14.3 Energy, Momentum, and Torque in Multi-Stage Problems

A Ball's Three Lives

[Video: A billiard ball rolls along a table at a steady pace. The camera follows from the side. The ball reaches the edge and rolls off. It arcs through the air --- spinning, falling, tracing a parabolic path. It strikes the floor, bounces once with a sharp crack, and skids forward before rolling away. Three labels appear in sequence, one for each phase: "Rolling (constrained rotation)," "Projectile (free flight)," and "Impact (collision)."]

Watch that again. One ball, one continuous motion --- but three completely different kinds of physics.

During the roll on the table, friction and the rolling constraint tie translation and rotation together. The moment the ball leaves the edge, that constraint vanishes. In the air, gravity acts on the center of mass, and nothing exerts a torque. When the ball hits the floor, forces spike to enormous values for a tiny fraction of a second, and momentum --- both linear and angular --- changes abruptly.

If you tried to write one set of equations that covers the entire motion from table to final rest, you would drown in complexity. But you don't have to. You already know how to handle each stage individually. The real skill is managing the transitions.

Prediction

Before you read on, commit to answers for each of the following. During the projectile phase (after the ball leaves the table, before it hits the floor):

Does the ball's spin rate ($\omega$) change?

Does the ball's angular momentum ($L$) change?

Does the ball's linear momentum change?

Write down your answers --- and your reasoning --- before continuing.

Let's take these one at a time.

Spin rate ($\omega$). What could change the ball's spin? A torque. Is there a torque on the ball while it flies through the air? Gravity acts at the center of mass. The ball is spherically symmetric, so gravity produces no torque about the center of mass. Air resistance? We are ignoring it (and even with air resistance on a smooth sphere, the torque is negligible). No torque means no angular acceleration: $\alpha = \tau/I = 0$. The spin rate stays constant throughout the flight.

Angular momentum ($L$). Since $\tau_{\text{net}} = dL/dt$ and $\tau_{\text{net}} = 0$, angular momentum is conserved during the projectile phase. For a rigid sphere, $L = I\omega$, and since both $I$ and $\omega$ are constant, $L$ doesn't change.

Linear momentum. This one is different. Gravity acts on the ball the entire time it's in the air. There is a net external force ($mg$ downward), so linear momentum is not conserved. The vertical component of momentum changes; the horizontal component stays constant (no horizontal force).

If you got all three right, good --- you are connecting the right tool to the right condition. If you missed one, pay attention to the pattern: the question "does this quantity change?" always comes down to "is there a net force/torque that would change it?"

The Guiding Question

How can one problem require different conservation laws and dynamics ideas at different stages?

This is not a new idea. In Section 2.4, you learned to handle piecewise acceleration: the rocket engine fires, then cuts off, and you stitch the stages together using continuity of position and velocity. That was segmentation and continuity with a limited toolkit --- just kinematics.

Now you have the full toolkit: Newton's laws, energy conservation, work-energy theorems, impulse-momentum, angular momentum, torque. The segmentation idea is the same, but the menu of tools is much richer --- and so is the question of which tool to reach for in each stage.

Exploration: Anatomy of a Multi-Stage Problem

[Interactive: Multi-Stage Timeline. A horizontal timeline bar spans the full motion of the ball-off-table problem. The timeline is divided into three color-coded stages:

Stage 1 (blue): Rolling on the table. Below this stage, cards show the relevant physics: rolling constraint ($v = R\omega$), friction force, normal force, energy conservation with both translational and rotational KE.

Stage 2 (green): Projectile flight. Cards show: projectile kinematics, $\omega = \text{const}$, angular momentum conserved, energy conservation (KE $\leftrightarrow$ PE, with rotational KE unchanged).

Stage 3 (red): Impact with the floor. Cards show: impulse-momentum theorem, coefficient of restitution, angular impulse, possible sliding/rolling transition.

At each boundary between stages, a gold connector strip asks the student to drag-and-drop which quantities carry over continuously: position, velocity (components), angular velocity, energy. Correct drops light up green; incorrect ones bounce back with a one-line explanation.]

Work through this interactive carefully. At each stage boundary, ask yourself two questions:

What quantities are continuous? (What value at the end of one stage equals the starting value of the next?)
What can change discontinuously? (What is different about the physics on the two sides of the boundary?)

Let's make this concrete.

Boundary 1: Table edge (rolling $\to$ projectile)

At the instant the ball leaves the table:

Quantity	Continuous?	Why?
Position ($x$, $y$)	Yes	The ball doesn't teleport
Velocity ($v_x$, $v_y$)	Yes	No infinite force at this transition
Angular velocity ($\omega$)	Yes	No impulsive torque at the edge
Rolling constraint ($v = R\omega$)	Ceases to apply	No contact surface in the air
Normal force	Drops to zero	No surface to push against
Friction force	Drops to zero	No contact, no friction

The ball leaves the edge with the same $v_x$, $v_y$, and $\omega$ it had the instant before. But the rules change: the rolling constraint is gone, friction is gone, and the only force is gravity.

Pause and think: The rolling constraint $v = R\omega$ was enforced by friction on the table. Once the ball is in the air, is the relationship $v_x = R\omega$ still true?

It is still numerically true at the instant of departure --- because continuity guarantees the values haven't changed. But it is no longer enforced. If something were to change $v_x$ without changing $\omega$ (or vice versa), there would be no friction to restore the relationship. In the air, $v_x$ and $\omega$ evolve independently. It just happens that, for a ball in free flight with no air resistance, $v_x$ stays constant and $\omega$ stays constant, so the numerical equality $v_x = R\omega$ persists throughout the flight. But this is a coincidence of the specific forces, not a constraint.

Boundary 2: Impact with the floor (projectile $\to$ collision)

This boundary is fundamentally different. During the collision:

Quantity	Continuous?	Why?
Position ($x$, $y$)	Yes	Ball doesn't teleport during impact
Velocity ($v_x$, $v_y$)	No	Impulsive forces act during the collision
Angular velocity ($\omega$)	Possibly not	Impulsive friction can exert angular impulse
Kinetic energy	No	Energy is lost in inelastic collisions

This is where things get interesting. During a collision, forces become extremely large for a very short time. The impulse $\vec{J} = \int \vec{F}\,dt$ is finite even though $\Delta t \to 0$, because $F \to \infty$ fast enough to compensate. This means velocity can change abruptly. So can angular velocity, if there is an impulsive torque (from friction at the contact point during impact).

You cannot use energy conservation across the collision (energy is lost). You cannot use $F = ma$ directly (the force is unknown and enormous). The right tool is the impulse-momentum theorem, both linear and angular:

$$\vec{J} = \Delta \vec{p} = m\Delta \vec{v}$$

$$\vec{\tau}_{\text{imp}} \cdot \Delta t = \Delta \vec{L} = I\Delta\vec{\omega}$$

After the collision, you re-establish initial conditions for whatever comes next (sliding, rolling, another flight).

Concept Reveal: The Multi-Stage Method

Here is the general structure. Every multi-stage problem follows the same architecture:

Step 1: Segment the motion

Read the problem and identify the stages. A new stage begins whenever:

A force turns on or off (engine ignition/cutoff, leaving a surface, entering a fluid)
A constraint begins or ends (rolling $\to$ sliding, string goes taut, object separates from a surface)
A collision or explosion occurs
The geometry changes (a ramp ends and a flat surface begins)

Step 2: Choose the right tool for each stage

This is the judgment call. For each stage, ask:

Question	If yes, use...
Do I need speed but not time?	Energy methods
Do I need time or acceleration?	Newton's second law ($F = ma$, $\tau = I\alpha$)
Is there a collision or explosion?	Impulse-momentum theorem
Are internal forces unknown?	Energy or momentum (to bypass those forces)
Is the system isolated (no external force/torque)?	Conservation of momentum / angular momentum
Is there rolling without slipping?	Rolling constraint $v = R\omega$, combined energy

You practiced this decision-making in earlier chapters. What is new here is that different stages of the same problem may require different tools.

Step 3: Solve each stage

Apply the chosen tool within each stage. This is just the physics you already know, applied to one piece of the problem.

Step 4: Enforce continuity at the boundaries

At each stage transition, identify what carries over:

Always continuous (for non-collision boundaries): position, velocity, angular velocity.
Always continuous (even across collisions): position.
Possibly discontinuous (across collisions): velocity, angular velocity, kinetic energy.
Changes at every boundary: the governing equations, the active forces, the applicable constraints.

The end-of-stage values become the initial conditions for the next stage. This is the thread that stitches the solution together.

Step 5: Assemble the answer

The question usually asks about something that spans multiple stages --- a final speed, a total distance, a landing position. Chain the stage solutions together to get there.

The multi-stage strategy in one sentence: Segment the problem at every change of physics, pick the best tool for each segment, and let the continuity conditions carry information across the boundaries.

Compare this to Section 2.4: the structure is identical. What has changed is the toolkit. In Section 2.4, every stage was "apply the kinematic equations." Now, one stage might use energy conservation, the next might use impulse-momentum, and the next might use torque and angular kinematics. The organizational framework is the same; the menu of tools is wider.

A Complete Example: Ball Rolling Off a Table

Let's work through the opening problem quantitatively.

Setup: A solid sphere of mass $m = 0.2$ kg and radius $R = 0.03$ m rolls without slipping along a table at height $h = 1.0$ m. Its center-of-mass speed on the table is $v_0 = 2.0$ m/s. It rolls off the edge. Find: (a) the ball's speed just before hitting the floor, and (b) how far from the base of the table it lands.

Before computing: Which tool will you use for the projectile phase --- energy or kinematics? Both are valid. Think about which is more efficient for each sub-question.

Stage 1: Rolling on the table

The ball rolls without slipping, so $v_0 = R\omega_0$, giving:

$$\omega_0 = \frac{v_0}{R} = \frac{2.0}{0.03} \approx 66.7 \text{ rad/s}$$

On the table, if the surface is level and we assume no energy loss, the ball maintains constant speed. The ball reaches the edge with:

$$v_{x} = v_0 = 2.0 \text{ m/s}, \quad v_{y} = 0, \quad \omega = \omega_0 \approx 66.7 \text{ rad/s}$$

Boundary 1: Leaving the table

All three quantities ($v_x$, $v_y$, $\omega$) carry over continuously. The rolling constraint ceases to be enforced, but $v_x = R\omega$ is still numerically true at this instant.

Stage 2: Projectile motion

Now the ball is in free flight. No torque acts, so $\omega$ stays at $66.7$ rad/s throughout the flight. We need the speed at impact and the horizontal distance.

Part (a): Speed at impact using energy.

The ball has both translational and rotational kinetic energy. During free flight, gravity converts potential energy into translational kinetic energy. The rotational KE doesn't change (no torque, so $\omega$ is constant):

$$\frac{1}{2}mv_0^2 + mgh + \underbrace{\frac{1}{2}I\omega_0^2}{\text{constant}} = \frac{1}{2}mv_f^2 + 0 + \underbrace{\frac{1}{2}I\omega_0^2}$$}

The rotational terms cancel:

$$\frac{1}{2}mv_0^2 + mgh = \frac{1}{2}mv_f^2$$

$$v_f = \sqrt{v_0^2 + 2gh} = \sqrt{(2.0)^2 + 2(9.8)(1.0)} = \sqrt{4 + 19.6} = \sqrt{23.6} \approx 4.86 \text{ m/s}$$

Notice something important: the rotational KE cancelled out of the energy equation because $\omega$ didn't change. The answer is the same as for a point particle sliding off the table. The spin is irrelevant to the translational speed at landing --- an insight you might not have predicted.

Part (b): Horizontal distance using projectile kinematics.

For the horizontal distance, energy methods are not ideal --- we need the time of flight, and energy doesn't give us time directly. Switch tools: use kinematics.

The ball leaves the edge at height $h = 1.0$ m with $v_y = 0$ (horizontal launch). The time to fall:

$$h = \frac{1}{2}gt^2 \implies t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2(1.0)}{9.8}} \approx 0.452 \text{ s}$$

Horizontal distance:

$$d = v_x \cdot t = 2.0 \times 0.452 \approx 0.90 \text{ m}$$

The ball lands about 90 cm from the base of the table.

Reflect on tool choice: We used energy for part (a) because it directly gives speed without needing time. We used kinematics for part (b) because we needed time, which energy methods don't provide. Within a single stage, different sub-questions can call for different tools.

A Richer Example: Ramp, Flight, and Collision

Here is a three-stage problem that exercises the full method.

Setup: A solid cylinder of mass $M = 0.5$ kg and radius $R = 0.05$ m starts from rest at the top of a ramp of height $H = 0.8$ m and rolls without slipping to the bottom. The bottom of the ramp is at the edge of a cliff of height $h = 2.0$ m. The cylinder rolls off the cliff, flies through the air, and hits the ground. (For a solid cylinder, $I = \frac{1}{2}MR^2$.)

[Interactive: Three-stage diagram. A ramp is shown at left, ending at a cliff edge. The cylinder's path is traced: a curved path down the ramp (Stage 1, blue), a parabolic arc through the air (Stage 2, green), and an impact point on the ground (Stage 3, red). Students can hover over each stage to see the active forces and applicable tools highlighted.]

Before solving: Identify the stages, name the best tool for each, and list what carries over at each boundary. Do this before reading further.

Stage 1: Rolling down the ramp

Tool: Energy conservation (we need speed at the bottom, not time).

The cylinder starts from rest at height $H$. At the bottom of the ramp, all potential energy has been converted to kinetic energy --- both translational and rotational:

$$MgH = \frac{1}{2}Mv_{\text{cm}}^2 + \frac{1}{2}I\omega^2$$

Using $I = \frac{1}{2}MR^2$ and the rolling constraint $\omega = v_{\text{cm}}/R$:

$$MgH = \frac{1}{2}Mv_{\text{cm}}^2 + \frac{1}{2}\left(\frac{1}{2}MR^2\right)\left(\frac{v_{\text{cm}}}{R}\right)^2 = \frac{1}{2}Mv_{\text{cm}}^2 + \frac{1}{4}Mv_{\text{cm}}^2 = \frac{3}{4}Mv_{\text{cm}}^2$$

$$v_{\text{cm}} = \sqrt{\frac{4gH}{3}} = \sqrt{\frac{4(9.8)(0.8)}{3}} = \sqrt{10.45} \approx 3.23 \text{ m/s}$$

And the angular velocity:

$$\omega = \frac{v_{\text{cm}}}{R} = \frac{3.23}{0.05} = 64.7 \text{ rad/s}$$

Boundary 1: Cliff edge

The cylinder leaves the ramp horizontally (assuming the ramp bottom is tangent to the horizontal). Continuity gives:

$$v_x = 3.23 \text{ m/s}, \quad v_y = 0, \quad \omega = 64.7 \text{ rad/s}$$

The rolling constraint ceases. From here, $v_x$ and $\omega$ evolve independently (though, in practice, both remain constant in free flight).

Stage 2: Projectile motion

Tool: Kinematics (we need time and landing position) and energy (for speed at impact).

Time to fall height $h$:

$$t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2(2.0)}{9.8}} = \sqrt{0.408} \approx 0.639 \text{ s}$$

Horizontal distance:

$$d = v_x \cdot t = 3.23 \times 0.639 \approx 2.06 \text{ m}$$

Speed at impact (using energy, with rotational KE cancelling as before):

$$v_f = \sqrt{v_{\text{cm}}^2 + 2gh} = \sqrt{(3.23)^2 + 2(9.8)(2.0)} = \sqrt{10.45 + 39.2} = \sqrt{49.65} \approx 7.05 \text{ m/s}$$

The cylinder arrives spinning at $\omega = 64.7$ rad/s, moving at $7.05$ m/s at some angle below the horizontal. Its velocity components are:

$$v_x = 3.23 \text{ m/s}, \quad v_y = gt = 9.8(0.639) = 6.26 \text{ m/s (downward)}$$

Boundary 2: Impact

This is where we would apply impulse-momentum methods if the problem asked about post-collision velocities. The collision involves impulsive normal and friction forces. Energy is not conserved across the impact. If we needed the post-bounce speed, we would use the coefficient of restitution for the normal direction and angular impulse for the spin change.

For this problem, we stop at the impact. But notice how the three-stage structure worked: energy gave us the speed off the ramp, kinematics gave us the flight trajectory, and each boundary handed off values to the next stage.

What Can and Cannot Cross a Boundary

This is worth organizing carefully. Different types of boundaries have different continuity rules.

Non-impulsive boundaries (no collision)

These include: leaving a surface, entering a new region, a force turning on/off smoothly.

Quantity	Continuous?
Position	Yes
Velocity (all components)	Yes
Angular velocity	Yes
Kinetic energy	Yes (value is continuous, though the expression may change)
Momentum	Yes (at the instant of transition)
Angular momentum	Yes (at the instant of transition)

At these boundaries, the values of all dynamical quantities are continuous. What changes is the equations of motion --- which forces act, which constraints apply.

Impulsive boundaries (collisions)

These include: impacts, explosions, sudden catches (ball caught by hand, bullet embedding in block).

Quantity	Continuous?
Position	Yes
Velocity	No --- impulse changes momentum in zero time
Angular velocity	No --- angular impulse can change angular momentum
Kinetic energy	No --- energy is typically lost
Momentum (system)	Yes, if no external impulse; No, if an external impulsive force acts
Angular momentum (system)	Yes, if no external impulsive torque; No, if one acts

A useful rule of thumb: At a non-collision boundary, everything is continuous and only the rules change. At a collision boundary, the values can jump, and you need impulse-momentum methods to find the new values.

Common Errors in Multi-Stage Problems

Students who understand each individual tool often stumble when combining them across stages. Here are the most frequent mistakes.

Error 1: Using energy conservation across a collision.

A student writes $\frac{1}{2}mv_{\text{before}}^2 = \frac{1}{2}mv_{\text{after}}^2$ for an inelastic collision. This is wrong --- kinetic energy is lost. Use momentum conservation (or coefficient of restitution) instead.

Error 2: Forgetting that the rolling constraint is stage-specific.

On the ramp, $v = R\omega$. In the air, this relationship is not enforced. After landing, it may or may not hold depending on whether the ball rolls or slides. Students who blindly apply $v = R\omega$ in every stage get incorrect results.

Error 3: Resetting initial conditions to zero.

A student starts Stage 2 with $v = 0$ and $x = 0$ instead of using the values inherited from Stage 1. This throws away all the information from the previous stage.

Error 4: Applying a conservation law when its conditions aren't met.

"Angular momentum is conserved" requires zero net external torque. "Momentum is conserved" requires zero net external force (or zero net external impulse). These conditions hold in some stages and not in others. You must check for each stage separately.

Error 5: Trying to write one equation for the whole problem.

The whole point of segmentation is that no single equation spans all stages. Resist the urge to find a shortcut that jumps from the initial state to the final state in one step. Sometimes energy conservation does let you skip intermediate stages (if no energy is lost anywhere), but this only works when you can verify that the conditions for energy conservation hold across all the stages you are skipping.

Connection to Earlier Sections

The segmentation idea appeared early in this course. Section 2.4 introduced piecewise acceleration: different kinematic rules in different time intervals, stitched together by continuity of position and velocity. The strategy there was:

Identify the stages.
Solve each stage with the kinematic equations.
Use continuity conditions at the boundaries.

This section is the same architecture with a fuller toolkit. Here is the comparison:

	Section 2.4	Section 14.3
What defines a stage	Change in acceleration	Change in forces, constraints, or physics type
Tool per stage	Kinematic equations only	Energy, momentum, torque, kinematics, impulse --- whatever fits
Continuous at boundaries	Position, velocity	Position; velocity and $\omega$ (unless collision); sometimes energy
Discontinuous at boundaries	Acceleration	Forces, constraints, applicable conservation laws
What's hard	Bookkeeping time intervals	Choosing the right tool for each stage

The organizational framework never changed. What grew was your toolkit.

Practice

Layer 1: Concrete --- Two-Stage Problem

A solid sphere ($I = \frac{2}{5}mR^2$) of mass $m = 0.3$ kg and radius $R = 0.04$ m rolls without slipping along a horizontal surface at $v_0 = 3.0$ m/s. It reaches the edge of a table of height $h = 1.2$ m and rolls off.

(a) What is the sphere's angular velocity $\omega_0$ as it leaves the table?

(b) What is the sphere's translational speed just before it hits the ground?

(c) What is the sphere's angular velocity just before it hits the ground?

(d) How far from the base of the table does it land?

Check your answer

**(a)** From the rolling constraint on the table: $$\omega_0 = \frac{v_0}{R} = \frac{3.0}{0.04} = 75 \text{ rad/s}$$ **(b)** During free flight, no torque acts, so $\omega$ is constant and the rotational KE doesn't change. Using energy conservation for the translational motion: $$v_f = \sqrt{v_0^2 + 2gh} = \sqrt{(3.0)^2 + 2(9.8)(1.2)} = \sqrt{9 + 23.52} = \sqrt{32.52} \approx 5.70 \text{ m/s}$$ **(c)** No torque in the air, so the angular velocity is unchanged: $$\omega_f = \omega_0 = 75 \text{ rad/s}$$ This is a key insight: the translational speed changes (gravity accelerates the ball downward) but the spin does not (no torque). **(d)** Time of flight from $h = \frac{1}{2}gt^2$: $$t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2(1.2)}{9.8}} \approx 0.495 \text{ s}$$ Horizontal distance: $$d = v_0 \cdot t = 3.0 \times 0.495 \approx 1.48 \text{ m}$$

Layer 2: Pattern --- Identify Stages and Tools

For each scenario below, identify the stages, name the best tool for each stage, and state what carries over at each boundary. Do not solve numerically.

(a) A block slides down a frictionless ramp and then onto a rough horizontal surface, where it decelerates and stops.

(b) A ball is thrown upward, rises, falls back down, and bounces off the ground (losing some kinetic energy), then rises again.

(c) A cylinder rolls without slipping down a ramp, reaches the bottom, and collides with a stationary block on a frictionless floor. The cylinder and block stick together and slide forward.

(d) A pendulum swings down from rest, and at the bottom of its arc the string is cut. The bob then undergoes projectile motion.

Check your answer

**(a)** Two stages. - **Stage 1 (frictionless ramp):** Energy conservation. $mgh = \frac{1}{2}mv^2$. No friction, so mechanical energy is conserved. - **Boundary:** Speed at the bottom carries over continuously. The direction of velocity changes from along the ramp to horizontal. - **Stage 2 (rough floor):** Work-energy theorem or Newton's second law. Friction does negative work: $\frac{1}{2}mv^2 = \mu_k mg \cdot d$. - **Tool choice reasoning:** Energy is efficient for Stage 1 (gives speed without needing time). Stage 2 also works well with energy (work done by friction equals loss of KE), or with $F = ma$ if you need the deceleration and stopping time. **(b)** Three stages. - **Stage 1 (upward flight):** Projectile kinematics or energy. Only gravity acts. - **Boundary 1:** Ball reaches peak with $v = 0$, then falls. - **Stage 2 (downward flight):** Same physics as Stage 1. By symmetry, it returns to the launch height at the same speed it was thrown. - **Boundary 2 (bounce):** Collision. Velocity changes discontinuously. Use coefficient of restitution: $v_{\text{after}} = e \cdot v_{\text{before}}$. Energy is lost. - **Stage 3 (second upward flight):** Projectile kinematics or energy, starting from $v_{\text{after}}$. **(c)** Three stages. - **Stage 1 (rolling down ramp):** Energy conservation with rolling: $MgH = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2$. - **Boundary 1:** $v$ and $\omega$ carry over. - **Stage 2 (collision):** Momentum conservation (perfectly inelastic). Energy is *not* conserved. The cylinder's rotational state complicates things: you need to consider whether friction during the collision affects the spin. For an idealized instant collision with the block, linear momentum is conserved: $Mv = (M + m_{\text{block}})v_f$. - **Boundary 2:** Post-collision velocity is the new initial condition. - **Stage 3 (sliding on frictionless floor):** Trivially constant velocity (no friction, no net force). Or, if the floor has friction, use $F = ma$ or work-energy. **(d)** Two stages. - **Stage 1 (pendulum swing):** Energy conservation. $mgh = \frac{1}{2}mv_{\text{bottom}}^2$, where $h$ is the initial height. - **Boundary:** At the bottom of the swing, the string is cut. The tension force disappears. Velocity carries over continuously (cutting a string is not an impulsive event --- no sudden force spike). The velocity at the bottom is horizontal. - **Stage 2 (projectile):** Standard projectile motion from the release point.

Layer 3: Structure --- Continuity at Boundaries

Answer the following questions about stage boundaries.

(a) What quantities are always continuous at stage boundaries, regardless of the type of boundary?

(b) What quantities can be discontinuous at a collision boundary but are continuous at a non-collision boundary?

(c) A ball rolls without slipping on a surface and then leaves the surface. At the instant of leaving, is $v = R\omega$ still true? Is it still a constraint? Explain the difference.

(d) A student claims: "If momentum is conserved across a collision, then kinetic energy must also be conserved." Is this correct?

Check your answer

**(a)** Position is always continuous --- an object cannot teleport, regardless of the type of boundary. Even in a collision, the object is at a definite location when the impact occurs. **(b)** Velocity (linear and angular) can be discontinuous at a collision boundary due to impulsive forces/torques. Kinetic energy can decrease at a collision boundary (inelastic collision). At a non-collision boundary, all of these are continuous. **(c)** At the instant of leaving, $v = R\omega$ is still numerically true (by continuity, the values haven't changed yet). But it is no longer a constraint --- there is no surface and no friction to enforce it. The distinction matters: a constraint is a *rule that the physics enforces going forward*. A numerical equality that happens to hold at one instant but is not enforced afterward is not a constraint. If a horizontal force were applied to the ball in the air (say, a brief gust of wind), $v_x$ would change but $\omega$ would not, and the equality $v_x = R\omega$ would break. **(d)** This is incorrect. Momentum conservation and energy conservation are independent conditions. In a perfectly inelastic collision (objects stick together), momentum is conserved but kinetic energy is maximally lost (as much as possible while still conserving momentum). Only in a perfectly elastic collision are *both* momentum and kinetic energy conserved. Momentum conservation tells you the final velocity; whether energy is also conserved is a separate physical fact about the collision.

Layer 4: Debug --- Find the Error

A student solves the following problem:

"A ball of mass $m$ rolls without slipping down a ramp of height $H$, then collides with a stationary block of mass $M$ on a frictionless floor. Find the speed of the block after the collision (assume perfectly inelastic)."

The student writes:

$$mgH = \frac{1}{2}(m + M)v_f^2$$

and solves for $v_f$. What did the student do wrong? What is the correct approach?

Check your answer

The student applied energy conservation across the entire problem, including the collision. This is wrong for two reasons: **Error 1:** The collision is perfectly inelastic --- kinetic energy is *not* conserved. The equation $mgH = \frac{1}{2}(m+M)v_f^2$ implicitly assumes that all the potential energy at the top of the ramp ends up as kinetic energy of the combined mass. But energy is lost during the collision (converted to heat, sound, deformation), so the right side is too large. **Error 2:** The student also forgot about rotational kinetic energy. Rolling down the ramp, some energy goes into rotation, so the ball's translational speed at the bottom is less than $\sqrt{2gH}$. **Correct approach (two stages):** *Stage 1 (rolling down ramp):* Energy conservation with rotation. $$mgH = \frac{1}{2}mv_1^2 + \frac{1}{2}I\omega_1^2 = \frac{1}{2}mv_1^2\left(1 + \frac{I}{mR^2}\right)$$ For a solid sphere, $I = \frac{2}{5}mR^2$, so: $$v_1 = \sqrt{\frac{2gH}{1 + 2/5}} = \sqrt{\frac{10gH}{7}}$$ *Stage 2 (perfectly inelastic collision):* Momentum conservation (not energy conservation). $$mv_1 = (m + M)v_f$$ $$v_f = \frac{m}{m + M}\sqrt{\frac{10gH}{7}}$$ The correct answer is smaller than the student's answer for two reasons: the ball is slower at the bottom (rotational energy), and the collision loses kinetic energy (inelastic). This problem illustrates why you must segment: energy conservation works for Stage 1 (no energy loss), but momentum conservation is required for Stage 2 (energy is lost, but momentum is conserved because the floor is frictionless and the collision forces are internal to the ball-block system).

A Practical Checklist for Multi-Stage Problems

When you face a problem that changes character partway through, use this procedure:

Read the whole problem first. Get the big picture before touching any equations.
Draw a timeline or diagram. Label the stages. Mark the boundaries. This is not optional decoration --- it is the most important step.
For each stage, identify:
What forces and torques act?
What constraints apply (rolling, string tension, contact)?
Is mechanical energy conserved?
Is momentum conserved? Angular momentum?
What tool gives the answer most efficiently?
For each boundary, identify:
Is it a collision (impulsive) or a smooth transition?
What quantities carry over continuously?
What can change?
Solve stage by stage, forward in time. Use the end-of-stage values as the beginning-of-next-stage values.
Check your answer. Do the units work? Do the limits make sense? (If $M \to 0$ in the collision problem, does the ball's speed approach what you'd expect? If $H \to 0$, does the final speed go to zero?)

Reflection

What is your personal strategy for decomposing a multi-stage problem?

Think about how you approached the problems in this section. Did you naturally segment first, or did you try to write equations immediately? Did you identify the tool before writing, or did you start with whatever formula came to mind?

The experts' habit is: draw first, segment second, choose tools third, compute last. If you find yourself computing before you have a clear picture of the stages, slow down. The time you invest in organization always pays for itself.

Also consider: What makes multi-stage problems feel hard? It is rarely the individual steps. Each stage is a problem you have solved before. The difficulty is organizational --- managing the transitions, choosing the right tool, tracking what carries over. Recognizing this can reduce the anxiety. You are not facing one hard problem; you are facing several easy problems linked by continuity conditions.