IB physics diagrams

Reading cue. The area under the triangle is the impulse; 40 N is the peak value at one instant, not a force that acts throughout, so nothing on this graph may simply be multiplied by 0.20 s.

1. aState State the magnitude of the resultant force acting at t = 0.050 s.

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   Check answer · 2 marks

   1. reads the rising straight line halfway between the origin and the peak
   2. F = 20 N

2. bSketch The object has mass 0.50 kg. Sketch the acceleration-time graph for the same 0.20 s, marking values on both axes.

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   Check answer · 3 marks

   1. same triangular shape: straight rise from the origin, straight fall to zero at t = 0.20 s
   2. peak acceleration = 40 / 0.50 = 80 m s⁻² at t = 0.10 s
   3. axes labelled a / m s⁻² and t / s with 80, 0.10 and 0.20 marked

3. cDetermine Determine the impulse delivered during the first 0.10 s, and hence determine the time at which the object, starting from rest, is moving at half of its final speed.

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   Check answer · 3 marks

   1. impulse to t = 0.10 s = ½ × 0.10 × 40 = 2.0 N s
   2. total impulse = ½ × 0.20 × 40 = 4.0 N s, so exactly half the momentum has been delivered by the peak
   3. mass is constant and the object started from rest, so half the momentum is half the speed: t = 0.10 s

4. dDetermine The same force pulse is now applied to a 0.50 kg object already moving at 8.0 m s⁻¹ in the direction opposite to the force. Determine its velocity at t = 0.20 s, and explain why the same area under the graph still applies.

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   Check answer · 4 marks

   1. area under the graph = ½ × 0.20 × 40 = 4.0 N s
   2. impulse gives the change in momentum, so Δv = 4.0 / 0.50 = 8.0 m s⁻¹ in the direction of the force
   3. taking the force direction as positive, v = −8.0 + 8.0 = 0, so the object is momentarily at rest
   4. the area fixes the change in velocity only, and is independent of the velocity the object started with

Reading cue. L runs from the jaws of the clamp to the centre of the bob, not to its top: the radius of the bob is part of the length, and the dashed arc shows the path, not an angle.

1. aState A student can only reach the string with a metre rule. State the two lengths that must be added together to give the L marked on the drawing.

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   Check answer · 2 marks

   1. the length of string from the jaws of the clamp down to the top of the bob
   2. plus the radius of the bob

2. bOutline Outline why the timing of each swing should be started and stopped as the bob passes the lowest point of the dashed arc rather than at either end of it.

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   Check answer · 3 marks

   1. the bob moves fastest at the lowest point, so the uncertainty in judging the instant of passing is smallest
   2. at the ends of the arc the bob is momentarily at rest, so the eye cannot fix the instant it turns
   3. a fiducial mark placed at the lowest point makes the judgement repeatable from swing to swing

3. cDetermine A student grips the string in the jaws, measures 0.788 m from the jaws to the top of a bob of radius 12 mm, and uses that as L with a measured period of 1.794 s. Determine the percentage error this introduces into g.

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   Check answer · 3 marks

   1. true length L = 0.788 + 0.012 = 0.800 m
   2. g = 4π²L / T², so with T fixed the fractional error in g equals the fractional error in L
   3. error = 0.012 / 0.800 = 1.5%, and g comes out as 9.67 m s⁻² instead of 9.81 m s⁻² — too low

4. dDiscuss The clamp arm is carried on a vertical rod in a heavy base. Discuss the effect on the measured period if the base were light enough to let the rod rock slightly in time with the swing.

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   Check answer · 4 marks

   1. the point of suspension would no longer be fixed, so the length governing the period is not the L that was measured
   2. rod and pendulum exchange energy, so the amplitude decays faster than air resistance alone would cause
   3. the period is shifted in the same direction on every trial, so the error is systematic and repeating the timing does not reduce it
   4. concludes that the heavy base, or clamping the base to the bench, is what makes L a valid measure of the pendulum length

Reading cue. The 12 m s⁻¹ arrow is horizontal and is the whole initial velocity, so the ball leaves the roof with zero vertical speed and 25 m is a free-fall drop, not a length of the dashed curve.

1. aState State the horizontal component of the ball's velocity at the instant it reaches the ground, and give a reason.

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   Check answer · 2 marks

   1. 12 m s⁻¹
   2. no horizontal force acts while air resistance is negligible, so the horizontal component is unchanged throughout

2. bDetermine Determine the horizontal distance from the foot of the building to the point where the dashed path meets the ground.

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   Check answer · 3 marks

   1. vertical: 25 = ½ × 9.81 × t², so t = 2.26 s
   2. horizontal motion is at constant velocity: x = 12 × 2.26
   3. x = 27 m

3. cDetermine Determine the angle below the horizontal at which the dashed path meets the ground.

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   Check answer · 3 marks

   1. vertical component on landing: v_y = 9.81 × 2.26 = 22.1 m s⁻¹
   2. horizontal component is still 12 m s⁻¹, so tan θ = 22.1 / 12 = 1.85
   3. θ = 62° below the horizontal

4. dDiscuss A second ball is thrown horizontally from the same point at 24 m s⁻¹. Discuss how its time of flight, its landing distance and its landing angle each compare with those of the first ball.

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   Check answer · 4 marks

   1. the vertical motion is unaffected by the horizontal velocity, so the time of flight is the same, 2.26 s
   2. the landing distance doubles to 54 m, since x = ut with the same t
   3. the vertical component on landing is still 22.1 m s⁻¹, so tan θ = 22.1 / 24 and θ = 43°
   4. concludes that throwing harder flattens the path and moves the landing point out, but does not keep the ball in the air any longer

Reading cue. The 25 N arrow points down the slope, so it is the resistance and not the drive; the 4.0° is measured from the dashed horizontal, so the weight component along the road uses sin, not cos.

1. aState State the resultant force acting on the cyclist and bicycle while they travel up the slope at the speed shown, and give a reason.

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   Check answer · 2 marks

   1. zero
   2. the velocity is constant, so by Newton's first law the forward force, the weight component and the 25 N resistance balance

2. bDetermine Determine the gravitational potential energy gained by the cyclist and bicycle during one minute of riding at the speed shown.

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   Check answer · 3 marks

   1. distance along the road = 5.5 × 60 = 330 m
   2. vertical rise = 330 × sin 4.0° = 23.0 m
   3. ΔE_p = 78 × 9.81 × 23.0 = 1.76 × 10⁴ J

3. cDetermine Determine the work done against the resistive force over the same minute, and hence determine the percentage of the cyclist's useful work that goes into raising her and the bicycle.

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   Check answer · 3 marks

   1. work against resistance = 25 × 330 = 8.25 × 10³ J
   2. total useful work = 1.76 × 10⁴ + 0.825 × 10⁴ = 2.59 × 10⁴ J
   3. fraction raising the cyclist = 1.76 × 10⁴ / 2.59 × 10⁴ = 68%

4. dDiscuss Further on, the road steepens to 8.0° and the cyclist holds the same useful power output of 431 W. Discuss how her steady speed changes, supporting your answer with a calculation, and whether the resistive force would still be 25 N.

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   Check answer · 4 marks

   1. component of weight down an 8.0° slope = 78 × 9.81 × sin 8.0° = 106 N
   2. forward force needed = 106 + 25 = 131 N, so v = 431 / 131 = 3.3 m s⁻¹
   3. the force needed has risen from 78.4 N only to 131 N, a factor of 1.7, so the speed falls by 1.7 and not by the factor of 2 the doubled angle might suggest — the 25 N resistance is unchanged while only the weight component doubles
   4. air resistance is part of the 25 N and falls as the speed falls, so the true steady speed is a little above 3.3 m s⁻¹ and this figure is a lower bound

Reading cue. The 1.2 m is measured from the dashed line through the axis, so it is the drop of the centre of mass and not the length of the slope; 0.15 m is a radius, not a diameter.

1. aState State the relationship between the translational speed of the cylinder's centre and its angular speed while it rolls without slipping, using the dimension marked on the drawing.

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   Check answer · 2 marks

   1. v = ωR
   2. with R = 0.15 m from the figure, v = 0.15ω

2. bDetermine The cylinder reaches the bottom of the slope with a translational speed of 3.96 m s⁻¹. Determine its angular speed there.

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   Check answer · 3 marks

   1. rolling without slipping, so ω = v / R
   2. ω = 3.96 / 0.15
   3. ω = 26 rad s⁻¹

3. cDetermine The slope is inclined at 25° to the horizontal. Determine the acceleration of the cylinder's centre down the slope, and the time it takes to reach the bottom from rest.

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   Check answer · 4 marks

   1. for a solid cylinder a = g sin θ / (1 + I/MR²) = g sin θ / 1.5 = (2/3)g sin θ
   2. a = (2/3) × 9.81 × sin 25° = 2.76 m s⁻²
   3. the distance travelled is along the slope, 1.2 / sin 25° = 2.84 m, not 1.2 m
   4. t = √(2 × 2.84 / 2.76) = 1.43 s, which checks against v = at = 2.76 × 1.43 = 3.96 m s⁻¹

4. dShow (that) Friction is the only force exerting a torque about the axis. Show that the coefficient of static friction must be at least (tan θ)/3 for the cylinder to roll without slipping, and evaluate this for the 25° slope.

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   Check answer · 4 marks

   1. torque about the axis: fR = Iα = ½MR²(a/R), so f = ½Ma
   2. substituting a = (2/3)g sin θ gives f = (1/3)Mg sin θ
   3. the normal force is N = Mg cos θ, so μ_min = f/N = (tan θ)/3
   4. for θ = 25°, μ_min = 0.466 / 3 = 0.16

Reading cue. The 2d across the end of the lower wire is a diameter, not an area: doubling it multiplies the cross-section by four, so wire 2 is twice as long but four times as wide in area.

1. aState State how the resistivity of wire 2 compares with that of wire 1, and give a reason.

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   Check answer · 2 marks

   1. it is the same
   2. resistivity is a property of the material and its temperature, not of the length or thickness of the specimen

2. bDetermine Wire 1 has resistance R. Determine the length of wire of the lower wire's thickness that would be needed to give a resistance of R, and state how it compares with the 2L drawn.

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   Check answer · 3 marks

   1. A ∝ d², so the lower wire's cross-sectional area is four times that of wire 1 and its resistance per unit length is R/(4L)
   2. for a resistance of R the length must be 4L
   3. that is twice the 2L drawn, so wire 2 as drawn has a resistance of only R/2

3. cDetermine The two wires are connected in turn across the same battery. Determine the ratio of the current in wire 2 to the current in wire 1, and the ratio of the electron drift speeds in them.

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   Check answer · 3 marks

   1. the same p.d. is applied to each, so I ∝ 1/R and I₂/I₁ = R/(R/2) = 2
   2. I = nAvq with the same n and q in both, so v = I/(nAq)
   3. A₂ = 4A₁ and I₂ = 2I₁, so v₂/v₁ = 2/4 = 0.5 — the electrons drift at half the speed in the thicker wire

4. dExplain The two wires are now joined end to end and the pair is connected across a supply. Explain which wire dissipates the greater power, and what happens to the electron drift speed as the electrons cross the junction.

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   Check answer · 4 marks

   1. in series the current is the same in both, so P = I²R and wire 1 dissipates twice the power of wire 2 (R against R/2)
   2. charge is conserved, so the same number of electrons per second passes every cross-section, including the junction
   3. I = nAvq with A four times larger in wire 2, so the drift speed there is one quarter of that in wire 1
   4. the electrons neither pile up nor speed up at the junction: the current is fixed and only the drift speed adjusts to the area

Reading cue. The thermometer bulb sits in the liquid and not against the coil, and 50.0 W is labelled on the supply, so it is the electrical input rather than the power reaching the liquid.

1. aState State what the 50.0 W on the power supply measures, and state one feature of the drawing that means less than 50.0 W warms the liquid.

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   Check answer · 2 marks

   1. the rate at which electrical energy is supplied to the heating element
   2. the container is open at the top, so energy escapes there by evaporation and convection (or: the leads, the heater and the container itself absorb energy)

2. bDetermine The liquid has specific heat capacity 2.4 × 10³ J kg⁻¹ K⁻¹. Determine how long the heater must run to raise the temperature of the liquid shown by 1.00 K, assuming all the electrical energy reaches it.

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   Check answer · 3 marks

   1. E = mcΔT = 0.500 × 2.4 × 10³ × 1.00 = 1.20 × 10³ J
   2. t = E / P = 1.20 × 10³ / 50.0
   3. t = 24.0 s

3. cDetermine The open top is the one surface the hatched insulation does not cover, and 0.42 g of liquid evaporates from it each minute. The specific latent heat of vaporisation is 8.5 × 10⁵ J kg⁻¹. Determine the power this carries away and the resulting error in c.

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   Check answer · 4 marks

   1. mass evaporating per second = 4.2 × 10⁻⁴ / 60 = 7.0 × 10⁻⁶ kg s⁻¹
   2. power carried away = 7.0 × 10⁻⁶ × 8.5 × 10⁵ = 6.0 W
   3. only 50.0 − 6.0 = 44.0 W actually warms the liquid, so a value calculated from 50.0 W is too large by a factor 50.0/44.0 = 1.14
   4. c comes out about 14% too high

4. dSuggest The student switches the supply off and keeps reading the thermometer as the liquid cools. Suggest how the cooling readings can be used to correct her value of c, and outline the assumption the correction rests on.

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   Check answer · 4 marks

   1. after switch-off nothing but the losses is acting, so the rate of fall measured at a given temperature is a direct measure of the loss at that temperature
   2. at that same temperature the heating run gives 50.0 = mc × (rate of rise) + mc × (rate of fall on cooling), so adding the two gradients yields c without needing the loss in watts first
   3. the corrected value of c is smaller than the uncorrected one, because part of the 50.0 W was never warming the liquid
   4. assumes the rate of loss depends only on the excess temperature over the surroundings, so it is the same whether the heater is on or off

Reading cue. The arrows differ in length as well as direction, so the drawing shows a spread of speeds; the rigid walls fix V, which is why warming the gas raises p rather than expanding it.

1. aState State the two quantities shown by or implied by the drawing that cannot change when the container is warmed, and give a reason for each.

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   Check answer · 2 marks

   1. the volume, 5.0 × 10⁻³ m³, because the container is rigid
   2. the amount of gas, because the container is sealed

2. bDetermine Determine the temperature to which the gas must be warmed for the pressure to reach 3.0 × 10⁵ Pa.

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   Check answer · 3 marks

   1. V and n are fixed, so p/T is constant: p₁/T₁ = p₂/T₂
   2. T₂ = 290 × (3.0 × 10⁵ / 2.4 × 10⁵) = 362.5 K
   3. T₂ = 363 K, that is about 89 °C

3. cDetermine The arrows represent a spread of molecular speeds. Determine the root mean square speed of a nitrogen molecule at the temperature shown. The molar mass of N₂ is 28.0 g mol⁻¹.

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   Show a hint

   The mass of one molecule is the molar mass divided by the Avogadro constant, not the molar mass itself.

   Check answer · 4 marks

   1. average kinetic energy = (3/2)k_BT = 1.5 × 1.38 × 10⁻²³ × 290 = 6.0 × 10⁻²¹ J
   2. mass of one molecule = 28.0 × 10⁻³ / 6.02 × 10²³ = 4.65 × 10⁻²⁶ kg
   3. ½mv²_rms = 6.0 × 10⁻²¹, so v²_rms = 2.58 × 10⁵ m² s⁻²
   4. v_rms = 5.1 × 10² m s⁻¹

4. dExplain A valve is opened briefly and exactly half the nitrogen escapes, the temperature being held at 290 K. Explain how the pressure, the average kinetic energy of a molecule, and the rate of collisions on a given wall each change.

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   Check answer · 4 marks

   1. p = nRT/V with n halved and V and T unchanged, so the pressure halves to 1.2 × 10⁵ Pa
   2. the average kinetic energy of a molecule is unchanged, because it depends only on the temperature
   3. each remaining molecule moves just as fast and hits just as hard, but there are half as many, so the rate of collisions on a given wall halves
   4. the pressure falls because the collisions are fewer, not because they are gentler

Reading cue. The dashed boundary is not a component: r = 0.75 Ω sits inside the cell, so 6.0 V is the e.m.f. of the whole loop and never the reading across the terminals while current flows.

1. aState The loop is broken at R and a voltmeter of very high resistance is connected across the cell's terminals. State its reading, and give a reason.

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   Check answer · 2 marks

   1. 6.0 V
   2. with no current there is no potential difference across r, so the terminal p.d. equals the e.m.f.

2. bDetermine The current marked I is 1.6 A. Determine the charge that passes through R in 2.0 minutes, and the chemical energy the cell converts in that time.

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   Check answer · 3 marks

   1. Q = It = 1.6 × 120 = 192 C
   2. the e.m.f. is the energy the cell gives to each coulomb, 6.0 J C⁻¹
   3. E = εQ = 6.0 × 192 = 1.2 × 10³ J

3. cDetermine A second 3.0 Ω resistor is connected in parallel with R. Determine the new terminal potential difference of the cell, and state what happens to the e.m.f.

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   Check answer · 4 marks

   1. external resistance becomes 1.5 Ω
   2. I = 6.0 / (1.5 + 0.75) = 2.67 A
   3. terminal p.d. = 2.67 × 1.5 = 4.0 V
   4. the e.m.f. stays at 6.0 V; only the lost volts Ir has changed, from 1.2 V to 2.0 V

4. dShow (that) Show that the power delivered to the external resistor is greatest when R equals r, determine that maximum power for the cell drawn, and state the power actually delivered to the 3.0 Ω resistor.

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   Show a hint

   No calculus is needed: try rewriting (R + r)²/R as (R − r)²/R + 4r.

   Check answer · 4 marks

   1. P = I²R = ε²R/(R + r)², which can be written P = ε²/[(R + r)²/R]
   2. (R + r)²/R = (R − r)²/R + 4r, so the denominator is least, and P greatest, when R = r
   3. maximum power = ε²/(4r) = 6.0² / (4 × 0.75) = 12 W
   4. with R = 3.0 Ω the circuit delivers only I²R = 1.6² × 3.0 = 7.7 W, because R is four times r

Reading cue. The hatched wall is a displacement node and the unblocked end an antinode; L = 0.85 m is the whole pipe, not the quarter wavelength it holds.

1. aState State how many displacement nodes and how many displacement antinodes the air column contains when the pipe sounds its lowest note, and say where each one lies on the drawing.

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   Check answer · 2 marks

   1. one node, at the closed (hatched) end
   2. one antinode, at the open end

2. bCalculate Calculate the wavelength and the frequency of the next standing wave this air column can support above its lowest note.

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   Check answer · 3 marks

   1. next mode fits three quarter-wavelengths into the pipe, so λ = 4L/3
   2. λ = 4 × 0.85 / 3 = 1.13 m
   3. f = v / λ = 3 × 340 / (4 × 0.85) = 300 Hz

3. cDetermine The open end is now sealed with a second wall, so that both ends of the 0.85 m pipe are closed. Determine the frequency of the lowest note the pipe can then sound, and state how it compares with the value for the pipe as drawn.

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   Check answer · 4 marks

   1. both ends are now displacement nodes, so the pipe holds half a wavelength: L = λ/2
   2. λ = 2 × 0.85 = 1.70 m
   3. f = 340 / 1.70 = 200 Hz
   4. this is twice the 100 Hz of the pipe as drawn

4. dSuggest When this pipe is built and blown, its lowest note is measured to be a few hertz below the value predicted from the 0.85 m marked on the drawing. Suggest a physical reason for the discrepancy.

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   Check answer · 3 marks

   1. the displacement antinode does not form exactly at the mouth but a little way outside it
   2. the effective length of the air column is therefore greater than the 0.85 m marked between the ticks
   3. a greater effective length gives a longer wavelength, and so a frequency below the predicted value

Reading cue. The arrow leaves the ambulance pointing at the observer, so 25 m s⁻¹ is a source closing on a still listener: it is subtracted from 340 m s⁻¹, not added.

1. aState State how the wavelength of the sound in the region between the ambulance and the observer compares with the wavelength of the sound behind the ambulance.

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   Check answer · 2 marks

   1. wavelength is shorter in front of the ambulance, on the side where the observer stands
   2. wavelength is longer behind the ambulance

2. bCalculate Calculate the wavelength of the sound in the air between the ambulance and the observer.

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   Check answer · 3 marks

   1. in one period the wavefronts advance 340 m s⁻¹ while the source advances 25 m s⁻¹ after them
   2. λ = (340 - 25) / 512
   3. λ = 0.615 m

3. cDetermine Determine the frequency the observer hears once the ambulance has passed and is travelling away along the same road, and hence determine the change in the frequency heard as the ambulance goes by.

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   Check answer · 4 marks

   1. receding source: the source speed is added, giving (340 + 25) in the denominator
   2. f = 512 × 340 / 365 = 477 Hz
   3. approaching frequency is 512 × 340 / 315 = 553 Hz
   4. change = 553 - 477 = 76 Hz, heard as a sudden drop in pitch as the ambulance passes

4. dDiscuss The ambulance is instead parked at the roadside with its siren sounding, and the observer runs towards it along the road at 25 m s⁻¹, the same speed as the arrow in the figure. Discuss whether the observer now hears the same frequency as in the situation drawn.

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   Check answer · 4 marks

   1. moving observer, stationary source: f = 512 x (340 + 25) / 340 = 550 Hz
   2. this is close to, but not equal to, the 553 Hz of the drawn moving-source case
   3. so the shift is not fixed by the relative motion alone
   4. the air is a preferred medium: a moving source alters the wavelength itself, while a moving observer meets unaltered wavefronts at a different rate

Reading cue. Only the middle ray enters the gap, so b is the gap and not the bar; the tall central band reaches the first minimum on both sides, so it spans twice each side band.

1. aState State, in terms of the wavelength and the slit width b marked on the figure, the angle at which the first dark band appears, and state the line on the drawing from which that angle is measured.

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   Check answer · 2 marks

   1. θ = λ / b
   2. measured from the dashed straight line continuing from the slit to the screen

2. bDetermine The slit width is b = 0.12 mm, the light has wavelength 590 nm, and the screen is 1.8 m from the slit. Determine the width of the tall central band drawn on the screen.

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   Check answer · 3 marks

   1. first minimum at θ = λ / b = 590 × 10⁻⁹ / 1.2 × 10⁻⁴ = 4.9 × 10⁻³ rad
   2. the central band runs from the first minimum on one side to the first minimum on the other, so its width is 2 λ D / b
   3. width = 2 × 590 × 10⁻⁹ x 1.8 / 1.2 × 10⁻⁴ = 1.8 × 10⁻² m (1.8 cm)

3. cShow (that) Show that the central band should be drawn twice as wide as each of the bands beside it.

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   Check answer · 4 marks

   1. minima occur at θ = λ/b, 2 λ/b, 3 λ/b, ... from the centre
   2. the central band lies between the minima at -λ/b and +λ/b, so it spans 2 λ/b
   3. each side band lies between consecutive minima, for example λ/b and 2 λ/b, so it spans λ/b
   4. hence the widths are in the ratio 2 : 1

4. dDiscuss The single slit is narrowed until b is smaller than the wavelength of the light. Discuss what then becomes of the row of bands drawn on the screen.

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   Check answer · 5 marks

   1. a minimum requires sin θ = λ / b
   2. with b < λ this gives λ/b > 1, which no angle can satisfy
   3. so no dark bands form and the screen is lit right across its width with no minima
   4. far less light passes through, so the illumination is much fainter
   5. the drawn row of separate bands is replaced by a single very broad, faint spread

Reading cue. The oscillation arrow is centred on the hanging position, not the spring's natural length: x is measured from there, where the spring already carries an extension mg/k.

1. aState State the point of the oscillation drawn at which the block's acceleration is zero, and the points at which its magnitude is greatest.

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   Check answer · 2 marks

   1. acceleration is zero at the hanging position, at the middle of the double-headed arrow
   2. acceleration is greatest at the two ends of the arrow, at maximum displacement

2. bShow (that) Show that, before it is set oscillating, the block stretches the spring by mg/k, and hence calculate this extension for m = 0.20 kg and k = 25 N m⁻¹.

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   Check answer · 3 marks

   1. at rest the upward spring tension balances the weight: ke = mg
   2. rearranging gives e = mg / k
   3. e = 0.20 × 9.8 / 25 = 7.8 × 10⁻² m (7.8 cm)

3. cDetermine For the same block and spring, determine the period of the oscillation, and determine the magnitude of the block's acceleration at the top of the arrow when the amplitude is 3.0 cm.

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   Check answer · 4 marks

   1. ω = sqrt(k/m) = sqrt(25 / 0.20) = 11.2 rad s⁻¹
   2. T = 2π / ω = 0.56 s
   3. using a = -ω² x, the magnitude at maximum displacement is ω² x0 = 125 × 0.030
   4. a = 3.8 m s⁻², directed downward, back towards the hanging position

4. dExplain Explain why, for this apparatus, the motion stops being simple harmonic once the amplitude exceeds the extension found earlier.

   top of the paper4 marks

   Show a hint

   What force can act on the block once the spring has returned to its natural length?

   Check answer · 4 marks

   1. at an amplitude equal to the static extension e, the maximum acceleration is ω² e = (k/m)(mg/k) = g
   2. at the top of that swing the spring has returned to its natural length and exerts no force on the block
   3. a larger amplitude would demand a downward acceleration greater than g, which gravity alone cannot supply
   4. the spring goes slack instead, the restoring force is no longer proportional to displacement, and the block briefly falls freely

Reading cue. d = 0.25 mm is centre-to-centre between the slits, and D = 2.40 m runs from the slit plane to the screen, not from the laser drawn further left.

1. aState State why the two slits are illuminated by the single laser drawn, rather than by two separate lasers of the same wavelength placed one behind each slit.

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   Check answer · 2 marks

   1. the two slits must act as coherent sources, with a constant phase difference between them
   2. one laser feeding both slits guarantees this; two independent lasers would drift in phase and the fringes would wash out

2. bDetermine The whole arrangement, from slits to screen, is immersed in water of refractive index 1.33. Determine the wavelength of the light in the water and the new separation of adjacent bright fringes.

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   Check answer · 4 marks

   1. the frequency is unchanged and the speed falls, so λ = 633 / 1.33 = 476 nm
   2. fringe separation s = λ D / d = 476 × 10⁻⁹ x 2.40 / 0.25 × 10⁻³
   3. s = 4.6 × 10⁻³ m (4.6 mm)
   4. the fringes move closer together than the 6.1 mm they are in air

3. cDetermine Keeping the laser and the 2.40 m screen distance as drawn, determine the slit separation that would be needed to space the bright fringes 1.0 cm apart, and state one disadvantage of working with the pattern this produces.

   demanding4 marks

   Check answer · 4 marks

   1. rearrange s = λ D / d to give d = λ D / s
   2. d = 633 × 10⁻⁹ x 2.40 / 0.010
   3. d = 1.5 × 10⁻⁴ m (0.15 mm), so the slits must be closer together than the 0.25 mm drawn
   4. disadvantage: the same light is spread over fewer, wider fringes, so each is fainter and fewer of them fall on the screen

4. dEvaluate A student uses this arrangement to find the laser's wavelength from λ = sd/D. The measurements are d = 0.25 +/- 0.01 mm, D = 2.40 +/- 0.01 m, and the width of ten fringe spacings is 61 +/- 1 mm. Evaluate which measurement limits the precision of the result.

   top of the paper6 marks

   Check answer · 6 marks

   1. for λ = sd/D the fractional uncertainties add: 0.01/0.25 = 4%, 1/61 = 1.6%, 0.01/2.40 = 0.4%
   2. total fractional uncertainty is about 6%
   3. s = 61/10 = 6.1 mm, so λ = 6.1 × 10⁻³ x 0.25 × 10⁻³ / 2.40 = 6.35 × 10⁻⁷ m (635 nm)
   4. 6% of that is 38 nm, so the result is (6.4 +/- 0.4) x 10² nm, which covers the true 633 nm
   5. the slit separation dominates, contributing two thirds of the total; it is the smallest length measured and is quoted to only two significant figures
   6. even a perfect D and a perfect fringe measurement would leave 4%, so only a better measurement of d improves the result appreciably

Reading cue. The horizontal axis is displacement, not time: kinetic energy is an inverted parabola peaking at x = 0 here, not a cosine curve, and the vertical axis carries no scale.

1. aState The axes are drawn to span the whole of the motion, from one extreme to the other. State the values of x, read from the horizontal scale, at which a potential-energy curve drawn on these axes would reach the total energy of the particle, and state the kinetic energy there.

   recall2 marks

   Check answer · 2 marks

   1. at x = -8.0 cm and x = +8.0 cm, the ends of the scale, which are the amplitude
   2. the kinetic energy is zero at those two displacements

2. bDetermine The total energy of the particle is 1.23 × 10⁻² J. Determine its potential energy and its kinetic energy at the gridline x = +4.0 cm.

   routine3 marks

   Check answer · 3 marks

   1. at that gridline x / x0 = 4.0 / 8.0 = 0.50, so x² / x0² = 0.25
   2. potential energy = 0.25 × 1.23 × 10⁻² = 3.1 × 10⁻³ J
   3. kinetic energy = 1.23 × 10⁻² - 3.1 × 10⁻³ = 9.2 × 10⁻³ J

3. cDetermine Determine the displacement, on the scale given, at which the kinetic and potential energies of the particle are equal, and state whether this falls on the +4.0 cm gridline drawn.

   demanding4 marks

   Check answer · 4 marks

   1. equal energies means each is half the total, so x² / x0² = 0.50
   2. x = x0 / sqrt(2) = 8.0 / sqrt(2)
   3. x = +/- 5.7 cm
   4. this lies outside the +/- 4.0 cm gridlines, so the curves do not cross at half the amplitude

4. dExplain The particle is restarted with amplitude 4.0 cm, the gridline on these axes, and the same period. Explain what happens to each of the two energy curves, giving the new total energy.

   top of the paper4 marks

   Check answer · 4 marks

   1. total energy is proportional to x0², so it falls to a quarter: 3.1 × 10⁻³ J
   2. the potential-energy curve keeps exactly the same shape, since potential energy depends on x and not on the amplitude; it is simply followed only out to +/- 4.0 cm
   3. the kinetic-energy curve is a new, lower inverted parabola, still peaking at x = 0 but now at 3.1 × 10⁻³ J
   4. the curves still cross where each is half the total, now at x = 4.0 / sqrt(2) = +/- 2.8 cm

Reading cue. The crosses mean B is into the page, and v is the tangent at the lowest point, so the force there points straight up towards the centre, at right angles to v.

1. aState State the direction of the magnetic force on the proton at the instant drawn, and state whether the proton travels clockwise or anticlockwise round the circle.

   recall2 marks

   Check answer · 2 marks

   1. the force is vertically upward at that instant, along the radius towards the centre of the circle
   2. the proton travels anticlockwise round the circle

2. bDetermine The field has magnitude B = 0.35 T and the radius marked is r = 4.0 cm. Determine the speed of the proton.

   routine3 marks

   Check answer · 3 marks

   1. the magnetic force supplies the centripetal force: qvB = mv² / r, so r = mv / (qB)
   2. v = qBr / m = 1.60 × 10⁻¹⁹ x 0.35 × 0.040 / 1.67 × 10⁻²⁷
   3. v = 1.3 × 10⁶ m s⁻¹

3. cShow (that) Show that the time taken for the proton to travel once round the circle drawn does not depend on the radius r, and calculate that time for B = 0.35 T.

   demanding4 marks

   Check answer · 4 marks

   1. from qvB = mv² / r, the radius is r = mv / (qB)
   2. the period is T = 2π r / v = 2π m / (qB), in which both v and r have cancelled
   3. T = 2π x 1.67 × 10⁻²⁷ / (1.60 × 10⁻¹⁹ x 0.35)
   4. T = 1.9 × 10⁻⁷ s

4. dDiscuss An electron enters the same field region at the same point, travelling to the right with the same speed as the proton. Discuss how the path drawn would change.

   top of the paper4 marks

   Check answer · 4 marks

   1. the charge is negative, so the force reverses: the centre of the circle now lies below the entry point and the electron is traced clockwise
   2. since r = mv / (qB) is proportional to mass, the radius shrinks by the mass ratio of about 1836
   3. r = 0.040 / 1836 = 2.2 × 10⁻⁵ m, far too small to show on the drawing at this scale
   4. the period shrinks by the same factor, to about 1.0 × 10⁻¹⁰ s

Reading cue. r runs from the planet's centre to the satellite, not from its surface, and v is tangential, at right angles to r, so gravity does no work and the speed is constant.

1. aState State the direction of the resultant force on the satellite at the instant drawn, and state its effect on the satellite's speed.

   recall2 marks

   Check answer · 2 marks

   1. the force acts along the line marked r, directed towards the centre of the planet
   2. it is perpendicular to v, so it changes the direction of the velocity but not the speed

2. bShow (that) Show that the orbital speed of the satellite is v = sqrt(GM/r), and hence show that the square of its orbital period is proportional to the cube of the radius marked.

   routine4 marks

   Check answer · 4 marks

   1. gravitational attraction supplies the centripetal force: GMm / r² = mv² / r
   2. the satellite mass m cancels, giving v² = GM / r and so v = sqrt(GM/r)
   3. the period is T = 2π r / v, so T² = 4π² r² / (GM/r) = 4π² r³ / (GM)
   4. hence T² is proportional to r³, with a constant that does not contain m

3. cDetermine The planet has mass M = 6.4 × 10²³ kg and the orbit radius marked is r = 9.4 × 10⁶ m. Determine the orbital speed of the satellite and the time it takes to complete one orbit.

   demanding4 marks

   Check answer · 4 marks

   1. v = sqrt(GM/r) = sqrt(6.67 × 10⁻¹¹ x 6.4 × 10²³ / 9.4 × 10⁶)
   2. v = 2.1 × 10³ m s⁻¹
   3. T = 2π r / v = 2π x 9.4 × 10⁶ / 2.1 × 10³
   4. T = 2.8 × 10⁴ s, about 7.7 hours

4. dDiscuss A thruster fires briefly along the direction of the arrow v, and the satellite settles into a new circular orbit of larger radius. Discuss why its orbital speed in the new orbit is lower than in the orbit drawn, even though the thruster did positive work on it.

   top of the paper4 marks

   Check answer · 4 marks

   1. in any circular orbit v = sqrt(GM/r), so a larger radius gives a smaller speed
   2. the kinetic energy GMm/(2r) therefore falls, while the potential energy -GMm/r rises, becoming less negative
   3. the potential energy gains twice as much as the kinetic energy loses
   4. so the total energy -GMm/(2r) still increases, and the thruster's work equals that increase: the satellite trades speed for height

Reading cue. The pole facing the ring is N, not the S drawn at the top; as the magnet falls the downward flux through the ring is rising, and that rise fixes the induced current.

1. aState State which pole of the magnet faces the ring, and state the direction of the magnetic field along the dashed line just below the magnet.

   recall2 marks

   Check answer · 2 marks

   1. the north pole faces the ring, the south pole being at the top, away from it
   2. the field just below the magnet points downward, away from the north pole, along the dashed line

2. bDetermine Determine the direction of the induced current in the copper ring as the magnet approaches, as seen by an observer looking down from above the magnet.

   routine3 marks

   Check answer · 3 marks

   1. the flux through the ring points downward and is increasing as the magnet falls
   2. by Lenz's law the induced current opposes the increase, so it must produce upward flux inside the ring
   3. the current therefore flows anticlockwise as seen from above, making the upper face of the ring behave as a north pole

3. cExplain As the magnet falls past the ring the induced current reverses direction, yet the force the ring exerts on the magnet stays upward throughout. Explain why the reversal does not reverse the force, and identify the one position of the magnet at which the ring exerts no force on it at all.

   demanding4 marks

   Show a hint

   Ask where the flux through the ring is greatest, not where the field is strongest.

   Check answer · 4 marks

   1. above the ring the downward flux is increasing, so the current opposes the increase and the upper face of the ring acts as a north pole, repelling the approaching magnet
   2. below the ring the downward flux is decreasing, so the current reverses and the lower face acts as a south pole, attracting the receding magnet
   3. the induced effect always opposes the change producing it, so the force acts against the motion in both phases, that is upward, and the acceleration is less than g
   4. the force is zero when the centre of the magnet is level with the plane of the ring: the flux is a maximum there, so its rate of change is momentarily zero and no current flows

4. dDiscuss Discuss what would change if the copper ring were cut so that a narrow gap ran through it, and separately what would change if the ring were replaced by one of identical dimensions made from a metal of higher resistivity.

   top of the paper4 marks

   Check answer · 4 marks

   1. with a gap the circuit is broken, so although an emf is still induced around the ring, no current can flow
   2. with no current there is no opposing magnetic field and no retarding force, so the magnet falls with acceleration g throughout
   3. with a ring of higher resistivity the same emf drives a smaller current, since I = emf / R, so the retarding force is smaller and the magnet is slowed less
   4. the retarding force is what converts the magnet's gravitational potential energy into resistive heating in the ring, so less current means less heating and a faster arrival

Reading cue. r is drawn from the wire to the near face of the probe, not to the sensor inside it, and the field being measured points perpendicular to the page, not along the line r.

1. aState State how the direction of the magnetic field at the probe is related to the plane of the drawing, and state whether the field points the same way once the probe is moved to the dashed position.

   recall3 marks

   Check answer · 3 marks

   1. Right-hand grip rule applied to the current arrow drawn on the wire
   2. Field at the probe is perpendicular to the plane of the drawing, not along the dimension line r
   3. Field at the dashed position points the same way, because that position lies on the same side of the wire; only the magnitude falls

2. bCalculate The wire carries a steady current of 4.0 A. The dimension r is 25 mm for the probe drawn in full and 100 mm for the dashed position. Calculate the magnetic flux density at each of these two positions.

   routine3 marks

   Check answer · 3 marks

   1. Use of B = μ₀I/(2πr), with r the perpendicular distance drawn from the wire
   2. At r = 0.025 m: B = (2 × 10⁻⁷ × 4.0)/0.025 = 3.2 × 10⁻⁵ T (32 μT)
   3. At r = 0.100 m: B = (2 × 10⁻⁷ × 4.0)/0.100 = 8.0 × 10⁻⁶ T (8.0 μT)

3. cDetermine The dimension r is drawn to the near face of the probe, but the sensing element sits 5.0 mm behind that face. Determine the percentage by which each recorded flux density falls below μ₀I/(2πr) when r is recorded as 25 mm and when it is recorded as 100 mm, and state which end of a graph of B against 1/r is distorted more.

   demanding4 marks

   Check answer · 4 marks

   1. True separation is r + 5.0 mm, so the recorded value is in the ratio r/(r + 5.0 mm) of the expected one
   2. At r = 25 mm: 25/30, so the reading is 17% below the expected value
   3. At r = 100 mm: 100/105, so the reading is 4.8% below the expected value
   4. Readings taken closest to the wire are affected most, so the graph bends below a straight line at the large-1/r end

4. dDiscuss The wire is drawn extending well above and well below the level of the probe. Discuss what happens to the readings at both probe positions if the wire is shortened until its ends lie only a few centimetres above and below that level.

   top of the paper4 marks

   Check answer · 4 marks

   1. μ₀I/(2πr) assumes an infinitely long wire, so that current elements at all distances along it contribute
   2. A shortened wire is missing those contributions, so every measured value of B is smaller than μ₀I/(2πr)
   3. The shortfall grows with r, since the shortened wire subtends a smaller angle at the probe, so the dashed far position is affected more than the near one
   4. The plotted line therefore falls away from a straight line at the small-1/r end, and its gradient no longer gives the true current

Reading cue. The crosses mean the field is into the page and the drawn arc bends downwards; the charge is negative, so check your rule reproduces that downward force before trusting any later step.

1. aState State the direction of the magnetic force on the electron at the instant it crosses the dashed boundary, and state the direction of the conventional current that the moving electron represents.

   recall2 marks

   Check answer · 2 marks

   1. Conventional current is directed to the left, opposite to the drawn motion of the electron
   2. Force is directed downwards, towards the bottom of the page, in agreement with the way the drawn arc bends

2. bDetermine Determine the time the electron would take to complete one full circle inside the field region, and state how that time would differ if the electron had been accelerated through 2000 V instead of 500 V.

   routine4 marks

   Check answer · 4 marks

   1. Use of T = 2πm/(qB)
   2. T = 2π × 9.11 × 10⁻³¹/(1.60 × 10⁻¹⁹ × 2.5 × 10⁻³)
   3. T = 1.4 × 10⁻⁸ s
   4. Time would be unchanged, because the period does not depend on the speed

3. cDetermine The dashed region is 5.0 cm wide in the direction of the electron's initial motion, and is tall enough that the electron never reaches its upper or lower edge. Determine the greatest distance the electron penetrates into the region, and determine where it leaves.

   demanding4 marks

   Show a hint

   The centre of the circular path lies on the perpendicular to the velocity drawn at the entry point.

   Check answer · 4 marks

   1. Speed on entry v = √(2eV/m) = 1.3 × 10⁷ m s⁻¹
   2. Radius r = mv/(eB) = 1.21 × 10⁻²³/(4.0 × 10⁻²²) = 3.0 × 10⁻² m
   3. Greatest penetration equals r, i.e. 3.0 cm after a quarter circle, which is less than the 5.0 cm width
   4. Electron turns through 180° and leaves through the boundary it entered, 2r = 6.0 cm below the entry point

4. dEvaluate The accelerating potential difference is raised from 500 V to 1000 V with everything else unchanged. Evaluate whether the electron can be made to leave through the far edge of the region drawn, either by this change or by raising the accelerating potential difference further.

   top of the paper4 marks

   Check answer · 4 marks

   1. r ∝ √V, so the radius rises by a factor of √2, from 3.0 cm to 4.3 cm
   2. Penetration still equals r, and 4.3 cm is less than the 5.0 cm width, so the electron still turns back
   3. Penetration reaches 5.0 cm when r = 5.0 cm, requiring V = 500 × (5.0/3.0)² ≈ 1.4 × 10³ V
   4. Above about 1.4 kV the electron does reach the far edge, so the behaviour drawn is a consequence of this particular accelerating voltage rather than a general feature of the arrangement

Reading cue. Two slip rings, not a split ring, so the output alternates; and it is the plane of the coil, not its normal, that the drawing tilts relative to the four field arrows.

1. aState State the frequency and the period of the potential difference appearing at the terminals marked as the output to the external circuit, and state whether that potential difference reverses in sign.

   recall3 marks

   Check answer · 3 marks

   1. Frequency = 50 Hz, taken from the 50 revolutions per second marked at the axis
   2. Period = 1/50 = 0.020 s (20 ms)
   3. Potential difference does reverse in sign each half revolution, because the coil is taken off through two slip rings rather than a split ring

2. bSketch Sketch a graph of the output potential difference against time for two complete revolutions of the coil, taking t = 0 at the instant when the plane of the coil contains the direction of the field arrows. Mark the period on the time axis and mark the two peaks as equal in size and opposite in sign.

   routine4 marks

   Check answer · 4 marks

   1. Sinusoidal curve, symmetrical about the time axis and taking both positive and negative values
   2. Curve at a maximum at t = 0, since the flux linkage is zero and changing fastest at that instant
   3. Two complete cycles drawn, each of period 20 ms, filling 40 ms of the time axis
   4. Positive and negative peaks marked equal in size, as the take-off is through two slip rings

3. cDetermine At the instant drawn, the plane of the coil makes an angle of 30° with the direction of the field arrows. Determine the e.m.f. induced at that instant.

   demanding4 marks

   Check answer · 4 marks

   1. ω = 2π × 50 = 314 rad s⁻¹
   2. Peak e.m.f. = NBAω = 250 × 0.085 × 3.2 × 10⁻³ × 314 = 21.4 V
   3. Angle between the normal to the coil and the field is 90° − 30° = 60°, so e.m.f. = 21.4 sin 60°
   4. e.m.f. = 18.5 V (19 V to two significant figures)

4. dDiscuss The arrangement is now altered so that the dashed axis of rotation lies along the field arrows instead of across them, the coil still turning at 50 revolutions per second about that axis. Discuss the output now obtained.

   top of the paper4 marks

   Check answer · 4 marks

   1. The plane of the coil contains the axis of rotation, so the normal to the coil stays perpendicular to the field throughout the turn
   2. The flux linkage is therefore zero at every instant of the rotation
   3. Since the flux linkage does not change, its rate of change is zero and no e.m.f. is induced
   4. No output is obtained however fast the coil is turned, showing that it is the rate of change of flux linkage, not the motion of the coil in the field, that generates the e.m.f.

Reading cue. The arrow length means nothing here, the diagram is not to scale: take every energy from the printed labels, and note every level is negative because 0 eV is the just-free electron.

1. aState State the energy that must be supplied to a hydrogen atom to remove its electron completely when that electron occupies the level marked n = 2, and state why every level below n = ∞ carries a negative value.

   recall2 marks

   Check answer · 2 marks

   1. 3.40 eV, the difference between the −3.40 eV level and the 0 eV level marked n = ∞
   2. Zero is defined as the electron free of the atom and at rest, so a bound electron has less energy than this and its value is negative

2. bCalculate Calculate the wavelength of the photon emitted when the electron falls from the level marked n = 4 to the level marked n = 2. Use hc = 1240 eV nm.

   routine3 marks

   Check answer · 3 marks

   1. ΔE = −0.85 − (−3.40) = 2.55 eV
   2. λ = 1240/2.55
   3. λ = 486 nm

3. cDetermine An electron begins in the level marked n = 4. Using only the levels drawn, determine how many different photon wavelengths could be emitted as it returns to the ground state, and determine the shortest of those wavelengths.

   demanding4 marks

   Check answer · 4 marks

   1. Every downward transition between the four bound levels is available: 4→3, 4→2, 4→1, 3→2, 3→1 and 2→1
   2. Six different wavelengths
   3. Shortest wavelength comes from the largest energy drop, 4→1: ΔE = −0.85 − (−13.6) = 12.75 eV
   4. λ = 1240/12.75 = 97.3 nm

4. dExplain Hydrogen atoms in the ground state are bombarded first with free electrons of kinetic energy 12.5 eV, and then with photons of energy 12.5 eV. Explain, using the levels drawn, why the electrons excite the atoms but the photons do not.

   top of the paper4 marks

   Check answer · 4 marks

   1. Excitation from the ground state requires exactly 10.20 eV, 12.09 eV or 12.75 eV to reach n = 2, n = 3 or n = 4
   2. A photon is absorbed whole or not at all, and 12.5 eV matches none of these gaps, so the photons pass through unabsorbed
   3. A free electron transfers energy by collision and need not give up all of it, so it can supply exactly 12.09 eV and raise the atom to n = 3
   4. The bombarding electron then moves on with the remaining 12.5 − 12.09 = 0.41 eV of kinetic energy

Reading cue. Follow the supply plus terminal: it reaches the collector, so electrons are being pulled across. As drawn the meter reads a collected current, not anything about a stopping potential.

1. aState State the direction of the conventional current in the wire joining the plate to the microammeter, and state why the tube must be evacuated.

   recall2 marks

   Check answer · 2 marks

   1. Conventional current flows from the plate through the microammeter towards the negative terminal of the supply, opposite to the electron flow in that wire
   2. Evacuated so that photoelectrons are not scattered or absorbed by gas molecules before reaching the collector

2. bDetermine The incident light has a wavelength of 400 nm and the plate has a work function of 2.30 eV. Determine the reverse potential difference that would have to be applied to the collector to bring the microammeter reading to zero, and determine the reading in μA when 1.2 × 10¹² electrons leave the plate each second and all are collected. Use hc = 1240 eV nm.

   routine4 marks

   Check answer · 4 marks

   1. Photon energy = 1240/400 = 3.10 eV
   2. Maximum kinetic energy = 3.10 − 2.30 = 0.80 eV
   3. Stopping potential difference = 0.80 V, with the collector made negative
   4. Current = 1.2 × 10¹² × 1.60 × 10⁻¹⁹ = 1.9 × 10⁻⁷ A, that is 0.19 μA

3. cSketch Sketch the variation of the microammeter reading with the potential difference applied to the collector, from a reverse value, through zero, to a forward value large enough for the reading to stop rising. On the same axes sketch the result of replacing the light with light of shorter wavelength delivering the same number of photons per second.

   demanding4 marks

   Check answer · 4 marks

   1. First curve rises from zero at a negative potential difference and levels off at a constant saturation current once the collector is positive
   2. The arrangement drawn, with the collector held positive, lies on the flat saturated part of that curve
   3. Second curve saturates at the same current, because the number of photons arriving each second is unchanged
   4. Second curve meets the potential-difference axis at a more negative value, because the photoelectrons now leave with greater maximum kinetic energy

4. dDiscuss The collecting electrode is drawn small and clear of the light beam. Discuss what the microammeter would record if it were enlarged so that it faced the whole plate and was itself illuminated.

   top of the paper5 marks

   Check answer · 5 marks

   1. Light reaching the enlarged collector ejects photoelectrons from the collector as well as from the plate, provided its work function is small enough for the wavelength used
   2. With the collector held positive, as drawn, the field between the electrodes returns those electrons to the collector, so the saturation reading is barely altered by them
   3. An electrode large enough to face the whole plate does stand in the path of the beam, so less light reaches the plate and the saturation current falls for that reason instead
   4. Once the potential difference is reversed to look for the stopping potential, the field then drives the collector's own photoelectrons across to the plate, giving a current in the opposite sense
   5. The reading therefore does not fall to zero at the true stopping potential difference, so any maximum kinetic energy obtained from it would be wrong

Reading cue. The vertical axis is ln (A / Bq) and starts at 5.2, not zero. Where the line meets t = 0 you read ln A0, so it must be exponentiated before it is an activity in Bq.

1. aDetermine The plotted crosses run from ln (A / Bq) = 6.4 at t = 0 down to ln (A / Bq) = 5.2 at t = 40 minutes. Determine the activity of the source at the moment the first reading was taken.

   recall2 marks

   Check answer · 2 marks

   1. Recognition that the intercept is ln A₀, so A₀ = e⁶.4 rather than 6.4
   2. A₀ = 6.0 × 10² Bq

2. bDetermine Determine the activity of the source, in Bq, 25 minutes after the first reading.

   routine3 marks

   Check answer · 3 marks

   1. Gradient = (5.2 − 6.4)/40 = −0.030 min⁻¹
   2. ln (A / Bq) at t = 25 min is 6.4 − 0.030 × 25 = 5.65
   3. A = e⁵.65 = 2.8 × 10² Bq

3. cDetermine Determine the number of undecayed nuclei of this nuclide present in the source at the moment the first reading was taken.

   demanding4 marks

   Check answer · 4 marks

   1. Use of A = λN
   2. Decay constant converted from the graph's minutes to seconds: 0.030/60 = 5.0 × 10⁻⁴ s⁻¹
   3. N₀ = 602/(5.0 × 10⁻⁴)
   4. N₀ = 1.2 × 10⁶ nuclei

4. dEvaluate The background in this laboratory corresponds to an activity of 0.50 Bq. A student extends the straight line through these five crosses in order to predict the time at which the corrected activity falls to that value. Evaluate the prediction.

   top of the paper4 marks

   Check answer · 4 marks

   1. Extending the line gives ln 0.50 = −0.69, so t = (6.4 + 0.69)/0.030 = 2.4 × 10² minutes, about four hours
   2. This is roughly six times beyond the last plotted point, so the prediction rests entirely on a single decay constant continuing to hold
   3. That assumption fails if the source contains a second nuclide or if the daughter is itself active, and nothing in the plotted range would reveal either
   4. As the corrected activity approaches the background, it becomes the small difference between two comparable and randomly fluctuating counts, so the prediction could not be tested with any precision anyway

Reading cue. Count what crosses the drawing: one neutron in, two fragments out, nothing else. The free neutrons that carry the chain on are not drawn, so the nucleon numbers as pictured do not balance.

1. aState State the form in which almost all of the 200 MeV appears immediately after the split, and state what the two arrows drawn on the fragments show about the momentum of the system.

   recall2 marks

   Check answer · 2 marks

   1. Almost all appears as kinetic energy of the two fragments, driven apart by the electrostatic repulsion between their positive charges
   2. The arrows point in opposite directions, so the fragments carry equal and opposite momenta and the total momentum stays essentially that of the slow incoming neutron, close to zero

2. bDetermine Determine the energy released in the single event drawn, in joules, and hence determine the number of such events needed each second to sustain a thermal output of 3.2 GW.

   routine4 marks

   Check answer · 4 marks

   1. 200 MeV = 200 × 10⁶ × 1.60 × 10⁻¹⁹ J
   2. = 3.2 × 10⁻¹¹ J
   3. Number per second = 3.2 × 10⁹/(3.2 × 10⁻¹¹)
   4. = 1.0 × 10²⁰ events per second

3. cDetermine The two fragments drawn have nucleon numbers 141 and 92, and about 170 MeV of the energy released appears as their kinetic energy. Determine the kinetic energy carried by each fragment.

   demanding4 marks

   Check answer · 4 marks

   1. Momentum conservation gives the two fragments equal and opposite momenta
   2. Since kinetic energy equals p²/2m, the energy is shared in inverse proportion to mass, so the shares are in the ratio 141 : 92 in favour of the lighter fragment
   3. Lighter fragment, nucleon number 92: 170 × 141/233 = 103 MeV
   4. Heavier fragment, nucleon number 141: 170 × 92/233 = 67 MeV

4. dDiscuss Nothing is drawn leaving the reaction except the two fragments. Discuss what else must leave the nucleus, and what decides whether the single event drawn grows into a self-sustaining chain.

   top of the paper4 marks

   Check answer · 4 marks

   1. Nucleon numbers must balance: 235 + 1 = 236, while 141 + 92 = 233, so three free neutrons must also be released
   2. A chain is sustained only if, on average, exactly one neutron from each fission goes on to cause a further fission
   3. The remaining neutrons are lost by escaping through the surface or by being absorbed without causing fission, so the mass and shape of the sample decide the outcome
   4. The released neutrons are fast and are captured by ²³⁵U far less readily until repeated collisions in a moderator have slowed them

Reading cue. Only the 420 nm label fixes the photon energy, the slant of the rays does not. The single drawn arrow is one electron among many, not necessarily the fastest one to leave.

1. aState State the minimum energy that must be given to a single electron for it to leave the surface drawn, and state how that energy reaches the electron.

   recall2 marks

   Check answer · 2 marks

   1. 2.1 eV, the work function labelled on the block
   2. From one photon of the incident light, absorbed whole in a single one-to-one interaction

2. bDetermine The rays drawn represent a beam delivering 1.5 mW to the surface. Determine the number of photons striking the surface each second. Use hc = 1240 eV nm.

   routine4 marks

   Check answer · 4 marks

   1. Photon energy = 1240/420 = 2.95 eV
   2. = 2.95 × 1.60 × 10⁻¹⁹ = 4.72 × 10⁻¹⁹ J
   3. Rate = 1.5 × 10⁻³/(4.72 × 10⁻¹⁹)
   4. = 3.2 × 10¹⁵ photons per second

3. cDetermine The photoelectron drawn leaves at 60° to the surface, carrying the maximum possible kinetic energy. A uniform retarding field of 500 V m⁻¹ is now applied at right angles to the surface. Determine how far from the surface this electron travels before it stops moving away from it.

   demanding4 marks

   Show a hint

   Resolve the velocity into components along and normal to the surface; the field acts on only one of them.

   Check answer · 4 marks

   1. Maximum kinetic energy = 2.95 − 2.1 = 0.85 eV
   2. Only the velocity component normal to the surface is retarded, so the energy to be removed is 0.85 sin²60° = 0.64 eV
   3. Distance = energy in eV divided by the field in V m⁻¹, d = 0.64/500
   4. d = 1.3 × 10⁻³ m (1.3 mm), the electron still moving parallel to the surface at that moment

4. dDiscuss The single electron is drawn leaving the surface at a point some distance along from where the rays land. Discuss how faithful this drawing is to the photon model of the effect.

   top of the paper4 marks

   Check answer · 4 marks

   1. In the photon model one photon is absorbed by one electron and emission follows with no measurable delay, so electrons leave from the illuminated region itself
   2. The drawn separation implies the energy travels along the surface before emission, which the model does not allow and which would introduce a delay that is not observed
   3. Only one arrow is drawn, whereas photoelectrons leave in all directions above the surface
   4. The drawn electron need not carry the maximum kinetic energy either: electrons freed below the surface lose energy on the way out, giving a spread of energies from zero up to 0.85 eV