A Level physics diagrams

Reading cue. Read the arrow as the whole of the initial velocity, drawn along one direction only, and the 45 m as the vertical drop — not as the length of the dashed path.

1. aState State the vertical component of the stone's velocity at the instant it leaves the cliff edge, and give the feature of Fig. 2.1 that tells you this.

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   Check answer · 2 marks

   1. vertical component of the initial velocity is zero
   2. because the velocity arrow at the cliff edge is drawn horizontal

2. bCalculate Calculate the speed of the stone at the instant it reaches the ground.

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   Check answer · 3 marks

   1. time to fall 45 m: t = √(2 × 45 / 9.81) = 3.03 s
   2. vertical component on landing: v = 9.81 × 3.03 = 29.7 m s⁻¹
   3. resultant speed = √(12² + 29.7²) = 32 m s⁻¹

3. cDetermine Determine the angle between the dashed path and the ground at the point where the stone lands.

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   Check answer · 3 marks

   1. uses horizontal component 12 m s⁻¹ with vertical component 29.7 m s⁻¹
   2. tan θ = 29.7 / 12
   3. θ = 68° to the horizontal

4. dShow (that) Taking the cliff edge as origin, with x measured horizontally and y measured vertically downwards, show that the dashed curve in Fig. 2.1 is a parabola, and state the numerical constant in your equation.

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   Show a hint

   Write x and y separately in terms of t, then eliminate t.

   Check answer · 4 marks

   1. x = 12t, so t = x/12
   2. y = ½ × 9.81 × t²
   3. substitution gives y = 9.81x² / (2 × 12²), which is of the form y ∝ x², a parabola
   4. y = 0.034x², with x and y in metres

Reading cue. Count the loops between the supports, not the humps of the dashed curve: the dashed line is the same wave half a period later, and the 1.2 m spans all four loops.

1. aState State the number of nodes and the number of antinodes shown on the string in Fig. 4.1, counting the two fixed ends.

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   Check answer · 2 marks

   1. 5 nodes
   2. 4 antinodes

2. bDetermine Determine the wavelength of the progressive waves on the string, and the distance along the string from a node to the nearest antinode.

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   Check answer · 3 marks

   1. each loop is half a wavelength: loop length = 1.2 / 4 = 0.30 m
   2. λ = 0.60 m
   3. node to nearest antinode = λ/4 = 0.15 m

3. cExplain Explain why the three points between the supports stay at rest while the rest of the string moves between the solid and dashed positions.

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   Check answer · 3 marks

   1. two progressive waves of equal frequency and amplitude travel in opposite directions along the string (incident and reflected)
   2. at those points the two waves always arrive with a phase difference of 180°
   3. the displacements cancel at all times, so the resultant displacement there is permanently zero

4. dDeduce The pattern drawn in Fig. 4.1 is produced at 150 Hz. Deduce the next frequency above 150 Hz at which a clear stationary wave pattern appears on this string, and describe what is seen between the two frequencies.

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   Check answer · 4 marks

   1. four loops means the string is vibrating in its fourth mode, so the lowest possible frequency is 150 / 4 = 37.5 Hz
   2. the next pattern has five loops, at 5 × 37.5 Hz
   3. = 188 Hz
   4. between the two the string is not resonating: the loops lose their definition and the amplitude is small

Reading cue. W is drawn from the dot at her centre of mass and F from the surface meeting the air; v is a velocity, not a third force, and W is drawn longer than F for a reason.

1. aDeduce Deduce from the relative lengths of the arrows W and F in Fig. 7.1 whether the skydiver has yet reached terminal velocity.

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   Check answer · 2 marks

   1. F is drawn shorter than W, so there is a resultant force downwards
   2. she is therefore still accelerating and has not reached terminal velocity

2. bCalculate At the instant drawn, the drag force F is 3.4 × 10² N. Calculate the weight of the skydiver and her acceleration at that instant.

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   Check answer · 3 marks

   1. W = 85 × 9.81 = 8.3 × 10² N
   2. resultant force = 834 − 340 = 4.9 × 10² N downwards
   3. a = 494 / 85 = 5.8 m s⁻² downwards

3. cDetermine The skydiver falls 250 m from rest and is then moving at 50 m s⁻¹. Determine the average drag force acting on her over that fall.

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   Check answer · 4 marks

   1. loss of gravitational potential energy = 85 × 9.81 × 250 = 2.08 × 10⁵ J
   2. gain in kinetic energy = ½ × 85 × 50² = 1.06 × 10⁵ J
   3. work done against drag = 2.08 × 10⁵ − 1.06 × 10⁵ = 1.02 × 10⁵ J
   4. average drag force = 1.02 × 10⁵ / 250 = 4.1 × 10² N

4. dExplain The skydiver is falling at constant velocity when she turns into a head-down dive, presenting a much smaller area to the airflow. Explain how the two arrows of Fig. 7.1 change, and describe her subsequent motion.

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   Check answer · 4 marks

   1. the drag is reduced, so F becomes shorter while W is unchanged
   2. there is now a resultant downward force, so she accelerates again
   3. as her speed rises the drag increases until F is once more equal to W
   4. she then falls at constant velocity at a higher terminal speed than before

Reading cue. The 45 N is the weight of the block, so it equals the tension only if the wire's own weight is ignored; 0.56 mm is a diameter, and 2.50 m is the length before loading.

1. aCalculate Calculate the cross-sectional area of the wire labelled in Fig. 8.1.

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   Check answer · 2 marks

   1. A = πd²/4 with d = 0.56 × 10⁻³ m
   2. A = 2.5 × 10⁻⁷ m²

2. bDetermine Steel of this type breaks at a tensile stress of 8.0 × 10⁸ Pa. Determine the greatest load this wire could support, and the factor by which the 45 N load could be increased before the wire breaks.

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   Check answer · 3 marks

   1. maximum load = stress × area = 8.0 × 10⁸ × 2.46 × 10⁻⁷
   2. = 2.0 × 10² N
   3. factor = 197 / 45 = 4.4

3. cDeduce The wire is replaced by one of the same steel and the same original length but of diameter 1.12 mm, and the same 45 N load is hung from it. Deduce the factor by which the extension changes.

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   Check answer · 3 marks

   1. doubling the diameter makes the cross-sectional area 4 times greater
   2. for the same load the stress, and hence the strain, is one quarter of its previous value
   3. the extension becomes one quarter of its previous value

4. dSuggest The density of steel is 7800 kg m⁻³. Suggest, with a supporting calculation, whether taking the tension in the wire to be 45 N along its whole length is justified.

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   Check answer · 4 marks

   1. volume of wire = 2.46 × 10⁻⁷ × 2.50 = 6.2 × 10⁻⁷ m³
   2. weight of wire = 7800 × 6.2 × 10⁻⁷ × 9.81 = 0.047 N
   3. this is about 0.1% of 45 N, and only the part of the wire below a given point adds to the tension there
   4. so treating the tension as 45 N throughout introduces no significant error

Reading cue. θ is measured from the straight-through n = 0 beam, not from the grating face, and 500 lines per mm must first be inverted to give the line spacing d.

1. aCalculate Calculate the distance between the centres of adjacent lines on the grating labelled in Fig. 9.1.

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   Check answer · 2 marks

   1. d = 1 mm / 500
   2. d = 2.0 × 10⁻⁶ m

2. bDetermine Determine the angle between the two first-order beams drawn either side of the straight-through direction.

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   Check answer · 3 marks

   1. sin θ = λ/d = 590 × 10⁻⁹ / 2.0 × 10⁻⁶ = 0.295
   2. θ = 17.2°
   3. the two n = 1 beams are symmetrical about n = 0, so the angle between them is 34.3°

3. cDetermine The source is replaced by one giving white light of wavelengths from 400 nm to 700 nm. Determine the angular width of the first-order spectrum.

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   Check answer · 3 marks

   1. sin θ = 400 × 10⁻⁹ / 2.0 × 10⁻⁶ = 0.200, giving θ = 11.5°
   2. sin θ = 700 × 10⁻⁹ / 2.0 × 10⁻⁶ = 0.350, giving θ = 20.5°
   3. angular width = 20.5 − 11.5 = 9.0°

4. dDeduce With the white-light source still in place, deduce whether the second-order and third-order spectra overlap.

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   Check answer · 4 marks

   1. second order at 700 nm: sin θ = 2 × 700 × 10⁻⁹ / 2.0 × 10⁻⁶ = 0.700, θ = 44.4°
   2. third order at 400 nm: sin θ = 3 × 400 × 10⁻⁹ / 2.0 × 10⁻⁶ = 0.600, θ = 36.9°
   3. the third-order violet leaves at a smaller angle than the second-order red, so the two spectra overlap
   4. the third order is incomplete: sin θ would exceed 1 beyond about 670 nm, so its red end is missing

Reading cue. r sits inside the dashed box with the cell, so read the box as the whole source, and read 12.0 V as labelling that source, not the p.d. across its terminals.

1. aState The ammeter reads 2.0 A. State the current in the resistor r inside the dashed box, and give the reason from Fig. 10.1.

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   Check answer · 2 marks

   1. 2.0 A
   2. the components form a single series loop, so the current is the same at every point in it

2. bCalculate With R set to 4.0 Ω the ammeter reads 2.0 A. Calculate the charge that passes through the ammeter in 5.0 minutes, and the energy delivered to R in that time.

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   Check answer · 3 marks

   1. Q = It = 2.0 × 300 = 6.0 × 10² C
   2. p.d. across R = 2.0 × 4.0 = 8.0 V
   3. energy = VQ = 8.0 × 600 = 4.8 × 10³ J

3. cDetermine The internal resistance of the battery is 2.0 Ω. Determine the setting of R that makes the ammeter read 1.5 A.

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   Check answer · 3 marks

   1. e.m.f. = I(R + r), so 12.0 = 1.5 × (R + 2.0)
   2. R + 2.0 = 8.0 Ω
   3. R = 6.0 Ω

4. dDeduce A second identical battery is connected in series with the first in the loop of Fig. 10.1, with R left at 4.0 Ω. Deduce whether the ammeter reading doubles.

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   Check answer · 4 marks

   1. total e.m.f. = 24.0 V
   2. total resistance = 4.0 + 2.0 + 2.0 = 8.0 Ω
   3. I = 24.0 / 8.0 = 3.0 A
   4. the reading rises from 2.0 A to 3.0 A, not to 4.0 A, because the internal resistance in the loop has doubled as well

Reading cue. Crosses mean B points into the page, and the particle is negative, so the conventional current is opposite to the arrow v — apply the left-hand rule to that reversed direction.

1. aState State the direction of the force on the electron at the instant it crosses the dashed boundary in Fig. 5.1.

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   Check answer · 2 marks

   1. vertically downwards in the plane of the page
   2. perpendicular to both v and B, from the left-hand rule applied with the conventional current opposite to v

2. bExplain Explain why the electron travels at constant speed inside the field region even though its velocity is changing.

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   Check answer · 3 marks

   1. the magnetic force is always perpendicular to the velocity
   2. so no work is done on the electron and its kinetic energy is unchanged
   3. a constant-magnitude force perpendicular to the motion changes only the direction of the velocity, giving a circular path

3. cDetermine The electron crosses the boundary at 2.4 × 10⁷ m s⁻¹ into a field of flux density 8.5 mT. Determine the radius of its path, and hence the least distance the field region must extend beyond the boundary if the electron is to leave the region travelling at right angles to its original direction.

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   Check answer · 3 marks

   1. r = mv/(eB) = (9.11 × 10⁻³¹ × 2.4 × 10⁷) / (1.60 × 10⁻¹⁹ × 8.5 × 10⁻³)
   2. r = 1.6 × 10⁻² m
   3. after turning through 90° the electron has advanced one radius along its original direction, so the region must extend at least 1.6 cm

4. dDeduce A proton crosses the boundary at the same point and with the same velocity as the electron. Deduce how its path differs from that of the electron.

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   Check answer · 3 marks

   1. the proton is positive, so the magnetic force acts in the opposite direction and the path curves upwards instead of downwards
   2. r = mv/(qB) and the charge magnitudes are equal, so the radius is greater in the ratio of the masses, about 1.8 × 10³
   3. r ≈ 29 m, so within the region drawn the proton's path is almost straight

Reading cue. The dashed line labelled r runs from the dot at the Earth's centre to the satellite, so it is the orbit radius, not the height above the surface, and it is not to scale.

1. aState State what provides the centripetal force on the satellite in Fig. 7.1, and state its direction on the figure.

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   Check answer · 2 marks

   1. the gravitational attraction of the Earth on the satellite
   2. directed along the dashed line towards the dot at the centre of the Earth

2. bCalculate The orbit radius r is 4.2 × 10⁷ m and the radius of the Earth is 6.4 × 10⁶ m. Calculate the height of the satellite above the Earth's surface.

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   Check answer · 2 marks

   1. height = r − radius of the Earth = 4.2 × 10⁷ − 6.4 × 10⁶
   2. = 3.6 × 10⁷ m

3. cDetermine The satellite has a mass of 1.5 × 10³ kg, and for the Earth GM = 3.99 × 10¹⁴ m³ s⁻². Determine the gravitational field strength of the Earth at this orbit radius and the force the Earth exerts on the satellite.

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   Check answer · 3 marks

   1. g = GM/r² = 3.99 × 10¹⁴ / (4.2 × 10⁷)²
   2. g = 0.23 N kg⁻¹
   3. F = mg = 1.5 × 10³ × 0.226 = 3.4 × 10² N

4. dDeduce A second satellite is placed in a circular orbit of radius 2.1 × 10⁷ m about the same centre, with GM = 3.99 × 10¹⁴ m³ s⁻² as before. Deduce its period, and deduce whether it can remain above one point on the equator.

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   Check answer · 4 marks

   1. GMm/r² = 4π²mr/T², so T = 2π√(r³/GM)
   2. T = 2π√((2.1 × 10⁷)³ / 3.99 × 10¹⁴) = 3.0 × 10⁴ s
   3. = 8.4 hours, so it circles the Earth almost three times each day
   4. its period is not 24 hours, so it cannot stay above a single point on the equator

Reading cue. Each 4.0 cm runs from the equilibrium line to one extreme, so it is the amplitude; the block travels the 8.0 cm between the outer dashed lines twice in each cycle.

1. aState State the total distance travelled by the block in one complete oscillation between the dashed lines of Fig. 9.1.

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   Check answer · 2 marks

   1. distance = 4 × amplitude
   2. = 16 cm (0.16 m)

2. bCalculate The period of the oscillation is 0.80 s. Calculate the maximum acceleration of the block, and state where on Fig. 9.1 it occurs.

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   Check answer · 3 marks

   1. ω = 2π/0.80 = 7.85 rad s⁻¹
   2. a₀ = ω²x₀ = 7.85² × 0.040 = 2.5 m s⁻²
   3. at the two outer dashed lines, directed towards the equilibrium line

3. cDetermine The block has a mass of 0.25 kg. Determine the spring constant of the spring and the maximum resultant force on the block.

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   Check answer · 3 marks

   1. for a mass on a spring ω² = k/m, so k = mω² = 0.25 × 61.7
   2. k = 15 N m⁻¹
   3. F = ma₀ = 0.25 × 2.47 = 0.62 N

4. dDeduce Deduce the extension of the spring when the block is at the equilibrium line marked in Fig. 9.1, and explain why that line is not level with the lower end of the unloaded spring.

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   Check answer · 4 marks

   1. at the equilibrium position the spring force balances the weight: kx = mg
   2. x = (0.25 × 9.81) / 15.4 = 0.16 m
   3. the spring is already stretched by 16 cm in supporting the block, so the equilibrium line lies 16 cm below the unloaded end
   4. the oscillation takes place about this stretched position, not about the natural length

Reading cue. 5.0 mm is the gap, not a plate length, and no field lines are drawn — the plus and minus signs on the plates are the only thing in the figure that fixes the field direction.

1. aState State the direction of the electric field in the gap in Fig. 10.1, and the direction of the force it exerts on an electron placed there.

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   Check answer · 2 marks

   1. the field points vertically downwards, from the positive upper plate to the negative lower plate
   2. the force on an electron is vertically upwards, towards the positive plate

2. bCalculate Calculate the work done on an electron that moves from the lower plate to the upper plate, and the speed with which it arrives.

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   Check answer · 3 marks

   1. W = eV = 1.60 × 10⁻¹⁹ × 2.0 × 10³ = 3.2 × 10⁻¹⁶ J
   2. ½mv² = 3.2 × 10⁻¹⁶ J
   3. v = 2.7 × 10⁷ m s⁻¹

3. cDetermine A charged dust particle of weight 1.28 × 10⁻¹³ N is held at rest midway between the plates. Determine the magnitude and sign of its charge, and the number of excess electrons it carries.

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   Check answer · 4 marks

   1. E = V/d = 2.0 × 10³ / 5.0 × 10⁻³ = 4.0 × 10⁵ V m⁻¹
   2. for equilibrium qE = weight, so q = 1.28 × 10⁻¹³ / 4.0 × 10⁵ = 3.2 × 10⁻¹⁹ C
   3. the electric force must act upwards while the field points downwards, so the charge is negative
   4. 3.2 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ = 2 excess electrons

4. dDeduce The battery p.d. is suddenly reduced to 1.0 kV while the particle is still midway between the plates. Deduce the acceleration of the particle, and calculate the time it takes to reach a plate.

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   Check answer · 4 marks

   1. new field = 2.0 × 10⁵ V m⁻¹, so the electric force = 3.2 × 10⁻¹⁹ × 2.0 × 10⁵ = 6.4 × 10⁻¹⁴ N upwards
   2. resultant = 1.28 × 10⁻¹³ − 6.4 × 10⁻¹⁴ = 6.4 × 10⁻¹⁴ N downwards
   3. mass = 1.28 × 10⁻¹³ / 9.81 = 1.30 × 10⁻¹⁴ kg, so a = 4.9 m s⁻² downwards, that is g/2
   4. falling the 2.5 mm to the lower plate: t = √(2 × 2.5 × 10⁻³ / 4.9) = 3.2 × 10⁻² s

Reading cue. The 250 V acts only between cathode and anode; beyond the anode gap there is no field, so the electrons meet the crystal at the speed they had on leaving it.

1. aExplain Explain why the cell in Fig. 11.1 is connected with its positive terminal to the anode.

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   Check answer · 2 marks

   1. electrons carry negative charge
   2. they are repelled from the negative cathode and attracted to the positive anode, so this connection accelerates them along the tube towards the screen

2. bDescribe Describe what is seen on the blank fluorescent screen once the electron beam has passed through the thin crystal.

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   Check answer · 3 marks

   1. a set of concentric bright rings
   2. centred on the point where the undeviated beam meets the screen
   3. the pattern is brightest at the centre, with the rings becoming fainter further out

3. cDetermine The accelerating p.d. is increased from 250 V to 1000 V. Determine the factor by which the de Broglie wavelength of the electrons changes, and describe the effect on the pattern on the screen.

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   Check answer · 3 marks

   1. eV = ½mv² and λ = h/mv, so λ ∝ 1/√V
   2. V is 4 times greater, so λ is halved: factor 0.50 (7.8 × 10⁻¹¹ m falls to 3.9 × 10⁻¹¹ m)
   3. the diffraction angles are smaller, so the rings close in towards the centre of the screen

4. dSuggest Protons are accelerated through the same 250 V and directed at the same crystal. Suggest why no ring pattern appears on the screen.

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   Check answer · 4 marks

   1. λ = h/√(2meV), so the proton wavelength is smaller than the electron wavelength by √(mp/me) ≈ 43
   2. λ ≈ 1.8 × 10⁻¹² m for the protons
   3. this is far smaller than the spacing of the atoms in the crystal, so the diffraction angles are too small to be seen
   4. (the much heavier protons would also be absorbed within the crystal)

Reading cue. The voltmeter is permanently across C, so it is part of the discharge path even before S closes; R joins the circuit only when S is closed.

1. aCalculate The capacitor in Fig. 12.1 has capacitance 470 μF. Calculate the time constant of the circuit after S is closed.

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   Check answer · 2 marks

   1. τ = RC = 100 × 10³ × 470 × 10⁻⁶
   2. τ = 47 s

2. bDetermine The voltmeter reads 9.0 V at the instant S is closed. Determine its reading 60 s later.

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   Check answer · 3 marks

   1. V = V₀e^(−t/RC), with t/RC = 60/47 = 1.28
   2. V = 9.0 × e^(−1.28)
   3. V = 2.5 V

3. cDetermine Determine the time taken for the voltmeter reading to halve, and state whether this time would differ if the reading at the instant of closing S were 4.5 V instead of 9.0 V.

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   Check answer · 3 marks

   1. ½ = e^(−t/RC), so t = RC ln 2
   2. t = 47 × 0.693 = 33 s
   3. the time to halve is independent of the starting p.d., so it would still be 33 s

4. dDeduce Deduce the energy transferred to R while the voltmeter reading falls from 9.0 V to 4.5 V, and state the assumption you make about the voltmeter.

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   Check answer · 4 marks

   1. energy stored at 9.0 V = ½CV² = ½ × 470 × 10⁻⁶ × 9.0² = 1.9 × 10⁻² J
   2. energy stored at 4.5 V = ½ × 470 × 10⁻⁶ × 4.5² = 4.8 × 10⁻³ J
   3. energy transferred to R = 1.9 × 10⁻² − 4.8 × 10⁻³ = 1.4 × 10⁻² J
   4. assumes no charge flows through the voltmeter branch, i.e. its resistance is effectively infinite