AP physics diagrams

Reading cue. Velocity is on the vertical axis, so a sloping line is a changing speed, not a change of direction; and the fall meets the axis at 9.0 s, not at the last gridline at 10 s.

1. aDetermine Determine the acceleration of the object during the first stage of the motion shown.

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   Check answer · 2 marks

   1. reads the gradient of the first segment as (8.0 − 0)/(4.0 − 0)
   2. a = 2.0 m/s², with unit

2. bCalculate Calculate the distance travelled by the object between t = 4.0 s and t = 9.0 s.

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   Check answer · 3 marks

   1. rectangle from 4.0 s to 7.0 s = 8.0 × 3.0 = 24 m
   2. triangle from 7.0 s to 9.0 s = ½ × 2.0 × 8.0 = 8.0 m
   3. total distance = 32 m

3. cDetermine Determine the time at which the object has travelled exactly 20 m from its starting point.

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   Check answer · 3 marks

   1. area under the first segment = ½ × 4.0 × 8.0 = 16 m, so the object is still short of 20 m at t = 4.0 s
   2. the remaining 4.0 m is covered at the constant 8.0 m/s, taking 4.0/8.0 = 0.50 s
   3. t = 4.5 s

4. dExplain A student looks at the last stage of the graph and says the object must be moving backwards, because the line is sloping downwards. Explain why the graph does not show this, and state what the drawing would look like if the object did reverse.

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   Check answer · 3 marks

   1. the plotted line stays above the time axis for the whole 9.0 s, so the velocity is positive throughout and the direction never changes
   2. a line falling towards the axis while still above it shows the speed decreasing in an unchanged direction
   3. a reversal would be drawn as the line crossing the time axis into negative velocity

Reading cue. The labels are speeds, not velocities — direction is carried only by the arrowheads — so fix a positive direction along the track before you write down a single momentum.

1. aCalculate Calculate the total kinetic energy of the two balls at the instant drawn.

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   Check answer · 2 marks

   1. ½(0.50)(4.0)² = 4.0 J and ½(1.5)(2.0)² = 3.0 J
   2. total = 7.0 J, with the directions of the arrows playing no part because kinetic energy is a scalar

2. bDetermine Determine the speed the 1.5 kg ball would need, with its direction of travel unchanged, for the total momentum of the pair to be zero.

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   Check answer · 3 marks

   1. momentum of the 0.50 kg ball = 0.50 × 4.0 = 2.0 kg m/s to the right
   2. for zero total the 1.5 kg ball must carry 2.0 kg m/s to the left
   3. v = 2.0/1.5 = 1.3 m/s, slower than the 2.0 m/s marked on the figure

3. cDetermine The collision is in fact perfectly elastic rather than sticking. Determine the velocity of each ball immediately afterwards.

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   Show a hint

   Two conservation statements give two equations. The quickest route is the elastic-collision result that the balls separate as fast as they approached — read the approach speed straight off the two arrows.

   Check answer · 4 marks

   1. momentum: 0.50v₁ + 1.5v₂ = 0.50(4.0) + 1.5(−2.0) = −1.0 kg m/s
   2. kinetic energy after must equal the 7.0 J before, or equivalently the separation speed equals the 6.0 m/s approach speed
   3. solving gives v₁ = −5.0 m/s for the 0.50 kg ball, that is 5.0 m/s to the left
   4. v₂ = +1.0 m/s for the 1.5 kg ball, that is 1.0 m/s to the right

4. dExplain Explain why the two arrows drawn on the figure cannot by themselves tell you whether the collision that follows is elastic, and describe one measurement made afterwards that would settle it.

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   Check answer · 3 marks

   1. the figure fixes only the masses and the velocities before contact, and an elastic and an inelastic collision start from exactly the same drawn state
   2. momentum is conserved either way, so the total 1.0 kg m/s to the left is no test of elasticity
   3. measuring both final speeds and comparing the total kinetic energy after with the 7.0 J before decides it: equal means elastic, less means kinetic energy was lost

Reading cue. h is the vertical dimension line at the side of the wedge, not the distance the block slides: along the face that is h/sin θ. μ is printed twice because both surfaces share it.

1. aDetermine Determine the normal force exerted on the block by the sloping face, in terms of m, θ and physical constants.

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   Check answer · 2 marks

   1. resolves the weight mg perpendicular to the sloping face
   2. N = mg cos θ, since the block has no acceleration perpendicular to the face

2. bCalculate Calculate the acceleration of the block down the sloping face for θ = 30° and μ = 0.25.

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   Check answer · 3 marks

   1. along the face, ma = mg sin θ − μmg cos θ, so a = g(sin θ − μ cos θ) with m cancelling
   2. a = 9.8(0.500 − 0.25 × 0.866) = 9.8 × 0.284
   3. a = 2.8 m/s², directed down the slope

3. cDetermine Determine the time the block takes to reach the foot of the slope when h = 1.5 m, for the same θ and μ.

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   Check answer · 4 marks

   1. the dimension line gives the vertical drop, so the distance along the face is L = h/sin θ = 1.5/0.500 = 3.0 m
   2. from rest with uniform acceleration, L = ½at², so t = √(2L/a)
   3. t = √(2 × 3.0/2.78) = √2.16
   4. t = 1.5 s

4. dJustify An identical block is released from the same height h on a steeper wedge carrying the same μ on both of its surfaces. Justify whether it stops nearer to or further from the foot of the slope than the block in the figure.

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   Check answer · 4 marks

   1. the friction force on the face is μmg cos θ and the sliding length is h/sin θ, so the energy lost on the slope is μmgh cot θ
   2. cot θ falls as θ increases, so the steeper wedge takes less energy from the block
   3. the block therefore reaches the foot with more kinetic energy, while the friction force on the floor is μmg and is unchanged by the wedge
   4. so it stops further from the foot of the slope, not nearer

Reading cue. The fall is one straight segment, so a single constant acceleration spans it; and the time axis is labelled from 1 s, so the start of the graph at t = 0 sits on an unlabelled origin.

1. aDetermine Determine the displacement of the cart between t = 5.0 s and t = 7.0 s.

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   Check answer · 2 marks

   1. the segment is a rectangle of height −3.0 m/s and width 2.0 s
   2. displacement = −6.0 m, that is 6.0 m in the negative direction

2. bCalculate Calculate the average velocity of the cart over the interval t = 2.0 s to t = 5.0 s.

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   Check answer · 3 marks

   1. area from 2.0 s to 3.5 s = ½(1.5)(3.0) = +2.25 m, and area from 3.5 s to 5.0 s = ½(1.5)(3.0) = −2.25 m
   2. net displacement over the interval is zero
   3. average velocity = 0/3.0 = 0 m/s, even though the cart is moving throughout and its average speed is 1.5 m/s

3. cDetermine Determine both the total distance travelled by the cart and its net displacement over the whole 7.0 s shown.

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   Check answer · 4 marks

   1. areas above the axis: +6.0 m from 0 to 2.0 s and +2.25 m from 2.0 s to 3.5 s
   2. areas below the axis: −2.25 m from 3.5 s to 5.0 s and −6.0 m from 5.0 s to 7.0 s
   3. net displacement = 6.0 + 2.25 − 2.25 − 6.0 = 0 m, so the cart ends where it started
   4. total distance = 6.0 + 2.25 + 2.25 + 6.0 = 16.5 m

4. dSketch Sketch on the same axes the velocity–time graph of a second cart that starts from the same point at t = 0, has one constant acceleration for the whole 7.0 s, and at t = 7.0 s is at the same position and has the same velocity as the cart drawn. Indicate its initial velocity and the value of its acceleration.

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   Show a hint

   For a constant acceleration the average velocity over any interval is just the mean of the velocities at its two ends. Zero net displacement then tells you at once how the initial and final velocities must be related.

   Check answer · 4 marks

   1. a single straight line of constant negative gradient across the whole 7.0 s
   2. for constant acceleration the average velocity is the mean of the initial and final values, and the net displacement must be zero as in part (c), so the initial velocity must be +3.0 m/s to match the final −3.0 m/s
   3. gradient = (−3.0 − 3.0)/7.0 = −0.86 m/s²
   4. the line crosses zero at t = 3.5 s, the same instant as the drawn cart

Reading cue. Six separate items, not an assembled apparatus: there is no protractor and no force sensor, so the stopwatch and the meter stick are the only route to anything the spring scale cannot give.

1. aDetermine Each of the known masses drawn in the stack is 0.50 kg, and the block itself weighs 4.9 N. Determine the largest normal force between the block and the level board that the equipment drawn can produce.

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   Check answer · 2 marks

   1. the stack in the figure holds three masses, so the greatest load is 3 × 0.50 = 1.5 kg, a weight of 14.7 N
   2. on a level board the normal force equals the total weight: N = 4.9 + 14.7 = 19.6 N

2. bExplain Explain why the spring scale must be read while the block is already sliding, and why the largest reading, taken at the instant the block first breaks away, is not the one wanted.

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   Check answer · 3 marks

   1. the quantity sought is the coefficient of kinetic friction, which applies only while the two surfaces are sliding over one another
   2. with the block moving at constant velocity the net force is zero, so the scale reading equals the friction force exactly
   3. the peak reading at break-away measures the maximum static friction, which is larger and would give too high a value

3. cDescribe The spring scale is difficult to hold at a steady reading. Describe how the stopwatch and the meter stick drawn in the figure could be used instead, with no use of the spring scale at all, and state the equation that would give μ from those measurements.

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   Check answer · 4 marks

   1. give the block a push so that it slides freely along the level board and comes to rest, so that friction is the only horizontal force acting during the slide
   2. measure with the meter stick the distance d from the point of release to the stopping point, and time that slide with the stopwatch as t
   3. for uniform deceleration ending at rest, d = ½at², so a = 2d/t²
   4. friction alone gives a = μg, so μ = 2d/(gt²); repeat the slide several times and average

4. dJustify Justify whether the value of μ obtained would change if the block were stood on its smallest face instead of its largest, and describe the measurement from this equipment that would test your answer.

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   Check answer · 4 marks

   1. the standard model gives f = μN with no area term at all, so the coefficient should come out unchanged
   2. the weight and therefore the normal force are unchanged, so the constant-speed scale reading should also be unchanged
   3. stand the block on its smallest face with the same loading masses on top and repeat the constant-speed pull, comparing the readings
   4. agreement within the spread of the repeats supports the model, while a consistent difference would show the model failing for these two surfaces

Reading cue. The dashed line exists to fix that both start at the same height h; the labels under the two ramps — rolls without slipping, frictionless — are the whole difference between the cases.

1. aIndicate Indicate whether the gravitational potential energy converted by the sphere on its way to the floor is greater than, less than, or equal to that converted by the block. No justification is required.

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   Check answer · 1 mark

   1. equal: both have mass M and both fall through the same height h from the dashed line, so each converts Mgh

2. bDetermine Determine what fraction of the sphere's total kinetic energy at the foot of the ramp is rotational.

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   Check answer · 3 marks

   1. rolling without slipping gives ω = v/R, so the rotational term is ½(2/5)MR²(v/R)² = (1/5)Mv²
   2. total kinetic energy = ½Mv² + (1/5)Mv² = (7/10)Mv²
   3. fraction rotational = (1/5)/(7/10) = 2/7, about 0.29

3. cDetermine Both objects cover the same distance along the face of their identical ramps, starting from rest with uniform acceleration. Determine the ratio of the time the sphere takes to reach the floor to the time the block takes.

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   Check answer · 4 marks

   1. for uniform acceleration from rest the average speed is half the final speed, so the face length L gives t = 2L/v for each object
   2. L is the same for both, so the ratio of times is the inverse ratio of the final speeds: t_sphere/t_block = v_block/v_sphere
   3. v_block = √(2gh) and v_sphere = √(10gh/7), so the ratio is √(2 ÷ 10/7) = √(7/5)
   4. t_sphere/t_block = √1.4 = 1.18, so the sphere takes about 18% longer

4. dDerive The sphere on the left ramp is replaced by a hollow spherical shell of the same mass and radius, for which I = (2/3)MR², released from rest on the same dashed line. Derive its speed at the floor and explain where it ranks against the two speeds the figure compares.

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   Check answer · 4 marks

   1. energy conservation with ω = v/R: Mgh = ½Mv² + ½(2/3)MR²(v/R)² = (5/6)Mv²
   2. v_shell = √(6gh/5) = 1.10√(gh), against 1.20√(gh) for the solid sphere and 1.41√(gh) for the block
   3. the shell is slowest: v_shell/v_sphere = √(0.84) = 0.92, so it is about 8% slower than the sphere
   4. all of the shell's mass sits at the rim, giving the largest moment of inertia for the same M and R, so the largest share of the Mgh goes into rotation and the least into translation

Reading cue. The crosses mean B points into the page, away from you, and the 0.25 m dimension line spans only the part of the wire inside the field region — that is the length that counts.

1. aDetermine Determine the direction of the magnetic force acting on the wire.

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   Check answer · 2 marks

   1. applies a right-hand rule with the current directed to the right and the field directed into the page
   2. the force is directed up the page, at right angles to both the current and the field

2. bDetermine The current in the wire is reversed and at the same instant the field is increased to 0.80 T. Determine the new magnitude and direction of the force on the wire.

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   Check answer · 3 marks

   1. F = BIL with L unchanged, so F = 0.80 × 3.0 × 0.25
   2. F = 0.60 N, double the original magnitude
   3. reversing the current reverses the force, so it now points down the page

3. cExplain The wire is turned to a new direction while still lying in the plane of the page, with the same 0.25 m of it inside the field region. Explain why the magnitude of the force is unchanged, and state the one change of orientation that would reduce that force to zero.

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   Check answer · 3 marks

   1. the crosses show that B is perpendicular to the page, so any wire lying in the page makes an angle of 90° with the field and sin θ = 1
   2. B, I and the length inside the field are all unchanged, so the magnitude BIL is identical for every direction that can be drawn in the page; only the direction of the force turns as the wire turns
   3. the force falls to zero only if the wire is turned out of the page to lie along the field itself, pointing straight into the page, where sin θ = 0

4. dExplain A student says that because the force on the wire is 0.30 N, the magnetic field must do 0.30 J of work on the wire for every metre the wire is pushed sideways. Explain why the magnetic field in fact does no work on the charges in the wire, and indicate what does supply the energy.

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   Check answer · 4 marks

   1. the magnetic force on any charge is qv × B, always perpendicular to that charge's own velocity, so it does no work on the charge
   2. once the wire moves sideways across the field there is a motional emf in it opposing the original current, by Faraday's law and Lenz's law
   3. the source driving the 3.0 A must therefore do extra work against this back-emf to hold the current steady
   4. the mechanical energy gained by the wire comes from that source, with the field acting only as the intermediary that redirects it

Reading cue. F is marked on both sides of the lens, and the object stands 10 cm away against f = 15 cm: it is inside the focal point, and that single fact decides the whole answer.

1. aSketch Sketch on a copy of the figure the ray that leaves the tip of the object travelling parallel to the principal axis, and indicate its path after it has passed through the lens.

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   Check answer · 2 marks

   1. a straight ray drawn from the tip of the object arrow, parallel to the dashed axis, as far as the lens
   2. after the lens the ray is bent towards the axis and drawn through the focal point F marked on the far side

2. bDetermine Determine the image distance for the object position marked on the figure.

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   Check answer · 3 marks

   1. 1/v = 1/f − 1/u = 1/15 − 1/10
   2. 1/v = (2 − 3)/30 = −1/30
   3. v = −30 cm, so the image lies 30 cm from the lens on the same side as the object

3. cCalculate The object arrow drawn on the figure represents an object 4.0 mm tall. Calculate the height of the image and determine the distance between the object and its image.

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   Check answer · 4 marks

   1. magnification m = −v/u = −(−30)/10 = +3.0
   2. the positive sign shows the image is upright, the same way up as the drawn object arrow
   3. image height = 3.0 × 4.0 = 12 mm
   4. both lie on the same side of the lens, so the separation is 30 − 10 = 20 cm

4. dDetermine Determine how far the object must be moved along the axis for the image to lie twice as far from the lens as it does now, and explain what happens as the object is moved all the way out to the focal point.

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   Check answer · 4 marks

   1. for the image at v = −60 cm, 1/u = 1/f − 1/v = 1/15 + 1/60
   2. 1/u = 5/60, so u = 12 cm
   3. the object must be moved 12 − 10 = 2.0 cm further from the lens, that is towards the near focal point
   4. as u approaches 15 cm, 1/v approaches zero and the image distance grows without limit: the rays leave the lens parallel and no image is formed at all

Reading cue. Both axes carry multipliers — A is 2.0 × 10⁵ Pa and 1.0 × 10⁻³ m³ — and the arrows run clockwise, which is why the enclosed area is net work done by the gas.

1. aDetermine Determine which of the four processes drawn involve no work done by or on the gas, and state why.

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   Check answer · 2 marks

   1. B → C and D → A are the vertical sides of the rectangle, so ΔV = 0 for both
   2. work is the area under the process line, and a vertical line encloses none, so W = PΔV = 0 in each case

2. bCalculate Calculate the ratio of the temperature of the gas at state C to its temperature at state A.

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   Check answer · 3 marks

   1. the quantity of gas is fixed, so T ∝ PV
   2. P_A V_A = 2.0 × 10⁵ × 1.0 × 10⁻³ = 200 J and P_C V_C = 1.0 × 10⁵ × 3.0 × 10⁻³ = 300 J
   3. T_C/T_A = 300/200 = 1.5

3. cDetermine Determine which of the four plotted states is the hottest and which is the coldest, and determine the thermal energy transferred to or from the gas during the process C → D.

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   Check answer · 4 marks

   1. PV is 600 J at B and 100 J at D, against 200 J at A and 300 J at C, so B is the hottest state and D the coldest
   2. work done by the gas on C → D = PΔV = 1.0 × 10⁵ × (−2.0 × 10⁻³) = −200 J
   3. ΔU = (3/2)Δ(PV) = (3/2)(100 − 300) = −300 J for a monatomic gas
   4. Q = ΔU + W = −300 − 200 = −500 J, so 500 J is removed from the gas

4. dDetermine Determine the efficiency of this cycle worked as a heat engine, and compare it with the ideal limit 1 − T_cold/T_hot set by the hottest and coldest states on the diagram.

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   Check answer · 4 marks

   1. heat enters only on A → B, where Q = (3/2)(600 − 200) + 400 = 1000 J, and on D → A, where Q = (3/2)(200 − 100) = 150 J with no work, giving Q_in = 1150 J
   2. net work per cycle is the enclosed rectangle, 1.0 × 10⁵ × 2.0 × 10⁻³ = 200 J
   3. efficiency = 200/1150 = 0.17, that is 17%
   4. T ∝ PV makes T_D/T_B = 100/600, so the ideal limit is 1 − 1/6 = 83%: this cycle reaches only about a fifth of what the same two temperatures would allow

Reading cue. The dashed line runs well beyond both charges on purpose, because not every answer lies between them; the origin sits at +2q, so measure every distance from there.

1. aDetermine Determine the magnitude and direction of the electrostatic force that the +2q charge exerts on the −q charge.

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   Check answer · 2 marks

   1. F = k(2q)(q)/d² = 2kq²/d², using the separation d marked by the dimension line
   2. the charges are opposite in sign, so the force on −q is attractive and points to the left, towards +2q

2. bDetermine Determine the magnitude and direction of the net electric field at the midpoint of the line joining the two charges.

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   Check answer · 3 marks

   1. each charge is d/2 from the midpoint; the field of +2q points away from it, to the right, and the field of −q points towards it, also to the right
   2. magnitudes are k(2q)/(d/2)² = 8kq/d² and kq/(d/2)² = 4kq/d²
   3. they add to 12kq/d², directed from +2q towards −q

3. cCalculate Calculate the work an external agent must do to bring a charge +q from far away to the midpoint of the line joining the two charges, taking the potential to be zero at infinity.

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   Check answer · 3 marks

   1. potential is a scalar sum: V = k(2q)/(d/2) + k(−q)/(d/2) = 4kq/d − 2kq/d
   2. V = 2kq/d at the midpoint
   3. W = qV = 2kq²/d, and being positive it must be supplied by the external agent

4. dDetermine There is a second point on the dashed line, other than the one between the charges, at which the electric potential is zero. Determine where it lies, and explain why no such point exists to the left of +2q.

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   Show a hint

   Potential is a scalar, so there is no direction for the two contributions to cancel along — only sign. Write V for a general point in each of the three regions of the dashed line and see which of the three equations has a solution that actually lies in its own region.

   Check answer · 4 marks

   1. for a point at distance x from the origin lying beyond −q, V = 2kq/x − kq/(x − d)
   2. setting V = 0 gives 2(x − d) = x, so x = 2d measured from the origin
   3. that places it a distance d beyond the −q charge, still on the dashed line
   4. to the left of +2q the larger charge is always the nearer one, so 2kq/s always exceeds kq/(s + d) and the sum can never reach zero there

Reading cue. Both dimension lines run to the centre of the lens, not to its mount or its glass face, so a misjudged centre raises u by exactly what it takes off v.

1. aDetermine With a sharp image on the screen, the object stands at the 12.0 cm mark on the rail, the lens at the 42.0 cm mark and the screen at the 102.0 cm mark. Determine u and v.

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   Check answer · 2 marks

   1. u is the object-to-lens distance: 42.0 − 12.0 = 30.0 cm
   2. v is the lens-to-screen distance: 102.0 − 42.0 = 60.0 cm

2. bCalculate Calculate the magnification of the image for that setting, and determine whether the image caught on the screen is upright or inverted.

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   Check answer · 3 marks

   1. magnification m = −v/u = −60.0/30.0
   2. m = −2.0, so the image is twice the height of the illuminated object
   3. an image that can be caught on a screen is real, and the negative sign shows it is inverted

3. cDetermine Leaving the object and the screen exactly where they are, determine the second position of the lens on the rail that also gives a sharp image, and state the magnification there.

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   Check answer · 4 marks

   1. from part (a), 1/f = 1/30.0 + 1/60.0 = 1/20.0, so f = 20.0 cm
   2. object and screen are fixed 90.0 cm apart, so u + v = 90.0 while uv = f(u + v) = 20.0 × 90.0 = 1800 cm²
   3. u and v are then the roots of t² − 90t + 1800 = 0, namely 60.0 cm and 30.0 cm, so the lens also focuses when set at the 72.0 cm mark
   4. the magnification there is −30.0/60.0 = −0.50, the reciprocal in size of the value in part (b)

4. dExplain The student now moves the screen so that object and screen are only 70.0 cm apart. Explain why no position of this lens on the rail will give a sharp image, and determine the smallest separation that does allow one.

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   Show a hint

   The two lens positions in part (c) came out of a quadratic. Ask what happens to the roots of that quadratic as the object and the screen are brought closer together.

   Check answer · 4 marks

   1. with a separation D the lens positions satisfy t² − Dt + fD = 0, whose roots are real only if D² ≥ 4fD, that is D ≥ 4f
   2. for this lens 4f = 80.0 cm, so 70.0 cm is too small a separation
   3. at D = 70.0 cm the discriminant is 70² − 4(20)(70) = 4900 − 5600 = −700, which is negative, so there is no real lens position at all
   4. the smallest workable separation is 80.0 cm, where the two positions merge into one at the midpoint with u = v = 40.0 cm and magnification 1 in size

Reading cue. Find the two junction dots first: everything before them carries the whole current, so X is not one of three equals — Y and Z each take half of what X carries.

1. aIndicate Indicate which bulb carries the largest current and which two carry equal currents, using only the way the wires meet at the junction dots.

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   Check answer · 3 marks

   1. X carries the largest current
   2. Y and Z carry equal currents
   3. reason: X carries the sum of the two branch currents, while Y and Z are identical resistances between the same pair of junctions

2. bDetermine The cell has negligible internal resistance. Determine the potential difference across bulb X and across bulb Y, in terms of ε.

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   Check answer · 4 marks

   1. parallel pair has combined resistance R/2
   2. total circuit resistance 3R/2, so current from the cell I = 2ε/3R
   3. V_X = IR = 2ε/3
   4. V_Y = ε − V_X = ε/3

3. cDetermine Taking the current from the cell to be I, determine the fraction of the total power delivered by the cell that is dissipated in each of the three bulbs, and show that the three fractions account for all of it.

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   Check answer · 5 marks

   1. current in X is I; current in each of Y and Z is I/2 by equal division between identical branches
   2. P_X = I²R and P_Y = P_Z = (I/2)²R = I²R/4
   3. total power delivered = εI = I²(3R/2), since the cell drives I through an effective resistance 3R/2
   4. fractions 2/3 in X and 1/6 in each of Y and Z
   5. the three add to exactly 1, as they must when the cell has no internal resistance

4. dDetermine Bulb Z is removed and replaced by a thick copper wire of negligible resistance joining the same two junction dots. Determine the new power dissipated in X and in Y, and determine the factor by which the total power delivered by the cell changes.

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   Check answer · 5 marks

   1. the wire short-circuits Y, so the potential difference across the parallel section is zero
   2. P_Y = 0 and bulb Y goes out
   3. circuit resistance becomes R, so the current becomes ε/R
   4. P_X = ε²/R, against 4ε²/9R before, so X is 2.25 times brighter
   5. total power rises from 2ε²/3R to ε²/R, a factor of 1.5

Reading cue. The dimension line runs from the axis symbol to the far end, so L is the whole rod and the axis is at the end, not the centre; the weight acts a distance L/2 from that axis.

1. aIndicate Indicate the point on the rod through which its weight acts, and state its distance from the axis marked on the figure.

   recall2 marks

   Check answer · 2 marks

   1. the midpoint of the rod, since the rod is uniform
   2. a distance L/2 from the marked axis

2. bDerive Taking the moment of inertia about the axis drawn to be ML²/3, derive the moment of inertia about a parallel axis through the centre of the rod.

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   Check answer · 3 marks

   1. parallel-axis theorem used in the form I_axis = I_cm + Md²
   2. d = L/2, so Md² = ML²/4
   3. I_cm = ML²/3 − ML²/4 = ML²/12

3. cDetermine Determine every position along the rod at which a parallel axis would give a moment of inertia exactly twice that about the centre, giving each as a distance from the axis drawn.

   demanding4 marks

   Show a hint

   There is more than one such axis.

   Check answer · 4 marks

   1. condition I_cm + Md² = 2I_cm, so Md² = I_cm
   2. d² = L²/12, giving d = L/(2√3) = 0.289L from the centre of mass
   3. two axes, at 0.5L ± 0.289L from the drawn axis
   4. 0.21L and 0.79L from the drawn axis, both lying on the rod

4. dDerive The rod is replaced by one of the same mass M and the same length L, but with a linear density that increases with distance x from the drawn axis as λ = λ₀x/L. Derive its moment of inertia about that axis and compare it with ML²/3.

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   Check answer · 4 marks

   1. M = ∫₀^L (λ₀x/L) dx = λ₀L/2, so λ₀ = 2M/L
   2. dI = x² dm with dm = λ dx = (2M/L²)x dx
   3. I = ∫₀^L (2M/L²)x³ dx = ML²/2
   4. 1.5 times the uniform value, because mass has been shifted towards the far end where the lever arm x is largest

Reading cue. The dashed outline is where the rod ends up, not a second rod: what falls is the centre of mass at L/2 from the pivot, and it falls L/2, not L.

1. aIndicate Indicate the point whose fall supplies the energy released as the rod swings to the dashed position, and state how far that point drops.

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   Check answer · 2 marks

   1. the centre of mass, at the midpoint of the rod
   2. it drops a vertical distance L/2, not L

2. bDetermine Determine the angular acceleration of the rod at the instant it reaches the dashed vertical position.

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   Check answer · 3 marks

   1. at the vertical the line of action of the weight passes through the pivot
   2. the moment arm, and hence the torque about the pivot, is zero
   3. α = 0 at that instant, even though the rod is turning fastest there

3. cDetermine Determine the horizontal and vertical components of the force the pivot exerts on the rod at the instant of release, in terms of M and g.

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   Check answer · 4 marks

   1. α = (MgL/2)/(ML²/3) = 3g/2L at release
   2. centre of mass acceleration a = αL/2 = 3g/4 downwards
   3. Newton's second law vertically: Mg − N = M(3g/4), so N = Mg/4 upwards
   4. ω = 0 at release, so no centripetal force is needed and the horizontal component is zero

4. dDerive A small coin rests on the rod a distance x from the pivot and is released with it. Derive the least value of x for which the coin loses contact with the rod immediately, and describe what is seen for a coin placed closer to the pivot.

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   Check answer · 4 marks

   1. the point of the rod under the coin has downward acceleration a = αx = 3gx/2L at release
   2. a coin acted on by gravity alone can accelerate downwards only at g, since the rod cannot pull it down
   3. contact is lost when 3gx/2L > g, that is x > 2L/3
   4. for x < 2L/3 the rod pushes up on the coin, the two accelerate together and the coin stays on the rod

Reading cue. The string leaves the small hub, not the wheel rim, so r is the moment arm and the block's acceleration is αr; the wheel's own radius enters nowhere in the mechanics.

1. aIndicate Indicate whether the moment arm of the string tension about the axle is the hub radius r or the radius of the large wheel, and state the relation between the block's linear acceleration and the wheel's angular acceleration.

   recall3 marks

   Check answer · 3 marks

   1. the hub radius r, because the string leaves the hub tangentially
   2. a = αr
   3. the outer radius of the wheel does not appear in the relation, because no force is applied at the rim

2. bDetermine The block accelerates downwards with acceleration a. Determine the tension in the string in terms of m, g and a, and explain why the tension must be smaller than the weight of the block.

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   Check answer · 3 marks

   1. Newton's second law on the block: mg − T = ma
   2. T = m(g − a)
   3. for a > 0 this requires T < mg; if T equalled mg the block could not accelerate downwards at all

3. cCalculate In one trial the hub radius is r = 4.0 cm and the block has mass 0.50 kg. From rest, the block falls 1.20 m in 3.0 s. Ignoring axle friction, calculate the moment of inertia of the wheel. Take g = 9.8 m s⁻².

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   Check answer · 4 marks

   1. a = 2h/t² = 2(1.20)/3.0² = 0.267 m s⁻²
   2. T = m(g − a) = 0.50(9.8 − 0.267) = 4.77 N
   3. α = a/r = 0.267/0.040 = 6.67 rad s⁻²
   4. I = Tr/α = (4.77 × 0.040)/6.67 = 2.9 × 10⁻² kg m²

4. dDetermine After the block lands, the string leaves the hub and the wheel is left to spin freely; it comes to rest 12 s later. Determine the frictional torque in the axle, determine a corrected moment of inertia, and determine the percentage error in the value found in the previous part.

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   Check answer · 5 marks

   1. speed at landing v = at = 0.80 m s⁻¹, so ω₀ = v/r = 20 rad s⁻¹
   2. while coasting, friction alone acts: α_f = ω₀/t = 20/12 = 1.67 rad s⁻² and τ_f = Iα_f
   3. during the fall Tr − τ_f = Iα, giving 0.191 = I(6.67 + 1.67)
   4. I = 2.3 × 10⁻² kg m² and τ_f = 3.8 × 10⁻² N m
   5. the frictionless value is 25% too high, because torque actually spent against friction was credited to the wheel's inertia

Reading cue. The arrow drawn from the block is labelled bv, not b: it shrinks as the block slows, so the acceleration is never constant and no constant-acceleration equation applies.

1. aDetermine Determine the magnitude and the direction of the block's acceleration at t = 0.

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   Check answer · 3 marks

   1. magnitude bv₀/m
   2. directed opposite to the velocity, that is to the left
   3. the surface contributes nothing, being frictionless, so this is the only horizontal force

2. bDetermine Determine the time at which the block's speed has fallen to half of v₀, in terms of m and b.

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   Check answer · 3 marks

   1. v = v₀e^(−bt/m) stated or derived
   2. ½ = e^(−bt/m)
   3. t = (m/b)ln2 = 0.69 m/b

3. cDerive Derive an expression for the speed of the block as a function of the distance x it has travelled, rather than of time, and state the shape of a graph of v against x.

   demanding4 marks

   Show a hint

   Acceleration can be written as v dv/dx when speed is wanted as a function of position.

   Check answer · 4 marks

   1. write the acceleration as v dv/dx, so m v dv/dx = −bv
   2. dv/dx = −b/m, a constant
   3. v = v₀ − bx/m
   4. a straight line of negative gradient b/m, reaching v = 0 at x = mv₀/b

4. dJustify Justify the claim that the block gives up three-quarters of its initial kinetic energy while covering the first half of the distance it will ever travel, and state how long that first half takes.

   top of the paper4 marks

   Check answer · 4 marks

   1. the total distance is mv₀/b, so the halfway point is x = mv₀/2b
   2. from v = v₀ − bx/m the speed there is exactly v₀/2
   3. kinetic energy is then ¼ of its initial value, so ¾ has been dissipated
   4. the speed reaches v₀/2 at t = (m/b)ln2, so the first half of the journey takes a finite 0.69 m/b while the second half takes forever

Reading cue. The switch is drawn open, so Q is the quantity held fixed and V is the one free to move; the slab fills the gap completely, so the whole field between the plates is affected.

1. aIndicate Indicate which of the charge on the plates and the potential difference across them is fixed while the slab is inserted, and state the feature of the drawing that settles it.

   recall3 marks

   Check answer · 3 marks

   1. the charge ±Q is fixed
   2. because the switch is drawn open, so no charge can reach or leave the plates
   3. the potential difference is therefore free to change

2. bDetermine The plates have area 2.0 × 10⁻² m² and separation 1.0 mm and carry ±4.0 nC, and the slab has κ = 3.0. Determine the potential difference and the electric field between the plates before and after the slab is inserted. Take ε₀ = 8.85 × 10⁻¹² F m⁻¹.

   routine4 marks

   Check answer · 4 marks

   1. C₀ = ε₀A/d = 1.77 × 10⁻¹⁰ F
   2. V₀ = Q/C₀ = 22.6 V and E₀ = V₀/d = 2.3 × 10⁴ V m⁻¹
   3. C = κC₀ = 5.3 × 10⁻¹⁰ F
   4. V = 7.5 V and E = 7.5 × 10³ V m⁻¹, each reduced by the factor κ

3. cDetermine Determine the energy stored before and after insertion, and hence determine the work done by the person holding the slab as it goes in. State what the sign of that work says about the force on the slab.

   demanding4 marks

   Check answer · 4 marks

   1. U₀ = Q²/2C₀ = 45 nJ
   2. U = U₀/κ = 15 nJ
   3. work done by the person equals ΔU = −30 nJ
   4. negative work means the field pulls the slab in, so the person has to hold it back rather than push it

4. dExplain Explain what would be different if the switch had been left closed, with the battery still connected, while the slab was inserted. Refer to the charge, the stored energy, and the energy supplied by the battery, and support your answer with values.

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   Check answer · 6 marks

   1. V is now held at 22.6 V, so the charge rises to κQ = 12 nC
   2. stored energy rises to κU₀ = 136 nJ, instead of falling
   3. the battery supplies ΔQ × V = 8.0 nC × 22.6 V = 181 nJ
   4. only 90 nJ of that appears as extra stored energy
   5. the other 90 nJ is the work the field does on the slab, which appears as kinetic energy and is dissipated as the slab is stopped
   6. the slab is pulled inwards either way, so the direction of the force does not depend on the switch

Reading cue. Crosses cover the page inside and outside the loop, but only those inside it carry flux; the field is into the page and growing, so the induced current opposes the growth, not the field.

1. aDetermine The loop has radius a and the field magnitude is B. Determine the magnetic flux through the loop, and state why the crosses drawn outside the loop make no difference to it.

   recall3 marks

   Check answer · 3 marks

   1. Φ = Bπa²
   2. the field is uniform and perpendicular to the plane of the loop, so no angle factor is needed
   3. flux counts only the field crossing the area enclosed by the loop

2. bCalculate The loop has radius 6.0 cm and resistance 0.25 Ω, and the field increases steadily from 0.20 T to 0.50 T in 1.5 s. Calculate the induced e.m.f. and the current in the loop.

   routine4 marks

   Check answer · 4 marks

   1. A = π(0.060)² = 1.13 × 10⁻² m²
   2. dB/dt = 0.30/1.5 = 0.20 T s⁻¹
   3. e.m.f. = A dB/dt = 2.3 × 10⁻³ V
   4. I = e.m.f./R = 9.0 × 10⁻³ A

3. cDetermine Determine the direction of the magnetic force on the wire at the top of the loop, and hence determine whether the wire of the loop is squeezed inwards or stretched outwards while the field is increasing.

   demanding5 marks

   Check answer · 5 marks

   1. the induced current is counterclockwise, opposing the increasing into-page flux
   2. at the top of the loop that current points to the left
   3. F = IL × B on that element points towards the centre of the loop
   4. the same argument holds all round, so the loop is squeezed inwards and the wire is in compression
   5. consistent with Lenz's law, since shrinking the area would reduce the flux that is increasing

4. dExplain The loop is cut so that a narrow gap is left, and an ideal voltmeter is connected across the gap. Explain which of the flux, the e.m.f., the current, the dissipated power and the force on the wire change, and state the voltmeter reading.

   top of the paper5 marks

   Check answer · 5 marks

   1. the flux and the e.m.f. are unchanged, since both depend on the field and the enclosed area and not on the wire
   2. the current falls to zero, because the conducting path is broken
   3. the dissipated power falls to zero
   4. the magnetic force on the wire disappears, since it required a current
   5. the voltmeter reads the full induced e.m.f., 2.3 × 10⁻³ V

Reading cue. The rings crowd towards the surface, so the density rises with radius: r runs from the centre out to P, and the charge inside r grows faster than r³.

1. aDetermine Determine the volume charge density at the centre of the sphere, at the point P marked at radius r, and at the surface, in terms of ρ₀.

   recall3 marks

   Check answer · 3 marks

   1. zero at the centre
   2. ρ₀r/R at P
   3. ρ₀ at the surface, which is why the rings are drawn closest together there

2. bDetermine Determine the fraction of the sphere's total charge that lies within a radius R/2 of the centre.

   routine4 marks

   Check answer · 4 marks

   1. q(r) = ∫₀^r (ρ₀s/R)4πs² ds = πρ₀r⁴/R
   2. so q ∝ r⁴, and the total charge is Q = πρ₀R³
   3. q(R/2)/Q = (1/2)⁴ = 1/16, that is 6.25%
   4. much less than the 1/8 a uniform sphere would give, because the charge is pushed outwards

3. cSketch Sketch a graph of the magnitude of the electric field against distance from the centre, from the centre out to 3R, marking the position and the value of the maximum.

   demanding5 marks

   Check answer · 5 marks

   1. for r < R, E = q(r)/4πε₀r² = ρ₀r²/4ε₀R, drawn as a curve rising from the origin with increasing gradient, not a straight line
   2. for r > R, E = Q/4πε₀r², falling as 1/r²
   3. the two expressions agree at r = R, so the curve has no step in it
   4. maximum at the surface, r = R, of value ρ₀R/4ε₀
   5. at r = 3R the field has fallen to one ninth of the maximum

4. dDerive Using an energy density of ½ε₀E², derive an expression for the total electric energy stored in the field outside the sphere, and explain why the same calculation cannot be carried through for a point charge.

   top of the paper5 marks

   Check answer · 5 marks

   1. dU = ½ε₀E²(4πr² dr) with E = Q/4πε₀r² for r > R
   2. U = (Q²/8πε₀)∫_R^∞ dr/r² = Q²/8πε₀R
   3. substituting Q = πρ₀R³ gives U = πρ₀²R⁵/8ε₀
   4. for a point charge the lower limit becomes r = 0 and the integral diverges
   5. so a point charge implies infinite field energy, which is why a finite size has to be assumed in such calculations

Reading cue. There are no junction dots: one current passes through R and C alike, and with C uncharged at t = 0 the capacitor acts momentarily like a plain wire, putting all of ε across R.

1. aDetermine Determine the current in the circuit immediately after the switch closes, and a long time afterwards.

   recall3 marks

   Check answer · 3 marks

   1. immediately after closing, the uncharged capacitor has no potential difference across it, so all of ε is across R and I₀ = ε/R
   2. after a long time the capacitor potential difference has risen to ε
   3. no potential difference is left across R, so the current is zero

2. bCalculate Take ε = 12 V, R = 47 kΩ and C = 22 μF. Calculate the potential difference across the resistor and across the capacitor, and the current in the circuit, at the moment t = RC.

   routine4 marks

   Check answer · 4 marks

   1. RC = 47 × 10³ × 22 × 10⁻⁶ = 1.03 s
   2. V_C = ε(1 − e⁻¹) = 12 × 0.632 = 7.6 V
   3. V_R = ε − V_C = 4.4 V
   4. I = V_R/R = 9.4 × 10⁻⁵ A

3. cDetermine For the interval from t = 0 to t = RC only, determine the energy delivered by the battery, the energy stored in the capacitor and the energy dissipated in the resistor, and comment on whether the battery's energy divides equally between the two over this interval.

   demanding5 marks

   Check answer · 5 marks

   1. charge delivered q = Cε(1 − e⁻¹) = 1.67 × 10⁻⁴ C
   2. energy from the battery = εq = 2.0 × 10⁻³ J
   3. energy stored = q²/2C = 6.3 × 10⁻⁴ J
   4. energy dissipated = 2.0 × 10⁻³ − 6.3 × 10⁻⁴ = 1.4 × 10⁻³ J
   5. only about 32% is stored over this interval, so the equal split holds for the whole charging process and not for a part of it

4. dDetermine A long time after closing, the switch is reopened. Determine what happens to the charge on the capacitor, describe the smallest change to the circuit that would let the capacitor discharge through R, and determine how long that discharge would take to reduce the charge to 10% of its initial value.

   top of the paper5 marks

   Check answer · 5 marks

   1. the figure is a single loop with the switch in series, so reopening it leaves no conducting path anywhere
   2. the charge Cε = 2.6 × 10⁻⁴ C simply stays on the capacitor, apart from slow leakage
   3. a path is needed that contains C and R but not the open switch, for example a two-way switch that replaces the battery branch with a plain wire
   4. discharge then follows q = Q e^(−t/RC)
   5. t = RC ln10 = 1.03 × 2.30 = 2.4 s

Reading cue. L is tick-marked between the rails, so it is the length of rod in the circuit, not a distance travelled; the loop's area is bounded by R at one end and the moving rod at the other.

1. aIndicate Indicate the direction, clockwise or counterclockwise as drawn, of the induced current around the loop while the rod moves to the right.

   recall3 marks

   Check answer · 3 marks

   1. the enclosed area is growing, so the outward flux through the loop is increasing
   2. the induced current opposes this by producing flux into the page inside the loop
   3. the current is therefore clockwise as drawn, passing down through the rod

2. bCalculate Take B = 0.45 T, L = 0.35 m, R = 0.80 Ω, m = 0.15 kg and v₀ = 4.0 m s⁻¹. Calculate the magnetic force on the rod and its deceleration at t = 0.

   routine4 marks

   Check answer · 4 marks

   1. e.m.f. = BLv₀ = 0.45 × 0.35 × 4.0 = 0.63 V
   2. I = e.m.f./R = 0.79 A
   3. F = BIL = 0.124 N
   4. a = F/m = 0.83 m s⁻², directed opposite to the motion

3. cDetermine Determine the speed of the rod 1.0 s after release, the distance it has travelled in that time, and the energy dissipated in R during that second.

   demanding4 marks

   Check answer · 4 marks

   1. time constant τ = mR/B²L² = (0.15 × 0.80)/(0.1575)² = 4.8 s
   2. v = v₀e^(−t/τ) = 4.0 × e^(−0.207) = 3.3 m s⁻¹
   3. distance = v₀τ(1 − e^(−t/τ)) = 4.0 × 4.84 × 0.187 = 3.6 m
   4. energy = ½m(v₀² − v²) = ½ × 0.15 × (16 − 10.6) = 0.41 J, all of it dissipated in R

4. dDerive Starting from rest, the rod is now pushed to the right by a constant applied force of 0.50 N. Derive an expression for its terminal speed, calculate its value, and show that at that speed every joule supplied by the applied force is dissipated in R.

   top of the paper5 marks

   Check answer · 5 marks

   1. equation of motion m dv/dt = F − B²L²v/R
   2. at terminal speed dv/dt = 0, so v_T = FR/B²L²
   3. v_T = (0.50 × 0.80)/(0.1575)² = 16 m s⁻¹
   4. mechanical power supplied = Fv_T = 8.1 W
   5. electrical power dissipated = (BLv_T)²/R = 8.1 W, equal because the kinetic energy is no longer changing