Mechanics Mastery

Q8: Comparing Gravitational Force on Satellites — Correct: (A) AP Standard 3.3

\(F_g = G M m / r^2\). For Satellite 1: \(m_0/R_0^2 = 1\). For Satellite 2: \(4m_0/(2R_0)^2 = 4/4 = 1\). For Satellite 3: \(6m_0/(3R_0)^2 = 6/9 = 0.67\). Therefore, \(F_1 = F_2 > F_3\).

Q9: Static Friction Threshold — Correct: (A) AP Standard 2.5

\(f_{s,max} = \mu_s F_N = 0.40(80\text{ N}) = 32\text{ N}\). The applied pull of \(24\text{ N}\) is less than \(f_{s,max}\), so the crate remains at rest and \(a = 0\text{ m/s}^2\). (Actual static friction is \(24\text{ N}\)).

Q10: Static Friction on Incline at Rest — Correct: (B) AP Standard 2.5

Because the block is at rest, \(\Sigma F_\parallel = mg\sin\theta - f_s = 0 \implies f_s = mg\sin\theta\). Static friction is an adjustable balancing force, not automatically \(\mu_s F_N\).

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Section I: Multiple Choice Answers & Rationales (Q11–Q20) AP Physics 1 Quarter 1

11: C

12: D

13: B

14: C

15: B

16: D

17: C

18: B

19: A

20: B

Q11: Block Sliding Up Incline — Correct: (C) AP Standard 2.3 & 2.5

As the block slides upward, the parallel component of gravity (\(Mg\sin\theta\)) acts down the incline, and kinetic friction (\(f_k = \mu_k Mg\cos\theta\)) also acts down the incline opposing velocity. Magnitudes add: \(F_{net,\parallel} = Mg\sin\theta + \mu_k Mg\cos\theta\).

Q12: Projectile Kinematic Vectors at Apex — Correct: (D) AP Standard 1.3

At the apex, vertical velocity is momentarily zero (\(v_y = 0\)), but horizontal velocity is non-zero (\(v_x = v_0\cos 35^\circ\)). Acceleration is constant downward due to gravity (\(a_y = -g \neq 0\)). Thus both velocity and acceleration are non-zero.

Q13: Force Components Along an Incline — Correct: (B) AP Standard 2.3

The pulling force is angled at \(\phi\) above the incline plane, so its component pulling up the incline is \(F\cos\phi\). Opposing it are gravity (\(mg\sin\theta\)) and friction (\(f\)). Net force is \(F\cos\phi - mg\sin\theta - f\).

Q14: Kinematic Turnaround on a Ramp — Correct: (C) AP Standard 1.1

At \(t = 5\text{ s}\), \(v(5) = +3.0 + (-1.0)(5.0) = -2.0\text{ m/s}\). Since velocity is negative (\(-x\)) and acceleration is negative (\(-x\)), the velocity and acceleration vectors point in the same direction, meaning the cart is moving in the \(-x\)-direction and speeding up.

Q15: Binary Orbit Center of Mass Acceleration — Correct: (B) AP Standard 2.1 & 3.1

Internal gravitational forces between the two stars cancel. The only net external force accelerating the center of mass of the binary system is the gravitational pull of the central black hole/star, which points radially toward the central body.

Q16: Newton's Third Law Invariance — Correct: (D) AP Standard 2.4

By Newton's Third Law, forces always occur in matched interaction pairs. If the tow truck exerts force \(F\) on any object, the object simultaneously exerts an equal and opposite force \(F\) on the tow truck, independent of mass, speed, or whether the object accelerates.

Q17: Average Acceleration Ranking — Correct: (C) AP Standard 1.2

\(|a_{avg}| = |\Delta v/\Delta t|\). Segment \(CD\): \(|\frac{0 - 8}{0.5}| = 16\text{ m/s}^2\). Segment \(AB\): \(8/1 = 8\text{ m/s}^2\). Segment \(DE\): \(4/1 = 4\text{ m/s}^2\). Segment \(BC\): \(0\text{ m/s}^2\). Thus \(a_{CD} > a_{AB} > a_{DE} > a_{BC}\).

Q18: Self-Supported Pulley Equilibrium — Correct: (B) AP Standard 2.2 & 2.3

The child + swing system is supported by two upward rope tensions: the rope attached to the seat and the rope held by the child's hands. Total upward force is \(2T\). For equilibrium, \(\Sigma F_y = 2T - Mg = 0 \implies 2T = Mg\).

Q19: Center of Mass Coordinate Shift — Correct: (A) AP Standard 2.1

\(\Delta x_{cm} = \frac{m(+\Delta x) + 4m(-\Delta x)}{5m} = -0.6\,\Delta x\). The center of mass shifts toward the heavier sphere (to the left/toward \(4m\)) by an amount less than \(\Delta x\).

Q20: Vertical Loop Critical Velocity — Correct: (B) AP Standard 3.2

At the top of the loop, \(\Sigma F_c = F_N + mg = m v^2/R\). At minimum speed, track normal force reaches zero (\(F_N = 0\)), giving \(mg = m v_{min}^2/R \implies v_{min} = \sqrt{gR}\).

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Section II: Constructed Response Rubrics • CR 1, 2 & 3 AP Point Allocations

CR 1: Chase & Catch-Up Kinematics (QQT) Total: 7 Points

Part (a) Derivations (3 pts):
• 1 pt: For using constant speed for truck: \(t_D = D / v_T\).
• 1 pt: For kinematics relation for car: \(D = \frac{1}{2} a_c t_D^2\).
• 1 pt: For algebraic substitution yielding \(a_c = \frac{2D}{t_D^2} = \frac{2v_T^2}{D}\).

Part (b) Argument Critique (4 pts):
• 1 pt: Identifying what is correct: both vehicles indeed have the same average speed over interval \(t_D\) (\(v_{avg} = D/t_D\)).
• 1 pt: Identifying what is incorrect: equal average speeds do not imply equal instantaneous final speeds.
• 1 pt: Explaining that since the car starts from rest (\(v_0 = 0\)), its final speed must be \(v_{c,final} = 2 v_{avg} = 2 v_T\) to cover equal area under the \(v-t\) curve.
• 1 pt: For consistent, coherent justification linking mathematical representation to conceptual reasoning.

CR 2: Multi-Trajectory Projectile Comparison Total: 6 Points

Part (a) Time in Air Ranking (2 pts):
• 1 pt: Correct ranking: \(t_1 > t_2 > t_3\).
• 1 pt: Justification based on initial vertical velocities: \(v_{y0,1} = +v_0\sin 45^\circ > v_{y0,2} = 0 > v_{y0,3} = -v_0\sin 45^\circ\). Greater upward initial velocity yields longer flight time.

Part (b) Impact Vertical Velocity (2 pts):
• 1 pt: Stating that \(|v_{y1}| = |v_{y3}|\) immediately before ground impact.
• 1 pt: Explaining symmetry: Rock 1 returns to its launch height with downward vertical speed \(v_0\sin 45^\circ\), identical to Rock 3's initial launch condition.

Part (c) Impact Speed Comparison (2 pts):
• 1 pt: Stating all three rocks strike the ground with the exact same total speed (\(v_1 = v_2 = v_3\)).
• 1 pt: Energy justification: \(E_i = \frac{1}{2}m v_0^2 + mgH = \frac{1}{2}m v_f^2 \implies v_f = \sqrt{v_0^2 + 2gH}\), independent of launch angle!

CR 3: Vertical Circular Motion & Rupture Dynamics Total: 7 Points

Part (a) Free-Body Diagrams (3 pts):
• Point P (1 pt): Upward velocity is immaterial to forces. Tension \(T_P\) points horizontally right (toward center); gravity \(Mg\) points straight down.
• Point Q (2 pts): Tension \(T_Q\) points straight up; gravity \(Mg\) points straight down. Vector \(T_Q\) must be drawn visibly LONGER than \(Mg\) to provide net upward centripetal acceleration (\(T_Q - Mg = Mv^2/R\)).

Part (b) Apex Minimum Speed Derivation (2 pts):
• 1 pt: Writing Newton's 2nd Law at top: \(T_Z + Mg = M\frac{v^2}{R}\).
• 1 pt: Setting threshold \(T_Z = 0\) to obtain \(v_{min} = \sqrt{gR}\).

Part (c) String Snapping Trajectory (2 pts):
• 1 pt: Stating the ball moves initially straight upward tangentially.
• 1 pt: Describing the 1D free-fall motion: decelerating upward at \(g\), reaching a peak, and falling vertically along a straight line.

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Section II: Constructed Response Rubrics • CR 4, 5 & 6 AP Point Allocations

CR 4: Modified Multi-Body Atwood Machine Total: 7 Points

Part (a) System Derivation (3 pts):
• 1 pt: Setting up system equation: \(\Sigma F_{ext} = m_2 g - m_1 g = (m_2 - m_1)g\).
• 1 pt: Identifying total accelerated system inertia: \(M_{total} = m_1 + m_2\).
• 1 pt: Solving: \(a_{sys} = \frac{(m_2 - m_1)g}{m_1 + m_2}\).

Part (b) Adding Middle Block \(m_3\) (4 pts):
• 1 pt: Correct claim: Acceleration of block 2 decreases.
• 1 pt: Stating net external driving force along the string line remains unchanged at \((m_2 - m_1)g\) because the table is frictionless (gravity on \(m_3\) is balanced by normal force).
• 1 pt: Explaining that total system inertia increases from \((m_1 + m_2)\) to \((m_1 + m_2 + m_3)\).
• 1 pt: Concluding \(a_{new} = \frac{(m_2 - m_1)g}{m_1 + m_2 + m_3} < a_{orig}\).

CR 5: Static Friction Against a Vertical Surface Total: 7 Points

Part (a) Free-Body Diagram (2 pts):
• 1 pt: Correct horizontal forces: Applied push \(F\) to the right; Normal force \(F_N\) from wall to the left (equal length).
• 1 pt: Correct vertical forces: Static friction \(f_s\) pointing UP; Gravity \(mg\) pointing DOWN (equal length).

Part (b) Minimum Force Derivation (3 pts):
• 1 pt: Identifying equilibrium: \(f_s = mg\) and \(F_N = F\).
• 1 pt: Applying inequality \(f_s \le \mu_s F_N \implies mg \le \mu_s F\).
• 1 pt: Solving: \(F_{min} = \frac{mg}{\mu_s} = \frac{(2.0)(10)}{0.40} = 50\text{ N}\).

Part (c) Doubling Applied Force (2 pts):
• 1 pt: Stating the friction force DOES NOT double (remains \(20\text{ N}\)).
• 1 pt: Explaining static friction is determined by equilibrium (\(\Sigma F_y = 0 \implies f_s = mg\)); increasing \(F\) only increases the maximum possible threshold \(f_{s,max}\), not the actual static friction required.

CR 6: Incline Multi-Body Dynamics & Severance Total: 6 Points

Part (a) Original System Acceleration (2 pts):
• 1 pt: Net force down incline is \(\Sigma F_\parallel = (M + M)g\sin\theta = 2Mg\sin\theta\).
• 1 pt: Dividing by total mass \(2M\) gives \(a_{cm,orig} = g\sin\theta\).

Part (b) Severed String with One Block Clamped (4 pts):
• 1 pt: Correct claim: \(a_{cm,new} = \frac{1}{2} a_{cm,orig}\) (it decreases).
• 1 pt: Stating block \(Y\) accelerates at \(g\sin\theta\) while block \(X\) has acceleration \(a_X = 0\).
• 1 pt: Applying definition of center of mass acceleration: \(a_{cm} = \frac{M a_X + M a_Y}{2M} = \frac{0 + M(g\sin\theta)}{2M} = \frac{1}{2}g\sin\theta\).
• 1 pt: Alternate external force rationale: System mass is \(2M\), but only mass \(Y\) has net gravitational driving force \(Mg\sin\theta\).

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Section II: Constructed Response Rubrics • CR 7 & 8 AP Point Allocations

CR 7: Experimental Design: Sprinter Kinematics Total: 7 Points

Part (a) Procedure & Setup (4 pts):
• 1 pt: Using measuring tape and chalk to mark 10 distinct, equidistant positions along the track (e.g., at \(10\text{ m}, 20\text{ m}, \dots, 100\text{ m}\)).
• 1 pt: Stationing one student timer with a stopwatch at each marked distance marker.
• 1 pt: Synchronizing timer starts using the visual flash/sound of the starter's horn/pistol.
• 1 pt: Recording the split time \(t_i\) when the runner's torso crosses each respective marker, and specifying multiple trials to reduce measurement uncertainty.

Part (b) Data Analysis & Graphing (3 pts):
• 1 pt: Plotting position \(x\) versus time \(t\) for all data points.
• 1 pt: Identifying \(t_u\) as the inflection point where the graph transitions from a parabolic curve (\(x \propto t^2\)) to a straight line (\(x \propto t\)).
• 1 pt: Calculating uniform acceleration: either fit \(x = \frac{1}{2}a_u t^2\) (graph \(x\) vs \(t^2\) for \(t \le t_u\) where slope \(= \frac{1}{2}a_u\)) or determine final constant speed \(v_{max}\) from linear slope and compute \(a_u = v_{max}/t_u\).

CR 8: Apparent Weight & Elevator Mechanics Total: 6 Points

Part (a) Quantitative Calculations (3 pts):
• Phase 1 (1 pt): Ascending & speeding up (\(a_y = +2\text{ m/s}^2\)):
\(\quad F_{N1} = M(g + a) = 60(10 + 2) = 720\text{ N}\).
• Phase 2 (1 pt): Ascending at constant velocity (\(a_y = 0\)):
\(\quad F_{N2} = M g = 60(10) = 600\text{ N}\).
• Phase 3 (1 pt): Ascending & slowing down (\(a_y = -3\text{ m/s}^2\)):
\(\quad F_{N3} = M(g - a) = 60(10 - 3) = 420\text{ N}\).

Part (b) Conceptual Explanation (3 pts):
• 1 pt: Explaining that a scale measures normal contact force \(F_N\), not the downward force of gravity \(F_g\).
• 1 pt: Explaining that gravity depends strictly on passenger mass and Earth's local gravitational field (\(F_g = Mg = 600\text{ N}\)), which is constant throughout.
• 1 pt: Connecting normal force variation directly to Newton's 2nd Law: \(\Sigma F_y = F_N - Mg = Ma_y \implies F_N = Mg + Ma_y\); to accelerate the passenger, normal force must differ from weight.

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Section II: Constructed Response Rubrics • CR 9 & 10 AP Point Allocations

CR 9: Gravitational Superposition & Null Point Total: 6 Points

Part (a) Null Point Algebraic Derivation (3 pts):
• 1 pt: Setting equal opposing gravitational fields: \(G\frac{M}{x^2} = G\frac{4M}{(D - x)^2}\).
• 1 pt: Canceling \(GM\) and taking square root: \(\frac{1}{x} = \frac{2}{D - x}\).
• 1 pt: Solving: \(D - x = 2x \implies 3x = D \implies x = \frac{D}{3}\) from Body 1.

Part (b) Stability of Equilibrium (3 pts):
• 1 pt: Stating the probe will accelerate AWAY from Point \(P\) toward Body 2 (unstable equilibrium).
• 1 pt: Explaining that moving toward Body 2 decreases distance \((D-x)\) and increases distance \(x\).
• 1 pt: By inverse-square law, the gravitational attraction from Body 2 increases while Body 1's attraction decreases, creating a net force in the direction of displacement.

CR 10: Incline Scale Tension vs Tilt Angle Total: 7 Points

Part (a) Function Derivation (3 pts):
• 1 pt: Drawing/identifying parallel forces: string tension \(F_T\) up incline, gravity component \(mg\sin\theta\) down incline.
• 1 pt: Setting up equilibrium: \(\Sigma F_\parallel = F_T - mg\sin\theta = 0\).
• 1 pt: Arriving at explicit function: \(F_T(\theta) = mg\sin\theta\).

Part (b) Graph Characteristics & Justification (4 pts):
• 1 pt: Graph sketch correctly starts at \((0^\circ, 0)\) and ends at \((90^\circ, mg)\).
• 1 pt: Curve is concave downward (rate of increase slows as angle approaches \(90^\circ\)).
• 1 pt: Mathematical reason: \(\frac{d^2}{d\theta^2}(mg\sin\theta) = -mg\sin\theta < 0\) for \(\theta \in (0, 90^\circ)\).
• 1 pt: Physical reason: At \(\theta = 90^\circ\), the ramp is vertical, normal force is zero, and the string supports the cart's full gravitational weight.

AP Reader Diagnostic Takeaways & Common Student Pitfalls

• Component drawing on FBDs: Remind students that drawing resolved components (e.g. \(mg\cos\theta\)) on an official FBD results in loss of points unless components are explicitly differentiated from actual physical forces.
• Equilibrium vs Maximum Static Friction: Students consistently default to \(f_s = \mu_s F_N\). Emphasize that static friction is an inequality (\(f_s \le \mu_s F_N\)) and equals the applied parallel force until slipping begins.
• Newton's Third Law: Reiterate that third-law pairs act on different objects and are equal under all kinematic conditions, including acceleration.

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