the exact centre is zero; indicated on the diagram by the absence (or equidistance) of field lines. 3 Diagram as per fig C but with the point charges labelled as negative; field line arrows pointing to electrons (i.e. opposite direction to fig C). 4 The charges are forced closest together on the spike, and hence they generate the strongest field at the point; indicated on the diagram by the closeness of the equipotential lines. 5 (a) V m 1 F = EQ = N Pearson Education Ltd Copying permitted for purchasing institution only. This material is not copyright free. 6

7 TOPIC 7 Electric and magnetic fields Coulomb s law N 2 (a) V m N 3 Separation must be measured to the centres of the spheres, which is difficult with spherical objects and may introduce uncertainties. 4 Students own estimates. With body surface area of 2 m 0.5 m, and weight 800 N: r = m 5 Graph of 1/r 2 against d gives a reasonable straight best-fit line that passes through the origin, proving proportionality, and verifying Coulomb s law, as F is proportional to d in this experiment. 7.1 Answers to Exam-style questions 1 B 2 B 3 C 4 (a) Space / area / region where a force acts on a charged particle (1) The force is the same at all points Or field strength is constant Or field lines equispaced (1) (accept diagram with a minimum of three equispaced parallel lines, with arrows for 2nd mark) Two parallel plates (accept wires for plates) (1) Connected to a potential difference Or potential difference is applied (1) Practical method to show force E.g. seeds in tray of glycerol Charged foil on end of rule, Charged pith ball on thread, Beam of electrons (in teltron tube) Charged oil drops (1) (do not credit charged object) (All 3 marks can be scored from a diagram. To score the third mark the set-up must be labelled.) Pearson Education Ltd Copying permitted for purchasing institution only. This material is not copyright free. 7

8 TOPIC 7 Electric and magnetic fields 5 (a) Weight / W / mg vertically down Tension / T parallel to thread and pointing away Electrical (force) horizontal to left Accept electrostatic (force), repulsive (force), coulomb (force) repelling (force). Do not accept just F or drag All three correct 2 marks Any two correct 1 mark The lines must start on the ball and have arrow heads to indicate direction. Minus 1 mark for each extra force line. (Candidates who draw forces on M correctly but also include forces on N score 1) (i) Use of T cos 35 = mg Or T sin 55 = mg (1) (c) g to kg and 9.81 (1) Tension = (N) (1) T cos 35 = mg T = ( kg 9.81 N kg 1 ) / cos 35 T = N (ii) Equates electric force to T sin 35 Or T cos 55 Or W tan 35 Or use of Pythagoras theorem (1) F E = Or (N) (1) (F E = N if show that value used. ecf T from (i)) F E = sin 35 F E = N (iii) Use of F = Q 2 /4πε or 2 Or F = kq 2 /r 2 (ecf value of F from (ii) (1) Conversion cm to m (1) Q = ( ) 10 7 C (1) (candidates who halve the value of r can score the first 2 marks) Q 2 = Fr 2 /k Q 2 = (0.020 N) ( m) 2 / ( N m 2 C 2 ) Q = C Both balls would move through the same angle/distance Or the balls are suspended at equal angles (to the vertical) (1) (Because) the force on both balls is the same (1) 6 (a) Use of E = V/d (1) Answer = V m 1 or N C 1 (1) E.g. E = 1.5 / Opposite forces (act on either end of molecule) (1) Molecule rotates / aligns with field (1) at top / + at bottom (1) 7 (a) At least three vertical lines spread over symmetrically over more than half of the plate length and touching both plates (1) (ignore edge lines that might curve) All lines equispaced and parallel [don t allow gaping to avoid oil drop] (1) Arrow pointing downwards (1) Negative / / ve (1) (negative and/or positive does not get the mark) Pearson Education Ltd Copying permitted for purchasing institution only. This material is not copyright free. 8

9 TOPIC 7 Electric and magnetic fields (c) Upward force labelled: electric (force) Or electrostatic (force) Or force due to electric field Or electromagnetic (force) (1) [do not accept repulsive/attractive force. If EQ used, the symbols must be defined] Downward force labelled: mg, weight, W, gravitational force (1) (for both marks the lines must touch the drop and be pointing away from it; ignore upthrust if drawn but one mark lost for each extra force added) (d) (i) E = 5100 V / 2 cm (1) Conversion of cm to m (1) Use of QE = mg ( kg) (1) Q = C (1) (E = (V m 1 ) scores MP1 and 2 Unit conversion missed Q = C scores MP1 and 3 If V is halved Q = C scores MP1, 2 and 3) E = V/d F = EQ = mg Q = mg/e = mgd/v Q = ( kg 9.81 m s m) / (5100 V) Q = C (ii) Answer to (d)(i) divided by e (1) 8 Max potential marks: 5 3 electrons Or sensible integer number less than 500 (1) (answers with very large numbers of electrons can get MP1 only) number of electrons = C / C number = 2.9, i.e. 3 electrons. Marks can be gained if the concept is seen on a diagram Radial field around electron Field line arrows pointing towards electron Potential lines are perpendicular to field lines Potential lines drawn as concentric circles (or spheres) around electron Distance of potential lines increases as 1/r Q Explanation that potential follows equation: V = 4π 0 r Diagram illustrating any of the above points 7.2 Capacitors Capacitor basics F 2 A 0.01 F capacitor is charged by and then isolated from an 8 V power supply. (a) J 0.08 C The charge will spread onto the other capacitor until they each have half of the original charge (0.04 C). This will result in a p.d. of 4V across each capacitor (and/or across the parallel combination). 4 E = J; so lights bulb for 4.86 ms Pearson Education Ltd Copying permitted for purchasing institution only. This material is not copyright free. 9

10 TOPIC 7 Electric and magnetic fields Charging and discharging capacitors 1 30 s 2 (a) Current: initially, the electric field from the supply sends a large surge of electrons onto the capacitor, so the current starts at its maximum. After a little time, the electrons already stored on the capacitor reduce the effective push (or electric field) from the supply, so the charge movement is reduced a smaller current. Eventually, the capacitor has as much charge as the 6 V supply can push onto it, and the mutual repulsion of electrons already stored stops any further charge moving onto the capacitor; the current falls to zero. (c) p.d.: initially, the capacitor has no charge on it, so its p.d. is zero. After a little time, the charge stored across the capacitor creates some p.d., which could be measured using a voltmeter. Eventually, the capacitor has as much charge as the 6 V supply can push onto it, which will be when it has also reached a voltage of 6 V, and it will remain at this value. Charge: initially, the capacitor has no charge on it. The current starts at its maximum so, after a little time, some charge is stored on the capacitor. The reduced current continues to add charge to the capacitor but at a slower rate. Eventually, when the current reaches zero, no more charge is added to the capacitor, and the amount on it remains constant at its maximum, CV. 3 Basic exponential decay curve as per fig B(a): initial current is 0.6 ma. This should fall to 0.22 ma in 0.5 s. At 1 s, the current has dropped to 0.08 ma, then at 1.5 s it is 0.03 ma, finishing at 0.01 ma by 2s. Depending on scale, it is likely that the final half second should look horizontal along zero ma Capacitor mathematics 1 (a) A Approx. 8 s (c) A 3 ln V = ln V 0 RC t Need RC = 74, so e.g. increase R to 740 In the experiment, take readings of p.d. against time. By plotting ln V against t, the log equation shows we will 1 get a straight-line graph. The gradient will be, so if R is known, this will give us C. Just plotting V against t RC would give the exponential curve, which is difficult to analyse accurately. 7.2 Answers to Exam-style questions 1 C 2 D 3 C 4 (a) (i) Capacitor charges up Or p.d. across capacitor becomes (equal to) p.d. of cell (1) Negative charge on one plate and positive charge on the other (ii) Or opposite charges on each plate Or movement of electrons from one plate and to the other (around the circuit) (1) (reference to positive charges moving or to charge moving directly between the plates negates the second mark) As capacitor charges current decreases Or as capacitor charges current drops to zero Or p.d. across capacitor becomes (equal to) p.d. of cell (1) No current through R (means no p.d.) Or V cell = V capacitor + V resistor (1) Pearson Education Ltd Copying permitted for purchasing institution only. This material is not copyright free. 10

11 TOPIC 7 Electric and magnetic fields * (QWC Work must be clear and organised in a logical manner using technical wording where appropriate) See Q = CV (1) As C increased then charge flowed (Or more charge stored) on capacitor (1) So p.d. across R (1) Charge flow / current /output signal reversed when plates move apart (1) Or See Q = CV (1) As C increased p.d. across capacitor decreased (1) So p.d. across R must increase (1) p.d. reverses when plates move apart (1) (c) Use of time constant = RC Or attempt to find half-life (1) Time constant = (s) Or t ½ = (s) (1) Use of T = 1/f (to give T = 0.05 s for the lowest audible frequency) (1) Capacitor completes discharging/charging during cycle of signal (1) (last mark can only be gained if supported by calculations) (f = 1/CR may be used to find the frequency of the microphone, rather than time. In this case candidates may just calculate f = 200 Hz rather than a time. Only first 3 marks are available) RC = Ω F RC = s f = 1/T = 1/20 = 0.05 s 5 (a) Method marks only Use of Q = CV with V = 16 V (1) Max value of C = (μf) (1) μf means 10 6 conversion of μf to F (1) C max = = F C max = F 16 V Q max = C Either use of ½ QV or ½ CV 2 (1) Energy = 1.5 J (1) W = ½ C 16 V energy = 1.54 J 6 (a) (i) Answer = 22 s (1) (ii) T = Ω F Time taken for the p.d. / charge / current to change by 63% Or to fall to 1/e of its original value Or to fall to (12/e) V Or to fall to 4.41 V (1) (i) To reduce / prevent charge / current through the voltmeter (1) (ii) Precision of stopwatch much less than reaction time Or uncertainty significantly less than 22 s (1) (iii) Repeat experiment and calculate mean value (1) Or use graphical method Pearson Education Ltd Copying permitted for purchasing institution only. This material is not copyright free. 11

12 TOPIC 7 Electric and magnetic fields (c) Correct expansion ln V = t/rc + ln V 0 (1) Compare with y = mx + c Or state that RC is constant (1) (d) ln values correct and to at least 2 SF (1) Labels axes with quantity and unit; y-axis must be labelled ln (V/V) (1) Scales (expect 1 cm : 0.1 vertically) (1) Points correctly plotted and best-fit straight line (1) (e) (i) Large triangle (at least half the drawn line) with correct values (1) Correct calculation for T in range 19.8 < T < 20.2 and at least 3 SF (no unit penalty) (1) (ii) Uses 22 s and graph value to calculate % difference correctly (1) ( )/22 = 2.0/22 = 9.1% (f) (i) ln 5 (= 1.609) used to find time from graph (1) Value in range s and to 3 SF (1) (ii) Answer = 150 kω (option C) (1) 220 (12/17.5) = Electromagnetic effects Magnetic fields 1 (a) Student sketch should include labels for geographic North and South Poles and field lines as per bar magnet, with arrows pointing along field lines from geographic South Pole to geographic North, e.g. The poles of a bar magnet have the field lines closest together, so the North and South Pole of the Earth would be where B is greatest. 2 (a) Estimate area of body perpendicular to field lines as 2 m 0.5 m, gives = Wb = Wb Wb-turns Pearson Education Ltd Copying permitted for purchasing institution only. This material is not copyright free. 12

13 TOPIC 7 Electric and magnetic fields Electric motors 1 Fleming s left-hand rule connects directions of magnetic field, electric current and force when determining effect on current-carrying conductor in a magnetic field. Thumb = force; first finger = magnetic field; second finger = conventional current. 2 (a) Down (c) Away from reader No force 3 With electromagnets the magnetic field can be varied easily, but the magnetic field is zero when the electric current is switched off Magnetic forces 1 (a) N (c) N N m s 1 ; faster than the speed of light. 3 v = t l and I = t Q F = BIl = B t Q l = B t l Q = BvQ where Q is a collection of charge Na + r = m; 22 Na + r = m. 0.3 mm is an easily detectable change. When the settings for B and V mean the radius of travel does not match exactly with the shape of the machine, the ions do not reach the detector at all, and the detectable differences are orders of magnitude less than the difference here Generating electricity 1 (a) Wb-turns (c) 0.14 V The current induced would feel a Fleming s left-hand rule force in the direction opposing the motion. 2 (a) The induced e.m.f. rises to a maximum of about 250 mv as the magnet enters the coil and reaches full flux linkage. As the magnet continues to fall, the magnetic field starts to leave the coil, so the induced e.m.f. is in the opposite direction, with a peak at about 400 mv higher than previously as gravity is constantly increasing the magnet s speed and hence its rate of change of flux linkage. The induced e.m.f. would drive a current through the bulb first in one direction and then in the opposite direction. If this resulted in a high enough current for it to be bright enough light to observe, it would look like a brief flash lasting perhaps 0.2 s, followed by a brighter flash of slightly shorter duration Alternating current 1 (a) 50 Hz 19.8 mv (c) (i) The trace becomes more squashed horizontally, so twice as many cycles appear on the screen. (ii) The trace extends up and down, twice as far, so the top and bottom sections of each cycle would be cut off by the edges of the screen turns 3 (a) 8.5 V I 0 = 15 ma; I rms = 11 ma 7.3 Answers to Exam-style questions 1 D 2 B 3 B 4 Use of W = mg (1) Pearson Education Ltd Copying permitted for purchasing institution only. This material is not copyright free. 13

14 TOPIC 7 Electric and magnetic fields Use of F = BIL (1) B = 0.04 T (1) 5 (a) (Magnetic) Flux linkage (1) (QWC (i and iii) spelling of technical terms must be correct and the answer must be organised in a logical sequence) Lenz s law / conservation of energy (1) induced current / e.m.f. (direction) (1) Opposes the change (that produced it) (1) 6 (a) Reference to magnetic flux (linkage) (1) Magnet vibrates / moves (1) Flux / field through the coil changes (1) Induces e.m.f. / p.d. (1) (i) Use of T = 2π/ω for a revolution (1) ω = 3.5 rad s 1 (1) ω = 33 2π rad/ 60 s ω = 3.5 rad s 1 (ii) ω / T / f remains constant (1) v = rω Or C = 2πr (1) So as the stylus moves towards the centre (tangential/linear) speed/velocity Or path length (per rotation) gets less (1) 7 (a)* (QWC Work must be clear and organised in a logical manner using technical wording where appropriate) Max 6 from Reference to changing / cutting of field / flux (1) Induced e.m.f. proportional to rate of change / cutting of flux (linkage) (1) (accept equation) Initial increase in e.m.f. as the magnet gets closer to the coil (1) Identifies region of negative gradient with magnet going through the coil (1) Indication that magnet s speed increases as it falls (1) Negative (max) value > positive (max) value (this mark is dependent on awarding MP5) (1) Time for second pulse shorter (this mark is dependent on awarding MP5) (1) The areas of the two parts of the graph will be the same (since NΦ constant) (1) Two sequential pulses (1) (if not two sequential pulses, scores zero) Pulses same height (± 3 mm squares) and width (by eye) (1) Pulses in opposite directions (1) Region of zero e.m.f. in the middle (1) Example (peaks could be in opposite directions) Pearson Education Ltd Copying permitted for purchasing institution only. This material is not copyright free. 14

15 TOPIC 8 Nuclear and particle physics 8.1 Probing matter A nuclear atom 1 Students own answers. 2 Main reason is the small size of atoms, experimentation needed to get to smaller and smaller scales, requiring detection technology at smaller and smaller scales. 3 Strengths: e.g. distance between nucleus and orbits is much larger than size of particles themselves, as with real Solar System; there is a force holding orbiting objects in place in both cases. Weaknesses: e.g. electrons do not all orbit in same plane, cf. the ecliptic; planets can have continuously variable orbital energies, electrons have only fixed energy orbits; electrons may not follow continuous path around their orbit, but follow probability function as to their location. 4 A = 9; B = Electrons from atoms 1 (a) m (c) (d) m m Students own answers of the order of m 2 The Davisson Germer experiment proved diffraction of electrons and measured their effective wavelength m s 1 which is faster than light. 4 Diagram with horizontal and vertical pairs of electric plates. Explanation that horizontal field will be used as time-base and vertical field connected to measuring electrodes in order to alter vertical position of trace. 8.1 Answers to Exam-style questions 1 (a) B 2 C 3 D B 4* (QWC (i and iii) Spelling of technical terms must be correct and the answer must be organised in a logical sequence.) Observations: Most alpha particles went straight through (1) Some deflected (1) (Very) few came straight back / large angle (1) Conclusions: Atom mainly (empty) space (1) Nucleus contains most of the mass (1) (Nucleus) very small / tiny (1) (Nucleus) charged / positive (1) 5 (a) Identifying the equations E k = p 2 /2m and λ = h/p Or λ = h/p, p = mv and E k = ½ mv 2 (1) Any combination or rearrangement (conditional mark) (1) (do not give 2nd mark just for quoting equation given in question) Pearson Education Ltd Copying permitted for purchasing institution only. This material is not copyright free. 15

16 TOPIC 8 Nuclear and particle physics (do not credit a reverse argument, i.e. starting with the given equation) Example of derivation p = 2mE k λ = h/ (2mE k) Correct substitution of h 2 and m (1) Use of E k = ev (1) λ = m (1) Or Use of E k = ½ mv 2 (to find v = (m s 1 )) (1) Use of λ = h/p with correct substitution for h and m (1) λ = m (1) λ = 2( ( J s) kg)(2500 V)( C) λ = m Or v = 2(2500 V)( kg 19 C) = λ = J s / ( kg)( m s 1 ) 6 (a) (i) Use of λ = h/p and p = mv Or v = h/mλ (1) Use of m = kg (1) v = m s 1 (1) λ = h/mv v = J s /( kg m) v = m s 1 (ii) Use of E k = ½ mv 2 Or E k = p 2 /2m Or see E k = J (1) Divided by (1) E k = 151 ev (accept values in range ev) (1) (ecf value of v from (a)) E k = ½ ( kg)( m s 1 ) 2 / ( J ev 1 ) E k = 151 ev The wavelength is similar in size to the nucleus (1) The wavelength/nucleus is (much) smaller / m / m (1) (if value is not given, wavelength is small or wavelength is very small is not sufficient) 7 (a) (i) Straight through, zero deflection, direction fired in. (ii) (Do not accept through or directly behind on its own) (1) (Atom consists) mainly/mostly of empty space Or Volume of atom very much greater than volume of nucleus (1) (do not credit if part of a list) Most of the mass is in the nucleus/centre (1) [it is not enough to say that the nucleus is dense/concentrated. Looking for idea that nearly all of the atom s mass is in the nucleus] Nucleus/centre is charged (1) Pearson Education Ltd Copying permitted for purchasing institution only. This material is not copyright free. 16

17 TOPIC 8 Nuclear and particle physics [ignore references to the charge being positive; just saying the nucleus is positive does not get the mark] (c) (i) Electrostatic / electromagnetic / electric / coulomb (1) (ii) Arrow starting on the path at closest point to the nucleus (1) Arrow pointing radially away from nucleus (1) (correct direction starting on the nucleus scores 2nd mark only) (iii) Deflection starts earlier (1) Final deflection is greater (1) (paths should diverge) 8 Marks can be gained if a concept is seen on a diagram Small central nucleus Most of atom is empty space Nucleus contains all positive charge in an atom Nucleus contains most of atom s mass Most alpha particles pass straight through with no deflection Some alpha particles pass through the nucleus and are deflected sideways A very few alpha particles are deflected by more than 90 Diagram illustrating any of the above points 8.2 Particle accelerators and detectors Particle accelerators 1 The particles are getting faster; the p.d. switches at a fixed frequency; to ensure p.d. switch at mid tube every time, tubes must get progressively longer. 2 Lawrence s second cyclotron, built with the assistance of Stanley Livingston, was 25 cm in diameter and accelerated protons to 1 MeV. (a) m s kg m s 1 (c) (d) 0.58 T 8.8 MHz 3 The Large Hadron Collider is a giant circular accelerator near Geneva in Switzerland. The LHC website has some facts about it: the circular ring has a circumference of 27 km each proton goes around over times a second each proton has ev of kinetic energy (a) m s 1 (c) (d) J J = ev which is several orders of magnitude smaller than 7 TeV. The high-speed protons have a significantly greater mass than kg because of the relativistic increase in mass at speeds near the speed of light Particle detectors 1 The ions created can be detected electrically; and causing ionisation is a common property of fast-moving ion particles; 2 Red = electron 3 radius of curvature of track is smaller above lead bar, so particle slower above, so must have travelled upwards through lead. Pearson Education Ltd Copying permitted for purchasing institution only. This material is not copyright free. 17

18 TOPIC 8 Nuclear and particle physics The Large Hadron Collider mv 1 r = so if we know the magnetic field strength and the speed of particle movement, we can find the charge Bq mass ratio for this particle. 2 If a particle track changes direction suddenly, that is a breach of conservation of momentum. This can only be reconciled if a particle not creating tracks is involved in a collision with the particle. 3 The collisions produce a huge variety of other particles, and so a large range of detectors is necessary. Each experiment concentrates on detectors of particular types. Also, it will not be possible for another lab elsewhere to verify the results, as building the LHC is such a monumental undertaking. Thus it must reproduce its own results, and the four separate experiments can independently verify each other. 4 Students own answers. 8.2 Answers to Exam-style questions 1 B 2 A 3 D 4 (a) (Magnetic) force acts at right angles to ion motion/current (1) Force is the centripetal force or causing centripetal acceleration or direction of acceleration/force is to centre (of circle) (1) See F = Bqv Or r = p/bq (1) F = mv 2 /r Or p = mv (1) f = v/2πr Or f = ω/2π) Or T = 2πr/v Or T = 2π/ω (1) (c) (i) Identifies positive (field) above and below (the ion) (1) which repels the ion (1) (ii) / (1) = (u) (1) (iii) Convert MeV to J (1) Convert J to kg (1) Convert kg to u (1) Mass loss = (u) (and this is more than u) (1) mass loss = 2.2 MeV J J to kg / kg kg to u / u 5 (a) Force on (charged) particles at right angles to motion (1) Causes circular motion [not spiral / curved] Or force/acceleration is centripetal (1) [credit first mark if clear from diagram] (i) Momentum: p = mv Or r = mv/be (1) v = 2πr/T Or v = rω Or ω = Be/m (1) Use of f = 1/T Or ω/2π (1) [allow q for e] Ber = mv Ber = m2πr/t Be = m2πf (ii) (Protons) accelerated / given energy, in the gaps / between dees / from one dee to the other (1) Pearson Education Ltd Copying permitted for purchasing institution only. This material is not copyright free. 18

19 TOPIC 8 Nuclear and particle physics Every half rotation/semicircle later (polarity of dees) needs a change (1) (iii) Relativistic effect / v approaching c / mass increases (1) so frequency decreases (1) [second mark consequent on first] (c) Must be accelerating due to circular motion (1) 6 Mark scheme: (Speed constant but) direction/velocity changing (1) Marks can be gained if a concept is seen on a diagram Charged particles can be accelerated by electric fields Both accelerators use fixed frequency alternating p.d. Linac acceleration tubes of increasing length Because particles move faster further along the accelerator Moving charged particles can be deflected in a circular path by a perpendicular magnetic field Cyclotron uses circular D-shaped electrodes Particles accelerate across gap between dees Magnetic field perpendicular to cyclotron dees makes particles accelerate in a circle Radius of circle travelled by particle increases with speed According to r = p Bq Cyclotron frequency is given by f = 2 Bq m Diagram illustrating any of the above points for linac Diagram illustrating any of the above points for cyclotron Magnetic fields can be applied to focus beams without deflecting sideways 8.3 The particle zoo Particle interactions 1 (a) 1.61 MeV = J Hz kg 3 (a) In order to conserve momentum kg = 1.00 MeV/c 2 4 (a) The lone blue track coming from the right. (c) It is the straightest track so moves the fastest. Also has no equal but opposite curvature partner track. The incoming anti-proton track is the straightest, so all the other smaller particles with their lesser kinetic energies can add up to a total of the p + /p initial mass/energy total. (d) Any red/green track pair with equal and opposite curvature, because their curvature is equal and opposite. (e) Each red track has a green counterpart, and these are opposite signs so cancel. Initial charge total of p + /p was zero. (f) As each pair of red and green tracks are mirror images, they start with opposite components of momentum which cancel, leaving only the initial momentum when all are added together The particle zoo 1 A positron has the same mass as an electron but is positively charged. 2 (a) Quarks feel the strong nuclear force but leptons do not. Opposite charges Pearson Education Ltd Copying permitted for purchasing institution only. This material is not copyright free. 19

20 TOPIC 8 Nuclear and particle physics (c) Muon has more mass (approx. 207 times) 3 (a) The predicted top quark has a much larger mass, so they needed more energy to be able to create this more massive particle. 175 GeV Mass = ; charge = 2 c Look at fig A and consider the three generations of neutrino, compared with the other fundamental particles in the Standard Model. (a) The neutrino masses are tiny. The neutrinos masses increase roughly an order of magnitude for each generation Particles and forces 1 Proton is uud, neutron is udd, so in beta minus decay, a down quark changes into an up quark. 2 (a) Baryon is qqq, meson is qq. Hadron can interact via strong nuclear force, lepton cannot. 3 The particle that carries the effect of a force between particles 4 It has a positive charge and the anti-particle would be negative, so different. 5 Photons are passed back and forth between the two protons 6 Their high mass values need high-energy experiments to create them, so we could only do that with the development of high-energy equipment Particle reactions 1 (a) Before: Q = +15; after: Q = +17 NOT permitted Before: Q = 0; after: Q = 0 permitted m p 0n e Demonstration that charge, baryon number and lepton number are conserved. None of the particles involved are strange, so strangeness is conserved at zero. 4 Proposed reaction 1: Q, B, L, S all conserved Proposed reaction 2: Q and B all conserved; but L is zero before and 2 (in total) after; and S is zero before and +1 (in total) after. 8.3 Answers to Exam-style questions 1 C 2 B 3 D 4* (QWC Work must be clear and organised in a logical manner using technical wording where appropriate) (After X) no tracks / track ceases (at X) / tracks cannot be seen (after X) (1) [allow lines for tracks] (so) uncharged/neutral particles produced Or only charged particles give tracks (1) At least one of the correct further events identified [i.e. at the V points] [in words or on diagram] (1) Both of the correct further events identified (1) 5* (a) 2/3 that of a proton or 2/ (C) (1) Mass = 80 MeV/c 2 (1) Charge = +1/3 (1) Pearson Education Ltd Copying permitted for purchasing institution only. This material is not copyright free. 20

21 TOPIC 8 Nuclear and particle physics (c) Recognition M means 10 6 (1) Convert ev to J or divide by c 2 (1) e.g Or / Answer (kg) (1) (d) (i) Kaon meson (1) Omega baryon (1) (ii) K + p (1) = K + + K 0 + Ω (1) [accept p or p + ; do not accept K for K 0 /K signs must be top right] (iii) Kaon plus = u s (1) (iv) Kaon neutral = d s or s d (1) [both marks can be inferred if equation in (d)(ii) is fully written in quark combinations] (QWC (i and iii) Spelling of technical terms must be correct and the answer must be organised in a logical sequence) Momentum conserved (1) Charge conserved (1) Energy / mass conserved (1) E = mc 2 (1) Kinetic energy (of kaon minus) is responsible (1) Momentum of three particles after = momentum of kaon before (1) Total charge 0 / charge before and after is 0 (1) Conservation of baryon number, quark number, strangeness (1) 6 (a) A sensible comment such as: A reference to symmetry Quarks in pairs (in the particle generations) 6 leptons known but only 5 quarks (1) (do not credit for each quark there has to be an anti-quark) (i) Same mass (1) Opposite charge (1) (ii) Conserve momentum (1) Initial (total) momentum is zero (1) (ignore reference to other conservation laws) (c) (i) Recognise (G)eV units of energy (1) (E = mc 2 so) E/c = mc = momentum (conditional mark) (1) Or recognise (G)eV/c 2 is unit of mass (1) Momentum is mass velocity (conditional mark) (1) (ii) Vectors added in sequence after μ 2 (1) Direction and magnitude of J3 and J4 accurate (1) Judge by eye and do not penalise missing arrows (iii) (GeV/c) (1) (iv) 7 values added together including the value from (iii) Or total length of vectors and 10 (1) (method mark) (v) Value in (iv) or 300 divided by 2 (1) (vi) Max 2 Large mass Or top quark (very) heavy (1) Large amount of energy required Or issue of providing sufficient energy (1) Availability of anti-matter is poor (1) Difficulty of storing anti-matter (1) Pearson Education Ltd Copying permitted for purchasing institution only. This material is not copyright free. 21

22 TOPIC 9 Thermodynamics 9.1 Heat and temperature Heat and temperature 1 (a) 300 K (c) 100 K 373 K (d) 15 C 2 The increase in temperature reduces the thermistor s resistance. This means it takes a smaller proportion of the supplied voltage, increasing the p.d. across the motor, so the motor (cooling fan) increases speed. 3 Change the motor for a voltmeter to record output voltages. Add a system for measuring the temperature of the thermistor, such as placing it (electrically insulated) in a water bath Internal energy 1 (a) 855 m s 1 29 K or 244 C 2 Estimate of 20 C (T = 293 K); assume all molecules are dinitrogen, so m = kg; gives 2 c = 509 m s 1. 3 Total KE of molecules in the sample 4 (a) Vertical line for c 0 reaching the peak of the curve Two more vertical lines slightly to the right of the peak, correctly labelled Heat transfer 1 (a) J 6780 kj needed, so 3080 s (over 51 minutes) 2 E = mc = = 8400 J E = L m = = E = mc = = J Total = = J C is the melting point of copper, so this additional energy was being absorbed to go towards melting the solid metal Ideal gas behaviour 1 (a) Students own answers, after fig A. Pearson Education Ltd Copying permitted for purchasing institution only. This material is not copyright free. 22

23 TOPIC 9 Thermodynamics (c) Measure ambient temperature throughout to confirm constant. Vary pressure as independent variable and for each pressure, measure volume from scale on glass tube. Plot p vs 1/V on graph and a straight best-fit line with positive gradient indicates that pressure is inversely proportional to volume. E.g. avoiding parallax errors on the glass tube. 2 Any refutation of an ideal gas assumption, e.g. Air molecules have a finite volume mol 4 Students own work. For example: Place a gas syringe in a water bath, with a thermocouple inside to measure the temperature. Keep at constant position (depth) underwater throughout, to confirm constant pressure. Vary temperature as independent variable and for each temperature, measure volume from scale on gas syringe. Plot V vs T on graph and a straight best-fit line with positive gradient indicates that volume is proportional to temperature Kinetic theory equations 1 The derivation considers the force on the wall of a container as it exerts a force on a molecule (Newton s third law) and how large this force must be to change the momentum of the molecule by a certain amount (Newton s second law) molecules 3 (a) 20.7 MPa 495 m s Answers to Exam-style questions 1 A 2 D 3 C 4 (a) Use of pv = NkT (1) * T = 870 (K) Or p = 12.4 (atmospheres) (1) If final pressure is given as Pa, then just use of mark pv N m m T = = = K Nk J K Or p = NkT J K 900 K = 6 3 V 3 10 m 6 p = Pa Pa = = Pa (QWC Work must be clear and organised in a logical manner using technical wording where appropriate) Atoms/molecules would gain energy (1) Atoms/molecules would escape from the liquid Or liquid propellant would vaporise / turn into gas Or the amount of gas in can would increase (1) Pressure would increase due to both temperature/energy increase and increase in amount of gas Or pressure would increase more for the same temperature increase Or pressure would be greater than 12 atmospheres before 900 K (1) The can would explode before 900 K reached (1) Pearson Education Ltd Copying permitted for purchasing institution only. This material is not copyright free. 23

24 TOPIC 9 Thermodynamics 5 (a) Any two from: Air behaves as an ideal gas (1) Temperature (in the lungs) stays constant (1) Implication of no change in mass of gas (1) (i) Use of ρ = m/v (1) Correct answer ( kg s 1 ) (1) m = Vρ = m kg m 3 = kg m = kg = kg s 1 t 60 s (ii) Use of ΔE = mcδθ (1) Correct answer (2.2 W) ecf (1) P = kg s J kg 1 K 1 ( ) K = 2.2 W 6 (a) Temperature (of gas) [treat references to oil/room as neutral] (1) Mass of air/gas Or number of atoms/molecules/moles of air/gas (1) [accept amount of air/gas, number of particles of air/gas] Assumption: idea that volume occupied by trapped air length of air in tube [e.g. volume = cross-sectional area length] (1) pl = a constant [accept pv = a constant] Or if p doubles, L halves (1) At least 2 pairs of p, L values correctly read from graph (1) Readings show that pl = 4500 (kpa cm) [± 100 kpa cm] (1) Or Readings show that p doubles when L is halved (1) [Accept references to V instead of L] p = 400 kpa, L = 11.0 cm pl = = 4400 p = 200 kpa, L = 23.0 cm pl = = 4600 (c) Use of pv = NkT [allow use of pv = nrt and N = nn A] (1) Conversion of temperature to kelvin (1) N = [accept answers in range to ] (1) [Answer in range but with an incorrect temperature conversion score max 2] N = Pa m m (d) (i) No change (1) J K ( ) K (ii) Similar curve (1) Shifted higher Or shifted to the right (1) = [an annotated diagram can score full marks] Pearson Education Ltd Copying permitted for purchasing institution only. This material is not copyright free. 24

25 TOPIC 9 Thermodynamics 7 (a) Circuit showing power supply unit (psu), heater, ammeter and voltmeter in parallel with heater (1) Max 6 Start below and finish above room temperature Measure the p.d. (voltage) and current at the start and at the end and find the average Switch off current and measure highest temperature reached Insulate block Measure mass of block (and heater) (Put oil into holes to help) Good thermal contact between heater, thermometer and block Use stop clock to measure time of current flow Calculate energy by multiplying voltage by current by time Draw appropriate graph and find gradient Use gradient correctly to find c Pearson Education Ltd Copying permitted for purchasing institution only. This material is not copyright free. 25

26 TOPIC 10 Nuclear radiation 10.1 Radioactivity Nuclear radiation 1 Students own answers: should include comparisons of penetrating power, ionising ability, hazards, structures Bq 3 Alpha radiation is blocked by a surface layer of dead skin so cannot penetrate to healthy tissue. Contamination on hand may be passed into mouth from hands and then is emitting alpha internally. 4 Cornwall generally has the highest background radiation level in Great Britain. 5 Description of experiment with different absorbers to selectively block each type of radiation and compare corrected count rates for each. Must mention background measurement and subtraction from each main reading Rate of radioactive decay 1 (a) s s 1 (c) s billion atoms 3 6 hours 4 Half-life is inversely proportional to decay constant. 5 The graph in fig D has a gradient of giving a half-life of 76 s Fission and fusion 1 Proton: J Neutron: J Electron: J J (124 MeV) 3 (a) 7.3 MeV Binding energy per nucleon goes up by 0.6 MeV 4 X = 95 (c) E k = J; rms speed = m s MeV MeV 7 Both fission and fusion products have higher binding energy per nucleon than the starting nuclei. This means that they are more tightly bound so there is less mass per nucleon. This drop in mass is released as energy Nuclear power stations 1 Costs to build; costs to decommission; fuel price; amount of fuel needed; power output; costs to dispose of radioactive waste; environmental impact; public opinion; and all these compared with the alternatives. 2 Students own answers 3 In case of an emergency, a mechanical system would not fail safe. 4 Australia 29%, Kazakhstan 12% and Russia 9% 5 Students own answers Pearson Education Ltd Copying permitted for purchasing institution only. This material is not copyright free. 26

27 TOPIC 10 Nuclear radiation 10.1 Answers to Exam-style questions 1 A 2 B 3 A 4 (a) Use of λ = ln 2/t 1/2 (1) λ = (yr 1 ) [λ = (s 1 ), λ = (min 1 )] (1) Use of A = A 0e λt (1) t = 950 (yr) [if λ = , then t = 960 (yr)] (1) [credit answers that use a constant ratio method to find the number of half-lives elapsed] λ = yr = yr s 1 = 16.5 s 1 e yr t s ln s t = = 947 yr yr Initial value of count rate should be bigger than 16.5 min 1 Or greater count rate from living wood in the past [e.g. A/A 0 smaller] Or initial value of count rate underestimated in the calculation Or initial number of undecayed atoms greater [e.g. N/N 0 smaller] Age of sample has been underestimated Or ship is older than 950 yr Or sample has been decaying for a longer time [If a calculation has been carried out to show that a greater value of initial activity leads to a greater age, then award both marks] 5 (a) (i) Use of m = kg (1) (ii) Use of ½ m<c 2 > = 3/2 kt (1) c rms = 2800(m s 1 ) (no ue) (1) <c 2 > = 3kT = m <c 2 > = m 2 s 2 c rms = U n J K 310 K = m 2 s kg m s = m s 1 U Cs Rb n Nucleon, proton numbers correct [236, 55] (1) Number of neutrons correct [2] (1) (iii) Attempt at calculation of mass defect (1) Use of 1 u = MeV (1) Use of fission rate = Fission rate = s 1 (1) power output energy per fission (1) Δm = ( ) Δm = kg = kg Pearson Education Ltd Copying permitted for purchasing institution only. This material is not copyright free. 27