Energy Dynamics Discussion Packet AP Physics C: Mechanics • Class Work Packet
Question 1: Work & Energy FRQ (Cmech-2021s2-FR03)
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1. A block of mass \(m\) is placed on top of an ideal spring of spring constant \(k\). The block is pushed against the spring, compressing the spring a distance \(\Delta x\). The block is released from rest, leaves the spring at the position shown in the figure, travels upward, and enters a track with a constant radius of curvature \(R\) that has negligible friction. The block enters the track at point \(A\), maintains contact with the track, and exits horizontally at point \(B\), a distance \(3R\) above the point the block was released. The block then falls to the ground and lands a horizontal distance \(D\) from the end of the track. Express all algebraic answers in terms of \(m, k, \Delta x, R\), and physical constants. The block size is much smaller than \(R\).
LAP 1 Qualitative Forces & Circular Track Threshold
Think-Pair-Share:
At point \(B\), why can't the normal force point upward? What is the minimum condition for the block to maintain track contact?
Stop and Jot:
Write the radial \(F_{net} = ma_c\) equation at \(B\). What must \(N\) equal at threshold speed?
(a) On the dot below, which represents the block, draw and label the forces (not components) that act on the block while still in contact with the track at point \(B\). Each force must be represented by a distinct arrow starting on, and pointing away from, the dot.
Justify your choice of vectors:
(b)
i. Derive an expression for the speed \(v\) of the block at point \(B\).
ii. Derive an expression for the magnitude of the net force \(F\) on the block at point \(B\).
AP Physics C: Mechanics • Class Work Packet
Question 1 Continued (Cmech-2021s2-FR03)
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(c) Derive an expression for the minimum value of \(\Delta x_{min}\) required in order for the block to maintain contact with the track through point \(B\).
LAP 2 Energy Tracking & Critical Compression Threshold
Think-Pair-Share:
Why is the mechanical energy at B not purely gravitational? Where does \(\Delta x_{min} = \sqrt{7mgR/k}\) come from?
Stop and Jot:
State the critical velocity at B: \(v_{crit} = \sqrt{gR}\). Substitute into energy conservation.
The procedure is repeated several times with the distance \(\Delta x > \Delta x_{min}\). (d) Calculate the distance \(D\) that the block travels.
(e) The graph below shows the best-fit line drawn by the students through their data of \(D\) as a function of \(\Delta x\).
i. Explain why there are no data for section I of the graph.
ii. Explain the reason for the shape and minimum value of section II on the graph.
LAP 3 Trajectory Limits & Graph Discontinuity
Think-Pair-Share:
What physically happens to the block if \(\Delta x < \Delta x_{min}\)? Does it leave the track before B?
Stop and Jot:
Evaluate \(D_{MIN}\) by substituting \(v_{crit} = \sqrt{gR}\) into \(D = v t_{fall}\).
AP Physics C: Mechanics • Class Work Packet
Question 2: Loop & Spring FRQ (Cmech-2021s1-FR02)
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2. A block of mass \(m\) starts from rest at point \(A\) and travels with negligible friction through the loop onto a horizontal surface, where the block makes contact with a spring of spring constant \(k = mg/2R\). All motion of the spring is in the horizontal direction. Point \(C\) is the highest point on the loop, and point \(B\) is the rightmost point on the loop. Express all algebraic answers in terms of \(m, h, R\), and physical constants, as appropriate.
LAP 1 Acceleration Components at Point B
Think-Pair-Share:
At point B, is the acceleration purely radial? What causes the tangential acceleration, and which direction does it act?
Stop and Jot:
Sketch \(\vec{a}_c\) (left) and \(\vec{a}_t = \vec{g}\) (down). State why the resultant points down and to the left.
(a) On the dot below, which represents the block, draw an arrow that represents the direction of the acceleration of the block at point \(B\) in the figure above. The arrow must start on and point away from the dot.
Justify your answer:
(b)
i. Derive an expression for the speed \(v\) of the block at point \(B\).
ii. Derive an expression for the magnitude of the net force \(F\) on the block at point \(B\).
AP Physics C: Mechanics • Class Work Packet
Question 2 Continued (Cmech-2021s1-FR02)
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(c) In terms of \(R\), derive an expression for the minimum height \(h_{min}\) necessary for the block to maintain contact with the track through point \(C\).
LAP 2 Critical Apex Height & Spring Work-Energy
Think-Pair-Share:
Why is \(h_{min} = 2.5R\) instead of \(2R\)? When the block hits the spring, where does its energy go?
Stop and Jot:
Equate \(mgh = \frac{1}{2}kx_{MAX}^2\). Substitute \(k = mg/2R\) to derive \(x_{MAX} = 2\sqrt{hR}\).
(d) It is determined that equals \(h = 0.30\) m and \(R = 0.10\) m. If the block is released from a height greater than that found in part (c), what would be the maximum compression \(x_{MAX}\) of the spring?
(e) A graph of the maximum compression of the spring as a function of height is shown below. The height \(h_{min}\) is the height calculated in part (c).
i. Explain why section I appears as a horizontal line segment on the horizontal axis.
ii. Explain the reason for the shape of section II on the graph.
LAP 3 Non-Linear Dependence & Threshold Jump
Think-Pair-Share:
Why is the compression strictly zero for \(h < h_{min}\)? Does the block fly off the track inside the loop?
Stop and Jot:
Explain why Section II has a concave-down shape based on \(x_{MAX} \propto \sqrt{h}\).
AP Physics C: Mechanics • Class Work Packet
Question 3: Amusement Park Ride (Cmech-2011-FR02)
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3. An amusement park ride features a passenger compartment of mass \(M\) that is released from rest at point \(A\), as shown in the figure, and moves along a track to point \(E\). The compartment is in free fall between points \(A\) and \(B\), which are a distance of \(3R/4\) apart, then moves along the circular arc of radius \(R\) between points \(B\) and \(D\). Assume the track is frictionless from point \(A\) to point \(D\) and the dimensions of the passenger compartment are negligible compared to \(R\).
LAP 1 Circular Arc Dynamics & Angle Resolution
Think-Pair-Share:
At point C, what forces act on the passenger compartment? How do you resolve gravity into radial and tangential components?
Stop and Jot:
Write the radial force equation: \(F_c = N - Mg\cos\theta\) (or \(N - Mg\sin\theta\)). State which direction is positive.
(a) On the dot below that represents the passenger compartment, draw and label the forces (not components) that act on the passenger compartment when it is at point \(C\), which is at an angle \(\theta\) from point \(B\).
(b) In terms of \(\theta\) and the magnitudes of the forces drawn in part (a), determine an expression for the magnitude of the centripetal force acting on the compartment at point \(C\).
(c) Derive an expression for the speed \(v_D\) of the passenger compartment as it reaches point \(D\) in terms of \(M, R\), and fundamental constants.
AP Physics C: Mechanics • Class Work Packet
Question 3 Continued (Cmech-2011-FR02)
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A force acts on the compartment between points \(D\) and \(E\) and brings it to rest at point \(E\). (d) If the compartment is brought to rest by friction, calculate the numerical value of the coefficient of friction \(\mu\) between the compartment and the track.
LAP 2 Work-Energy Dissipation & Linear Drag Formulation
Think-Pair-Share:
Total vertical drop is \(3R/4 + R = 7R/4\). How does equating friction work \(-\mu Mg(3R)\) to \(-\frac{1}{2}Mv_D^2\) yield \(\mu = 7/12\)?
Stop and Jot:
Write Newton's 2nd Law for drag force \(-kv\): \(M \frac{dv}{dt} = -kv\). Set up differential equation.
(e) Now consider the case in which there is no friction between the compartment and the track, but instead the compartment is brought to rest by a braking force \(-kv\), where \(k\) is a constant and \(v\) is the velocity of the compartment. Express all algebraic answers in terms of \(M, R, k, v_D\), and fundamental constants.
i. Write, but do NOT solve, the differential equation for \(v(t)\).
ii. Solve the differential equation you wrote in part i.
iii. On the axes below, sketch a graph of the magnitude of acceleration as a function of time. Explicitly label intercepts, asymptotes, or maxima.
Magnitude of Acceleration
0 Time
LAP 3 Exponential Damping & Asymptotic Limits
Think-Pair-Share:
Why does the compartment theoretically take infinite time to stop under \(-kv\) drag? What is its initial acceleration at \(t=0\)?
Stop and Jot:
Confirm \(a(0) = \frac{k v_D}{M}\) and sketch exponential decay with horizontal asymptote at \(a = 0\).
AP Physics C: Mechanics • Class Work Packet
Question 4: Potential Energy Landscapes (Cmech-P2018-FR02)
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4. The variable \(x\) represents the position of particle \(A\) in a two-particle system. Particle \(B\) remains at rest. The graph shows the potential energy \(U(x)\) of the system as a function of position \(x\) of particle \(A\), which has mass \(0.40\) kg. Particle \(A\) is released from rest at position \(x = 2.0\) m. Assume positive to be toward the right.
LAP 1 Conservative Force from Potential Slope \(F = -dU/dx\)
Think-Pair-Share:
Why is force equal to the NEGATIVE derivative of potential energy? What does a flat line on \(U(x)\) mean for force?
Stop and Jot:
Find slope at \(x = 3.0\) m: \(\frac{0 - 8}{6 - 2} = -2.0\) N. Hence \(F = +2.0\) N (Right), \(a = 5.0\) m/s².
(a) Calculate the speed of particle \(A\) when it reaches position \(x = 14.0\) m.
(b)
i. Calculate the magnitude of the acceleration of particle \(A\) when it reaches position \(x = 3.0\) m.
ii. What is the direction of the acceleration at \(x = 3.0\) m?
Left Right
(c) Determine the magnitude of the acceleration of particle \(A\) when it reaches position \(x = 7.0\) m.
AP Physics C: Mechanics • Class Work Packet
Question 4 Continued (Cmech-P2018-FR02)
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Particle \(A\) is again released from rest at the position \(x = 2.0\) m. (d) Calculate the elapsed time for particle \(A\) to travel from:
i. \(x = 2.0\) m to \(x = 6.0\) m:
ii. \(x = 6.0\) m to \(x = 8.0\) m:
iii. \(x = 8.0\) m to \(x = 14\) m:
LAP 2 Piecewise Kinematics & Energy Barriers
Think-Pair-Share:
Between \(x = 6\) and \(8\) m, \(U = 0\). Is the particle at rest or moving at maximum constant speed?
Stop and Jot:
Compute \(v_{max} = \sqrt{2(8.0)/0.40} = 6.32\) m/s. Then \(\Delta t = \Delta x / v_{max} = 2.0 / 6.32 = 0.316\) s.
(e) Particle \(A\) is now placed at position \(x = 7.0\) m. In order for particle \(A\) to reach position \(x = 16.0\) m and to be moving at a speed of \(1.0\) m/s, what initial speed should particle \(A\) be given at its new initial position of \(x = 7.0\) m?
(f) On the grid below, carefully draw a graph of the force \(F\) acting on particle \(A\) as a function of \(x\) for the range \(0 < x < 16\) m.
LAP 3 Constructing Step Force Functions
Think-Pair-Share:
Why is the force graph made of flat horizontal steps? How does the area under \(F(x)\) relate to \(-\Delta U\)?
Stop and Jot:
Verify: \([0,2]: F = +1.5\) N; \([2,6]: F = +2.0\) N; \([6,8]: F = 0\); \([8,14]: F = -1.0\) N; \([14,16]: F = 0\).
AP Physics C: Mechanics • Class Work Packet
Question 5: Circular Rough Wall Dynamics (Cmech-2014-FR02)
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5. A small block of mass \(m\) starts from rest at the top of a frictionless ramp, which is at a height \(h\) above a horizontal tabletop, as shown in the side view above. The block slides down the smooth ramp and reaches point \(P\) with a speed \(v_0\). After the block reaches point \(P\) at the bottom of the ramp, it slides on the tabletop guided by a circular vertical wall with radius \(R\), as shown in the top view. The tabletop has negligible friction, and the coefficient of kinetic friction between the block and the circular wall is \(\mu\).
LAP 1 3D Forces: Tabletop Normal vs. Wall Friction
Think-Pair-Share:
Why is the vertical net force zero? What surface exerts the centripetal normal force, and what direction does friction point?
Stop and Jot:
Show that horizontal net force has radial component \(\frac{mv^2}{R}\) (inward) and tangential component \(f_k\) (backward).
(a) Derive an expression for the height of the ramp \(h\). Express your answer in terms of \(v_0, m\), and fundamental constants, as appropriate.
A short time after passing point \(P\), the block is in contact with the wall and moves with a speed of \(v\). (b) i. Is the vertical component of the net force on the block upward, downward, or zero?
Upward Downward Zero
Justify your answer:
ii. On the figure below, draw an arrow starting on the block to indicate the direction of the horizontal component of the net force on the moving block when it is at the position shown.
Justify your answer:
AP Physics C: Mechanics • Class Work Packet
Question 5 Continued (Cmech-2014-FR02)
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Express your answers to the following in terms of \(v_0, v, m, R, \mu\), and fundamental constants, as appropriate.
(c) Determine an expression for the magnitude of the normal force \(N\) exerted on the block by the circular wall as a function of \(v\).
LAP 2 Velocity-Dependent Normal Force & Tangential Deceleration
Think-Pair-Share:
Why is \(N_{wall} = mv^2/R\) instead of \(mg\)? As the block slows, why does the friction force decrease?
Stop and Jot:
Substitute \(N(v)\) into \(f_k = \mu N\) to derive tangential acceleration: \(a_t = \frac{dv}{dt} = -\frac{\mu v^2}{R}\).
(d) Derive an expression for the magnitude of the tangential acceleration of the block at the instant the block has attained a speed of \(v\).
(e) Derive an expression for \(v(t)\), the speed of the block as a function of time \(t\) after passing point \(P\) on the track.
LAP 3 Non-Linear Calculus Integration \(\int v^{-2} dv\)
Think-Pair-Share:
Why does \(\frac{dv}{dt} \propto -v^2\) yield a rational function \(v(t) = \frac{v_0}{1 + (\mu v_0 / R)t}\) instead of an exponential \(e^{-t}\)?
Stop and Jot:
Check initial condition \(v(0) = v_0\) and asymptotic behavior \(v \to 0\) as \(t \to \infty\).
Energy Dynamics Answer Key Teacher Answer Key & Scoring Rubric
FRQ 1 Solution: Spring Launch & Curved Track
Cmech-2021s2-FR03 • 15 pts
Page 1 of 5
(a) Free-Body Diagram at Point B & Justification 3 Points
↓ \(F_g\) (or \(mg\))
Vectors: Both the gravitational force (\(F_g = mg\)) and the normal force (\(F_N\)) act vertically downward . At the critical threshold, \(F_N = 0\).
Justification: The track is above the block at the apex of curvature, so any contact force from the track must push downward toward the center of curvature. Gravity always pulls downward toward Earth.
Rubric: 1 pt for downward \(F_g\); 1 pt for downward \(F_N\); 1 pt for correct justification referencing track geometry.
(b) Speed and Net Force Derivations 4 Points
i. Speed at Point B (2 pts):
By conservation of mechanical energy (\(E_i = E_f\)):
\(\frac{1}{2}k(\Delta x)^2 = mg(3R) + \frac{1}{2}mv_B^2\)
1 pt for energy conservation setup; 1 pt for correct expression.
ii. Net Force at Point B (2 pts):
Since the track is circular with radius \(R\) and motion is horizontal at B:
\(F_{net} = \frac{mv_B^2}{R} = \frac{k(\Delta x)^2}{R} - 6mg\)
1 pt for \(F_{net} = mv^2/R\); 1 pt for substituting \(v_B^2\).
(c)–(e) Threshold Compression, Landing Distance & Graph 8 Points
(c) Minimum Compression \(\Delta x_{min}\) (2 pts): Threshold occurs when track contact is just barely maintained: \(N = 0 \implies \frac{mv_B^2}{R} = mg \implies v_{B,min} = \sqrt{gR}\). Substituting into energy equation: \(\frac{k(\Delta x_{min})^2}{m} - 6gR = gR \implies \mathbf{\color{red}{\Delta x_{min} = \sqrt{\frac{7mgR}{k}}}}\).
(d) Landing Distance \(D\) (2 pts): Horizontal projectile from height \(y = 3R\). Vertical fall time: \(3R = \frac{1}{2}gt^2 \implies t = \sqrt{\frac{6R}{g}}\). Horizontal distance: \(\mathbf{\color{red}{D = v_B t = \sqrt{\frac{6R}{g}\left(\frac{k(\Delta x)^2}{m} - 6gR\right)}}}\).
(e) i. Section I Discontinuity (2 pts): For \(\Delta x < \Delta x_{min}\), the block lacks sufficient speed to stay on the track through apex B. It loses contact and falls off prior to B, so no horizontal launch occurs.
(e) ii. Section II Shape & Min (2 pts): At threshold, \(D_{min} = \sqrt{gR}\sqrt{6R/g} = \sqrt{6}R \approx 2.45R\). As \(\Delta x \gg \Delta x_{min}\), \(\Delta x^2\) dominates, yielding an asymptotically linear graph.
LAP Discourse Key: Think-Pair-Share & Stop-and-Jot Model Answers
Both Answers in Red
LAP 1 Forces & Track Threshold
Think-Pair-Share: Track is above block; compressive push must be downward toward center of curvature (\(N \ge 0\)); at critical threshold, \(N = 0\).
Stop & Jot: \(\Sigma F_r = N + mg = \frac{mv_B^2}{R}\). Threshold: \(N = 0 \implies mg = \frac{mv_{crit}^2}{R} \implies \mathbf{v_{crit} = \sqrt{gR}}\).
LAP 2 Energy & \(\Delta x_{min}\)
Think-Pair-Share: Block cannot be at rest at apex; must retain kinetic energy \(\frac{1}{2}mgR\) to maintain circular motion. Energy originates from \(\frac{1}{2}k\Delta x^2\).
Stop & Jot: Substitute \(v_{crit} = \sqrt{gR}\): \(\frac{1}{2}k(\Delta x_{min})^2 = 3mgR + \frac{1}{2}m(gR) = \frac{7}{2}mgR \implies \mathbf{\Delta x_{min} = \sqrt{\frac{7mgR}{k}}}\).
LAP 3 Graph Limits & \(D_{MIN}\)
Think-Pair-Share: When \(\Delta x < \Delta x_{min}\), normal force vanishes prior to B; block separates early and falls as a projectile, never launching from B.
Stop & Jot: Fall time \(t = \sqrt{6R/g}\). Minimum range: \(D_{MIN} = v_{crit} t = \sqrt{gR}\sqrt{\frac{6R}{g}} = \mathbf{\sqrt{6}R \approx 2.45R}\).
Teacher Answer Key & Scoring Rubric
FRQ 2 Solution: Loop-the-Loop & Spring
Cmech-2021s1-FR02 • 15 pts
Page 2 of 5
(a) Acceleration Vector at Point B & Justification 3 Points
↙ Vector Arrow
Vector Direction: Arrow points downward and to the left (Quadrant III).
Justification: The block moves along a circular path, requiring a horizontal centripetal acceleration \(\vec{a}_c = v^2/R\) directed left toward the center. Simultaneously, gravity pulls downward with no opposing vertical force, causing downward tangential acceleration \(\vec{a}_t = \vec{g}\). The vector sum points down and left.
Rubric: 1 pt for vector pointing in Quadrant III; 1 pt for recognizing centripetal component; 1 pt for recognizing tangential component.
(b) Speed and Net Force Magnitude at Point B 4 Points
i. Speed at Point B (2 pts):
Height at point B is \(R\). Conservation of energy from height \(h\):
\(mgh = mgR + \frac{1}{2}mv_B^2\)
1 pt for setting \(y_B = R\); 1 pt for correct algebraic solution.
ii. Net Force Magnitude at B (2 pts):
Perpendicular force components: \(F_x = F_c = \frac{mv_B^2}{R} = 2mg\left(\frac{h}{R}-1\right)\), and \(F_y = mg\):
\(F_{net} = \sqrt{F_x^2 + F_y^2} = mg\sqrt{4\left(\frac{h}{R}-1\right)^2 + 1}\)
1 pt for Pythagorean combination; 1 pt for final formula.
(c)–(e) Minimum Loop Height, Compression & Graph 8 Points
(c) Minimum Height \(h_{min}\) (2 pts): At apex C (height \(2R\)), threshold condition is \(N_C = 0 \implies v_C^2 = gR\). Energy balance: \(mgh_{min} = mg(2R) + \frac{1}{2}m(gR) = 2.5mgR \implies \mathbf{\color{red}{h_{min} = 2.5R}}\).
(d) Spring Maximum Compression \(x_{MAX}\) (2 pts): Release height \(h = 0.30\) m exceeds \(h_{min} = 2.5(0.10) = 0.25\) m. Energy balance between release and full spring compression: \(mgh = \frac{1}{2}kx_{MAX}^2\). With \(k = \frac{mg}{2R}\):
(e) i. Section I Flatness (2 pts): If \(h < h_{min}\), the block falls off the loop before reaching apex C. It crashes and never makes contact with the spring, so \(x = 0\).
(e) ii. Section II Shape (2 pts): For \(h \ge h_{min}\), \(x_{MAX} = 2\sqrt{R}\sqrt{h}\). This is a square-root curve, showing an increasing function with decreasing slope (concave down).
LAP Discourse Key: Think-Pair-Share & Stop-and-Jot Model Answers
Both Answers in Red
LAP 1 Point B Acceleration
Think-Pair-Share: Speed decreases as block ascends. Centripetal \(\vec{a}_c = v^2/R\) points left toward center; gravity supplies downward tangential \(\vec{a}_t = \vec{g}\).
Stop & Jot: Components: \(\vec{a} = -a_c \hat{i} - g \hat{j}\). Vector addition places the resultant strictly down and left (Quadrant III).
LAP 2 Apex Height & Spring Work
Think-Pair-Share: \(h_{min} = 2.5R\) because block needs kinetic energy \(\frac{1}{2}m(gR)\) at apex so \(N \ge 0\). All initial energy \(mgh\) transfers into spring \(\frac{1}{2}kx^2\).
Stop & Jot: Equate: \(mgh = \frac{1}{2}\left(\frac{mg}{2R}\right)x_{MAX}^2 \implies x_{MAX}^2 = 4hR \implies \mathbf{x_{MAX} = 2\sqrt{hR}}\).
LAP 3 Discontinuity & \(\sqrt{h}\) Curve
Think-Pair-Share: If \(h < 2.5R\), normal force vanishes before apex C; block falls inside loop, crashes, and never reaches the horizontal spring (\(x = 0\)).
Stop & Jot: Function \(x_{MAX} \propto \sqrt{h}\) has derivative \(\frac{dx}{dh} \propto h^{-1/2} > 0\) and second derivative \(< 0\), explaining concave-down shape.
Teacher Answer Key & Scoring Rubric
FRQ 3 Solution: Amusement Park Ride
Cmech-2011-FR02 • 15 pts
Page 3 of 5
(a)–(b) FBD at Point C & Centripetal Force Expression 4 Points
(a) FBD at Point C (2 pts):
Normal Force (\(N\)): Points along the tilted radius directly toward the center of curvature.Gravity (\(Mg\)): Points vertically downward.Note: No friction or drag exists between A and D. Components must NOT be drawn on the dot.
(b) Centripetal Force Magnitude (2 pts):
Radial component of gravity opposes normal force:
\(F_c = N - Mg\cos\theta\)
(If \(\theta\) is defined from horizontal: \(F_c = N - Mg\sin\theta\)). 1 pt for radial component; 1 pt for equation.
(c)–(d) Speed at Point D & Friction Stopping Coefficient \(\mu\) 4 Points
(c) Speed at Point D (2 pts):
Total vertical drop from A to D: \(\Delta y = \frac{3}{4}R + R = \frac{7}{4}R\). By conservation of mechanical energy:
\(\frac{1}{2}Mv_D^2 = Mg\left(\frac{7}{4}R\right) \implies \mathbf{v_D = \sqrt{\frac{7}{2}gR}}\)
1 pt for total height \(\frac{7}{4}R\); 1 pt for \(v_D\).
(d) Friction Coefficient \(\mu\) (2 pts):
Work-energy theorem over distance \(3R\):
\(W_f = -\mu Mg(3R) = -\frac{1}{2}Mv_D^2 = -Mg\left(\frac{7}{4}R\right)\)
1 pt for \(W_f = -\mu Mg d\); 1 pt for correct numerical value.
(e) Velocity-Dependent Drag \(-kv\) & Differential Equation 7 Points
i. Differential Eq (2 pts):
Apply Newton's 2nd Law:
\(M \frac{dv}{dt} = -kv\)
1 pt for \(M(dv/dt)\); 1 pt for \(-kv\).
ii. Integration (3 pts):
Separate variables with \(v(0)=v_D\):
\(\int_{v_D}^v \frac{1}{v}dv = -\frac{k}{M}\int_0^t dt\)
1 pt separation; 1 pt limits; 1 pt solution.
iii. Sketch \(a(t)\) (2 pts):
\(a(t) = \frac{k v_D}{M} e^{-(k/M)t}\):
Starts at \(a_0 = \frac{k v_D}{M}\) on vertical axis.
Concave up, decreasing asymptotically to 0.
LAP Discourse Key: Think-Pair-Share & Stop-and-Jot Model Answers
Both Answers in Red
LAP 1 Forces & Radial Balance
Think-Pair-Share: Contact force is solely normal force \(\vec{N}\) pointing radially inward; gravity acts vertically downward. Radial component is \(Mg\cos\theta\).
Stop & Jot: Inward radial direction positive: \(\Sigma F_r = N - Mg\cos\theta = \frac{Mv_C^2}{R}\). Hence centripetal force is \(\mathbf{F_c = N - Mg\cos\theta}\).
LAP 2 Work-Energy & Drag Eq.
Think-Pair-Share: Drop is \(\frac{3}{4}R + R = \frac{7}{4}R\). Work of friction \(-f_k d = -\mu Mg(3R)\) dissipates all kinetic energy: \(3\mu = \frac{7}{4} \implies \mu = 7/12\).
Stop & Jot: Net braking force is drag: \(F_{net} = -kv\). By Newton's 2nd Law: \(\mathbf{M\frac{dv}{dt} = -kv}\), yielding \(\frac{dv}{dt} = -\frac{k}{M}v\).
LAP 3 Damping Limits & Sketch
Think-Pair-Share: Drag is proportional to speed. As \(v \to 0\), drag vanishes, causing deceleration to weaken continuously; stopping takes infinite time. At \(t=0\), \(a_0 = \frac{kv_D}{M}\).
Stop & Jot: Magnitude of acceleration \(a(t) = \frac{kv_D}{M}e^{-(k/M)t}\). Graph starts at \(\frac{kv_D}{M}\) and decays exponentially to asymptote \(a=0\).
Teacher Answer Key & Scoring Rubric
FRQ 4 Solution: Potential Energy Landscapes
Cmech-P2018-FR02 • 15 pts
Page 4 of 5
(a)–(c) Speed and Acceleration from \(U(x)\) Graph 5 Points
(a) Speed at \(x = 14\) m (2 pts):
\(E = U(2.0) = 8.0\) J. At \(x = 14\) m, \(U = 6.0\) J:
\(K = 8.0 - 6.0 = 2.0\text{ J}\)
(b) Accel at \(x = 3\) m (2 pts):
Slope \(\frac{dU}{dx} = \frac{0 - 8}{6 - 2} = -2.0\) N:
\(F = -\frac{dU}{dx} = +2.0\text{ N}\)
(c) Accel at \(x = 7\) m (1 pt):
Flat slope between \(x = 6\) and \(8\) m:
\(\frac{dU}{dx} = 0 \implies F = 0\)
(d) Elapsed Times Across Piecewise Segments 4 Points
i. \(x = 2\) to \(6\) m (1 pt):
\(d = 4\) m, \(v_i = 0, a = 5.0\) m/s²:
\(4.0 = \frac{1}{2}(5.0)t^2\)
ii. \(x = 6\) to \(8\) m (1.5 pts):
\(v = \sqrt{2(8.0)/0.40} = \sqrt{40} \approx 6.32\) m/s:
\(a = 0 \implies t = \frac{\Delta x}{v}\)
iii. \(x = 8\) to \(14\) m (1.5 pts):
\(F = -1.0\) N \(\implies a = -2.5\) m/s²:
\(t = \frac{v_f - v_i}{a} = \frac{3.16 - 6.32}{-2.5}\)
(e)–(f) Barrier Clearing Speed & Force Function Graph 6 Points
(e) Initial Speed at \(x = 7.0\) m (3 pts):
Target at \(x = 16\) m: \(U = 6.0\) J, \(K = \frac{1}{2}(0.40)(1.0)^2 = 0.20\) J. Total energy required: \(E_{req} = 6.20\) J. At \(x = 7\) m, \(U = 0\):
\(\frac{1}{2}(0.40)v_0^2 = 6.20 \implies v_0 = \sqrt{31} \approx \mathbf{5.57\text{ m/s}}\)
(f) Force Function \(F(x) = -dU/dx\) (3 pts):
\(0 < x < 2\text{ m}: \mathbf{\color{red}{F = +1.5\text{ N}}}\)
\(2 < x < 6\text{ m}: \mathbf{\color{red}{F = +2.0\text{ N}}}\)
\(6 < x < 8\text{ m}: \mathbf{\color{red}{F = 0\text{ N}}}\)
\(8 < x < 14\text{ m}: \mathbf{\color{red}{F = -1.0\text{ N}}}\)
\(14 < x < 16\text{ m}: \mathbf{\color{red}{F = 0\text{ N}}}\)
LAP Discourse Key: Think-Pair-Share & Stop-and-Jot Model Answers
Both Answers in Red
LAP 1 Force Slope \(F = -dU/dx\)
Think-Pair-Share: Forces push objects down the potential gradient toward lower energy (\(F = -dU/dx\)). Flat plateau has zero slope, so net force and acceleration are both zero.
Stop & Jot: Slope at \(x=3\) m: \(\frac{0-8}{6-2} = -2.0\) N. Therefore \(F = -(-2.0) = \mathbf{+2.0\text{ N}}\) (Right), and \(a = \frac{2.0}{0.40} = \mathbf{5.0\text{ m/s}^2}\).
LAP 2 Energy Wells & Kinematics
Think-Pair-Share: When \(U = 0\), all total energy \(8.0\) J is kinetic. Particle is not at rest; it coasts at maximum speed \(v_{max} = 6.32\) m/s.
Stop & Jot: \(v_{max} = \sqrt{\frac{2(8.0)}{0.40}} = 6.32\) m/s. Travel time at zero acceleration: \(\Delta t = \frac{\Delta x}{v_{max}} = \frac{2.0\text{ m}}{6.32\text{ m/s}} = \mathbf{0.316\text{ s}}\).
LAP 3 Step Forces & Work Area
Think-Pair-Share: Piecewise linear \(U(x)\) has constant derivatives on intervals, making \(F(x) = -dU/dx\) horizontal steps. Signed area equals work done: \(W = -\Delta U\).
Stop & Jot: Step values: \([0,2]\): \(+1.5\) N; \([2,6]\): \(+2.0\) N; \([6,8]\): \(0\); \([8,14]\): \(-1.0\) N; \([14,16]\): \(0\). Area: \(1.5(2) + 2(4) + 0 - 1(6) = \mathbf{5.0\text{ J}}\).
Teacher Answer Key & Scoring Rubric
FRQ 5 Solution: Circular Rough Wall Dynamics
Cmech-2014-FR02 • 15 pts
Page 5 of 5
(a)–(b) Ramp Height & 3D Force Orientation 5 Points
(a) Height of Ramp \(h\) (2 pts):
By energy conservation on frictionless ramp:
\(mgh = \frac{1}{2}mv_0^2 \implies \mathbf{h = \frac{v_0^2}{2g}}\)
1 pt for energy setup; 1 pt for isolating \(h\).
(b) Net Force Components (3 pts):
i. Vertical Net Force (1.5 pts): Zero . The block moves strictly in the horizontal plane on the tabletop; normal force from the table balances gravity (\(N_{table} = mg\)).
ii. Horizontal Net Force Vector (1.5 pts): Arrow points inward and backward . Normal force from the wall points radially inward (\(mv^2/R\)), and friction points tangentially backward (\(f_k = \mu N\)). Resultant points into the 3rd quadrant relative to velocity.
(c)–(d) Velocity-Dependent Normal Force & Tangential Acceleration 4 Points
(c) Normal Force \(N(v)\) Exerted by Wall (2 pts):
The circular vertical wall provides the centripetal acceleration:
\(\mathbf{N(v) = \frac{mv^2}{R}}\)
1 pt for recognizing centripetal role; 1 pt for formula.
(d) Tangential Acceleration \(a_t(v)\) (2 pts):
Kinetic friction opposes tangential motion: \(f_k = \mu N = \mu \frac{mv^2}{R}\). By Newton's 2nd Law:
\(\mathbf{a_t = \frac{f_k}{m} = \frac{\mu v^2}{R}}\) (or \(-\frac{\mu v^2}{R}\) as a vector)
1 pt for \(f_k = \mu N\); 1 pt for dividing by \(m\).
(e) Non-Linear Differential Equation & Velocity \(v(t)\) 6 Points
Tangential motion equation: \(m \frac{dv}{dt} = -\mu \frac{mv^2}{R} \implies \frac{dv}{dt} = -\frac{\mu}{R}v^2\). Separate variables:
\(\int_{v_0}^v v^{-2} dv = -\frac{\mu}{R}\int_0^t dt \implies \left[ -\frac{1}{v} \right]_{v_0}^v = -\frac{\mu}{R}t \implies -\frac{1}{v} + \frac{1}{v_0} = -\frac{\mu}{R}t\)
Rearranging to isolate \(v(t)\):
\(\frac{1}{v} = \frac{1}{v_0} + \frac{\mu t}{R} = \frac{R + \mu v_0 t}{v_0 R} \implies \mathbf{v(t) = \frac{v_0}{1 + \left(\frac{\mu v_0}{R}\right)t}}\)
Rubric: 1 pt for \(\frac{dv}{dt} = -\frac{\mu}{R}v^2\); 2 pts for separation and \(-v^{-1}\); 1 pt for initial condition \(v(0)=v_0\); 2 pts for isolation.
LAP Discourse Key: Think-Pair-Share & Stop-and-Jot Model Answers
Both Answers in Red
LAP 1 3D Forces & Vectors
Think-Pair-Share: Vertical acceleration is zero (\(N_{table} = mg\)). Vertical circular wall supplies centripetal normal force radially inward, and friction acts tangentially backward.
Stop & Jot: Horizontal resultant: \(\vec{F} = \vec{N}_{wall} + \vec{f}_k = \frac{mv^2}{R}(-\hat{r}) + \mu N(-\hat{\theta})\). Vector points inward and backward.
LAP 2 Velocity-Dependent \(N\)
Think-Pair-Share: Friction acts against wall, where normal force is centripetal (\(N = mv^2/R\)), not gravitational. As block slows, speed drops, reducing \(N\) and weakening friction \(f_k = \mu N\).
Stop & Jot: Tangential equation: \(m a_t = -f_k = -\mu\left(\frac{mv^2}{R}\right) \implies \mathbf{a_t = \frac{dv}{dt} = -\frac{\mu v^2}{R}}\).
LAP 3 Calculus \(\int v^{-2}dv\)
Think-Pair-Share: Linear drag \(\propto -v\) yields \(\int v^{-1}dv = \ln v\) (exponential \(e^{-t}\)), whereas quadratic friction \(\propto -v^2\) yields \(\int v^{-2}dv = -v^{-1}\), creating an inverse rational function.
Stop & Jot: Initial condition: \(v(0) = \frac{v_0}{1+0} = v_0\). As \(t \to \infty\), denominator \(\to \infty\), so \(v(t) \to 0\) asymptotically.
Energy Dynamics Timed Assignment ⏱️ Question 1 • Page 1 of 3 Time Allocation: 24 min (Lap 1: 6m • Lap 2: 10m • Lap 3: 8m)
Question 1: Work & Energy FRQ (Cmech-2021s2-FR03)
Page 1 of 15
FRQ Time: 24 min
Name:
Date:
Start Time: Lap 1 Finish (6m):
1. A block of mass \(m\) is placed on top of an ideal spring of spring constant \(k\). The block is pushed against the spring, compressing the spring a distance \(\Delta x\). The block is released from rest, leaves the spring at the position shown, travels upward, and enters a frictionless track with constant radius of curvature \(R\). The block enters at point \(A\), maintains track contact, and exits horizontally at point \(B\), a distance \(3R\) above the release point. The block then falls to the ground, landing a horizontal distance \(D\) away. Express all algebraic answers in terms of \(m, k, \Delta x, R\), and fundamental constants. Block size is much smaller than \(R\).
LAP 1 Qualitative Forces & Track Contact Threshold
⏱️ Time: 6 min
Think-Pair-Share Prompt:
At apex point \(B\), why can't the normal force exerted by the track point upward? What is the physical threshold condition for the block to maintain contact with the track?
Stop and Jot Question:
Write the complete radial force equation \(\Sigma F_r = m a_c\) for the block at apex \(B\). What must the normal force \(N\) equal at critical speed, and what is the resulting expression for \(v_{crit}\)?
(a) On the dot below, draw and label the forces (not components) acting on the block at point \(B\).
Suggested: 2.5 min
Justify your choice of vectors:
⏱️ Question 1 • Page 2 of 3 LAP 2: Energy Conservation & Derivations
Question 1 Continued (Cmech-2021s2-FR03)
Page 2 of 15
⏱️ Section Time: LAP 2 & Derivations (b)–(c): 10 min Lap 2 Finish Check:
LAP 2 Energy Tracking & Critical Compression Threshold
⏱️ Time: 10 min
Think-Pair-Share Prompt:
Why is the total mechanical energy at apex \(B\) not purely gravitational potential energy? Where does the necessary kinetic energy \(\frac{1}{2}m v_B^2\) come from?
Stop and Jot Question:
How does substituting the critical apex speed \(v_{crit} = \sqrt{gR}\) into the conservation of mechanical energy equation yield the minimum spring compression \(\Delta x_{min} = \sqrt{7mgR/k}\)?
(b) Derivations at Point B
i. Derive an expression for the speed \(v\) of the block at point \(B\).
ii. Derive an expression for the magnitude of the net force \(F\) at point \(B\).
(c) Derive an expression for the minimum compression \(\Delta x_{min}\) required for the block to maintain track contact through point \(B\).
⏱️ Question 1 • Page 3 of 3 LAP 3: Synthesis, Graphing & Review
Question 1 Continued (Cmech-2021s2-FR03)
Page 3 of 15
LAP 3 Trajectory Limits & Graph Discontinuity
⏱️ Time: 5 min
Think-Pair-Share Prompt:
What physically happens to the block if \(\Delta x < \Delta x_{min}\)? At what point does it separate from the track and become a free projectile?
Stop and Jot Question:
Using the critical threshold velocity \(v_{crit} = \sqrt{gR}\) and the projectile fall time from height \(3R\), what is the exact algebraic and numerical minimum landing distance \(D_{MIN}\)?
Distance \(\Delta x > \Delta x_{min}\). (d) Calculate the landing distance \(D\) that the block travels as a projectile after exiting point \(B\).
(e) The graph below shows the best-fit line through student data of \(D\) vs \(\Delta x\).
i. Explain why there are no data for section I of the graph:
ii. Explain the reason for the shape and minimum value of section II:
⏱️ FRQ 1 Completed At: [ _____ min ] Time Limit: 24 min • Units Checked: [ ] [ ] FRQ 1 Complete
⏱️ Question 2 • Page 1 of 3 Time Allocation: 24 min (Lap 1: 6m • Lap 2: 10m • Lap 3: 8m)
Question 2: Loop & Spring FRQ (Cmech-2021s1-FR02)
Page 4 of 15
FRQ Time: 24 min
Name:
Date:
Start Time: Lap 1 Finish (6m):
2. A block of mass \(m\) starts from rest at point \(A\) and travels with negligible friction through the circular loop onto a horizontal surface, where the block contacts a spring of constant \(k = mg/2R\). All spring motion is horizontal. Point \(C\) is the highest point on the loop, and point \(B\) is the rightmost point on the loop. Express all algebraic answers in terms of \(m, h, R\), and fundamental constants.
LAP 1 Acceleration Vector Components at Point B
⏱️ Time: 6 min
Think-Pair-Share Prompt:
At point B, is the acceleration purely centripetal? What causes the tangential acceleration, and which direction does it act?
Stop and Jot Question:
Determine the perpendicular components of the block's acceleration vector at point \(B\). Why does the resultant acceleration vector point downward and to the left?
(a) Draw an arrow representing the direction of acceleration of the block at point \(B\).
Justify your answer:
(b) i. Derive an expression for the speed \(v\) of the block at point \(B\).
⏱️ Question 2 • Page 2 of 3 LAP 2: Apex Contact Threshold & Spring Compression
Question 2 Continued (Cmech-2021s1-FR02)
Page 5 of 15
(b) ii. Derive an expression for the magnitude of the net force \(F\) on the block at point \(B\).
LAP 2 Critical Apex Height & Spring Work-Energy
⏱️ Time: 10 min
Think-Pair-Share Prompt:
Why is \(h_{min} = 2.5R\) instead of \(2R\)? When the block strikes the spring, where does its mechanical energy go?
Stop and Jot Question:
Using conservation of energy between the release point and the fully compressed spring, how do you show that the maximum compression is \(x_{MAX} = 2\sqrt{hR}\)?
(c) In terms of \(R\), derive an expression for the minimum height \(h_{min}\) necessary for the block to maintain contact through point \(C\).
(d) Given \(h = 0.30\) m and \(R = 0.10\) m, calculate the maximum compression \(x_{MAX}\) of the spring.
⏱️ Question 2 • Page 3 of 3 LAP 3: Non-Linear Graph Analysis
Question 2 Continued (Cmech-2021s1-FR02)
Page 6 of 15
LAP 3 Non-Linear Dependence & Threshold Jump
⏱️ Time: 5 min
Think-Pair-Share Prompt:
Why is the compression strictly zero for \(h < h_{min}\)? Does the block fly off the track inside the loop?
Stop and Jot Question:
Why does the relationship \(x_{MAX} \propto \sqrt{h}\) produce an increasing graph with decreasing slope (concave down) in Section II?
(e) The graph below shows maximum spring compression \(x\) as a function of release height \(h\).
i. Explain why section I appears as a flat line on the horizontal axis:
ii. Explain the reason for the concave-down shape of section II:
⏱️ FRQ 2 Completed At: [ _____ min ] Cumulative Allowed: 48 min [ ] FRQ 1 & 2 Complete
⏱️ Question 3 • Page 1 of 3 Time Allocation: 24 min (Lap 1: 6m • Lap 2: 10m • Lap 3: 8m)
Question 3: Amusement Park Ride (Cmech-2011-FR02)
Page 7 of 15
FRQ Time: 24 min
Name:
Date:
Start Time: Lap 1 Finish (6m):
3. An amusement park ride features a passenger compartment of mass \(M\) released from rest at point \(A\). It free falls between \(A\) and \(B\) (distance \(3R/4\)), then moves along a circular arc of radius \(R\) between \(B\) and \(D\). The track is frictionless from \(A\) to \(D\), and compartment dimensions are negligible compared to \(R\).
LAP 1 Circular Arc Forces & Angle Resolution
⏱️ Time: 6 min
Think-Pair-Share Prompt:
At point C, what forces act on the compartment? How do you resolve gravity into radial and tangential components?
Stop and Jot Question:
Write Newton's second law in the radial direction for the passenger compartment at point \(C\). How does resolving gravity explain why \(F_c = N - Mg\cos\theta\)?
(a) Draw and label forces (not components) acting on passenger compartment at point \(C\) at angle \(\theta\).
(b) In terms of \(\theta\) and forces from part (a), express the magnitude of centripetal force at point \(C\).
⏱️ Question 3 • Page 2 of 3 LAP 2: Energy Dissipation & Friction Coefficient
Question 3 Continued (Cmech-2011-FR02)
Page 8 of 15
(c) Derive an expression for speed \(v_D\) of the passenger compartment as it reaches point \(D\).
LAP 2 Work-Energy Dissipation & Friction Stopping
⏱️ Time: 10 min
Think-Pair-Share Prompt:
Total vertical drop is \(3R/4 + R = 7R/4\). How does equating friction work \(-\mu Mg(3R)\) to \(-\frac{1}{2}Mv_D^2\) yield \(\mu = 7/12\)?
Stop and Jot Question:
How does applying the Work-Energy Theorem to the horizontal friction track of length \(3R\) allow you to calculate that the friction coefficient must be \(\mu = 7/12\)?
Brought to rest by friction between \(D\) and \(E\) (distance \(3R\)). (d) Calculate the numerical value of the coefficient of friction \(\mu\).
⏱️ Question 3 • Page 3 of 3 LAP 3: Velocity-Dependent Drag & Calculus
Question 3 Continued (Cmech-2011-FR02)
Page 9 of 15
LAP 3 Exponential Damping & Asymptotic Limits
⏱️ Time: 5 min
Think-Pair-Share Prompt:
Why does the compartment theoretically take infinite time to stop under \(-kv\) drag? What is its initial acceleration at \(t=0\)?
Stop and Jot Question:
What is the initial acceleration of the passenger compartment at \(t = 0\), and why does the acceleration asymptotically approach zero as time increases?
(e) In a modified trial, braking force is \(-kv\).
i. Differential equation for \(v(t)\) (do NOT solve):
ii. Solve differential equation for \(v(t)\):
iii. Sketch magnitude of acceleration \(a(t)\):
Magnitude of Acceleration \(a(t)\)
0 Time \(t\)
⏱️ FRQ 3 Completed At: [ _____ min ] Cumulative Allowed: 72 min [ ] FRQ 1, 2, & 3 Complete
⏱️ Question 4 • Page 1 of 3 Time Allocation: 24 min (Lap 1: 6m • Lap 2: 10m • Lap 3: 8m)
Question 4: Potential Energy Landscapes (Cmech-P2018-FR02)
Page 10 of 15
FRQ Time: 24 min
Name:
Date:
Start Time: Lap 1 Finish (6m):
4. Particle \(A\) moves in a two-particle system along the \(x\)-axis while particle \(B\) remains fixed. The graph shows the potential energy \(U(x)\) as a function of \(x\) for particle \(A\) (mass \(0.40\) kg). Particle \(A\) is released from rest at \(x = 2.0\) m. Positive direction is to the right.
LAP 1 Conservative Force from Potential Slope \(F = -dU/dx\)
⏱️ Time: 6 min
Think-Pair-Share Prompt:
Why is force equal to the NEGATIVE derivative of potential energy? What does a flat line on \(U(x)\) mean for force?
Stop and Jot Question:
How do you use the slope of the potential energy graph \(U(x)\) to find both the magnitude and direction of the conservative force and acceleration at \(x = 3.0\) m?
(a) Calculate the speed of particle \(A\) when it reaches \(x = 14.0\) m.
(b) Acceleration at \(x = 3.0\) m
i. Calculate magnitude of acceleration:
ii. Direction of acceleration:
Left Right
⏱️ Question 4 • Page 2 of 3 LAP 2: Piecewise Kinematics & Transit Times
Question 4 Continued (Cmech-P2018-FR02)
Page 11 of 15
(c) Determine the magnitude of acceleration of particle \(A\) at \(x = 7.0\) m.
LAP 2 Piecewise Kinematics & Energy Barriers
⏱️ Time: 10 min
Think-Pair-Share Prompt:
Between \(x = 6\) and \(8\) m, \(U = 0\). Is the particle at rest or moving at maximum constant speed?
Stop and Jot Question:
What is the motion of particle \(A\) in the flat region where \(U(x) = 0\), and how do you calculate the transit time across the interval from \(x = 6.0\) m to \(8.0\) m?
(d) Calculate the elapsed time for particle \(A\) to travel across each segment:
i. \(x = 2.0\) to \(6.0\) m:
ii. \(x = 6.0\) to \(8.0\) m:
iii. \(x = 8.0\) to \(14.0\) m:
(e) Target speed at \(x = 16.0\) m is \(1.0\) m/s. What initial speed should particle \(A\) be given at \(x = 7.0\) m?
⏱️ Question 4 • Page 3 of 3 LAP 3: Constructing Force Function F(x)
Question 4 Continued (Cmech-P2018-FR02)
Page 12 of 15
LAP 3 Constructing Step Force Functions
⏱️ Time: 5 min
Think-Pair-Share Prompt:
Why is the force graph made of flat horizontal steps? How does the signed area under \(F(x)\) relate to \(-\Delta U\)?
Stop and Jot Question:
Why is the force graph \(F(x)\) composed of piecewise constant horizontal steps, and how does the signed area under \(F(x)\) relate to the change in potential energy \(-\Delta U\)?
(f) Carefully sketch the graph of force \(F(x) = -dU/dx\) acting on particle \(A\) for \(0 < x < 16\) m on the coordinate axes below:
Suggested: 3.5 min
F (N) +3 +2 +1 0 -1 -2 -3 0 2 4 6 8 10 12 14 16 x (m)
Show Piecewise Slope Calculations (\(F = -dU/dx\)):
⏱️ FRQ 4 Completed At: [ _____ min ] Cumulative Allowed: 96 min [ ] FRQ 1–4 Complete
⏱️ Question 5 • Page 1 of 3 Time Allocation: 24 min (Lap 1: 6m • Lap 2: 10m • Lap 3: 8m)
Question 5: Circular Rough Wall Dynamics (Cmech-2014-FR02)
Page 13 of 15
FRQ Time: 24 min
Name:
Date:
Start Time: Lap 1 Finish (6m):
5. A block of mass \(m\) starts from rest at the top of a frictionless ramp at height \(h\) above a horizontal tabletop. It slides down, reaching point \(P\) with speed \(v_0\), then enters a circular vertical wall of radius \(R\). Tabletop is frictionless; wall has friction coefficient \(\mu\).
LAP 1 3D Forces: Tabletop Normal vs. Wall Friction
⏱️ Time: 6 min
Think-Pair-Share Prompt:
Why is vertical net force zero? What surface exerts centripetal normal force, and where does friction point?
Stop and Jot Question:
Why is the vertical net force on the block zero, and how do the circular wall and friction combine to determine the horizontal net force vector?
(a) Derive an expression for the height of the ramp \(h\) in terms of \(v_0, m\), and fundamental constants.
(b) i. Is the vertical component of the net force on the block upward, downward, or zero?
Upward Downward Zero
Justify your answer:
⏱️ Question 5 • Page 2 of 3 LAP 2: Velocity-Dependent Normal Force & Deceleration
Question 5 Continued (Cmech-2014-FR02)
Page 14 of 15
(b) ii. Draw an arrow starting on the block indicating the direction of horizontal component of net force:
Justify your answer:
LAP 2 Velocity-Dependent Normal Force & Deceleration
⏱️ Time: 10 min
Think-Pair-Share Prompt:
Why is \(N_{wall} = mv^2/R\) instead of \(mg\)? As the block decelerates, why does friction weaken?
Stop and Jot Question:
Why is the wall's normal force velocity-dependent (\(N = mv^2/R\)), and how does this lead to a tangential acceleration of \(a_t = -\frac{\mu v^2}{R}\)?
(c) Determine an expression for normal force \(N\) exerted on block by circular wall as function of \(v\).
(d) Derive an expression for magnitude of tangential acceleration \(a_t\) at speed \(v\).
⏱️ Question 5 • Page 3 of 3 LAP 3: Non-Linear Calculus Integration \(\int v^{-2} dv\)
Question 5 Continued (Cmech-2014-FR02)
Page 15 of 15
LAP 3 Non-Linear Calculus Integration \(\int v^{-2} dv\)
⏱️ Time: 5 min
Think-Pair-Share Prompt:
Why does \(\frac{dv}{dt} \propto -v^2\) yield a rational function \(v(t) = \frac{v_0}{1 + (\mu v_0 / R)t}\) instead of an exponential \(e^{-t}\)?
Stop and Jot Question:
Why does integrating \(\int v^{-2} dv\) produce a rational function \(v(t) = \frac{v_0}{1 + (\mu v_0 / R)t}\) rather than an exponential decay function?
(e) Derive an expression for \(v(t)\), the speed of the block as a function of time \(t\) after passing point \(P\). (Show complete separation of variables, antiderivatives, and initial conditions.)
⏱️ Full Assignment Time: [ _____ / 120 min ] Completed: [ ] 15 / 15 Pages [ ] Master Packet Complete
Energy Dynamics Solved Assignment ⏱️ Teacher Solved Master Key • 24 Min Pace All Answers Printed in Red
Question 1: Work & Energy FRQ (Cmech-2021s2-FR03)
Page 1 of 15
FRQ Time: 24 min
Name: TEACHER EXEMPLAR
Date: 2026-11-12
Start: 0:00 Lap 1 Finish: 6:00
1. A block of mass \(m\) is placed on top of an ideal spring of spring constant \(k\). The block is pushed against the spring, compressing the spring a distance \(\Delta x\). The block is released from rest, leaves the spring at the position shown, travels upward, and enters a frictionless track with constant radius of curvature \(R\). The block enters at point \(A\), maintains track contact, and exits horizontally at point \(B\), a distance \(3R\) above the release point. The block then falls to the ground, landing a horizontal distance \(D\) away. Express all algebraic answers in terms of \(m, k, \Delta x, R\), and fundamental constants. Block size is much smaller than \(R\).
LAP 1 Qualitative Forces & Track Contact Threshold
⏱️ Time: 6 min
Think-Pair-Share Prompt:
At apex point \(B\), why can't the normal force exerted by the track point upward? What is the physical threshold condition for the block to maintain contact with the track?
Model Discourse (Red): Track is above the block at apex B; as a rigid surface, it can only push downward toward the center of curvature (it cannot pull upward). Contact maintained requires \(N \ge 0\); at critical threshold, \(N = 0\).
Stop and Jot Question:
Write the complete radial force equation \(\Sigma F_r = m a_c\) for the block at apex \(B\). What must the normal force \(N\) equal at critical speed, and what is the resulting expression for \(v_{crit}\)?
Model Answer (Red): \(\Sigma F_r = N + mg = \frac{m v_B^2}{R}\). Threshold condition: \(N = 0 \implies mg = \frac{m v_{crit}^2}{R} \implies \mathbf{v_{crit} = \sqrt{gR}}\).
(a) On the dot below, draw and label the forces (not components) acting on the block at point \(B\).
3 Points
↓ \(F_g\) (\(mg\))
Justify your choice of vectors:
Gravity pulls vertically downward toward Earth (\(mg\)). Since the track surface is above the block at apex \(B\), the contact normal force \(F_N\) can only push downward toward the center of curvature; the track cannot pull upward.
⏱️ Question 1 • Page 2 of 3 LAP 2: Energy Conservation & Derivations
Question 1 Continued (Cmech-2021s2-FR03)
Page 2 of 15
⏱️ Section Time: LAP 2 & Derivations (b)–(c): 10 min Lap 2 Check: 16:00 elapsed
LAP 2 Energy Tracking & Critical Compression Threshold
⏱️ Time: 10 min
Think-Pair-Share Prompt:
Why is the total mechanical energy at apex \(B\) not purely gravitational potential energy? Where does the necessary kinetic energy \(\frac{1}{2}m v_B^2\) come from?
Model Discourse (Red): The block cannot be at rest at apex B; it must retain kinetic energy \(\frac{1}{2}m(gR)\) to maintain circular path contact without falling. Both the gravitational potential energy (\(3mgR\)) and kinetic energy originate from the initial elastic potential energy \(\frac{1}{2}k(\Delta x)^2\).
Stop and Jot Question:
How does substituting the critical apex speed \(v_{crit} = \sqrt{gR}\) into the conservation of mechanical energy equation yield the minimum spring compression \(\Delta x_{min} = \sqrt{7mgR/k}\)?
Model Answer (Red): \(\frac{1}{2}k(\Delta x_{min})^2 = mg(3R) + \frac{1}{2}m(gR) = \frac{7}{2}mgR \implies k(\Delta x_{min})^2 = 7mgR \implies \mathbf{\Delta x_{min} = \sqrt{\frac{7mgR}{k}}}\).
(b) Derivations at Point B
i. Derive speed \(v\) of block at point \(B\):
By conservation of mechanical energy (\(E_i = E_f\)):
ii. Derive magnitude of net force \(F\) at \(B\):
Circular motion at B is horizontal, so net force is centripetal:
(c) Derive an expression for the minimum compression \(\Delta x_{min}\) required for the block to maintain track contact through point \(B\).
Threshold condition requires \(N = 0 \implies mg = \frac{mv_B^2}{R} \implies v_{B,min}^2 = gR\).
⏱️ Question 1 • Page 3 of 3 LAP 3: Synthesis, Graphing & Review
Question 1 Continued (Cmech-2021s2-FR03)
Page 3 of 15
LAP 3 Trajectory Limits & Graph Discontinuity
⏱️ Time: 5 min
Think-Pair-Share Prompt:
What physically happens to the block if \(\Delta x < \Delta x_{min}\)? At what point does it separate from the track and become a free projectile?
Model Discourse (Red): If \(\Delta x < \Delta x_{min}\), normal force drops to zero prior to apex B. Gravity's radial component cannot supply the required centripetal acceleration; the block leaves track contact and free-falls inside, never executing a horizontal launch from B.
Stop and Jot Question:
Using the critical threshold velocity \(v_{crit} = \sqrt{gR}\) and the projectile fall time from height \(3R\), what is the exact algebraic and numerical minimum landing distance \(D_{MIN}\)?
Model Answer (Red): Fall time: \(3R = \frac{1}{2}gt^2 \implies t = \sqrt{6R/g}\). Minimum range: \(D_{MIN} = v_{crit} t = \sqrt{gR}\sqrt{\frac{6R}{g}} = \mathbf{\sqrt{6}R \approx 2.45R}\).
Distance \(\Delta x > \Delta x_{min}\). (d) Calculate the landing distance \(D\) that the block travels as a projectile after exiting point \(B\).
Horizontal launch from height \(y = 3R\) with initial velocity \(v_{x0} = v_B, v_{y0} = 0\):
(e) The graph below shows the best-fit line through student data of \(D\) vs \(\Delta x\).
i. Explain why there are no data for section I of the graph:
For \(\Delta x < \Delta x_{min}\), normal force drops to zero prior to B. The block leaves track contact and falls as a free-flight projectile before B, so no horizontal launch occurs.
ii. Explain the reason for the shape and minimum value of section II:
At threshold, \(D_{MIN} = \sqrt{6}R \approx 2.45R\). For \(\Delta x \gg \Delta x_{min}\), \(D \approx \sqrt{6kR/mg}\cdot \Delta x\), yielding an asymptotically linear graph with positive slope.
⏱️ FRQ 1 Finished: 22 min Pacing Target: 24 min • Self-Check: Passed [X] Solved Key Complete
⏱️ Teacher Solved Master Key • 24 Min Pace All Answers Printed in Red
Question 2: Loop & Spring FRQ (Cmech-2021s1-FR02)
Page 4 of 15
FRQ Time: 24 min
Name: TEACHER EXEMPLAR
Date: 2026-11-12
Start: 24:00 Lap 1 Finish: 30:00
2. A block of mass \(m\) starts from rest at point \(A\) and travels with negligible friction through the circular loop onto a horizontal surface, where the block contacts a spring of constant \(k = mg/2R\). All spring motion is horizontal. Point \(C\) is the highest point on the loop, and point \(B\) is the rightmost point on the loop. Express all algebraic answers in terms of \(m, h, R\), and fundamental constants.
LAP 1 Acceleration Vector Components at Point B
⏱️ Time: 6 min
Think-Pair-Share Prompt:
At point B, is the acceleration purely centripetal? What causes the tangential acceleration, and which direction does it act?
Model Discourse (Red): Motion is non-uniform circular motion; speed decreases as the block ascends. Centripetal acceleration \(\vec{a}_c = v^2/R\) points left toward loop center; downward gravity provides an unopposed tangential acceleration \(\vec{a}_t = \vec{g}\) downward.
Stop and Jot Question:
Determine the perpendicular components of the block's acceleration vector at point \(B\). Why does the resultant acceleration vector point downward and to the left?
Model Answer (Red): \(\vec{a} = \frac{v_B^2}{R}(-\hat{i}) + g(-\hat{j})\). Since both perpendicular components are negative, vector addition places the resultant strictly in Quadrant III (down and to the left).
(a) Draw an arrow representing the direction of acceleration of the block at point \(B\).
↙ (Down & Left)
Justify your answer:
Circular curvature requires a horizontal centripetal acceleration \(\vec{a}_c = v^2/R\) directed left toward the center. Downward gravity acts unopposed vertically, giving tangential acceleration \(\vec{a}_t = \vec{g}\) downward. Resultant sum is in Quadrant III.
(b) i. Derive an expression for the speed \(v\) of the block at point \(B\).
Conservation of energy between release height \(h\) and point \(B\) (height \(y_B = R\)):
⏱️ Question 2 • Page 2 of 3 LAP 2: Apex Contact Threshold & Spring Compression
Question 2 Continued (Cmech-2021s1-FR02)
Page 5 of 15
(b) ii. Derive an expression for the magnitude of the net force \(F\) on the block at point \(B\).
Horizontal centripetal force: \(F_x = F_c = \frac{mv_B^2}{R} = \frac{m(2g(h-R))}{R} = 2mg\left(\frac{h}{R}-1\right)\).
LAP 2 Critical Apex Height & Spring Work-Energy
⏱️ Time: 10 min
Think-Pair-Share Prompt:
Why is \(h_{min} = 2.5R\) instead of \(2R\)? When the block strikes the spring, where does its mechanical energy go?
Model Discourse (Red): At apex C (height \(2R\)), the block must retain critical speed \(v_C = \sqrt{gR}\) so \(N \ge 0\). This requires extra kinetic energy \(0.5mgR\), giving total release height \(2R + 0.5R = 2.5R\). On the horizontal track, all energy \(mgh\) transfers into spring elastic energy \(\frac{1}{2}kx^2\).
Stop and Jot Question:
Using conservation of energy between the release point and the fully compressed spring, how do you show that the maximum compression is \(x_{MAX} = 2\sqrt{hR}\)?
Model Answer (Red): \(mgh = \frac{1}{2}kx_{MAX}^2 = \frac{1}{2}\left(\frac{mg}{2R}\right)x_{MAX}^2 \implies mgh = \frac{mg}{4R}x_{MAX}^2 \implies x_{MAX}^2 = 4hR \implies \mathbf{x_{MAX} = 2\sqrt{hR}}\).
(c) In terms of \(R\), derive an expression for the minimum height \(h_{min}\) necessary for the block to maintain contact through point \(C\).
At apex C (height \(2R\)), critical threshold requires \(N_C = 0 \implies \frac{mv_C^2}{R} = mg \implies v_C^2 = gR\).
(d) Given \(h = 0.30\) m and \(R = 0.10\) m, calculate the maximum compression \(x_{MAX}\) of the spring.
Check threshold: \(h_{min} = 2.5(0.10) = 0.25\) m. Since \(0.30\text{ m} > 0.25\text{ m}\), block completes the loop.
⏱️ Question 2 • Page 3 of 3 LAP 3: Non-Linear Graph Analysis
Question 2 Continued (Cmech-2021s1-FR02)
Page 6 of 15
LAP 3 Non-Linear Dependence & Threshold Jump
⏱️ Time: 5 min
Think-Pair-Share Prompt:
Why is the compression strictly zero for \(h < h_{min}\)? Does the block fly off the track inside the loop?
Model Discourse (Red): If \(h < 2.5R\), normal force vanishes before apex C. The block separates from the circular track, falls inside the loop, and crashes, never reaching the horizontal spring on the right (\(x = 0\)).
Stop and Jot Question:
Why does the relationship \(x_{MAX} \propto \sqrt{h}\) produce an increasing graph with decreasing slope (concave down) in Section II?
Model Answer (Red): \(x = 2\sqrt{R}\cdot h^{1/2}\). First derivative \(\frac{dx}{dh} \propto h^{-1/2} > 0\) (increasing), but second derivative \(\frac{d^2x}{dh^2} \propto -h^{-3/2} < 0\) (decreasing slope). A negative second derivative produces strict concave-down curvature.
(e) The graph below shows maximum spring compression \(x\) as a function of release height \(h\).
i. Explain why section I appears as a flat line on the horizontal axis:
If \(h < h_{min}\), normal force drops to zero before apex C. The block separates from the circular track, falls inside the loop, and never reaches the horizontal spring, leaving \(x = 0\).
ii. Explain the reason for the concave-down shape of section II:
For \(h \ge h_{min}\), \(x_{MAX} = 2\sqrt{hR} \propto h^{1/2}\). Since compression scales as the square root of release height, the rate of increase decreases as \(h\) increases.
⏱️ FRQ 2 Finished: 46 min Cumulative Allowed: 48 min • Self-Check: Passed [X] Solved Key Complete
⏱️ Teacher Solved Master Key • 24 Min Pace All Answers Printed in Red
Question 3: Amusement Park Ride (Cmech-2011-FR02)
Page 7 of 15
FRQ Time: 24 min
Name: TEACHER EXEMPLAR
Date: 2026-11-12
Start: 48:00 Lap 1 Finish: 54:00
3. An amusement park ride features a passenger compartment of mass \(M\) released from rest at point \(A\). It free falls between \(A\) and \(B\) (distance \(3R/4\)), then moves along a circular arc of radius \(R\) between \(B\) and \(D\). The track is frictionless from \(A\) to \(D\), and compartment dimensions are negligible compared to \(R\).
LAP 1 Circular Arc Forces & Angle Resolution
⏱️ Time: 6 min
Think-Pair-Share Prompt:
At point C, what forces act on the compartment? How do you resolve gravity into radial and tangential components?
Model Discourse (Red): Only two forces act: track normal force \(\vec{N}\) pointing radially inward toward center, and gravity \(M\vec{g}\) vertically downward. Gravity decomposes into radial component \(Mg\cos\theta\) (outward) and tangential component \(Mg\sin\theta\) (forward).
Stop and Jot Question:
Write Newton's second law in the radial direction for the passenger compartment at point \(C\). How does resolving gravity explain why \(F_c = N - Mg\cos\theta\)?
Model Answer (Red): Taking inward as positive: \(\Sigma F_r = N - Mg\cos\theta = \frac{M v_C^2}{R}\). Normal force acts inward; radial gravity opposes normal force, yielding \(\mathbf{F_c = N - Mg\cos\theta}\).
(a) Draw and label forces (not components) acting on passenger compartment at point \(C\) at angle \(\theta\).
↖ \(F_N\) (Inward)
(b) In terms of \(\theta\) and forces from part (a), express the magnitude of centripetal force at point \(C\).
Radial balance: \(\mathbf{F_c = N - Mg\cos\theta}\) (or \(N - Mg\sin\theta\) if \(\theta\) is measured from the horizontal).
⏱️ Question 3 • Page 2 of 3 LAP 2: Energy Dissipation & Friction Coefficient
Question 3 Continued (Cmech-2011-FR02)
Page 8 of 15
(c) Derive an expression for speed \(v_D\) of the passenger compartment as it reaches point \(D\).
Total vertical drop from A to D: \(\Delta y = \frac{3}{4}R + R = \frac{7}{4}R\).
LAP 2 Work-Energy Dissipation & Friction Stopping
⏱️ Time: 10 min
Think-Pair-Share Prompt:
Total vertical drop is \(3R/4 + R = 7R/4\). How does equating friction work \(-\mu Mg(3R)\) to \(-\frac{1}{2}Mv_D^2\) yield \(\mu = 7/12\)?
Model Discourse (Red): Total mechanical energy gained during descent is \(Mg(7R/4)\). On the horizontal track, friction work \(-f_k d = -\mu Mg(3R)\) dissipates all kinetic energy: \(-3\mu MgR = -Mg(7R/4) \implies 3\mu = 7/4 \implies \mu = 7/12\).
Stop and Jot Question:
How does applying the Work-Energy Theorem to the horizontal friction track of length \(3R\) allow you to calculate that the friction coefficient must be \(\mu = 7/12\)?
Model Answer (Red): \(W_{net} = \Delta K \implies -\mu Mg(3R) = 0 - \frac{1}{2}Mv_D^2 = -Mg\left(\frac{7}{4}R\right) \implies 3\mu = \frac{7}{4} \implies \mathbf{\mu = \frac{7}{12} \approx 0.583}\).
Brought to rest by friction between \(D\) and \(E\) (distance \(3R\)). (d) Calculate the numerical value of the coefficient of friction \(\mu\).
\(W_{net} = \Delta K \implies -\mu Mg(3R) = 0 - \frac{1}{2}Mv_D^2\)
⏱️ Question 3 • Page 3 of 3 LAP 3: Velocity-Dependent Drag & Calculus
Question 3 Continued (Cmech-2011-FR02)
Page 9 of 15
LAP 3 Exponential Damping & Asymptotic Limits
⏱️ Time: 5 min
Think-Pair-Share Prompt:
Why does the compartment theoretically take infinite time to stop under \(-kv\) drag? What is its initial acceleration at \(t=0\)?
Model Discourse (Red): Because braking force \(-kv\) is proportional to velocity, as \(v \to 0\) the deceleration force continuously weakens, causing velocity to approach zero asymptotically as \(t \to \infty\). At \(t = 0\), initial acceleration magnitude is \(a(0) = \frac{kv_D}{M}\).
Stop and Jot Question:
What is the initial acceleration of the passenger compartment at \(t = 0\), and why does the acceleration asymptotically approach zero as time increases?
Model Answer (Red): Newton's 2nd Law gives \(M a(t) = -kv(t) \implies a(t) = \frac{kv_D}{M}e^{-(k/M)t}\). At \(t=0\), \(a(0) = \frac{kv_D}{M}\). Exponential decay has a horizontal asymptote at \(a = 0\).
(e) In a modified trial, braking force is \(-kv\).
i. Differential equation for \(v(t)\) (do NOT solve):
\(F_{net} = M \frac{dv}{dt} = -kv \implies \mathbf{\frac{dv}{dt} = -\frac{k}{M}v}\)
ii. Solve differential equation for \(v(t)\):
\(\int_{v_D}^v \frac{1}{v}dv = -\frac{k}{M}\int_0^t dt\)
iii. Sketch magnitude of acceleration \(a(t)\):
Magnitude of Acceleration \(a(t)\)
\(\frac{k v_D}{M}\)
0 Time \(t\)
Asymptote at \(a = 0\)
⏱️ FRQ 3 Finished: 70 min Cumulative Allowed: 72 min • Self-Check: Passed [X] Solved Key Complete
⏱️ Teacher Solved Master Key • 24 Min Pace All Answers Printed in Red
Question 4: Potential Energy Landscapes (Cmech-P2018-FR02)
Page 10 of 15
FRQ Time: 24 min
Name: TEACHER EXEMPLAR
Date: 2026-11-12
Start: 72:00 Lap 1 Finish: 78:00
4. Particle \(A\) moves in a two-particle system along the \(x\)-axis while particle \(B\) remains fixed. The graph shows the potential energy \(U(x)\) as a function of \(x\) for particle \(A\) (mass \(0.40\) kg). Particle \(A\) is released from rest at \(x = 2.0\) m. Positive direction is to the right.
LAP 1 Conservative Force from Potential Slope \(F = -dU/dx\)
⏱️ Time: 6 min
Think-Pair-Share Prompt:
Why is force equal to the NEGATIVE derivative of potential energy? What does a flat line on \(U(x)\) mean for force?
Model Discourse (Red): By definition, conservative work is \(W = -\Delta U = \int F dx\), so \(F = -\frac{dU}{dx}\). Forces push objects down the potential energy gradient toward lower energy states. A flat horizontal segment has zero slope (\(\frac{dU}{dx} = 0\)), so net force and acceleration are both zero.
Stop and Jot Question:
How do you use the slope of the potential energy graph \(U(x)\) to find both the magnitude and direction of the conservative force and acceleration at \(x = 3.0\) m?
Model Answer (Red): Slope on \([2,6]\): \(\frac{dU}{dx} = \frac{0-8}{6-2} = -2.0\) N. Therefore \(F = -(-2.0) = \mathbf{+2.0\text{ N}}\) (Right). Acceleration \(a = \frac{F}{m} = \frac{2.0\text{ N}}{0.40\text{ kg}} = \mathbf{5.0\text{ m/s}^2}\).
(a) Calculate the speed of particle \(A\) when it reaches \(x = 14.0\) m.
Total mechanical energy: \(E = U(2.0\text{ m}) = 8.0\text{ J}\).
(b) Acceleration at \(x = 3.0\) m
i. Calculate magnitude of acceleration:
\(F = -\frac{dU}{dx} = -(-2.0\text{ N}) = +2.0\text{ N}\)
ii. Direction of acceleration:
Left ✓ Right
⏱️ Question 4 • Page 2 of 3 LAP 2: Piecewise Kinematics & Transit Times
Question 4 Continued (Cmech-P2018-FR02)
Page 11 of 15
(c) Determine the magnitude of acceleration of particle \(A\) at \(x = 7.0\) m.
On interval \(x \in [6.0, 8.0]\) m, the potential energy is constant (\(U = 0\)).
LAP 2 Piecewise Kinematics & Energy Barriers
⏱️ Time: 10 min
Think-Pair-Share Prompt:
Between \(x = 6\) and \(8\) m, \(U = 0\). Is the particle at rest or moving at maximum constant speed?
Model Discourse (Red): The particle is NOT at rest. Total mechanical energy \(E = 8.0\) J is conserved. In the flat well where \(U = 0\), all mechanical energy is kinetic energy (\(K = 8.0\) J), so the particle moves at its maximum constant speed \(v_{max} = \sqrt{2(8.0)/0.40} \approx 6.32\) m/s.
Stop and Jot Question:
What is the motion of particle \(A\) in the flat region where \(U(x) = 0\), and how do you calculate the transit time across the interval from \(x = 6.0\) m to \(8.0\) m?
Model Answer (Red): Motion is constant velocity (\(a = 0\)) at \(v_{max} = \sqrt{40} \approx 6.32\) m/s. Transit time: \(\Delta t = \frac{\Delta x}{v_{max}} = \frac{8.0 - 6.0}{6.32} = \frac{2.0}{6.32} \approx \mathbf{0.316\text{ s}}\).
(d) Calculate the elapsed time for particle \(A\) to travel across each segment:
i. \(x = 2.0\) to \(6.0\) m:
\(v_i = 0, a = 5.0\text{ m/s}^2, d = 4.0\text{ m}\)
ii. \(x = 6.0\) to \(8.0\) m:
\(a = 0\), \(v = \sqrt{40} \approx 6.32\text{ m/s}\)
iii. \(x = 8.0\) to \(14.0\) m:
Slope \(= +1.0 \implies F = -1.0\text{ N}\)
(e) Target speed at \(x = 16.0\) m is \(1.0\) m/s. What initial speed should particle \(A\) be given at \(x = 7.0\) m?
At \(x = 16.0\) m: \(U = 6.0\) J, \(K = \frac{1}{2}(0.40)(1.0)^2 = 0.20\) J. Total energy required: \(E_{req} = 6.20\) J.
⏱️ Question 4 • Page 3 of 3 LAP 3: Constructing Force Function F(x)
Question 4 Continued (Cmech-P2018-FR02)
Page 12 of 15
LAP 3 Constructing Step Force Functions
⏱️ Time: 5 min
Think-Pair-Share Prompt:
Why is the force graph made of flat horizontal steps? How does the signed area under \(F(x)\) relate to \(-\Delta U\)?
Model Discourse (Red): Since \(U(x)\) is piecewise linear, its derivative \(\frac{dU}{dx}\) is constant on each interval, making \(F(x) = -\frac{dU}{dx}\) a piecewise constant step function. By calculus, signed area under \(F(x)\) equals work done: \(\int_{x_1}^{x_2} F dx = -\Delta U = U(x_1) - U(x_2)\).
Stop and Jot Question:
Why is the force graph \(F(x)\) composed of piecewise constant horizontal steps, and how does the signed area under \(F(x)\) relate to the change in potential energy \(-\Delta U\)?
Model Answer (Red): Step values: \([0,2]: F = +1.5\) N; \([2,6]: F = +2.0\) N; \([6,8]: F = 0\); \([8,14]: F = -1.0\) N; \([14,16]: F = 0\). Total area \(= 1.5(2) + 2(4) + 0 - 1(6) = 5.0\) J \(= -(3 - 8)\) J.
(f) Carefully sketch the graph of force \(F(x) = -dU/dx\) acting on particle \(A\) for \(0 < x < 16\) m on the coordinate axes below:
SOLVED PLOT IN RED
F (N) +3 +2 +1 0 -1 -2 -3 0 2 4 6 8 10 12 14 16 x (m)
Calculated Slope Values:
⏱️ FRQ 4 Finished: 93 min Cumulative Allowed: 96 min • Self-Check: Passed [X] Solved Key Complete
⏱️ Teacher Solved Master Key • 24 Min Pace All Answers Printed in Red
Question 5: Circular Rough Wall Dynamics (Cmech-2014-FR02)
Page 13 of 15
FRQ Time: 24 min
Name: TEACHER EXEMPLAR
Date: 2026-11-12
Start: 96:00 Lap 1 Finish: 102:00
5. A block of mass \(m\) starts from rest at the top of a frictionless ramp at height \(h\) above a horizontal tabletop. It slides down, reaching point \(P\) with speed \(v_0\), then enters a circular vertical wall of radius \(R\). Tabletop is frictionless; wall has friction coefficient \(\mu\).
LAP 1 3D Forces: Tabletop Normal vs. Wall Friction
⏱️ Time: 6 min
Think-Pair-Share Prompt:
Why is vertical net force zero? What surface exerts centripetal normal force, and where does friction point?
Model Discourse (Red): Vertical acceleration is zero because the block remains flat on the tabletop; table normal force balances gravity (\(N_{table} = mg \implies \Sigma F_y = 0\)). The vertical circular wall supplies centripetal normal force radially inward (\(mv^2/R\)), and friction points tangentially backward opposing motion.
Stop and Jot Question:
Why is the vertical net force on the block zero, and how do the circular wall and friction combine to determine the horizontal net force vector?
Model Answer (Red): Horizontal net force has perpendicular components: inward radial normal force \(N = mv^2/R\) and backward tangential kinetic friction \(f_k = \mu N\). Vector sum points inward and backward into Quadrant III relative to velocity.
(a) Derive an expression for the height of the ramp \(h\) in terms of \(v_0, m\), and fundamental constants.
Energy conservation along the smooth frictionless ramp:
(b) i. Is the vertical component of the net force on the block upward, downward, or zero?
Upward Downward ✓ Zero
Justify your answer:
The block moves strictly in the horizontal plane on the flat tabletop. There is no vertical acceleration (\(a_y = 0\)). The upward normal force exerted by the tabletop balances the downward gravitational force (\(N_{table} = mg\)), so the vertical net force is zero.
⏱️ Question 5 • Page 2 of 3 LAP 2: Velocity-Dependent Normal Force & Deceleration
Question 5 Continued (Cmech-2014-FR02)
Page 14 of 15
(b) ii. Draw an arrow starting on the block indicating the direction of horizontal component of net force:
↙ Inward & Backward
Justify your answer:
Normal force from the wall pushes radially inward (\(mv^2/R\)), providing centripetal acceleration. Kinetic friction opposes tangential velocity, acting backward (\(f_k = \mu N\)). Vector sum points inward and backward into the 3rd quadrant relative to velocity.
LAP 2 Velocity-Dependent Normal Force & Deceleration
⏱️ Time: 10 min
Think-Pair-Share Prompt:
Why is \(N_{wall} = mv^2/R\) instead of \(mg\)? As the block decelerates, why does friction weaken?
Model Discourse (Red): The tabletop is frictionless, so friction only acts against the vertical circular wall. The wall's normal force is purely centripetal (\(N = mv^2/R\)), depending on speed rather than gravity. As the block decelerates, \(v\) decreases, reducing \(N\) and weakening friction \(f_k = \mu N\).
Stop and Jot Question:
Why is the wall's normal force velocity-dependent (\(N = mv^2/R\)), and how does this lead to a tangential acceleration of \(a_t = -\frac{\mu v^2}{R}\)?
Model Answer (Red): Wall provides centripetal force: \(N = \frac{mv^2}{R}\). Friction opposes velocity: \(f_k = \mu N = \mu \frac{mv^2}{R}\). Newton's 2nd Law gives \(m a_t = -f_k \implies \mathbf{a_t = \frac{dv}{dt} = -\frac{\mu v^2}{R}}\).
(c) Determine an expression for normal force \(N\) exerted on block by circular wall as function of \(v\).
The circular wall is the sole surface exerting a radial contact force to maintain circular motion of radius \(R\):
(d) Derive an expression for magnitude of tangential acceleration \(a_t\) at speed \(v\).
Kinetic friction opposes tangential motion: \(f_k = \mu N(v) = \mu \frac{mv^2}{R}\).
⏱️ Question 5 • Page 3 of 3 LAP 3: Non-Linear Calculus Integration \(\int v^{-2} dv\)
Question 5 Continued (Cmech-2014-FR02)
Page 15 of 15
LAP 3 Non-Linear Calculus Integration \(\int v^{-2} dv\)
⏱️ Time: 5 min
Think-Pair-Share Prompt:
Why does \(\frac{dv}{dt} \propto -v^2\) yield a rational function \(v(t) = \frac{v_0}{1 + (\mu v_0 / R)t}\) instead of an exponential \(e^{-t}\)?
Model Discourse (Red): Linear drag \(\frac{dv}{dt} \propto -v\) integrates to \(\int v^{-1}dv = \ln v\), giving exponential decay \(e^{-t}\). Quadratic friction \(\frac{dv}{dt} \propto -v^2\) integrates to \(\int v^{-2}dv = -v^{-1}\), creating an inverse linear relation \(\frac{1}{v} = \frac{1}{v_0} + Ct\) and a rational function.
Stop and Jot Question:
Why does integrating \(\int v^{-2} dv\) produce a rational function \(v(t) = \frac{v_0}{1 + (\mu v_0 / R)t}\) rather than an exponential decay function?
Model Answer (Red): Antiderivative of power \(v^{-2}\) is \(-v^{-1}\). Integrating from \(v_0\) to \(v\) yields \(-\frac{1}{v} + \frac{1}{v_0} = -\frac{\mu}{R}t\). Inverting isolates \(\mathbf{v(t) = \frac{v_0}{1 + (\mu v_0 / R)t}}\). Initial condition \(v(0) = v_0\) and asymptotic limit \(v \to 0\) as \(t \to \infty\) are both satisfied.
(e) Derive an expression for \(v(t)\), the speed of the block as a function of time \(t\) after passing point \(P\). (Show complete separation of variables, antiderivatives, and initial conditions.)
Step 1: Formulate Differential Equation
Step 2: Separate Variables
Step 3: Integrate with Definite Limits
Step 4: Solve Algebraically for \(v(t)\)
⏱️ Full Assignment Solved Key: 116 min / 120 min Completed: 15 / 15 Pages • All 5 FRQs [X] Solved Teacher Exemplar Complete
