14.4 Strategy Selection Across Force, Energy, and Momentum Methods

Three Students, One Problem

Here is a problem:

A 0.50 kg ball hangs from a 1.2 m string. You pull it to an angle of 60 degrees from vertical and release it. At the bottom of its swing, the ball collides with a 1.5 kg block sitting on a rough horizontal surface ($\mu_k = 0.30$). The collision is perfectly inelastic. How far does the block slide before stopping?

Three students solve it. All three get the correct answer.

Student A uses forces throughout. She decomposes gravity along the arc, writes Newton's second law in polar coordinates to find the speed at the bottom of the swing, then models the collision with forces (estimating impulse duration and contact forces), then applies kinetic friction and kinematics to find the stopping distance. Her solution covers two pages and takes twenty minutes. She gets the right number, but she had to estimate the collision duration, and her polar-coordinate work on the pendulum required careful attention to centripetal terms.

Student B uses energy conservation for the pendulum swing, then momentum conservation for the collision, then the work-energy theorem for the sliding block. Three clean stages, three clean equations. Her solution fits on half a page and takes three minutes.

Student C uses energy conservation for the entire problem. She writes $mgh = \frac{1}{2}(m + M)v_f^2 + \mu_k(m + M)gd$ --- treating the whole process as a single energy equation. She gets the wrong answer. The collision destroyed kinetic energy, and she missed it.

Before you read on: Why did Student B finish in three minutes while Student A took twenty? And what specifically went wrong with Student C's approach?

Think about what each method needs as input and what it provides as output. Then continue.

Student B was fast because she matched each stage of the problem to the tool designed for that stage. The pendulum swing conserves mechanical energy (no friction, no collision), so energy gives the speed in one equation. The collision involves large unknown internal forces over a short time, so momentum conservation bypasses those forces entirely. The rough-surface slide dissipates energy through friction, so the work-energy theorem connects the post-collision kinetic energy to the stopping distance. Each tool was chosen for a reason. None was forced to do something it was not built to do.

Student A was slow because she used forces everywhere --- even in stages where forces are unknown or unnecessary. The collision stage was especially painful: she needed the contact force as a function of time, which she did not have and had to estimate.

Student C was wrong because she ignored the fact that the collision destroys kinetic energy. Energy is not conserved across a perfectly inelastic collision. She applied a conservation law across a boundary where it does not hold.

The difference between these three students is not knowledge. All three knew the same formulas. The difference is method selection --- the ability to read a problem, identify its structure, and choose the right tool for each part.

That is the subject of this section. And it is arguably the most important skill in this entire course.

Prediction

Commit to a strategy before you see the solution.

A uniform solid ball rolls without slipping along a horizontal surface at speed $v_0$. It reaches the bottom of a ramp and rolls up. At the top of the ramp, the ball launches off the edge and becomes a projectile, eventually landing on the ground below.

To find the ball's speed when it hits the ground, which method(s) do you need? In what order? Write down your proposed strategy --- which tools, applied to which stages --- before reading further.

[Interactive: Strategy Planner. Students see the problem scenario illustrated with three labeled stages: (1) rolling up the ramp, (2) projectile flight, (3) finding the landing speed. For each stage, students drag and drop method cards --- "Forces/Torques," "Energy Conservation," "Momentum Conservation," "Kinematics" --- into the stage box. After submitting, the optimal strategy is revealed with explanations.]

Here is the expert strategy:

Stage 1 (rolling up the ramp): Energy conservation. The ball rolls without slipping, so no energy is lost to friction. The question asks for speed at the top of the ramp --- not time, not acceleration, not friction. Energy gives you $v_{\text{top}}$ directly: $\frac{1}{2}mv_0^2 + \frac{1}{2}I\omega_0^2 = \frac{1}{2}mv_{\text{top}}^2 + \frac{1}{2}I\omega_{\text{top}}^2 + mgh$.

Stage 2 (projectile flight): Once the ball leaves the ramp, there is no surface contact and therefore no rolling constraint. Gravity is the only force. Here, you need to find the landing speed. You could use kinematics (resolving into components and tracking $v_x$ and $v_y$), or you could use energy conservation again: $\frac{1}{2}mv_{\text{top}}^2 + mgh_{\text{launch}} = \frac{1}{2}mv_{\text{land}}^2 + mgh_{\text{ground}}$. But notice: during the flight, the ball still spins at the same $\omega$ it had at the top of the ramp (no torque acts on it in the air). That rotational kinetic energy is still there. If the problem asks for the translational speed at landing, the rotational KE just comes along for the ride but does not convert into translational KE (there is no rolling constraint in free flight). You need to track translational and rotational energy separately.

What does not work here: Momentum conservation is useless --- there are external forces (gravity) at every stage. Forces and torques could solve stage 1, but with much more algebra. Applying energy across the entire problem works only if you carefully track what happens to rotational kinetic energy during the transition from rolling to free flight.

The point: the strategy matters as much as the execution.

The Guiding Question

How do experts decide which mechanics tool to reach for first?

This question has been building across the entire course. In Section 7.7, you compared forces and energy for translational problems. In Section 11.6, you added rotational tools to the comparison. In Section 14.3, you learned to decompose multi-stage problems and choose tools for each stage.

This section pulls it all together. The goal is not to learn new physics. There is none. The goal is to build a reliable, explicit decision process that you can apply to any mechanics problem.

Exploration: The Decision Tree

Here is a structured set of questions that, when asked in order, guides you toward the best method for any mechanics problem or problem-stage.

[Interactive: Strategy Decision Tree. The student sees a flowchart with yes/no branch points. At each node, a question is displayed. The student clicks "Yes" or "No" and the tree branches. At the leaves, a recommended method appears with a short justification. The student can input a problem description and walk through the tree, then compare their tree-recommended method to the one an expert would choose. Below the tree, five practice problems are listed; students walk each one through the tree and check against the expert recommendation.]

The decision process has five questions. Ask them in this order:

Question 1: What does the problem ask for?

This is always the first question. The answer immediately eliminates some methods and favors others.

If the problem asks for...	Then reach for...	Because...
Speed or distance	Energy methods	Energy directly connects force, displacement, and speed --- no time needed
Time	Forces + kinematics	Energy and momentum methods do not involve time
Acceleration	Forces ($\sum F = ma$) or torques ($\sum \tau = I\alpha$)	Acceleration is the direct output of Newton's second law
A specific force (normal, tension, friction)	Forces	Energy and momentum methods bypass individual forces
Velocity after a collision	Momentum conservation	Collisions involve unknown forces over short times; momentum handles them cleanly
Angular velocity after a reconfiguration	Angular momentum conservation	When the system changes shape with no external torques

Question 2: Is there a collision or explosion?

If yes, that stage almost certainly requires momentum conservation. Collisions involve forces that are huge, brief, and unknown in detail. Neither the force method (you do not know the force) nor the energy method (kinetic energy is typically not conserved) can handle them directly. Momentum conservation works precisely because it does not require knowing the forces --- only that they are internal.

After the collision, you will typically switch to energy or forces for the next stage.

Question 3: Is mechanical energy conserved in this stage?

Check: - Are the only forces conservative (gravity, springs)? - If friction acts, is it static friction during rolling (which does no work)? - Is there no collision in this stage?

If energy is conserved, and the problem asks for speed or distance, energy conservation is almost always the fastest path.

If energy is not conserved (kinetic friction does work, or a collision dissipates energy), you can still use the work-energy theorem --- but you must account for the nonconservative work: $W_{\text{nc}} = \Delta KE + \Delta PE$. This is still often faster than forces, especially for multi-phase paths.

Question 4: Is angular momentum conserved?

Check: is the net external torque about some axis zero? This is true when: - The system is isolated from external torques (e.g., spinning objects in space) - All external forces pass through the chosen axis (so they produce no torque about it) - Internal forces reconfigure the system (skater pulling arms in, disks locking together)

If angular momentum is conserved and the problem asks for an angular velocity, this is your method. One equation, one unknown, done.

Question 5: Is rotation involved?

If the object rolls, spins, or rotates: - Include $\frac{1}{2}I\omega^2$ in any energy accounting - Use the rolling constraint ($v = R\omega$) to link translational and rotational quantities - Consider whether you need $\sum \tau = I\alpha$ for angular acceleration - Check whether the rolling constraint holds at every stage (it breaks in free flight or when slipping occurs)

The Heuristics, Stated Plainly

Here are the method-selection rules distilled to their sharpest form:

Need speed but not time? Use energy.

Energy conservation ($KE_i + PE_i = KE_f + PE_f$) or the work-energy theorem ($W_{\text{net}} = \Delta KE$) connects speeds and positions without ever involving time. If the problem asks "how fast" or "how far," energy is your first candidate.

Need time or acceleration? Use forces.

Newton's second law ($\sum F = ma$, $\sum \tau = I\alpha$) plus kinematics is the only tool that gives you access to time. The kinematic equations --- $v = v_0 + at$, $x = x_0 + v_0 t + \frac{1}{2}at^2$, etc. --- require knowing the acceleration, and that comes from forces.

Collision or explosion? Use momentum.

When objects collide, stick together, or fly apart, the internal forces are unknown. Momentum conservation ($\vec{p}_i = \vec{p}_f$ for isolated systems, or angular momentum $L_i = L_f$ for systems with no external torque) bypasses those forces entirely.

Need to bypass unknown internal forces? Use momentum or energy.

Both methods are derived from Newton's second law by integrating --- momentum by integrating over time, energy by integrating over displacement. The integration eliminates the need to know the forces in detail. Whenever a problem involves forces you cannot write down (collision forces, internal tension in a breaking rope, friction between locking disks), these integral methods let you skip past the unknown.

Rotation involved? Check for rolling constraints and use rotational analogs.

If the object rolls, include both translational and rotational kinetic energy. If the problem asks for angular acceleration or torque, use $\sum \tau = I\alpha$. If there is no external torque, angular momentum may be conserved. Always check whether the rolling constraint $v = R\omega$ holds throughout the problem or only during some stages.

Putting the Heuristics to Work

Pause and think: Before reading the worked examples below, go back to the five questions in the decision tree and apply them to the opening problem (ball on string, collision with block, block slides to stop). Do your tree-recommended methods match Student B's three-stage approach?

Worked Example: Ramp, Collision, and Slide

A 2.0 kg solid sphere ($I = \frac{2}{5}mR^2$, $R = 0.10$ m) starts from rest at the top of a frictionless ramp of height $h = 1.5$ m and rolls without slipping to the bottom. At the bottom, it collides with a 3.0 kg block that is free to slide on a rough horizontal surface ($\mu_k = 0.25$). The collision is perfectly inelastic, and after the collision the sphere does not roll --- it and the block move together as a single sliding object. Find the distance the combined object slides before stopping.

Step 0: Read the problem. Identify the stages.

Stage 1: Sphere rolls down the ramp (rolling without slipping, no energy loss). Stage 2: Sphere collides with block (perfectly inelastic collision). Stage 3: Combined object slides on rough surface to a stop.

Step 1: Choose the method for each stage.

Stage 1 asks for speed at the bottom --- energy conservation. Energy is conserved because the sphere rolls without slipping on a frictionless ramp.

Stage 2 involves a collision with unknown internal forces --- momentum conservation. Kinetic energy is not conserved (perfectly inelastic).

Stage 3 asks for a distance with friction present --- work-energy theorem.

Step 2: Solve each stage.

Stage 1: Energy conservation.

$$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{2}{5}mR^2\right)\frac{v^2}{R^2} = \frac{7}{10}mv^2$$

$$v_1 = \sqrt{\frac{10gh}{7}} = \sqrt{\frac{10(9.8)(1.5)}{7}} = \sqrt{21.0} = 4.58 \text{ m/s}$$

Stage 2: Momentum conservation.

Before the collision, the sphere has translational momentum $mv_1$. The block is at rest. After the collision, they move together at speed $v_2$.

$$mv_1 = (m + M)v_2$$

$$v_2 = \frac{m}{m + M}v_1 = \frac{2.0}{5.0}(4.58) = 1.83 \text{ m/s}$$

Note: the sphere's rotational kinetic energy is lost in the collision (it stops rolling and the combined object slides). This is part of the energy dissipated. Momentum conservation handles the collision without needing to track where the energy went.

Stage 3: Work-energy theorem.

The combined object has kinetic energy $\frac{1}{2}(m + M)v_2^2$ and slides to a stop. Friction does negative work: $W_{\text{fric}} = -\mu_k(m + M)g \cdot d$.

$$0 - \frac{1}{2}(m + M)v_2^2 = -\mu_k(m + M)g \cdot d$$

$$d = \frac{v_2^2}{2\mu_k g} = \frac{(1.83)^2}{2(0.25)(9.8)} = \frac{3.35}{4.90} = 0.68 \text{ m}$$

The block slides 0.68 m.

Three stages, three methods, three equations. Each method was chosen because its outputs matched what that stage required. No wasted effort. No unknown forces to estimate.

Check your intuition: Go back to your prediction from the start of this section. Did you identify these three stages and these three methods? If not, at which point did your strategy diverge from the expert approach?

Common Method-Selection Mistakes

Students who have learned all the individual tools sometimes still misapply them. Here are the most common errors and how to avoid them.

Mistake 1: Applying energy conservation across a collision

Student C from the opening example made this error. It is the most common strategy mistake in multi-stage problems.

The rule: Mechanical energy is not conserved in inelastic collisions. If your problem includes a collision, you must use momentum conservation for that stage and restart your energy accounting after the collision with the post-collision kinetic energy.

Even in elastic collisions, you typically use momentum conservation together with the elastic condition ($KE_i = KE_f$) as a system of equations. Energy alone does not determine the individual velocities after the collision.

Mistake 2: Using forces when the forces are unknown

In a collision, the contact forces are enormous and last for milliseconds. You do not know the force as a function of time, so $\sum F = ma$ is useless in practice (even though it is correct in principle). Momentum conservation was invented precisely for this situation.

Similarly, if a rope breaks and an internal tension disappears, the force method requires knowing the tension history. Momentum conservation does not.

Mistake 3: Using energy when the problem asks for time

Energy methods connect speeds and positions. They never involve time. If the problem says "how long," you need forces and kinematics. No shortcut exists. You must find the acceleration first.

Mistake 4: Forgetting rotational kinetic energy

When a rolling object is involved, its total kinetic energy is $\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$. Forgetting the rotational term gives a speed that is too high (as you saw in Section 11.6, Practice Problem 4). For a hoop, the error is about 41%. That is not a rounding issue --- it is a conceptual error about where the energy goes.

Mistake 5: Assuming the rolling constraint holds everywhere

A ball that rolls on a surface and then launches into the air keeps spinning at the same $\omega$ it had on the surface. But the rolling constraint $v = R\omega$ no longer holds in free flight, because there is no surface to enforce it. Similarly, after a collision, a rolling object may start sliding instead of rolling if the collision changes $v$ and $\omega$ in incompatible ways.

The Full Toolkit

Here is the complete method-selection table, bringing together everything from Chapters 4 through 14.

Method	Core equation	What it gives you	What it cannot give you	Use when...
Newton's 2nd (translation)	$\sum \vec{F} = m\vec{a}$	Acceleration, specific forces, time (via kinematics)	Speed from complex paths efficiently	You need $a$, time, or a specific force value
Newton's 2nd (rotation)	$\sum \vec{\tau} = I\vec{\alpha}$	Angular acceleration, specific torques	Speed from complex rotation efficiently	You need $\alpha$, angular timing, or a specific torque
Energy conservation	$KE_i + PE_i = KE_f + PE_f$	Speed, distance, height	Time, acceleration, individual forces	No energy lost, asking for speed/distance
Work-energy theorem	$W_{\text{net}} = \Delta KE$	Speed, distance (even with friction)	Time, acceleration, individual forces	Nonconservative forces present, asking for speed/distance
Momentum conservation	$\vec{p}_i = \vec{p}_f$	Velocity after collision/explosion	Anything about the collision duration or forces	System isolated during the interaction, or impulse is known
Angular momentum conservation	$L_i = L_f$	Angular velocity after reconfiguration	Torques during the reconfiguration	No external torque about the chosen axis

Every method in this table is a repackaging of Newton's second law. Forces give you the raw equation. Energy comes from integrating $\sum F = ma$ over displacement. Momentum comes from integrating $\sum F = ma$ over time. The rotational versions come from $\sum \tau = I\alpha$ with the same two integrations. Each integration trades detailed information (the force as a function of time or position) for a simpler equation that skips directly to the quantity you want.

The expert's advantage is not knowing more equations. It is knowing which integration has already been done for you and what it costs.

Connection: Completing the Toolkit

This section completes a thread that has been running since Section 7.7.

In Section 7.7, you had two tools --- forces and energy --- and learned that energy wins when the problem asks for speed, while forces win when it asks for time. That was the first time the course asked you to choose rather than just apply.

In Section 11.6, the toolkit doubled. Rotational kinetic energy, angular momentum conservation, and the force/torque approach for rotation gave you four methods. The decision logic stayed the same --- match the method to the output the problem demands --- but the number of options grew.

In Section 14.3, you learned to decompose multi-stage problems. The key insight: different stages of the same problem may require different methods. The transitions between stages are where the bookkeeping matters most.

This section brings all of that together into a single decision framework. There are no new physics concepts here. The entire value is organizational. You are learning to see the structure of a problem before you solve it, and to make the strategic choice before you write the first equation.

This is the meta-skill of mechanics. Everything you have learned since Chapter 1 is a tool. This section teaches you how to decide which tool to pick up.

Teaching Emphasis: Method Selection Is Not an Afterthought

In many courses, students learn forces first, then energy, then momentum, and each tool is practiced in isolation. Method selection --- if it is taught at all --- comes at the end of the semester as a review topic. The implicit message is that the formulas are the main content, and choosing between them is a minor organizational detail.

That message is wrong. In novel problems --- problems you have never seen before, which is every problem on an exam and every problem in the real world --- the hardest part is not executing the algebra. It is deciding what to do in the first place. The student who can follow a worked example but freezes on an unfamiliar problem is not missing a formula. They are missing the strategy layer.

The ability to choose the right method is what separates a student who can follow worked examples from one who can solve novel problems.

Treat the decision process in this section as a first-class skill, not a summary to glance at before an exam. Practice it explicitly. Make it automatic.

Spaced Retrieval

Before moving to practice, recall the key ideas from the chapters that built these tools.

Recall prompt 1: What is the work-energy theorem? State it in words and in an equation. (Section 7.3)

Recall prompt 2: When is linear momentum conserved? State the precise condition. (Section 9.2)

Recall prompt 3: When is angular momentum conserved? State the precise condition. (Section 11.5)

Recall prompt 4: What quantities can the force method provide that the energy method cannot? (Section 7.7)

Recall prompt 5: For a rolling object, what is the total kinetic energy? What constraint links $v$ and $\omega$? (Section 11.1)

Practice Layers

Layer 1: Concrete --- Solve Problems with Deliberate Method Choice

For each problem below, first state which method(s) you will use and why, then solve.

Problem 1. A 0.40 kg ball is dropped from rest at a height of 2.0 m onto a vertical spring ($k = 500$ N/m) resting on the floor. How far does the spring compress?

Before solving: Does this problem involve a collision? Is energy conserved? What does the problem ask for?

Strategy check

No collision. The ball-spring-Earth system conserves mechanical energy (gravity is conservative, the spring force is conservative, and we ignore air resistance). The problem asks for a distance. **Energy conservation** is the right tool.

Full solution

Choose the reference level for gravitational PE at the top of the uncompressed spring. If the spring compresses by $x$, the ball falls a total distance of $h + x = 2.0 + x$. Energy conservation (taking the compressed position as the final state, where $v = 0$): $$mg(h + x) = \frac{1}{2}kx^2$$ $$(0.40)(9.8)(2.0 + x) = \frac{1}{2}(500)x^2$$ $$3.92(2.0 + x) = 250x^2$$ $$250x^2 - 3.92x - 7.84 = 0$$ Using the quadratic formula: $$x = \frac{3.92 + \sqrt{(3.92)^2 + 4(250)(7.84)}}{2(250)} = \frac{3.92 + \sqrt{15.4 + 7840}}{500} = \frac{3.92 + 88.6}{500} = 0.185 \text{ m}$$ The spring compresses about 0.19 m (19 cm). Note that using forces here would require solving a differential equation (the spring force changes with compression). Energy made it a single algebraic equation.

Problem 2. Two ice skaters, initially at rest, push off each other. Skater A (mass 60 kg) moves to the left at 1.5 m/s. Skater B has mass 80 kg. Find Skater B's velocity.

Before solving: Is this a collision, an explosion, or neither? Are the push forces known? What is conserved?

Strategy check

This is an "explosion" (two objects initially together, separating). The push forces are internal and unknown. The system is isolated horizontally (ice is frictionless). **Momentum conservation** is the right tool.

Full solution

Initial momentum: $\vec{p}_i = 0$ (both at rest). Final momentum: $\vec{p}_f = m_A v_A + m_B v_B$. Taking leftward as negative: $$0 = (60)(-1.5) + (80)v_B$$ $$v_B = \frac{90}{80} = 1.125 \text{ m/s (to the right)}$$ One equation, one unknown. Attempting this with forces would require knowing the push force as a function of time --- which we do not have.

Problem 3. A solid cylinder ($I = \frac{1}{2}mR^2$) of mass 3.0 kg and radius 0.15 m rolls without slipping down a ramp of height 2.5 m and then across a rough horizontal surface ($\mu_k = 0.20$). At the bottom of the ramp, the cylinder transitions from rolling to sliding (the rough surface has insufficient friction to maintain rolling). How far does the cylinder slide before stopping?

Before solving: How many stages does this problem have? What method fits each stage? Be careful about what happens to the rotational kinetic energy.

Strategy check

Two stages. Stage 1 (rolling down the ramp): **energy conservation** gives the speed at the bottom. Stage 2 (sliding on the rough surface): the cylinder is now sliding, not rolling, so the rotational kinetic energy is gradually dissipated by friction. The work-energy theorem must account for both the translational and rotational kinetic energy being converted to heat by friction. Actually, this requires careful thought. When the cylinder transitions from rolling to sliding, both translational and rotational friction act. If we treat the entire post-ramp motion as one stage, the **work-energy theorem** gives: friction does work equal to the total kinetic energy (translational plus rotational) that the cylinder had at the bottom. The total KE at the bottom of the ramp equals $mgh$ (from stage 1). So friction must dissipate $mgh$ worth of energy: $\mu_k mg \cdot d = mgh$, giving $d = h / \mu_k$. Wait --- that is only correct if all kinetic energy (translational and rotational) is eventually dissipated by the same friction force doing work over distance $d$. The sliding friction decelerates the translational motion and simultaneously exerts a torque that decelerates the rotation. Both are dissipated over the sliding distance, so the total work done by friction equals the total KE.

Full solution

*Stage 1: Rolling down the ramp.* $$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = \frac{3}{4}mv^2$$ (using $I = \frac{1}{2}mR^2$ and $v = R\omega$). Total KE at bottom $= mgh = (3.0)(9.8)(2.5) = 73.5$ J. *Stage 2: Sliding to a stop.* On the rough surface, kinetic friction acts on the cylinder. It decelerates the translational motion and simultaneously applies a torque that decelerates the spin. All kinetic energy --- translational and rotational --- is eventually converted to thermal energy by friction. The friction force is $f_k = \mu_k mg$. The total energy dissipated over distance $d$ is $f_k \cdot d$ (for the translational part) plus the work done by the friction torque on the rotation. But here is the key: the friction force acting over the distance the center of mass travels, combined with the friction torque acting over the angle the cylinder rotates, together dissipate all the kinetic energy. By the work-energy theorem applied to the whole system: $$\mu_k mg \cdot d_{\text{cm}} + \mu_k mg R \cdot \Delta\theta = KE_{\text{total, initial}}$$ This gets complicated. A cleaner approach: use energy. The total mechanical energy at the bottom of the ramp is $73.5$ J. All of it is eventually dissipated by friction. The friction force on the sliding cylinder is $\mu_k mg$. The distance the contact point travels relative to the surface determines the total energy dissipated. For pure sliding, the energy dissipated equals $f_k$ times the distance the center of mass moves (the relative sliding at the contact point accounts for both translational and rotational dissipation). This gives: $$\mu_k mg \cdot d = mgh$$ $$d = \frac{h}{\mu_k} = \frac{2.5}{0.20} = 12.5 \text{ m}$$ The cylinder slides 12.5 m. Notice that the mass cancelled, and the answer depends only on the height and the friction coefficient.

Layer 2: Pattern --- Identify the Best Method Without Solving

For each problem description below, identify the best method (or combination of methods). Write your choice and a one-sentence justification. Do not solve the problem.

Problem 4a. A bullet embeds itself in a wooden block hanging from a string. The block swings upward. Find the bullet's initial speed.

Problem 4b. A yo-yo unwinds from rest under gravity. Find its acceleration.

Problem 4c. A disk spinning at 10 rad/s has a ring dropped onto it. No external torques act. Find the final angular velocity.

Problem 4d. A ball is thrown upward from the edge of a cliff. Find the total time it is in the air.

Problem 4e. A roller coaster car starts from rest at the top of a 40 m hill and passes through a loop of radius 10 m. Find its speed at the top of the loop.

Check your answers

**4a. Momentum (collision) + Energy (swing).** The bullet embedding in the block is a perfectly inelastic collision --- use momentum conservation to relate the bullet's speed to the block+bullet speed after impact. Then use energy conservation for the swing upward to relate the post-collision speed to the height. Two stages, two methods. This is the classic "ballistic pendulum." **4b. Forces and torques.** The problem asks for *acceleration*, which is the direct output of $\sum F = ma$ and $\sum \tau = I\alpha$. Energy can give you speed at a given height but not acceleration. You need the force/torque approach with the constraint that the string unwinds ($a = R\alpha$). **4c. Angular momentum conservation.** No external torques, internal reconfiguration. $I_1\omega_1 = (I_1 + I_2)\omega_f$. One equation, done. Energy is not conserved (friction between disk and ring dissipates KE). **4d. Forces + kinematics.** The problem asks for *time*. Energy methods do not involve time. You need $a = -g$ and the kinematic equation $y = y_0 + v_0 t + \frac{1}{2}at^2$ with the appropriate initial conditions. **4e. Energy conservation.** The problem asks for *speed* after a complex curved path. Energy conservation (no friction assumed) gives $\frac{1}{2}mv_{\text{top}}^2 + mg(2R) = mg(40)$, yielding $v_{\text{top}}$ in one step. The force method would require tracking the normal force and centripetal acceleration around the loop. **The pattern:** match the output of the method to what the problem asks for. Collision? Momentum. Speed? Energy. Time or acceleration? Forces.

Layer 3: Structure --- Build Your Own Decision Guide

Problem 5. Create a one-page decision guide for method selection in mechanics. Your guide should:

(a) List the key questions a solver should ask before writing any equations.

(b) For each question, explain which method(s) it points toward and why.

(c) Include at least one example problem for each branch of your decision tree.

(d) Address what to do when a problem has multiple stages that require different methods.

Compare your guide to the decision tree presented earlier in this section. What did you include that the tree does not? What did the tree include that you initially missed?

Problem 6. Consider the following claim: "You can always solve any mechanics problem using forces and kinematics. Energy and momentum are just shortcuts."

(a) Is this claim technically true? Why or why not?

(b) Even if it is technically true, why is it practically misleading?

(c) Give a specific example of a problem where the "forces only" approach is essentially impossible in practice, even though it is correct in principle.

Discussion

**(a)** In principle, yes. Energy conservation and momentum conservation are both derived from $\sum F = ma$. Any result you can get from energy or momentum methods, you could in theory obtain from force analysis with sufficient information and computation. **(b)** It is misleading because "in theory" and "in practice" are very different. In a collision, you do not know the force as a function of time. You cannot apply $F = ma$ without that function. Momentum conservation bypasses this unknown entirely. Similarly, for a ball rolling through a complex multi-stage path, the energy method handles the entire journey in one equation; the force method requires solving separate problems for each stage. The "shortcut" framing suggests that energy and momentum are nice but optional. In practice, they are often the *only* feasible approach. **(c)** A lump of clay collides with and sticks to a wooden block. The contact force during the collision depends on the materials, deformation, and collision geometry in ways that are unknown and unmeasurable at this level. The force method would require knowing $F(t)$ during the few-millisecond collision. Momentum conservation solves the problem in one line without ever needing that force.

Layer 4: Creation --- Design Problems That Test Method Selection

Problem 7. Design a problem where two methods (your choice) are equally efficient --- both reach the answer in about the same number of steps.

Then modify your problem so that one method is clearly better than the other. Explain what you changed and why it shifted the balance.

Problem 8. Design a two-stage problem that requires different methods in each stage. Write the full problem statement, identify the stages, and explain why each stage requires its particular method.

Bonus: design a three-stage problem where each stage uses a different method (energy, momentum, and forces).

Hints and examples

**Problem 7 example.** Start with: "A block slides down a frictionless ramp of height $h$. Find its speed at the bottom." Both energy ($mgh = \frac{1}{2}mv^2$) and forces ($a = g\sin\theta$, $v^2 = 2aL$) solve this in about three steps. Now modify: "A block slides down a frictionless ramp of height $h$, then across a rough flat surface of length $L$, then up a second frictionless ramp. Find its speed at the top of the second ramp (height $h_2 < h$)." Now energy handles three phases in one equation, while forces require three separate sub-problems. Energy is clearly better. **Problem 8 example.** "A pendulum swings down from angle $\theta$ and the bob collides with a block on a rough surface. Find the distance the block slides." Stage 1: energy (pendulum swing gives speed at the bottom). Stage 2: momentum (collision gives post-collision speed). Stage 3: work-energy theorem (friction dissipates KE over the sliding distance). Three stages, three methods.

Reflection

After working through this section, think about how your problem-solving process has changed.

Before this chapter, when you saw a mechanics problem, what was your first instinct? Did you default to drawing an FBD and writing $F = ma$? Did you look for the "right formula" in your notes?

Now, consider the expert approach: read the problem, identify the stages, ask "what does it want?" and "what is conserved?", choose the method, and then write equations. The first equation you write should be the right one, not the first one you happen to remember.

Reflect on these questions:

• When you see a new problem, what questions do you now ask before writing any equations?
• Can you recall a problem earlier in this course where you used the wrong method (or a harder method) and now see the better choice?
• What is the single most important factor in choosing between forces, energy, and momentum?

The metacognitive shift --- from "what formula should I use?" to "what structure does this problem have?" --- is the most important outcome of this section. If you leave this section with a reliable decision process that you trust, you are ready for the synthesis problems in Section 14.5.