AP Prep — Physics 2 Practice

Question 1 of 10

MCQU10Topic 10.2Medium No calc Word

An initially neutral metal sphere rests on an insulating stand. A student brings a negatively charged rod near (but not touching) the right side of the sphere, then momentarily touches a grounding wire to the LEFT side of the sphere, then removes the wire, and finally removes the rod. Which statement correctly describes the sphere afterward, and why?

AThe sphere is neutral, because charge is conserved, so disconnecting the ground wire restores the sphere to exactly the state it had before the rod was ever brought near it.

BThe sphere is neutral overall but permanently polarized, with positive charge locked on the right side and negative charge locked on the left side even after the rod is taken away.

CThe sphere is negatively charged, because the strong field of the negative rod drove some of the rod's own excess electrons across the air gap and onto the surface of the metal.

DThe sphere is positively charged, because electrons repelled by the rod escaped through the ground wire and cannot return.

Explanation

This is charging by induction. The negative rod repels the sphere's free electrons to the far (left) side; touching a ground wire there gives those electrons a path to Earth, so they leave the sphere. Removing the wire first, and only then the rod, strands the sphere with an electron deficit — net POSITIVE. The first choice treats charge as jumping off an insulating rod (the rod's excess charge is not mobile, and no spark is described), and it also predicts the wrong sign. The second choice misapplies conservation: charge is conserved for the sphere + Earth system, not for the sphere alone once electrons have flowed away. The third choice describes only what happens while the rod is present; once the rod is removed the free electrons in a conductor redistribute, so an isolated neutral conductor cannot hold a permanent polarization.

Question 2 of 10

MCQU12Topic 12.2Medium No calc Word

A proton (charge +1.60×10⁻¹⁹ C) travels at 2.5×10⁵ m/s through a uniform magnetic field of magnitude 0.40 T. The proton's velocity makes an angle of 30° with the direction of the magnetic field. What is the magnitude of the magnetic force on the proton?

A1.6×10⁻¹⁴ N, because the full product qvB gives the force on any moving charge

B1.39×10⁻¹⁴ N, using the component of v along B in F = qvB cos θ

C8.0×10⁻¹⁵ N

D0 N, because a magnetic field can never exert a force on a single moving charge

Explanation

The magnetic force on a moving charge is F = qvB sin θ, where θ is the angle between v and B. F = (1.60×10⁻¹⁹ C)(2.5×10⁵ m/s)(0.40 T)(sin 30°) = (1.6×10⁻¹⁴ N)(0.50) = 8.0×10⁻¹⁵ N. The 1.6×10⁻¹⁴ N choice is what a student gets by dropping the sin θ factor entirely (that value is correct only at θ = 90°). Using cos θ is also wrong: the force depends on the component of v PERPENDICULAR to B, which is v sin θ, so the force is largest when v ⊥ B and zero when v is parallel to B.

Question 3 of 10

MCQU13Topic 13.4Medium Calc Word Diagram

A thin diverging lens has a focal length of magnitude 20 cm. A 6.0 cm tall arrow is placed 30 cm from the lens on the axis. What is the height and orientation of the image?

A12 cm tall and inverted

B2.4 cm tall and inverted, since the image distance works out negative and a negative magnification means the arrow is flipped over.

C6.0 cm tall and upright, because a lens changes only where the image forms, not how big it is.

D2.4 cm tall and upright

Explanation

A diverging lens has a NEGATIVE focal length, f = −20 cm. Then 1/d_i = 1/f − 1/d_o = −1/20 − 1/30 = −5/60, so d_i = −12 cm — the image is virtual and on the same side as the arrow. The magnification is m = −d_i/d_o = −(−12)/30 = +0.40, so the image height is 0.40 × 6.0 = 2.4 cm and, because m is positive, upright. A diverging lens always gives a reduced, upright, virtual image of a real object. The 12 cm answer comes from taking f = +20 cm (the sign error of making a diverging lens's focal length positive, which gives d_i = +60 cm and m = −2); calling the image inverted misreads the sign of m after the double negative; and a lens certainly does change image size whenever d_i ≠ d_o.

Question 4 of 10

MCQU9Topic 9.3Easy No calc Word

A hot steel bolt is dropped into a cup of cool oil, and the sealed, insulated cup is left alone until nothing further changes. Which statement best describes the final state?

AThe bolt and the oil are at the same temperature, and there is no further net transfer of energy between them

BThe bolt and the oil now contain equal amounts of heat, because heat spreads itself evenly between objects in contact

CThe bolt and the oil now contain equal amounts of internal energy, since energy flowed until the two amounts matched

DThe bolt still contains more heat than the oil, but the transfer has stopped because the bolt has run out of heat to give

Explanation

Thermal equilibrium is defined by equal TEMPERATURES, which makes the net rate of energy transfer between the objects zero (microscopic exchanges continue in both directions but cancel). Internal energies are not equal — those depend on mass and specific heat as well as temperature, so the larger heat capacity holds more internal energy at the same T. The two 'equal amounts of heat' and 'contains more heat' options treat heat as a fluid-like substance stored inside an object. Heat is energy IN TRANSIT because of a temperature difference; once the temperatures match, there is no heat at all, so an object cannot 'contain' any.

Question 5 of 10

MCQU11Topic 11.1Medium Calc Word

A wire carries a steady current of 0.80 A. Approximately how many electrons pass a given point in the wire in 5.0 s? (The magnitude of the electron charge is 1.60×10⁻¹⁹ C.)

A4.0×10¹⁹ electrons

B2.5×10¹⁹ electrons

C1.28×10⁻¹⁹ electrons

D5.0×10¹⁸ electrons

Explanation

First find the total charge: ΔQ = I·Δt = (0.80 A)(5.0 s) = 4.0 C. Then divide by the charge per electron: N = ΔQ/e = 4.0 C / 1.60×10⁻¹⁹ C = 2.5×10¹⁹ electrons. The choice 5.0×10¹⁸ is what you get from N = I/e, forgetting to multiply by the 5.0 s of elapsed time. 4.0×10¹⁹ comes from rounding e to 1.0×10⁻¹⁹ C, and 1.28×10⁻¹⁹ comes from multiplying by e instead of dividing, which also gives a physically impossible count.

Question 6 of 10

MCQU10Topic 10.3Hard Calc Word

A uniform electric field of magnitude 250 N/C points vertically DOWNWARD in the region between two large horizontal plates. A small bead carrying a charge of −2.0 μC is placed in this region. What bead mass would allow the bead to hang motionless, with the electric force exactly balancing gravity?

A4.9×10⁻³ kg

B5.1×10⁻⁵ kg

C1.0×10⁻⁴ kg

D5.0×10⁻⁴ kg

Explanation

The force on a NEGATIVE charge is opposite the field, so with E downward the electric force on the bead is UPWARD, magnitude F = |q|E = (2.0×10⁻⁶)(250) = 5.0×10⁻⁴ N. Equilibrium requires |q|E = mg, so m = |q|E/g = 5.0×10⁻⁴/9.8 = 5.1×10⁻⁵ kg. The 5.0×10⁻⁴ choice reports the FORCE in newtons as if it were a mass — the units do not balance. The 4.9×10⁻³ choice multiplies by g instead of dividing (|q|E·g). The 1.0×10⁻⁴ choice uses 4.0 μC, doubling the charge magnitude. Note that if the charge were positive the electric force would point down with gravity and no equilibrium would exist at all.

Question 7 of 10

MCQU9Topic 9.1Medium No calc Word

A sealed flask contains neon gas, a monatomic ideal gas, at a uniform temperature of 27 °C. What is the average translational kinetic energy of a single neon atom in the flask?

A5.6×10⁻²² J

B6.2×10⁻²¹ J

C4.1×10⁻²¹ J

D3.7×10³ J

Explanation

Temperature is a measure of the AVERAGE translational kinetic energy per particle: K_avg = (3/2)k_B·T, with T in kelvin. Convert first: T = 27 + 273 = 300 K. Then K_avg = 1.5 × (1.38×10⁻²³ J/K) × (300 K) = 6.21×10⁻²¹ J. The key distractor 5.6×10⁻²² J comes from plugging in 27 directly (Celsius) instead of 300 K — the kinetic-theory relation is built on absolute temperature, where zero really means zero molecular kinetic energy, so Celsius values are meaningless here. 4.1×10⁻²¹ J drops the 3/2 factor (that is k_B·T), and 3.7×10³ J uses R in place of k_B, which gives the energy of a whole MOLE, not one atom.

Question 8 of 10

MCQU13Topic 13.4Hard No calc Word Diagram

A student mounts a thin converging lens of focal length 12 cm on an optical bench. A small illuminated arrow 2.0 cm tall is placed upright on the axis 8.0 cm from the lens, on the side the light comes from. Taking the standard sign convention (object distance positive, image distance positive on the far side of the lens, positive magnification = upright), which of the following correctly gives the image distance, the nature of the image, and the magnification?

Ad_i = -4.8 cm; the image is virtual and upright, with m = +0.60

Bd_i = -24 cm; the image is virtual and upright, with m = +3.0

Cd_i = +24 cm; the image is real and inverted, with m = -3.0

Dd_i = -24 cm; the image is virtual and inverted, with m = -3.0

Explanation

Principle: the thin-lens equation 1/f = 1/d_o + 1/d_i with the magnification relation m = -d_i/d_o. Steps: a converging lens has f = +12 cm, and the object sits in front of it, so d_o = +8.0 cm. Then 1/d_i = 1/12 - 1/8 = (2 - 3)/24 = -1/24, giving d_i = -24 cm. A negative image distance means the image lies on the same side of the lens as the object, so it is VIRTUAL (the outgoing rays diverge and only appear to come from that point). The magnification is m = -d_i/d_o = -(-24)/8.0 = +3.0, so the image is upright and 3 times as tall, 6.0 cm. This is the magnifying-glass case: the object is INSIDE the focal point (8.0 cm < 12 cm), so no real image can form. Why the distractors are wrong: the option giving d_i = +24 cm with a real, inverted image comes from assuming that an object in front of a converging lens always produces a real image and simply dropping the minus sign; the algebra genuinely gives -24 cm, and a real image is impossible when d_o < f. The option giving d_i = -4.8 cm comes from wrongly using f = -12 cm for a converging lens: 1/d_i = -1/12 - 1/8 = -5/24 yields -4.8 cm. The option keeping d_i = -24 cm but calling the image inverted uses m = +d_i/d_o without the minus sign, which flips the reported orientation; a single converging lens forming a virtual image always forms it upright.

Question 9 of 10

MCQU14Topic 14.1Medium No calc Word

Two triangular pulses of identical shape and width travel toward each other along the same taut rope. One pulse displaces the rope upward with a peak height of 4.0 cm; the other displaces it downward with a peak depth of 3.0 cm. At the single instant when the two pulses exactly overlap, what is the maximum displacement of the rope from its equilibrium position?

A0 cm, because the two pulses cancel completely

B3.0 cm downward

C1.0 cm upward

D7.0 cm upward

Explanation

Waves obey the principle of superposition: the displacement of the rope at any point is the algebraic sum of the displacements each pulse would produce alone. Taking upward as positive, the sum at the overlap peak is (+4.0 cm) + (-3.0 cm) = +1.0 cm, so the rope rises 1.0 cm above equilibrium. The 7.0 cm choice comes from adding the magnitudes without regard to sign, which is what happens only when both pulses are on the same side of the rope. Complete cancellation (0 cm) would require pulses of equal magnitude and opposite sign, which is not the case here.

Question 10 of 10

MCQU11Topic 11.5Hard No calc Word Diagram

Three identical bulbs X, Y, and Z (each of fixed resistance R) are connected to an ideal battery: bulb X is in series with the battery, and bulbs Y and Z are connected in parallel with each other, that parallel pair being in series with X. Which statement correctly ranks the brightness of the bulbs?

AX is brightest; Y and Z are equally bright, each dissipating one quarter of X's power

BX is dimmest, because the current is partly used up in X before reaching Y and Z, leaving less charge flowing through the first bulb in the line

CAll three bulbs are equally bright, since they are identical and the same current that leaves the battery passes through every one of them

DY and Z are each brighter than X, because the parallel section has the smaller equivalent resistance and current takes the easier path

Explanation

Brightness is power dissipated, P = I²R. The full battery current I passes through X, but it splits equally between the identical parallel bulbs, so Y and Z each carry I/2. Thus P_X = I²R while P_Y = P_Z = (I/2)²R = I²R/4: X is brightest and the other two are equal at one quarter of X's power. The 'all equal' option assumes series and parallel elements carry the same current. The Y-and-Z-brighter option ranks brightness by resistance alone rather than by I²R with the actual branch currents. The last option uses the false idea that current is 'used up' in a bulb; current is conserved at every junction, and X in fact carries the most.

Free Response 1 · Section II

FRQMathematical RoutinesU9 Calc

A vertical cylinder of uniform cross-sectional area A = 0.0120 m² is closed at the bottom and sealed at the top by a frictionless piston of mass m = 15.0 kg that can slide without leaking. The cylinder contains a monatomic ideal gas. Atmospheric pressure outside the cylinder is P₀ = 1.0 × 10⁵ Pa, and the cylinder walls conduct heat only through a small heater at the base. Initially the gas is in equilibrium with the piston at rest, the gas volume is V₁ = 3.00 × 10⁻³ m³, and the gas temperature is T₁ = 300 K. The heater is then switched on and the gas is heated slowly until the gas volume has doubled, with the piston free to rise the whole time.

(a) Starting from a force analysis of the piston, derive a symbolic expression for the absolute pressure P of the gas in terms of P₀, m, g, and A. Explain why this pressure stays constant while the gas is being heated, and evaluate P numerically.

(b) Calculate the number of moles of gas sealed in the cylinder.

(c) Derive a symbolic expression for the total energy Q added to the gas during the heating, in terms of P and the volume change ΔV only, and then evaluate ΔU, the work done by the gas, and Q numerically for this process.

(d) Calculate the distance the piston rises, and determine what fraction of the energy Q added by the heater ends up as gravitational potential energy of the piston.

Free response is self-scored — work it out, then reveal the model answer and scoring checklist to compare.

Model answer

(a) Free-body diagram of the piston (in equilibrium, frictionless, no leakage): P·A (up, from the gas below) = P₀·A (down, from the atmosphere above) + m·g (weight) P = P₀ + mg/A Because P₀, m, g, and A are all fixed, the equilibrium gas pressure is fixed: the process is isobaric. If heating momentarily raises the pressure, the piston accelerates upward, the gas expands, and the pressure falls right back to this same value. P = 1.0 × 10⁵ + (15.0)(9.8)/(0.0120) = 1.0 × 10⁵ + 1.225 × 10⁴ = 1.1225 × 10⁵ Pa

(b) PV = nRT ⇒ n = PV₁/(R·T₁) n = (1.1225 × 10⁵)(3.00 × 10⁻³)/[(8.31)(300)] = 336.75/2493 = 0.135 mol

(c) Isobaric: W_by = P·ΔV. For a monatomic ideal gas ΔU = (3/2)nRΔT, and at constant pressure nRΔT = PΔV, so ΔU = (3/2)PΔV. First law: Q = ΔU + W_by = (3/2)PΔV + PΔV = (5/2)PΔV ΔV = 6.00 × 10⁻³ − 3.00 × 10⁻³ = 3.00 × 10⁻³ m³ W_by = (1.1225 × 10⁵)(3.00 × 10⁻³) = 337 J ΔU = (3/2)(336.75) = 505 J Q = (5/2)(1.1225 × 10⁵)(3.00 × 10⁻³) = 842 J (Consistency check: since V doubles at constant P, T₂ = 2T₁ = 600 K, and ΔU = (3/2)nRΔT = 1.5(0.13508)(8.31)(300) = 505 J.)

(d) Δx = ΔV/A = (3.00 × 10⁻³)/(0.0120) = 0.250 m ΔU_grav = mgΔx = (15.0)(9.8)(0.250) = 36.75 J ≈ 36.8 J Fraction = 36.75/841.9 = 0.0437, i.e. about 4.4 % of the heater's energy lifts the piston. The rest raises the internal energy of the gas (505 J) and pushes the atmosphere back (P₀ΔV = 300 J). Note 505 + 36.75 + 300 = 842 J, so the energy budget closes.

Scoring · 6 points

(a) Derives P = P₀ + mg/A from an explicit piston force balance rather than quoting it.

(a) Argues the process is isobaric because P₀, m, g, and A are all constant, AND evaluates P ≈ 1.12 × 10⁵ Pa.

(b) Applies PV = nRT with the part-(a) pressure and T₁ = 300 K to obtain n ≈ 0.135 mol.

(d) Obtains piston rise Δx = ΔV/A = 0.250 m and the fraction mgΔx/Q ≈ 0.044 (about 4 %).

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