AP Physics C · Mechanics

A particle moves along the x-axis with acceleration a(t) = 6t, where a is in m/s² and t is in seconds. At t = 0 the particle is at rest at the origin. What is its position at t = 2.0 s?

1. A4.0 m
2. B8.0 m
3. C12 m
4. D24 m

An object of mass m falls from rest through a fluid that exerts a resistive force of magnitude bv, where v is the speed. What is the terminal speed?

1. Ab/(mg)
2. Bmg/b
3. C√(mg/b)
4. Dmgb

A single force F(x) = 4x² acts on an object as it moves along the x-axis, where F is in newtons and x in metres. How much work does this force do on the object as it moves from x = 0 to x = 3.0 m?

1. A12 J
2. B36 J
3. C54 J
4. D108 J

A uniform thin rod of mass M and length L rotates about an axis perpendicular to the rod through one end. What is its moment of inertia about that axis?

1. AML²/12
2. BML²/6
3. CML²/3
4. DML²

A figure skater spinning with her arms extended pulls her arms in close to her body. Which of the following is true?

1. AHer angular momentum and her kinetic energy both stay the same.
2. BHer angular momentum stays the same and her kinetic energy increases.
3. CHer angular momentum increases and her kinetic energy stays the same.
4. DHer angular momentum and her kinetic energy both increase.

A particle of mass m moves in a potential U(x) = ½kx⁴. Which statement about its motion for small oscillations about x = 0 is correct?

1. AThe motion is simple harmonic with angular frequency √(k/m).
2. BThe motion is simple harmonic with angular frequency √(2k/m).
3. CThe motion is periodic but not simple harmonic, and its period depends on the amplitude.
4. DThe motion is not periodic.

A uniform thin rod of mass M and length L is free to rotate in a vertical plane about a frictionless horizontal pivot at one end. The rod is held horizontal and released from rest.

1. (a)

   Derive Using integration, derive the moment of inertia of the rod about the pivot. Show the mass element you use.

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   1. 1

      1 point: takes a mass element dm = (M/L)dx at distance x from the pivot

   2. 2

      1 point: writes I = ∫x² dm = ∫₀^L x²(M/L)dx

   3. 3

      1 point: evaluates the integral to (M/L)(L³/3)

   4. 4

      1 point: I = ML²/3

2. (b)

   Determine Determine the angular acceleration of the rod at the instant it is released, while it is still horizontal.

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   1. 1

      1 point: the weight acts at the centre of mass, a distance L/2 from the pivot, giving τ = MgL/2

   2. 2

      1 point: α = τ/I = (MgL/2)/(ML²/3)

   3. 3

      1 point: α = 3g/(2L)

3. (c)

   Derive Derive an expression for the angular speed of the rod when it reaches the vertical position.

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   1. 1

      1 point: the centre of mass falls a distance L/2, so the energy released is MgL/2

   2. 2

      1 point: sets MgL/2 = ½Iω² with I = ML²/3

   3. 3

      1 point: ω = √(3g/L)

4. (d)

   Explain Determine the linear acceleration of the free end of the rod at the instant of release, and explain how a point on the rod can have a downward acceleration greater than g.

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   1. 1

      1 point: a = αL = 3g/2, which is 1.5g

   2. 2

      1 point: the rod is a rigid body, not a free particle — the pivot exerts a force on it, so the net force on the rod is not simply its weight

   3. 3

      1 point: the pivot pushes down on the rod (or, equivalently, the inner part of the rod is accelerating more slowly than g and the rod's rigidity transmits this as a downward force on the outer part), allowing the far end to accelerate faster than free fall

A particle of mass m moves along the x-axis in a conservative field described by the potential energy function U(x) = ax⁴ − bx², where a and b are positive constants.

1. (a)

   Derive Derive an expression for the force on the particle as a function of x.

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   1. 1

      1 point: uses F = −dU/dx

   2. 2

      1 point: F = −4ax³ + 2bx

2. (b)

   Determine Determine the positions of all equilibrium points, and state for each whether it is stable or unstable.

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   1. 1

      1 point: equilibrium requires F = 0, so 2bx = 4ax³

   2. 2

      1 point: x = 0 and x = ±√(b/2a)

   3. 3

      1 point: x = 0 is unstable — U has a local maximum there, since d²U/dx² = −2b < 0

   4. 4

      1 point: x = ±√(b/2a) are stable, since d²U/dx² = 12ax² − 2b = 4b > 0 at those points

3. (c)

   Sketch Sketch a graph of U(x) against x, marking the equilibrium positions, and on the same axes indicate a total energy E for which the particle is confined to a region on one side of the origin only.

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   1. 1

      1 point: symmetric double-well curve, with U → +∞ as x → ±∞ and a local maximum of U = 0 at x = 0

   2. 2

      1 point: minima marked at x = ±√(b/2a), with U negative there

   3. 3

      1 point: a horizontal line drawn at a negative energy E, between the minimum value of U and zero, with the two turning points on one side identified

4. (d)

   Derive Derive an expression for the angular frequency of small oscillations of the particle about one of the stable equilibrium positions.

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   1. 1

      1 point: for small displacements about a minimum, the effective spring constant is k_eff = d²U/dx² evaluated at that point

   2. 2

      1 point: k_eff = 12a(b/2a) − 2b = 4b

   3. 3

      1 point: ω = √(k_eff/m) = 2√(b/m)

A student wants to determine experimentally the moment of inertia I of a wheel that is free to rotate about a fixed horizontal axle. A light string is wound around a hub of known radius r on the axle, and a hanging mass m is attached to the free end. When released, the mass falls and the wheel rotates. Friction in the axle is not negligible.

1. (a)

   Describe Describe a procedure for measuring the linear acceleration of the falling mass, and state the quantity the student should deliberately vary between trials.

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   1. 1

      1 point: release the mass from rest from a measured height h and time the fall with a stopwatch or photogate

   2. 2

      1 point: obtain the acceleration from h = ½at², repeating each trial and averaging the time

   3. 3

      1 point: vary the hanging mass m between trials, keeping the hub radius and the wheel unchanged

   4. 4

      1 point: use a photogate or video rather than a hand-operated stopwatch where possible, since the fall times are short and reaction time is a significant fraction of them

2. (b)

   Derive Ignoring friction for this part, derive a relationship between the acceleration a of the hanging mass and the moment of inertia I, and describe how the student should plot the data to obtain a straight line.

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   1. 1

      1 point: for the hanging mass, mg − T = ma; for the wheel, Tr = Iα with α = a/r

   2. 2

      1 point: eliminating T gives mg = ma + Ia/r², so a = mg/(m + I/r²)

   3. 3

      1 point: rearranges to a linear form — for example 1/a = (I/r²)(1/(mg)) + 1/g

   4. 4

      1 point: plot 1/a against 1/m; the slope is I/(gr²) and the vertical intercept is 1/g

3. (c)

   Explain Explain how the presence of friction in the axle would affect the value of I obtained from the slope, and describe a modification to the analysis that would account for it.

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   1. 1

      1 point: a frictional torque opposes the rotation, so every measured acceleration is smaller than the frictionless model predicts

   2. 2

      1 point: the analysis attributes that reduction entirely to rotational inertia, so the value of I obtained is too large

   3. 3

      1 point: include a constant frictional torque τ_f in the wheel's equation, Tr − τ_f = Iα, which adds a constant term to the linearised relationship and can be found from the intercept — or measure the torque needed to keep the wheel turning at constant speed and subtract it

A block of mass m slides on a frictionless horizontal surface with initial speed v₀ at t = 0. It experiences a resistive force from the surrounding air of magnitude bv, directed opposite to its velocity, where b is a positive constant.

1. (a)

   Derive Derive an expression for the speed of the block as a function of time.

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   1. 1

      1 point: Newton's second law gives m dv/dt = −bv

   2. 2

      1 point: separates variables, dv/v = −(b/m)dt

   3. 3

      1 point: integrates from v₀ at t = 0 to v at t, giving ln(v/v₀) = −bt/m

   4. 4

      1 point: v(t) = v₀e^(−bt/m)

2. (b)

   Determine Determine the total distance travelled by the block before it comes to rest.

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   1. 1

      1 point: x = ∫₀^∞ v₀e^(−bt/m) dt

   2. 2

      1 point: evaluates to v₀(m/b)[−e^(−bt/m)]₀^∞

   3. 3

      1 point: x = mv₀/b, a finite distance

3. (c)

   Sketch Sketch graphs of the speed and of the position of the block as functions of time, on separate axes.

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   1. 1

      1 point: speed starts at v₀, decreases, is concave up, and approaches zero asymptotically without reaching it

   2. 2

      1 point: position starts at zero with initial slope v₀ and increases

   3. 3

      1 point: position approaches the horizontal asymptote x = mv₀/b

4. (d)

   Explain Explain how the block can travel a finite total distance even though your expression predicts that its speed never reaches exactly zero.

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   1. 1

      1 point: as the block slows, the resistive force also falls in proportion, so the deceleration decreases and the block never quite stops within the model

   2. 2

      1 point: the distance covered in each successive interval of time falls off exponentially

   3. 3

      1 point: the sum of those ever-smaller contributions converges, so the total distance is finite even though the time taken is unbounded — a physical situation in which "never stops" and "travels a finite distance" are both true