Unit 2 / Force and Translational Dynamics

Unit 2: worked solutions

Compare the model and reasoning at each step. A matching final number is useful; a reason for every step is what lets you solve the next problem.

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Question 1 · AP topic 2.1

1. Topic 2.1 question 4

Total mass is M = 2 + 1 + 1 = 4 kg.
Why this step: Every mass contributes to the normalization of each coordinate, including a mass whose coordinate happens to be zero.
Compute x_CM = (2·0 + 1·4 + 1·0)/4 = 1 m.
Why this step: Only the particle at x = 4 m contributes a nonzero x moment; the mass at y = 4 m still counts in M.
Compute y_CM = (2·0 + 1·0 + 1·4)/4 = 1 m.
Why this step: Apply the same weighting separately to y; averaging radial distances would lose the coordinate directions.
The center is (1 m, 1 m), inside the 0–4 m coordinate bounds.
Why this step: A weighted mean with positive masses cannot fall beyond the smallest or largest constituent coordinate.

Total mass is 4 kg. x_CM = (2·0 + 1·4 + 1·0)/4 = 1 m; y_CM = (2·0 + 1·0 + 1·4)/4 = 1 m. Both lie between 0 and 4 m.

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Question 2 · AP topic 2.2

2. Topic 2.2 question 4

Draw Earth’s gravitational force down and the hand’s applied push horizontally toward the wall.
Why this step: These are the stated source interactions; neither motion nor a generic “holding force” is an additional interaction.
Draw the wall’s normal force horizontally away from the wall.
Why this step: A normal force is perpendicular to the contact surface; a vertical wall therefore has a horizontal normal direction.
Draw static friction from the wall upward.
Why this step: Without friction, gravity would make the stationary book slip downward relative to the wall, so static friction must oppose that tendency.
Start all four actual force arrows at the book’s dot and label their sources.
Why this step: The selected object is the book; forces the book exerts on the hand or wall belong on those other objects’ diagrams.

Gravity from Earth points downward, the hand's push points toward the wall, the wall's normal force points away from the wall, and static friction from the wall points upward. All arrows originate at the book's dot.

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Question 3 · AP topic 2.3

3. Topic 2.3 question 4

Take the 10 N force on the 5 kg cart as positive. Its interaction-only acceleration is 10/5 = +2 m/s².
Why this step: The stated force acts on the 5 kg cart; dividing by that cart’s own mass gives its response.
The 2 kg cart receives the opposite force, −10 N, and acceleration −10/2 = −5 m/s².
Why this step: Newton’s third law reverses the force but does not equalize masses, so this cart has the larger acceleration magnitude.
The interaction-only mass-weighted force sum is 5·2 + 2·(−5) = 0 N.
Why this step: This checks internal cancellation for both carts together; any separately applied external force would have to be added to its recipient’s equation.

The 5 kg cart feels 10 N from the 2 kg cart and accelerates at 2 m/s². The 2 kg cart feels the opposite 10 N and accelerates at 5 m/s² in the opposite direction. Other forces, if present, would also enter the net force.

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Question 4 · AP topic 2.4

4. Topic 2.4 question 4

For the stationary lamp, T − mg = 0, giving T = 4·10 = 40 N.
Why this step: The one vertical cable supplies the only stated upward force, and zero acceleration requires it to match weight.
For constant upward velocity, a is still zero, so the same two-arrow FBD and T = 40 N apply.
Why this step: Newton’s law depends on velocity change rather than on whether the lamp is already moving upward; constant motion adds no force of motion.

T − mg = 0 gives T = 40 N using g = 10 m/s². The FBD and force magnitudes are the same for constant upward velocity because acceleration remains zero.

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Question 5 · AP topic 2.5

5. Topic 2.5 question 4

Take right positive and add the horizontal forces: F_net = 15 − 3 = +12 N.
Why this step: Friction points left, so its component opposes rather than supplements the pull.
Use a = F_net/m = 12/4 = +3 m/s².
Why this step: The 4 kg cart is the selected system, and the constant resultant gives constant acceleration over the stated interval.
Find Δv = aΔt = 3·2 = +6 m/s.
Why this step: Two seconds of the same acceleration add a fixed positive velocity change; this does not require the cart to start from rest.
Final velocity would be v_f = v_i + 6 m/s.
Why this step: Initial velocity was not provided, so reporting +6 m/s as a final velocity would assume an unstated starting condition.

Net force is 15 − 3 = 12 N, so a = 12/4 = +3 m/s². The velocity change is aΔt = +6 m/s. Initial velocity is still needed to determine final velocity.

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Question 6 · AP topic 2.6

6. Topic 2.6 question 4

Compute gravitational weight: mg = (50 kg)(10 m/s²) = 500 N downward.
Why this step: The stated mass and local field determine weight independently of elevator acceleration.
For upward acceleration, N − 500 = 50(+1.5), so N = 575 N.
Why this step: The support must exceed gravity by 75 N to create the specified upward resultant.
For downward acceleration, N − 500 = 50(−1.5), so N = 425 N.
Why this step: Keep the same upward-positive axis; the negative acceleration now requires the support to be 75 N less than gravity.
Weight stays 500 N; only the scale reading changes.
Why this step: Neither elevator case changes the mass or local gravitational field, but each demands a different contact force.

Weight is 50·10 = 500 N. Upward acceleration gives N = 50(10 + 1.5) = 575 N; downward acceleration gives N = 50(10 − 1.5) = 425 N. Weight remains 500 N in both cases.

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Question 7 · AP topic 2.7

7. Topic 2.7 question 4

On the level floor, vertical balance gives N = mg = 2·10 = 20 N.
Why this step: The pulls are horizontal and no vertical acceleration is stated, so weight is the only downward load.
The static maximum is μ_sN = 0.40·20 = 8 N.
Why this step: This threshold decides whether a box initially at rest can remain in the no-slip regime.
For a 5 N pull, static friction is 5 N and acceleration is zero.
Why this step: The required opposing force is below 8 N, so static friction can adjust to balance the pull exactly.
For a 10 N pull from rest, the 8 N static capacity is exceeded. Once sliding, f_k = 0.25·20 = 5 N.
Why this step: The onset of slip changes the appropriate coefficient; the static maximum cannot be used as the continuing sliding force.
Then a = (10 − 5)/2 = 2.5 m/s² in the pull direction.
Why this step: The kinetic force opposes motion, leaving a positive 5 N resultant acting on the 2 kg mass.

N = 20 N, so the static maximum is 8 N. A 5 N pull gives f_s = 5 N and zero acceleration. A 10 N pull initiates sliding; f_k = 5 N and a = (10 − 5)/2 = 2.5 m/s².

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Question 8 · AP topic 2.8

8. Topic 2.8 question 4

Initial extension is x = 0.21 − 0.15 = 0.06 m.
Why this step: The 0.15 m relaxed length contributes no Hooke-law force; the spring’s full held length is not its deformation.
At rest the holding force balances the spring magnitude, so k = F_hold/x = 9/0.06 = 150 N/m.
Why this step: Force divided by deformation determines stiffness because the spring is assumed linear in this range.
At length 0.18 m, extension is 0.18 − 0.15 = 0.03 m.
Why this step: Both states must use the same relaxed-length reference, rather than taking the second length relative to the first.
The new holding-force magnitude is kx = 150·0.03 = 4.5 N.
Why this step: The deformation is half as large, so a linear ideal spring needs half the original 9 N supporting pull.

Extension is 0.21 − 0.15 = 0.06 m, so k = 9/0.06 = 150 N/m. The new extension is 0.03 m and holding force is 150(0.03) = 4.5 N.

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Question 9 · AP topic 2.9

9. Topic 2.9 question 4

Find inward acceleration a_c = v²/r = 4²/2 = 8 m/s².
Why this step: Circular curvature fixes the radial acceleration at the stated instantaneous speed and radius, independent of which real force supplies it.
The inward net-force component is ma_c = 0.50·8 = 4 N.
Why this step: Newton’s second law uses the given mass to convert radial acceleration into the radial resultant, not an additional force arrow.
If the first case maintains speed 4 m/s throughout a revolution, P = 2πr/v = 2π·2/4 = π s.
Why this step: A complete lap has length 2πr; dividing by one speed gives its duration only if that speed stays constant.
With tangential acceleration 6 m/s² at this instant, total magnitude is √(8² + 6²) = 10 m/s².
Why this step: Radial and tangential components are perpendicular, so Pythagoras rather than scalar addition combines them.
If the speed changes over a lap, the instantaneous 4 m/s alone cannot determine that lap’s period.
Why this step: The added tangential acceleration changes future speed; a full time history would be needed for a nonuniform revolution.

a_c = 4²/2 = 8 m/s², F_inward = 0.50·8 = 4 N, and P = 2π·2/4 = π s. With a_t = 6 m/s², total a = √(8² + 6²) = 10 m/s².

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Question 10 · AP topic 2.1

10. Topic 2.1 question 5

Choose both carts and their spring as one system. The spring exerts opposite internal forces on the carts.
Why this step: Including the spring keeps its interactions inside the boundary, while still allowing each cart to feel its own force.
Those internal forces cancel in the whole-system sum, so M a_CM = F_external = 0.
Why this step: Newton’s third-law pairing removes internal forces from the center-of-mass equation, not from each cart’s equation.
If v_CM initially equals zero, it remains zero even as the carts separate.
Why this step: Zero external force fixes the center-of-mass velocity; it does not require each constituent velocity to be zero.
Unequal opposing track-friction forces could give a nonzero net external force and accelerate the center of mass.
Why this step: The track is outside the chosen system; unequal friction magnitudes need not cancel when the carts move in opposite directions.

Internal spring forces change the individual motions while canceling in the whole-system force sum. With zero net external force, the center-of-mass velocity stays zero. Unequal track friction could supply a net external force.

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Question 11 · AP topic 2.2

11. Topic 2.2 question 5

Weight is mg = 3·10 = 30 N downward, and the added push is 8 N downward.
Why this step: The push is a second real interaction; it cannot be absorbed into the definition of the block’s gravitational weight.
Stationarity gives N − 30 − 8 = 0, so N = 38 N upward.
Why this step: The floor must balance both downward forces to maintain zero vertical acceleration, not gravity alone.
Weight remains 30 N while the contact normal force rises to 38 N.
Why this step: Changing an applied push does not change mass or local g; normal force adjusts to the contact constraint.

N − mg − 8 = 0, so N = 30 + 8 = 38 N. Normal force responds to the contact condition and must balance both downward forces.

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Question 12 · AP topic 2.3

12. Topic 2.3 question 5

On the hanging mass, draw string-on-mass upward and Earth-on-mass downward.
Why this step: The mass FBD shows only forces received by the mass; the source labels distinguish the two interactions.
On the ideal string, draw mass-on-string downward at the lower end and support-on-string upward at the upper end.
Why this step: The downward string force is the third-law partner of string-on-mass, so it acts on the string rather than on the mass.
For a negligible-mass string with finite acceleration, the endpoint force magnitudes balance, giving the same tension at both ends.
Why this step: Applying ΣF = m_string a with m_string ≈ 0 leaves no finite net force on the string; a massive chain would need a different model.
If the mass is also at rest, its own equation gives T = mg.
Why this step: Equal tension along an ideal string does not alone establish T = mg; that additional equality requires zero acceleration of the mass.

The mass diagram includes string-on-mass upward and Earth-on-mass downward. Mass-on-string acts on the string. For the ideal string at rest, its endpoint forces balance, consistent with uniform tension.

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Question 13 · AP topic 2.4

13. Topic 2.4 question 5

Add east-west components: 5 − 5 = 0 N. Add vertical components: 8 − 8 = 0 N.
Why this step: The opposing forces cancel within each axis, so the full vector resultant is zero rather than just one component.
The cart has zero acceleration, so its velocity stays at its initial value.
Why this step: In an inertial frame, zero resultant stops the velocity from changing but does not reset that value to zero.
You cannot determine that initial velocity from the given forces.
Why this step: A resting cart and a steadily moving cart can have the same balanced force configuration; a position-time or initial-velocity measurement is still needed.

The net force is zero, so velocity stays constant. Its value could be zero or any initially established constant vector; the force information alone doesn't set that initial condition.

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Question 14 · AP topic 2.5

14. Topic 2.5 question 5

For both carts, the external horizontal force is F and total mass is m₁ + m₂, so F = (m₁ + m₂)a.
Why this step: The track is frictionless, and the connecting tension is internal to the combined system; the taut ideal string gives a common acceleration.
Solve a = F/(m₁ + m₂).
Why this step: The external pull must accelerate both masses, so using just the front mass would overestimate the shared acceleration.
For the rear cart alone, T = m₂a.
Why this step: Only the string supplies its horizontal force in this frictionless model, so its own mass determines the required tension.
Substitute the shared acceleration: T = m₂F/(m₁ + m₂).
Why this step: This connects the whole-system result to the rear-cart force without reclassifying tension as an external force on both carts.

Both carts give F = (m₁ + m₂)a. The rear cart alone gives T = m₂a. Substituting the first result produces the requested tension expression.

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Question 15 · AP topic 2.6

15. Topic 2.6 question 5

Write F_old = Gm₁m₂/r² and F_new = G(2m₁)(2m₂)/(3r)².
Why this step: Universal gravitation multiplies the two masses and squares center separation, so each changed quantity must appear in its own factor.
Divide: F_new/F_old = 4/9.
Why this step: Both mass doublings supply a factor of four; tripling separation supplies a factor of nine that partly offsets it.
If h is surface altitude instead, center separation is R + h, not h.
Why this step: A planet’s gravitational model is centered at its mass center; tripling h does not generally triple R + h.
For fixed radius R, changing h to 3h would instead give the distance factor (R + h)²/(R + 3h)².
Why this step: Keeping the planet radius in the ratio prevents misapplying the 1/9 distance factor to an altitude change.

F_new/F_old = (2·2)/(3²) = 4/9. The inverse-square equation uses center separation, so surface altitude must first be added to the body's radius.

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Question 16 · AP topic 2.7

16. Topic 2.7 question 5

For a block at rest, N = mg cos θ and required uphill friction is mg sin θ.
Why this step: The incline constraint removes perpendicular acceleration, while the downhill gravity component would cause slip without an uphill contact force.
No slip requires mg sin θ ≤ μ_smg cos θ.
Why this step: Static friction can supply the required value only up to its normal-force-dependent maximum.
At impending slip, equality gives tan θ_max = μ_s, or θ_max = arctan μ_s.
Why this step: Dividing by positive mg cos θ converts the two component factors into a tangent; equality identifies the largest allowable angle.
The mass cancels, so changing mass alone does not change θ_max in this model.
Why this step: The extra mass increases both the downhill demand and friction capacity in the same proportion, assuming the surface coefficient remains unchanged.

Compare mg sin θ with μ_smg cos θ. At the threshold, tan θ_max = μ_s. Both required friction and available maximum scale with mass, so the mass cancels.

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Question 17 · AP topic 2.8

17. Topic 2.8 question 5

Take downward deformation x from relaxed length as positive. The total force is mg − kx.
Why this step: Gravity points along the positive axis while the stretched spring pulls upward; both interactions must be counted separately.
At equilibrium, mg − kx_eq = 0, giving x_eq = mg/k.
Why this step: Zero net force fixes the extension where spring support equals weight; it does not make the spring force itself zero.
If y is displacement below that equilibrium, the actual spring deformation is x = x_eq + y.
Why this step: Changing coordinate origin subtracts the equilibrium offset from position, but Hooke’s law still needs the relaxed-length deformation.
Substitute: F_net = mg − k(x_eq + y) = mg − kx_eq − ky = −ky.
Why this step: The equilibrium weight and spring-offset terms cancel because kx_eq = mg, leaving a restoring net force about the shifted position.
The spring force alone is −k(x_eq + y), not −ky.
Why this step: Only the combined spring-plus-gravity equation loses the constant offset; omitting it from the spring force would incorrectly predict no support at equilibrium.

With downward positive, mg − kx_eq = 0 gives x_eq = mg/k. At x = x_eq + y, net force is mg − kx_eq − ky = −ky. Spring force alone remains −k(x_eq + y).

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Question 18 · AP topic 2.9

18. Topic 2.9 question 5

The tilted normal force has vertical component N cos θ and inward horizontal component N sin θ.
Why this step: For bank angle θ, the normal tilts by θ from vertical; the frictionless assumption leaves no other contact force to supply horizontal acceleration.
Write N cos θ = mg and N sin θ = mv²/r.
Why this step: The car has no vertical acceleration but does have inward horizontal acceleration while rounding the level-radius curve.
Divide the second equation by the first: tan θ = v²/(rg).
Why this step: The unknown normal force and mass cancel in the component ratio, directly linking design angle, speed, and radius.
For 0 < θ < 90°, N = mg/cos θ > mg.
Why this step: Only the vertical component balances weight; since cos θ is less than one, the full tilted force must be larger than that component.

Vertical balance gives N cos θ = mg; inward balance gives N sin θ = mv²/r. Dividing cancels N and m. Since cos θ is less than one for a nonzero bank, N = mg/cos θ exceeds mg.

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Question 19 · AP topic 2.1

19. Topic 2.1 question 6

Treat the board’s 2 kg as located at its known center, x = 0.50 m, for the position calculation.
Why this step: The board’s distributed mass has the same total mass-position moment as that equivalent point; its internal details are not needed here.
Calculate x_CM = (2·0.50 + 1·0.80)/(2 + 1) = 1.80/3 = 0.60 m.
Why this step: The new point combines the board and washer, with the board carrying twice the weight in the average.
For a variable washer position x_w, x_CM = (1.00 + x_w)/3.
Why this step: Only the washer position changes; keeping both masses and the board center fixed makes the relationship linear.
The graph has slope 1/3, so a 0.30 m washer shift moves the center by 0.10 m.
Why this step: The washer is one third of the total mass, so its displacement produces one third as much center-of-mass displacement.

x_CM = (2·0.50 + 1·0.80)/3 = 0.60 m. As washer position grows, center position rises linearly with slope 1/3; moving 0.30 m shifts the balance point 0.10 m.

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Question 20 · AP topic 2.2

20. Topic 2.2 question 6

Choose x downhill along the incline and y perpendicular outward.
Why this step: The straight-track constraint makes perpendicular acceleration zero, leaving the changing motion in the parallel direction.
Draw gravity vertically down, normal force outward perpendicular, and kinetic friction uphill.
Why this step: The block slides downhill relative to the rough surface, so friction opposes that slip; the gravity arrow is not redrawn along the slope.
Write mg sin θ − f_k = ma_x.
Why this step: The gravitational parallel component points along positive x, while friction points against it; a_x is not set to zero without a constant-speed condition.
Write N − mg cos θ = 0, with f_k = μ_kN if that coefficient model applies.
Why this step: Contact on a straight incline prevents perpendicular acceleration, and the normal force sets the kinetic-friction magnitude.

Choose downhill x and outward-perpendicular y. Actual arrows are gravity down, normal outward, and kinetic friction uphill. Then mg sin θ − f_k = maₓ and N − mg cos θ = 0, if the block remains on a straight incline.

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Question 21 · AP topic 2.3

21. Topic 2.3 question 6

Reverse the sign of the second reading to compare paired magnitudes: 12.0 N and 11.7 N differ by 0.3 N.
Why this step: The third law predicts opposite signed forces, so comparing F₁ with −F₂ tests equal magnitudes rather than incorrectly demanding the same sign.
The first magnitude range is 11.7–12.3 N; the second is 11.4–12.0 N. They overlap from 11.7 to 12.0 N.
Why this step: Each reading allows ±0.3 N; at least one common magnitude is compatible with both measurements, so the mismatch is not resolved by these data.
The predicted signed sum is zero; the measured sum is +0.3 N with a conservative combined interval of −0.3 to +0.9 N.
Why this step: Adding the stated uncertainty bounds gives ±0.6 N for the sum without assuming an unprovided statistical model; that interval includes zero.
These readings are not clear evidence against the third law; compare calibrated, simultaneous readings and their uncertainties.
Why this step: The prediction remains within the measurement bounds, and time mismatch or sensor offset could also imitate unequal forces.

No. The magnitude difference is 0.3 N, compatible with the stated uncertainties. Compare calibrated simultaneous readings and their uncertainty ranges before making a claim of disagreement.

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Question 22 · AP topic 2.4

22. Topic 2.4 question 6

By symmetry, both cables have tension T. Their horizontal components are equal and opposite.
Why this step: The symmetric stationary arrangement cannot have an uncanceled sideways resultant.
Vertical balance is T sin θ + T sin θ − mg = 0.
Why this step: Angles are measured above horizontal, so sine selects each upward component; both cables share the supporting load.
Solve T = mg/(2 sin θ), for a supporting angle with sin θ > 0.
Why this step: Dividing by the nonzero upward-component factor isolates the full cable force; a horizontal cable has no finite supporting component.
At fixed θ, doubling m changes T to 2mg/(2 sin θ) = 2T.
Why this step: Only the weight factor changes, while the two-cable geometry stays fixed, making tension directly proportional to mass.

Horizontal components cancel; vertical balance gives 2T sin θ = mg. Hence T = mg/(2 sin θ). Doubling m doubles each tension when θ is fixed.

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Question 23 · AP topic 2.5

23. Topic 2.5 question 6

For correctly calculated net force, a = (1/M)F_net, so the fitted slope is 1/M.
Why this step: The graph places acceleration vertically and net force horizontally; its slope has inverse-mass units.
Invert the measured slope: M = 1/(0.25 kg⁻¹) = 4 kg.
Why this step: The reciprocal converts inverse kilograms to kilograms; multiplying by the slope would return the wrong dimensions.
If the horizontal value actually omits constant opposing friction, it is F_pull, and a = F_pull/M − f/M.
Why this step: The true net force is F_pull − f for motion in the positive direction; the omitted force appears as a constant negative offset.
That mislabeled applied-force graph has vertical intercept −f/M; a true F_net graph retains zero intercept in this model.
Why this step: An intercept shift diagnoses unaccounted force or calibration rather than a different mass slope, provided friction stays constant in the fitted interval.

Since slope is 1/M, M = 1/0.25 = 4 kg. If plotted force is the applied pull rather than net force, a = (F_pull − f)/M has a negative vertical intercept −f/M for positive-direction motion.

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Question 24 · AP topic 2.6

24. Topic 2.6 question 6

For r ≥ R, sketch the magnitude g = GM/r² as a positive decreasing curve.
Why this step: Outside a spherical source, attraction gives an inward vector field whose magnitude follows this center-distance dependence.
Mark g(R) = g₀ and g(2R) = g₀/4.
Why this step: The two labeled distances provide a quantitative scale check on the inverse-square graph.
For a small height h, g(R + h)/g₀ = 1/(1 + h/R)² is close to one when h/R is tiny.
Why this step: Near-surface classroom changes are small compared with the whole planet radius, so the relative denominator change is small.
Treat g as constant only over that short interval, not over the full R-to-2R range.
Why this step: The quarter-strength point shows that extending the approximation to planet-scale distances would hide a substantial change.

Draw a positive decreasing inverse-square curve: g(R) = g₀ and g(2R) = g₀/4. Over a tiny fraction of R, the fractional change is small; the constant-field approximation has a stated spatial limit.

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Question 25 · AP topic 2.7

25. Topic 2.7 question 6

Use the example’s 4 kg box, N = 40 N, μ_s = 0.50, and μ_k = 0.30.
Why this step: The given graph must refer to this box’s fixed contact conditions, not an unspecified surface or changing vertical load.
While the box remains at rest, draw f_s equal to pull magnitude up to f_s,max = 20 N.
Why this step: Zero acceleration requires matching horizontal forces, and the 0.50·40 limit is the point where that adjustment stops being possible.
After slip begins, draw the kinetic branch at f_k = 12 N.
Why this step: Sliding uses μ_kN = 0.30·40, which is lower than the preceding static maximum even as the applied pull continues increasing.
Label the first branch static and the second kinetic, with pull horizontally and friction magnitude vertically.
Why this step: The branch change is a contact-motion change; the graph shows friction magnitude, not the net force or signed acceleration.

While static, friction follows the pull up to 20 N. Just after slip, the ideal model drops to 12 N, then remains at that magnitude while the same contact conditions and sliding regime apply. Mark which part is static and which is kinetic.

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Question 26 · AP topic 2.8

26. Topic 2.8 question 6

Plot extension on x and holding-force magnitude on y; the three points lie on one straight line.
Why this step: The stated force rises by 2 N for every 0.01 m deformation, so constant slope is testable across more than one load.
Use k = ΔF/Δx = (6 − 2)/(0.03 − 0.01) = 200 N/m.
Why this step: This holding-force graph has positive slope k; dividing force change by length change gives the correct stiffness units.
The extrapolated intercept is 2 − 200·0.01 = 0 N.
Why this step: A zero intercept agrees with measuring extension from a zero-force relaxed length rather than from the spring’s total length.
Measure additional larger extensions without damaging the spring, then unload it and check its relaxed length.
Why this step: Curvature would show a changing stiffness, while failure to recover relaxed length would show that the assumed linear elastic model no longer applies.

The line through the origin has slope 200 N/m. Measure additional larger extensions, still below damage limits, and check for systematic departures from the fitted line or failure to return to the relaxed length.

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Question 27 · AP topic 2.9

27. Topic 2.9 question 6

Set GMm/r² = mv²/r, then cancel m to obtain v² = GM/r.
Why this step: Gravity is the only inward force in the circular-orbit model; satellite mass occurs on both sides and does not set orbital speed.
For uniform circular orbit, P = 2πr/v. Square it: P² = 4π²r²/v².
Why this step: One revolution covers the circumference at constant speed; squaring prepares the expression for substitution of v².
Substitute v² = GM/r to get P² = 4π²r³/(GM).
Why this step: Dividing by GM/r introduces one extra factor of r, giving the cubic-radius rather than quadratic-radius dependence.
Plot P² against r³: the line has zero intercept and slope 4π²/(GM), with units s²/m³.
Why this step: The same planet keeps GM fixed, so the coefficient is constant; the vertical-to-horizontal axis-unit ratio checks its dimensions.
A larger central mass gives a smaller slope.
Why this step: GM is in the denominator: at the same orbital radius, stronger gravity allows a faster orbit and therefore a shorter period.

Use v² = GM/r in P = 2πr/v. Squaring gives P² = 4π²r³/(GM). The graph is a line through the origin with slope 4π²/(GM), in s²/m³. Different central masses give different slopes.

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Question 28 · AP topic 2.1

28. Topic 2.1 question 1

Use x_CM = (m₁x₁ + m₂x₂)/(m₁ + m₂).
Why this step: The two masses differ, so a simple midpoint would give the lighter object too much influence.
Substitute x_CM = (1·0 + 3·4)/(1 + 3) = 12/4 = 3 m.
Why this step: The numerator is the sum of mass-position products; dividing by all 4 kg leaves a position in meters.
The answer lies nearer the 3 kg mass at 4 m.
Why this step: The heavier mass has three times the position weight, so a center at 2 m would not match the mass distribution.

3 m

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Question 29 · AP topic 2.2

29. Topic 2.2 question 1

List the block’s interactions: Earth exerts gravity, and the incline exerts a normal force.
Why this step: Frictionless contact excludes friction, and no rope, hand, or other contact force is stated.
Draw mg vertically down and N perpendicular outward from the incline, both starting at the block’s dot.
Why this step: Force direction comes from its source interaction; choosing inclined axes does not rotate gravity away from vertical.
Use mg sin θ and mg cos θ only when writing component equations.
Why this step: They are projections of the same gravitational force; drawing them as extra forces would count gravity more than once.

Gravity and normal force

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Question 30 · AP topic 2.3

30. Topic 2.3 question 1

Identify the pair: force of truck on car and force of car on truck.
Why this step: Both forces come from the same collision interaction, with the source and receiving object exchanged.
Newton’s third law gives equal magnitudes and opposite directions at each instant.
Why this step: The pair equality does not depend on equal masses, rest, or equal collision damage.
The lighter object can still have larger acceleration because a = F/m.
Why this step: The force magnitude can match while division by different masses produces different responses, so damage is not evidence of unequal paired forces.

They exert equal and opposite forces on each other

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Explain and apply: worked solutions

Give credit for equivalent correct reasoning. Use the rubric to find the next step to improve, even when your final number matches.

Decide how much substructure to keep

A system contains a 3 kg module initially at (0, 0) m and a 1 kg module initially at (4, 0) m on a straight horizontal frictionless track. Neglect connection mass and external horizontal forces except a 12 N rightward pull applied to the 3 kg module along the line through both modules. Compare three versions of the same system: the modules are loose, are fixed together by a rigid connection, or are joined by an initially relaxed spring along their separation. Both modules start at rest; no rotation occurs in this collinear setup.

Part a

Total mass is 4 kg; x_CM = [3(0) + 1(4)]/4 = 1.0 m and y_CM = 0 m. Represent it by one 4 kg point object at (1.0, 0) m when only center-of-mass translation under the known external resultant matters. Individual shapes, relative motions and internal forces may then be ignored. The replacement cannot predict which module moves first, the spring deformation, or internal stresses/rotation in a more general setup.

Calculate each mass-weighted coordinate using the same total mass.
Assign the replacement the total mass and put it at the center of mass.
Match the neglected information to the macroscopic question and state an omitted structural question.

Scoring criteria

Obtains (1.0, 0) m and 4 kg.
Explicitly places the replacement object at that coordinate.
Explains both when constituent properties/interactions can be neglected and a concrete limitation.

Part b

The common center-of-mass acceleration is 12/4 = 3 m/s² rightward. For loose modules, the 3 kg module initially accelerates at 12/3 = 4 m/s² rightward while the 1 kg module has zero initial acceleration. The rigid connection transfers force so their translational accelerations are both 3 m/s² in this collinear setup. For the initially relaxed spring, its initial force is zero, so initially the same 4 and 0 m/s² values hold; as the left module approaches the right one, the spring compresses and produces forces that alter the individual accelerations. The point model captures the common center-of-mass acceleration and cannot capture those internal differences.

Sum external forces for the whole-system acceleration.
Use whether the connection is absent, rigidly constrained or initially relaxed to determine initial constituent forces.
Explain how the spring’s later deformation creates internal forces despite the same external resultant.

Scoring criteria

Obtains 3 m/s² common center-of-mass acceleration and initial loose/spring constituent values 4 and 0 m/s².
Explains common 3 m/s² rigid translation in the collinear setup.
Connects differing constituent motions and spring compression to substructure and identifies what the replacement misses.

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Track transfers and structural changes

Choose a stationary cart, an attached electric heater and its contents as a fixed system. An external supply powers the heater at 10 W for 4 s. In a separate operation 0.20 kg of sand is poured into the initially 0.60 kg system. In another experiment, two modules joined by a 150 N/m ideal spring are held at rest by equal outward external pulls on the two modules, first 3 N each and then 9 N each.

Part a

The external supply transfers electrical energy to the included heater: E = Pt = 10(4) = 40 J. The incoming sand carries mass across the boundary: final system mass is 0.60 + 0.20 = 0.80 kg. The supply stays outside the system; sand becomes a constituent after entry. The selected system permits environment interactions, so a drawn boundary does not prohibit energy or mass transfer. No variable-mass motion calculation is implied.

List the included objects and locate the supply and incoming sand outside the boundary.
Calculate electrical energy transfer from power and elapsed time.
Update included mass after the sand crosses into the system.

Scoring criteria

Identifies the supply-to-heater energy interaction and obtains 40 J.
Identifies the sand boundary crossing and obtains 0.80 kg.
Explains that openness is determined by actual transfers, not by drawing a boundary.

Part b

Each spring endpoint balances one external pull, so extensions are 3/150 = 0.020 m and 9/150 = 0.060 m. Increasing the external loading changes the internal separation by 0.040 m. The opposite external pulls still sum to zero, but zero resultant constrains center-of-mass acceleration, not internal deformation. The spring stays in its assumed linear range.

Balance each module separately to relate spring force to one pull rather than the sum of magnitudes.
Use extension = force/k for each static configuration.
Separate the zero whole-system resultant from the nonzero force on an internal connection.

Scoring criteria

Obtains 0.020 and 0.060 m extensions without using twice the endpoint pull.
Names external pull magnitude and internal separation/extension as the changed variables.
Explains zero net external force with changed substructure.

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Find the source of every push

A person sits in a wheeled cart on a horizontal frictionless track and pushes the cart’s front wall. Choose person plus cart as the system. Later a rope outside the system pulls it rightward with 6 N. Total mass is 3 kg and there is no vertical acceleration. Use g = 10 m/s².

Part a

The person exerts a force on the wall/cart; the wall/cart exerts an equal opposite force on the person. Both recipients are inside the chosen system, so the pair cancels in its vector force sum. The system cannot exert a net force on itself; without an external horizontal resultant its center-of-mass velocity remains unchanged. The internal forces act on different constituents, so they can change relative motion or deformation even though their system sum is zero.

Name both forces by source and recipient.
Place both recipients inside the same boundary and add their vectors.
Distinguish a whole-system force sum from each constituent’s separate force equation.

Scoring criteria

Identifies person-on-cart and cart-on-person as the internal pair.
Explains equal/opposite cancellation and the absence of a net self-force.
Explains possible constituent motion with unchanged center-of-mass motion.

Part b

The objects touch, and electric interactions among atoms in the contacting materials resist compression and produce the observed pushes. The wall force and track normal force are macroscopic effects of these interatomic electric forces. Their microscopic origin is already represented by the contact-force arrow; adding another electric arrow would count the same interaction twice. AP Physics 1 uses the macroscopic contact model without calculating atom-level electric forces.

Identify the two pairs of touching objects.
Connect their atom-level electric interactions with the measured contact forces.
Keep one macroscopic representation for each actual interaction.

Scoring criteria

Explicitly states touching and interatomic electric origin.
Applies that cause to both wall push and track normal force.
Rejects double counting as separate macroscopic electric/contact arrows.

Part c

The FBD has rope-on-system 6 N right, Earth-on-system gravity 30 N down, and track-on-system normal 30 N up, each starting at the system dot. Internal person/wall forces are omitted from this whole-system diagram. Horizontal resultant is 6 N, so a_CM = 6/3 = 2 m/s² rightward. Vertical forces balance because vertical acceleration is zero.

Cross the chosen boundary to identify rope, Earth and track as external sources.
Use vertical balance to obtain N = mg = 30 N.
Divide the external horizontal resultant by total system mass.

Scoring criteria

Draws and names only the three actual external forces with correct directions and dot origins.
Excludes the internal person/cart pair from the whole-system FBD.
Obtains 2 m/s² rightward center-of-mass acceleration.

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Follow tension through an ideal apparatus

A 2 kg mass and a 3 kg mass hang on opposite sides of a taut unbranched string over a pulley. The string does not slip on the pulley. Use the ideal string and ideal pulley models, neglect air resistance, and use g = 10 m/s². Neither mass has reached a stop.

Part a

The ideal string has negligible mass and does not stretch under tension. The ideal pulley has negligible mass and rotates about an axle through its center of mass with negligible friction. Negligible string mass and pulley rotational inertia/axle friction allow equal tension in this setup; a nonstretching taut string over the fixed pulley constrains the masses to equal acceleration magnitudes in opposite directions. The stated no-slip condition links string and pulley motion. A massive or frictional pulley can require unequal tensions.

Separate the string model from the pulley model.
Name the center-of-mass axle geometry as well as mass and friction assumptions.
Connect force-uniformity assumptions and the geometric length constraint to different conclusions.

Scoring criteria

States negligible string mass and nonstretching.
States negligible pulley mass, axle through its center of mass and negligible axle friction.
Correctly distinguishes uniform tension from the equal-acceleration constraint.

Part b

Each segment is pulled by its neighbors, and the endpoint segments are pulled by the external masses. These successive interactions give the macroscopic tension. A negligible-mass short straight segment requires equal opposing longitudinal pulls for finite acceleration; an ideal pulley adds no required tension difference. Thus all specified locations, and intervening points, have the same tension magnitude T. The string direction and hence tension direction change over the pulley. For a massive hanging chain, upper portions must support more chain weight and tension is greater near the top, so the negligible-mass argument fails.

Identify force sources on a segment as its adjacent segments or an external endpoint load.
Apply the negligible-mass argument successively along the string, with ideal-pulley assumptions at the bend.
Distinguish equal magnitude from changing vector direction and compare with a heavy chain.

Scoring criteria

Explains transmission through adjacent segments in response to external loading.
Explicitly compares both ends and all interior portions as equal magnitude.
Explains the top-heavy-chain tension increase qualitatively.

Part c

Each mass has upward tension and downward gravity from the dot. The 3 kg side accelerates down and the 2 kg side up. With positive directions along each acceleration: T − 20 = 2a and 30 − T = 3a. Adding gives 10 = 5a, so a = 2.0 m/s². Then T = 20 + 2(2) = 24 N, which is the common magnitude at all points. The equal tensions are not third-law partners on the two masses; each string–mass interaction has its partner on the string.

Choose opposite physical positive directions for the two masses consistent with the fixed string length.
Write one force equation per mass and eliminate the common tension.
Use one equation to recover tension and check it in the other.

Scoring criteria

Draws only upward tension and downward gravity on each mass.
Finds the heavier side downward and acceleration 2.0 m/s².
Finds 24 N at every ideal-string point and avoids calling the two mass tensions a third-law pair.

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Use force-free motion to test a frame

A released puck has negligible horizontal external force. A ground observer measures its positions as 0, 4, 8 and 12 m at t = 0, 1, 2 and 3 s. Observer A’s origin has ground position x_A = (1 m/s)t and observer B’s origin has x_B = (1 m/s²)t² with t in seconds. Both use axes parallel to the ground axis and synchronized clocks. The ground frame is approximately inertial.

Part a

Subtract A’s origin positions 0, 1, 2, 3 m to obtain 0, 3, 6, 9 m. Interval velocities are +3 m/s throughout. An inertial frame is one in which an observer verifies Newton’s first law: zero net force gives constant velocity. A moves at constant velocity relative to the stipulated inertial ground frame and its force-free puck record verifies that law. A single noisy trial cannot prove every possible first-law test, but these assumptions establish A as inertial in this model.

Subtract the observer-origin position at each synchronized time.
Compare displacements over equal 1 s intervals.
Link unchanged velocity for zero force and constant relative frame velocity to the inertial definition.

Scoring criteria

Obtains positions 0, 3, 6, 9 m and velocities +3 m/s.
Defines inertial status through verification of the first law.
Justifies A’s status from the force-free result and constant velocity relative to ground.

Part b

Subtract B’s origins 0, 1, 4, 9 m from the puck record to get positions 0, 3, 4, 3 m. Interval velocities are +3, +1, −1 m/s. They change despite negligible horizontal external force, so B’s accelerating frame fails the inertial first-law test. The ground origin relation x_B = t² corresponds to +2 m/s² acceleration; the force-free puck’s apparent acceleration is −2 m/s² relative to B. This coordinate effect does not introduce a real horizontal contact interaction.

Use the same event times for puck and observer-origin coordinates.
Calculate each difference of successive relative positions over 1 s.
Explain the accelerating-origin cause and distinguish it from real external forces.

Scoring criteria

Obtains positions 0, 3, 4, 3 m and interval velocities +3, +1, −1 m/s.
Concludes that B is noninertial using the zero-force first-law condition.
Explains the coordinate acceleration without inventing a contact force.

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Represent attraction and measure a field

A spherical planet has GM = 1.20 × 10¹⁴ m³/s². In separate trials, 0.20 kg and 0.60 kg small test objects are at center distance r = 2.00 × 10⁶ m outside the planet. Their measured gravitational force magnitudes are 6.0 N and 18.0 N. Neglect other astronomical bodies, drag and the test objects’ effect on the planet’s field.

Part a

Gravity is attractive. The force on the test object starts at its modeled center of mass and points toward the planet center; the reciprocal force starts at the planet center of mass and points toward the test object. The arrows lie along their center line and have equal magnitudes and opposite directions by the third law (6.0 N for the 0.20 kg trial; 18.0 N for the 0.60 kg trial). The planet is spherical and the small test object is modeled as a point outside it. For a general asymmetric extended mass distribution the simple center-line model need not suffice.

Locate the centers of mass before choosing force directions or center separation.
Draw one attractive force on each recipient rather than placing both forces on one FBD.
Use equal/opposite pair symmetry and state the spherical/point-object approximation.

Scoring criteria

Describes gravity as attractive and draws each arrow toward the other center along the center line.
Places each arrow at the receiving system’s modeled center of mass.
States equal magnitude/opposite direction and the model’s spherical exterior-field scope.

Part b

The ratios are 6.0/0.20 = 30 N/kg and 18.0/0.60 = 30 N/kg. GM/r² = (1.20 × 10¹⁴)/(2.00 × 10⁶)² = 30 m/s² = 30 N/kg. Tripling test mass triples force while the ratio is unchanged; source mass and center distance set the field. These separate trials do not add the two test masses’ fields to the planet’s field.

Divide the measured gravitational force by the corresponding test-object mass for each trial.
Square the center distance, not surface altitude, in the source-field expression.
Compare numerator and denominator scaling under the test-mass change.

Scoring criteria

Calculates both force/mass ratios as 30 N/kg.
Calculates GM/r² as 30 m/s² and equates the units.
Explains test-mass independence rather than attributing a stronger field to a larger force.

Part c

Both accelerate at 30 m/s² toward the planet. This is numerically equal to 30 N/kg because N = kg·m/s², so N/kg = m/s². With a support exactly balancing gravity, net force and acceleration are zero even though field strength remains 30 N/kg. The equality a = g applies to gravity-only motion, not supported motion.

Set net force equal to gravitational force only under the stated free-fall condition.
Divide by each mass and infer direction from attractive gravity.
Recompute the net force when a balancing support is added.

Scoring criteria

Gives 30 m/s² for both masses with direction toward the planet.
Explicitly states numerical equality to 30 N/kg and explains the unit conversion.
States the gravity-only limit and zero acceleration under balanced support.

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Test an apparent-weight claim

A 5.0 kg block remains in contact with a horizontal scale in a region where g = 10 m/s². Compare upward acceleration 3.0 m/s², horizontal acceleration 3.0 m/s² produced by a separate horizontal pull with zero vertical acceleration, and no acceleration. Separately compare small sealed laboratories at rest in an approximately uniform 10 N/kg field and far from gravity with a floor accelerating at 10 m/s². A 1.5 kg test object rests relative to each laboratory floor.

Part a

Weight is mg = 50 N in all three cases. Upward normal-direction acceleration gives N − 50 = 5(3), so N = 65 N. With purely horizontal acceleration, a_y = 0, so N = 50 N; with no acceleration N = 50 N too. The horizontal case accelerates but has equal apparent and gravitational weight, disproving the unqualified claim. In this setup N differs from mg when the acceleration component along the supporting normal is nonzero, subject to other vertical forces and maintained contact.

Separate gravitational force from the support normal force measured by the scale.
Write the vertical equation N − mg = ma_y, using vertical acceleration rather than total acceleration magnitude.
Use the horizontal case as a counterexample and state the normal-direction condition.

Scoring criteria

Obtains 50 N weight and readings 65, 50 and 50 N.
Identifies purely horizontal acceleration as a numerical counterexample.
States a normal-direction force-balance qualification without promoting the broad source claim to universal validity.

Part b

Each floor provides 15 N. In the gravitational laboratory N = mg = 1.5(10); far from gravity the accelerating floor supplies N = ma = 1.5(10). The second laboratory is noninertial, and the local apparent-weight effect is indistinguishable from an equivalent uniform gravitational effect by this small-scale support-force experiment. This is local: tidal/spatial field variations across a large apparatus or external observations can distinguish the situations.

Use gravitational equilibrium for the first floor and inertial-frame acceleration for the second.
Compare the identical contact-force readings rather than assuming identical external causes.
State local indistinguishability and a concrete spatial or external-observation limit.

Scoring criteria

Calculates 15 N in both laboratories from the correct force balances.
Explains the noninertial apparent-weight/gravitational-effect equivalence.
Identifies a local-experiment limit such as tidal variation.

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Measure inertial and gravitational mass

For one object, a corrected net horizontal force of 1.8 N produces acceleration 0.60 m/s² on a low-friction cart. An independent gravitational force measurement gives 30 N in a known 10 N/kg field. The resulting inertial mass uncertainty is ±0.10 kg and gravitational mass uncertainty is ±0.06 kg. Cart/sensor mass has been accounted for.

Part a

Inertial mass is the proportionality in F_net = m_i a and measures resistance to acceleration under interactions. A larger m_i gives a smaller acceleration for the same net force. Gravitational mass sets attraction, represented locally by F_g = m_g g and universally by the mass factors in Gm_gM_g/r². A larger m_g feels greater gravitational force at the same field. These describe different measurement procedures, even though experiments establish equivalent mass values.

Distinguish a net-force/acceleration measurement from a gravitational-force/known-field measurement.
Use inverse acceleration dependence on inertial mass at fixed net force.
Use direct attraction-force dependence on gravitational mass at fixed source field.

Scoring criteria

Describes both mass concepts and their correct force relations.
Explains larger inertia as greater resistance and smaller acceleration.
Explains gravitational mass through stronger attraction at a fixed field.

Part b

m_i = 1.8/0.60 = 3.0 kg and m_g = 30/10 = 3.0 kg. The intervals 2.90–3.10 kg and 2.94–3.06 kg overlap, so this trial supports agreement within uncertainty. Inertial and gravitational masses have been experimentally verified to be equivalent; equality is not their conceptual definition, and one trial alone does not prove universal equivalence. A force-sensor zero offset, uncorrected friction, acceleration calibration, or including different cart masses can cause a spurious disagreement.

Infer inertial mass from net force/acceleration and gravitational mass from gravitational force/field.
Construct or compare the stated uncertainty intervals.
Distinguish experimental support from a definition and investigate model/calibration errors.

Scoring criteria

Obtains 3.0 kg from both independent ratios.
Uses the uncertainties to establish compatible measurements.
Explicitly states experimentally verified equivalence and names a relevant systematic check.

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Choose friction direction in a moving-surface frame

A 3 kg box moves east at 2 m/s while a horizontal belt beneath it moves east at 5 m/s. Their surfaces slide relative to one another. The kinetic coefficient is 0.20; no other horizontal force acts on the box and it has no vertical acceleration. Use g = 10 m/s².

Part a

Box relative to belt is 2 − 5 = −3 m/s, west. Belt relative to box is +3 m/s, east. Kinetic friction on the box is east, opposite its westward sliding relative to the belt; friction on the belt is west, opposite its eastward sliding relative to the box. The two forces are equal/opposite interaction partners. The box’s eastward ground velocity does not require westward friction.

Subtract the surface velocities in both orders.
Oppose each surface’s motion relative to the other surface.
Check the two interaction-force directions against Newton’s third law.

Scoring criteria

Finds −3 and +3 m/s relative velocities with directions.
Gives eastward friction on box and westward friction on belt.
Justifies both from relative surface motion.

Part b

The force arrows from the dot are gravity 30 N downward, normal force 30 N upward away from the horizontal belt, and kinetic friction 6 N east. Vertical balance gives N = mg = 30 N; f_k = 0.20(30) = 6 N; a = 6/3 = 2 m/s² east while this relative sliding condition lasts. The normal is the perpendicular component of the belt’s contact force and would point perpendicular outward from a tilted belt, not remain vertical. Normal and friction are the two contact components; no extra combined contact arrow is added alongside them.

Use zero vertical acceleration to find normal load.
Use the established sliding direction and f_k = μ_kN.
Divide the horizontal resultant by box mass and keep the direction tied to the surface orientation.

Scoring criteria

Draws gravity down, normal perpendicular outward and friction east from the dot.
Obtains N = 30 N, f_k = 6 N and acceleration 2 m/s² east.
Describes the tilted-surface normal as perpendicular outward.

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Separate area, material and friction regime

A rectangular block is tested with a fixed 50 N normal load. On the same dry material pair it slides on a 200 cm² face and then a 100 cm² face; μ_k = 0.30. The narrow face is then tested on a different surface with μ_k = 0.50. Keep speed in the ideal dry-friction regime and surface condition fixed within each pair. For the original pair a measured μ_s is 0.45.

Part a

Both orientations give f_k = 0.30(50) = 15 N. Halving apparent contact area does not change friction in the ideal dry-friction model. Holding normal load and the same material pair/condition fixed isolates area; changing either could change force for another reason. This is an approximation for the stated regime, not a universal result for adhesive or highly deformable contacts.

Use the same coefficient and normal load for both orientations.
Compare the results while area alone changes.
State what the controls isolate and the model domain.

Scoring criteria

Finds 15 N for both faces.
Explicitly explains area independence at equal load.
Names material/load controls and avoids universalizing the approximation.

Part b

On the replacement pair f_k = 0.50(50) = 25 N. The coefficient depends on the contacting materials and their surface condition, so replacing a surface can change μ_k even though area and load stay fixed. Area independence within a controlled pair does not mean material independence.

Keep the 50 N load and use the new material-pair coefficient.
Connect the coefficient change to material properties at contact.
Separate the controlled-area conclusion from the material comparison.

Scoring criteria

Obtains 25 N.
States dependence of μ_k on material properties/condition of both contacting surfaces.
Explains why the equal-area or equal-load condition does not force equal coefficients.

Part c

Typically μ_s > μ_k. Here maximum static friction is 0.45(50) = 22.5 N and kinetic friction is 0.30(50) = 15 N. Actual static friction adjusts to the force needed to prevent relative slip and can be below 22.5 N, including zero; 22.5 N applies at impending slip. “Typically” allows exceptions and does not assert every possible surface pair obeys the inequality.

State the typical coefficient inequality before applying the supplied values.
Multiply each coefficient by the same load to compare forces.
Distinguish an adjustable static force from its limit.

Scoring criteria

Explicitly states μ_s is typically greater than μ_k.
Obtains 22.5 N maximum static and 15 N kinetic friction.
Explains the nonautomatic maximum and retains the typical qualification.

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Distinguish the two restoring-force origins

An ideal spring has relaxed length 0.18 m. A 12 N holding force stretches it to 0.24 m within the linear range. The spring is then used vertically to suspend a 0.60 kg mass. Use g = 10 m/s², measure downward spring deformation x from relaxed length and displacement y from the hanging equilibrium.

Part a

The spring has negligible mass and its force magnitude is proportional to the change in length from relaxed length. Initial extension is 0.24 − 0.18 = 0.060 m, so k = 12/0.060 = 200 N/m. At 0.21 m length the extension is 0.030 m, giving 6.0 N. Total length includes the unstressed relaxed portion; only deformation produces the ideal spring force.

State negligible mass and identify the relaxed-length reference.
Subtract relaxed length from each loaded length.
Use the same positive stiffness for both deformations.

Scoring criteria

States negligible spring mass and proportionality to deformation.
Obtains k = 200 N/m and predicted holding force 6.0 N.
Explicitly distinguishes total length from extension.

Part b

At equilibrium mg − kx_eq = 0, so x_eq = 6/200 = 0.030 m. Above equilibrium y = −0.010 m, hence x = 0.020 m. F_s = −200(0.020) = −4.0 N upward. F_net = +6 − 4 = +2.0 N downward toward equilibrium. Spring force points away from equilibrium in this state while the net force restores it. A horizontal spring with no other horizontal force has equilibrium at relaxed length and spring force directed toward that position. A hanging spring uses F_s = −k(x_eq + y), while the combined restoring net force is −ky.

Balance gravity and spring force at equilibrium to derive its shifted extension.
Convert displacement from equilibrium to deformation from relaxed length.
Compare actual spring and net-force directions and identify which one restores the shifted system.

Scoring criteria

Derives x_eq = 0.030 m.
Obtains −4.0 N spring force and +2.0 N net force with correct directions.
Uses the above-equilibrium state as a counterexample and states the correct horizontal/vertical force distinction.

Part c

The spring-force line is F_s = −6 − 200y N, with slope −200 N/m and intercept −6 N. The net-force line is F_net = −200y N, with the same slope and zero intercept. Gravity adds +6 N to the spring force, shifting the line upward. At y = 0 only net force is zero; the spring force remains 6 N upward and balances gravity. Graphs apply while the spring remains in its linear range.

Express spring deformation as x_eq + y.
Add gravity to the spring-force expression to obtain the net line.
Mark both y = 0 intercepts and explain the vertical offset.

Scoring criteria

Sketches both decreasing lines with slope −200 N/m.
Labels intercepts −6 N for spring and 0 N for net.
Explains gravity’s offset and identifies zero net force at equilibrium.

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Construct acceleration and a contact threshold

A bead is at the rightmost point of a radius 3.0 m circle moving counterclockwise. Its speed increases at a constant rate from 4.0 to 8.0 m/s over 2.0 s; consider the instant its speed is 6.0 m/s. Separately, a 2.0 kg object moves on the inside of a vertical circular track of radius 2.5 m. Neglect drag and use g = 10 m/s².

Part a

Velocity points upward tangent to the path. Centripetal acceleration points left toward the center and is 6²/3 = 12 m/s². Tangential acceleration points upward along the local tangent while speeding up and is (8 − 4)/2 = 2.0 m/s². The perpendicular components give total magnitude √(12² + 2²) = √148 ≈ 12.2 m/s², pointing up and left. The tangential component changes speed; the radial component changes velocity direction.

Use counterclockwise motion at the rightmost point to establish the tangent direction.
Obtain the speed-change rate from the record and the inward turning rate from v²/r.
Add the perpendicular acceleration components as vectors.

Scoring criteria

Draws upward velocity/tangential acceleration and leftward centripetal acceleration.
Calculates 2.0 m/s² from Δspeed/Δt and 12 m/s² from v²/r.
Gives approximately 12.2 m/s² total and explains the different physical roles.

Part b

During slowing, tangential acceleration points down, opposite the upward velocity; centripetal acceleration remains left toward the center. At constant speed the tangential acceleration is zero because speed-change rate is zero, while centripetal acceleration remains nonzero because velocity direction changes. With nonzero speed, constant speed on a circle does not mean constant velocity.

Oppose the instantaneous velocity tangent when speed decreases.
Keep the radial direction set by the center rather than the sign of speed change.
Separate zero speed-change rate from nonzero vector-velocity change.

Scoring criteria

Gives downward slowing tangential direction and inward leftward radial direction.
Identifies zero tangential acceleration at constant speed.
Explains remaining acceleration from direction change.

Part c

At the top inward is down, so mg + N = mv²/r with N ≥ 0 for an inside track. The minimum occurs at N = 0: mg = mv_min²/r, hence v_min = √(gr) = √(10·2.5) = 5.0 m/s. Gravity alone supplies the radial acceleration at threshold. At 4.0 m/s, N = 2(16/2.5 − 10) = −7.2 N, which would require the track to pull; contact is lost. This is the top speed condition, not a launch condition at the bottom.

Identify the inward direction and actual top forces.
Apply nonnegative normal force and set N = 0 at the contact limit.
Evaluate the proposed speed and reject the negative normal force.

Scoring criteria

Derives v_min = √(gr) and calculates 5.0 m/s.
States gravity alone causes centripetal acceleration at N = 0.
Uses −7.2 N to explain loss of contact at 4.0 m/s and retains the local-top limit.

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Explain friction’s contribution on a bank

A circular road is banked so that its outer edge is higher than its inner edge. Tires do not slip relative to the road when the required static friction lies within its available limit. The frictionless design speed is the speed at which normal force alone provides the needed horizontal radial force. Analyze friction qualitatively; no numerical frictional-bank calculation is requested.

Part a

The FBD has gravity downward and normal force perpendicular outward from the sloping road, tilted inward and upward. Its vertical component balances mg; its horizontal inward component supplies mv²/r. Because only N cos θ balances weight for bank angle θ, N = mg/cos θ exceeds mg at nonzero angle. Components are constructions for the force equation, not additional actual-force arrows.

Draw gravity and the tilted normal from the selected car dot.
Resolve the one normal force into upward and horizontal inward components separately.
Use vertical balance to distinguish N from its vertical component.

Scoring criteria

Shows normal force perpendicular outward with inward/upward direction and gravity downward.
Identifies the inward normal component as the centripetal-force contribution.
Explains N > mg at a nonzero angle and avoids double-counting components.

Part b

Below design speed the tendency is down the bank; static friction acts up the bank, with outward and upward components. Its outward component subtracts from the inward normal contribution, and its upward component helps balance weight. Above design speed the tendency is up the bank; static friction acts down the bank, with inward and downward components. Its inward component adds to the inward normal contribution, while its downward component requires a larger upward normal contribution. These directions oppose impending relative slip. If the required static force exceeds μ_sN, no-slip circular motion at that speed cannot be maintained; the model does not promise unlimited friction.

Compare the required radial turning with the frictionless design-speed case.
Oppose each up/down-bank slipping tendency with static friction.
Resolve the static force qualitatively in radial and vertical directions and apply its finite limit.

Scoring criteria

Gives down-bank tendency/up-bank friction below design speed and the reverse above it.
Correctly states outward subtraction versus inward addition to the normal radial contribution.
Correctly states upward/downward vertical roles and the finite static-friction condition.

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Resolve tension in a conical pendulum

A 0.80 kg mass travels at constant speed in a horizontal circle on a taut 2.0 m ideal string. The string makes 60° with vertical, the support is fixed, and air resistance is negligible. Use g = 10 m/s². T denotes tension and P denotes period.

Part a

Actual forces are gravity 8.0 N vertically down and tension along the string toward the support. Vertical balance is T cos 60° = mg; horizontal inward balance is T sin 60° = mv²/r. The horizontal component of tension supplies the inward net force; tension’s vertical component balances gravity. Centripetal force names that inward resultant and is not a third interaction to add to the FBD. Draw components separately from the actual-force diagram.

Identify only the string and Earth as force sources.
Resolve the tension using the angle measured from vertical.
Separate vertical balance from horizontal radial acceleration.

Scoring criteria

Draws only gravity and tension with correct directions.
Identifies T sin 60° as the inward contribution and writes both equations.
Explains why there is no extra centripetal interaction.

Part b

r = L sin 60° = √3 m ≈ 1.73 m. T = 8.0/cos 60° = 16 N. The inward component is 16 sin 60° = 8√3 N ≈ 13.9 N. Then v² = (8√3)(√3)/0.80 = 30 m²/s², so v = √30 ≈ 5.48 m/s. P = 2πr/v = 2π/√10 ≈ 1.99 s. The circle radius is the string’s horizontal projection, not its full 2.0 m length.

Obtain radius from the horizontal projection and tension from vertical balance.
Use only the horizontal tension component in the radial equation.
Convert tangential speed into time for one circumference.

Scoring criteria

Obtains radius about 1.73 m, tension 16 N and inward force about 13.9 N.
Obtains speed about 5.48 m/s and period about 1.99 s.
Explains the radius projection and avoids substituting full string length.

Part c

T = mg/cos θ doubles, as does T sin θ. But a_c = T sin θ/m = g tan θ is unchanged; r = L sin θ is unchanged, so v² = rg tan θ and speed are unchanged. A greater tension is required for the greater mass without increasing the same constrained circular motion’s acceleration.

Use vertical equilibrium to scale total tension with mass.
Divide the horizontal component by mass to see the cancellation.
Keep the geometric radius fixed when relating acceleration to speed.

Scoring criteria

Predicts doubled tension and unchanged radial acceleration/speed.
Shows cancellation in T sin θ/m or v² = rg tan θ.
Connects the result to fixed angle and string length.

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