AP Physics 2 · Algebra-Based

An ideal gas expands at a constant pressure of 2.0 × 10⁵ Pa from a volume of 1.0 × 10⁻³ m³ to 3.0 × 10⁻³ m³. How much work is done by the gas?

1. A200 J
2. B400 J
3. C600 J
4. D800 J

Two point charges, +Q and −Q, are held a distance d apart. What are the electric field and the electric potential at the midpoint between them, taking the potential to be zero at infinity?

1. Afield zero, potential zero
2. Bfield zero, potential non-zero
3. Cfield non-zero, potential zero
4. Dfield non-zero, potential non-zero

Three identical light bulbs are connected to a battery of negligible internal resistance: bulb X is in series with a parallel combination of bulbs Y and Z. Bulb Z then burns out, breaking its branch. What happens to the brightness of bulb X?

1. AIt becomes brighter.
2. BIt becomes dimmer.
3. CIt is unchanged.
4. DIt goes out.

A straight wire of length 0.25 m carries a current of 3.0 A at right angles to a uniform magnetic field of magnitude 0.40 T. What is the magnitude of the magnetic force on the wire?

1. A0.030 N
2. B0.30 N
3. C3.0 N
4. D4.8 N

An object is placed 10 cm in front of a converging lens of focal length 15 cm. Which describes the image?

1. Areal, inverted, and 30 cm from the lens
2. Breal, inverted, and 6.0 cm from the lens
3. Cvirtual, upright, and 30 cm from the lens
4. Dvirtual, upright, and 6.0 cm from the lens

A nucleus of ²³⁸₉₂U decays to ²³⁴₉₀Th. What particle is emitted?

1. Aan alpha particle
2. Ba beta-minus particle
3. Ca beta-plus particle
4. Da gamma photon

A fixed quantity of a monatomic ideal gas is taken around the closed cycle A → B → C → D → A. State A is at pressure 2.0 × 10⁵ Pa and volume 1.0 × 10⁻³ m³. The gas expands at constant pressure to state B at 3.0 × 10⁻³ m³. It is then cooled at constant volume to state C at 1.0 × 10⁵ Pa. It is compressed at constant pressure to state D at 1.0 × 10⁻³ m³, and finally warmed at constant volume back to state A.

1. (a)

   Calculate Calculate the work done by the gas during the process A → B.

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   1. 1

      1 point: W = PΔV for a constant-pressure process

   2. 2

      1 point: W = 2.0 × 10⁵ × 2.0 × 10⁻³ = 400 J, done by the gas

2. (b)

   Determine Determine the change in internal energy of the gas during A → B, and hence the thermal energy added to the gas.

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   1. 1

      1 point: for a monatomic ideal gas ΔU = (3/2)nRΔT = (3/2)Δ(PV)

   2. 2

      1 point: Δ(PV) = 2.0 × 10⁵ (3.0 × 10⁻³ − 1.0 × 10⁻³) = 400 J, so ΔU = 600 J

   3. 3

      1 point: applies the first law, Q = ΔU + W_by

   4. 4

      1 point: Q = 600 + 400 = 1000 J added to the gas

3. (c)

   Determine Determine the net work done by the gas in one complete cycle.

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   1. 1

      1 point: net work is the area enclosed by the cycle on the PV diagram

   2. 2

      1 point: area = (2.0 × 10⁵ − 1.0 × 10⁵)(3.0 × 10⁻³ − 1.0 × 10⁻³)

   3. 3

      1 point: net work = 200 J, done by the gas because the cycle is traversed clockwise

4. (d)

   Explain Explain why the change in internal energy of the gas over one complete cycle is zero, and state what this implies about the net thermal energy exchanged with the surroundings.

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   1. 1

      1 point: internal energy is a function of state and depends only on the temperature

   2. 2

      1 point: the gas returns to state A, so its temperature — and hence its internal energy — is the same as at the start, giving ΔU = 0 for the cycle

   3. 3

      1 point: by the first law the net thermal energy added must equal the net work done by the gas, 200 J

Two point charges are fixed on a horizontal line: a charge +2q at the origin, and a charge −q at a distance d to the right of it.

1. (a)

   Sketch Sketch the electric field lines in the region around the two charges. Indicate the direction of each line with an arrow.

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   1. 1

      1 point: lines begin on the positive charge and end on the negative charge, with arrows pointing away from +2q and towards −q

   2. 2

      1 point: twice as many lines leave +2q as arrive at −q, with the surplus continuing outward to large distances

   3. 3

      1 point: lines meet both charges radially and never cross one another

2. (b)

   Determine Determine the location on the line through the two charges at which the net electric field is zero. Indicate whether it lies to the left of +2q, between the charges, or to the right of −q, and justify that region before calculating.

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   1. 1

      1 point: the null point must lie beyond the smaller charge, to the right of −q, because only there do the two fields oppose and the weaker charge is closer

   2. 2

      1 point: sets k(2q)/(d + x)² = kq/x², where x is measured from −q

   3. 3

      1 point: 2x² = (d + x)², so x√2 = d + x

   4. 4

      1 point: x = d/(√2 − 1) = 2.4d to the right of −q

3. (c)

   Determine Determine whether there is a point on the line between the two charges at which the electric potential is zero, and if so, where.

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   1. 1

      1 point: potentials are scalars, so a zero requires k(2q)/r₁ = kq/r₂ with r₁ + r₂ = d

   2. 2

      1 point: 2r₂ = r₁, so r₂ = d/3

   3. 3

      1 point: the potential is zero at a distance d/3 to the left of −q (that is, 2d/3 from +2q)

4. (d)

   Explain Explain why the point where the field is zero and the point where the potential is zero are in different places.

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   1. 1

      1 point: the field is a vector sum, so it vanishes where two contributions are equal in magnitude and opposite in direction — which can only happen outside the pair

   2. 2

      1 point: the potential is a scalar sum, so it vanishes where two contributions of opposite sign are equal in magnitude, which happens between the charges; the two conditions are different, so the points differ

A student is given a converging lens of unknown focal length, an illuminated object, a screen, an optical bench and a meter stick. The student measures the object distance u and the corresponding image distance v for several positions of the lens.

1. (a)

   Describe Describe how the student should obtain each value of v so that the measurement is as accurate as possible.

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   1. 1

      1 point: move the screen until the image of the object is as sharp as possible, then measure from the centre of the lens to the screen

   2. 2

      1 point: approach the sharpest position from both directions and take the midpoint of the range over which the image looks focused

   3. 3

      1 point: measure with the meter stick parallel to the bench and the eye directly above the scale, to avoid parallax

2. (b)

   Describe The thin lens equation is 1/u + 1/v = 1/f. Describe how the student should plot the data to obtain a straight line, and state what the slope and the intercepts of that line represent.

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   1. 1

      1 point: plot 1/v on the vertical axis against 1/u on the horizontal axis

   2. 2

      1 point: rearranged, 1/v = −(1/u) + 1/f, which has the form y = mx + c

   3. 3

      1 point: the slope is −1

   4. 4

      1 point: both intercepts equal 1/f

3. (c)

   Determine Use the data to determine the focal length of the lens.

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   1. 1

      1 point: adds 1/u and 1/v for any row — for example 0.0500 + 0.0167 = 0.0667 cm⁻¹

   2. 2

      1 point: confirms the sum is the same for every row, so 1/f = 0.0667 cm⁻¹

   3. 3

      1 point: f = 15.0 cm

4. (d)

   Explain The student notices that the point at u = 20.0 cm lies slightly further from the best-fit line than the others. Explain why the measurement of v is least precise for small object distances, and describe one change that would improve it.

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   1. 1

      1 point: at small u the image distance is large, the image is highly magnified and dim, and the range of screen positions over which it looks sharp is widest — so the uncertainty in v is greatest

   2. 2

      1 point: improvement — use a brighter object, work in a darkened room, or take repeated determinations of the sharpest position and average them

Three identical bulbs, each of resistance R, are connected to a battery of e.m.f. ε and negligible internal resistance. Bulb X is connected in series with a parallel combination of bulbs Y and Z. The brightness of a bulb is determined by the power it dissipates.

1. (a)

   Indicate Indicate which bulb or bulbs are brightest in the original circuit. No justification is required in this part.

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   1. 1

      1 point: bulb X

2. (b)

   Derive Derive expressions for the power dissipated by bulb X and by bulb Y in the original circuit, in terms of ε and R.

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   1. 1

      1 point: parallel pair has resistance R/2, so the total resistance is 3R/2

   2. 2

      1 point: current through X is I = ε/(3R/2) = 2ε/(3R)

   3. 3

      1 point: P_X = I²R = 4ε²/(9R)

   4. 4

      1 point: current through Y is half of I, so P_Y = (ε/(3R))²R = ε²/(9R)

3. (c)

   Determine Bulb Z now burns out, breaking its branch of the circuit. Determine the new power dissipated by bulb X, and state whether X becomes brighter, dimmer, or stays the same.

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   1. 1

      1 point: with Z's branch broken, X and Y are in series and the total resistance is 2R

   2. 2

      1 point: I = ε/(2R), so P_X = ε²/(4R)

   3. 3

      1 point: ε²/(4R) is less than 4ε²/(9R), so X becomes dimmer

4. (d)

   Explain Explain, without using equations, why bulb X becomes dimmer while bulb Y becomes brighter when bulb Z burns out.

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   1. 1

      1 point: removing one of two parallel branches raises the resistance of that section, and so raises the total resistance of the circuit

   2. 2

      1 point: a higher total resistance means a smaller current from the battery, and bulb X carries the whole of that current, so X dissipates less power and dims

   3. 3

      1 point: bulb Y previously carried only half of the current through X, because the current divided between two identical branches

   4. 4

      1 point: now Y carries all of the current, and although that total current is smaller than before, it is larger than the half-share Y used to have — so Y brightens