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Momentum, Impulse, and Conservation Laws

8.5 Two-Dimensional Collision Analysis

Scattering on a Table

A cue ball rolls across the felt and strikes an object ball off-center. The object ball flies toward a corner pocket. The cue ball veers sideways, taking a path you did not expect. Two objects entered the collision along one line; they left along two different lines in two different directions.

This is not a one-dimensional problem anymore. In every collision we studied in Sections 8.3 and 8.4, the objects moved along a single line --- head-on, back-and-forth, start to finish. That simplification was useful, but the real world rarely cooperates. Billiard balls scatter at angles. Cars collide at intersections. Subatomic particles ricochet off each other in every conceivable direction.

To handle these situations, you need to think about momentum as a vector --- not just a number with a sign, but a quantity with components in each direction. The good news: you already have every tool you need. The principle of conservation of momentum still holds. The only difference is that now it gives you two equations instead of one.

Prediction

Before you read on: Two balls of equal mass collide on a frictionless surface. Ball A is moving to the right; ball B is initially at rest. The collision is perfectly elastic and off-center (a glancing blow, not head-on).

After the collision, both balls are moving. What angle separates their two paths? Is it less than 90 degrees, exactly 90 degrees, or more than 90 degrees?

Commit to a prediction. If you have played pool, think about what you have seen at the table.

[Interactive: A top-down view of two equal-mass discs. Ball A approaches from the left. Ball B sits at rest. Students drag a slider to set the impact parameter (how far off-center the collision is). Before clicking "collide," they predict the angle between the outgoing paths.]

This is one of the most elegant results in classical mechanics. For any equal-mass elastic collision with one ball initially at rest --- regardless of the impact angle, regardless of how glancing or how nearly head-on the blow is --- the two outgoing velocity vectors are separated by exactly 90 degrees. Always. We will prove this below and then explore why it works.

The Guiding Question

How do conservation ideas scale when direction matters as much as magnitude?

In one dimension, momentum conservation gave you a single equation: total momentum before equals total momentum after. One equation, and the unknowns were speeds along a line. Now we move to two dimensions, and the unknowns include angles. You need more equations. Where do they come from?

The answer is not a new law. It is the same principle you used in Section 3.4 when you decomposed projectile motion into independent horizontal and vertical problems: component independence. Each component of momentum is conserved separately. The $x$-direction has its own conservation equation. The $y$-direction has its own conservation equation. Two dimensions, two equations, one principle.

Exploration: The 2D Collision Simulator

[Interactive: 2D Billiards Collision. A top-down view shows two discs on a frictionless surface. Ball A moves to the right with initial velocity $v_0$; ball B is at rest. Students control two parameters: (1) the impact parameter $b$ --- the vertical offset between the centers, which determines how glancing the collision is --- and (2) a mass ratio slider ($m_B / m_A$, ranging from 0.5 to 3.0). A restitution slider (elastic to perfectly inelastic) is also available.

After each collision, the simulator shows: - The outgoing velocity vectors of both balls (arrows showing direction and magnitude). - Component bar charts displaying $p_x$ and $p_y$ for each ball, before and after the collision. The totals are displayed and remain unchanged. - A readout of the angle between the two outgoing velocity vectors. - A kinetic energy bar showing total $KE$ before and after.

Guided prompts:

Prompt 1: Set the masses equal and the collision to elastic. Choose a moderate impact parameter. Run the collision. What is the angle between the two outgoing paths? Now change the impact parameter. What happens to the angle?

Prompt 2: Watch the component bars. Is total $p_x$ conserved? Is total $p_y$ conserved? Check both at several different impact parameters.

Prompt 3: Keep the collision elastic but change the mass ratio to $m_B / m_A = 2$. Is the 90-degree rule still true? What angle do you observe now?

Prompt 4: Set the masses back to equal but switch the collision to perfectly inelastic (the balls stick together). What happens to the angle now? Why?

Prompt 5: For the elastic equal-mass case, try an almost head-on collision (small $b$) and an extremely glancing collision (large $b$). How do the speeds of the two outgoing balls change? Does the angle between them stay the same?]

Spend time with this simulator. The key patterns you should notice:

  • Component-wise conservation holds exactly. No matter what you do with the sliders, the total $p_x$ before equals the total $p_y$ after, and the same for $p_y$. This is not approximate --- it is exact.

  • For equal-mass elastic collisions, the outgoing paths are always perpendicular. Change the impact parameter from nearly head-on to extremely glancing. The individual angles change dramatically, but their sum is always 90 degrees. This is remarkable and deserves an explanation.

  • Unequal masses break the 90-degree rule. When the target is heavier, the angle between the paths shrinks below 90 degrees. When the target is lighter, it opens beyond 90 degrees.

  • Inelastic collisions break the 90-degree rule too. When energy is lost, the angle between outgoing paths changes.

The 90-degree result is special: it requires both equal masses and an elastic collision. Change either condition, and it fails.

Concept Reveal: Momentum Conservation in Components

The Core Equations

In two dimensions, momentum is a vector: $\vec{p} = m\vec{v} = (mv_x, \, mv_y)$. Conservation of momentum for an isolated two-body system means:

$$\vec{p}{1,i} + \vec{p}} = \vec{p{1,f} + \vec{p}$$

This single vector equation is really two scalar equations, one for each component:

$$m_1 v_{1x,i} + m_2 v_{2x,i} = m_1 v_{1x,f} + m_2 v_{2x,f}$$

$$m_1 v_{1y,i} + m_2 v_{2y,i} = m_1 v_{1y,f} + m_2 v_{2y,f}$$

Each component is conserved independently. The $x$-momentum doesn't care what happens in the $y$-direction. The $y$-momentum doesn't care what happens in the $x$-direction. This is exactly the same independence of components you relied on in projectile motion (Section 3.4) --- the horizontal and vertical motions evolve independently.

Setting Up a Typical Problem

Consider a collision where ball 1 moves to the right with speed $v_0$ and ball 2 is at rest. Choose coordinates so the initial velocity is along the $x$-axis.

Before the collision: - Ball 1: $v_{1x,i} = v_0$, $\; v_{1y,i} = 0$ - Ball 2: $v_{2x,i} = 0$, $\; v_{2y,i} = 0$

After the collision, ball 1 moves at speed $v_1$ at angle $\theta_1$ above the $x$-axis, and ball 2 moves at speed $v_2$ at angle $\theta_2$ below the $x$-axis. The component equations become:

$$m_1 v_0 = m_1 v_1 \cos\theta_1 + m_2 v_2 \cos\theta_2 \qquad (x\text{-component})$$

$$0 = m_1 v_1 \sin\theta_1 - m_2 v_2 \sin\theta_2 \qquad (y\text{-component})$$

The $y$-equation tells you something immediately: whatever upward momentum ball 1 gains, ball 2 must gain an equal amount of downward momentum. The $y$-momenta must cancel, because there was no $y$-momentum to begin with.

The Counting Problem

Here is a crucial point. After the collision, there are four unknowns: $v_1$, $v_2$, $\theta_1$, $\theta_2$. But momentum conservation only gives you two equations ($x$ and $y$ components). Two equations, four unknowns --- you cannot solve the problem without more information.

What additional information could you have?

  • The collision is elastic. Then kinetic energy conservation adds a third equation: $\frac{1}{2}m_1 v_0^2 = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2$. That gets you to three equations, three unknowns (if one angle is given) --- or you can find relationships between the unknowns.

  • The collision is perfectly inelastic. The balls stick together, so $v_1 = v_2$ and $\theta_1 = \theta_2$. This removes two unknowns, and the problem is fully determined.

  • One outgoing angle is measured. If you know the direction of one ball after the collision (from observation or experiment), that specifies one unknown, and the system can be solved.

This is why 2D collisions require more information than 1D collisions. In one dimension, momentum conservation (plus energy conservation for elastic cases) was often enough. In two dimensions, the additional directional freedom introduces more unknowns.

The 90-Degree Rule: Proof and Geometry

The most beautiful special case deserves its own treatment. Set $m_1 = m_2 = m$ and ball 2 initially at rest. For an elastic collision, the three conservation equations become:

$x$-momentum: $\quad v_0 = v_1 \cos\theta_1 + v_2 \cos\theta_2$

$y$-momentum: $\quad 0 = v_1 \sin\theta_1 - v_2 \sin\theta_2$

Kinetic energy: $\quad v_0^2 = v_1^2 + v_2^2$

Look at the energy equation. It says $v_0^2 = v_1^2 + v_2^2$. That is the Pythagorean theorem. The three speeds form a right triangle, with $v_0$ as the hypotenuse.

Now look at the momentum equations. They say that the vector $\vec{v}_0$ (pointing along $x$) equals the vector sum $\vec{v}_1 + \vec{v}_2$. This means $\vec{v}_1$ and $\vec{v}_2$ are the two legs of a vector triangle whose hypotenuse is $\vec{v}_0$.

The Pythagorean relation from the energy equation tells you this vector triangle is a right triangle. Since $\vec{v}_1$ and $\vec{v}_2$ are the legs, the right angle is between them. Therefore:

$$\theta_1 + \theta_2 = 90°$$

The two outgoing velocity vectors are perpendicular. Always. Regardless of the impact parameter. A nearly head-on collision sends ball 1 barely deflected and ball 2 nearly straight ahead, but the angle between them is still 90 degrees. A very glancing collision sends ball 1 sideways and ball 2 barely moving forward, but the angle is still 90 degrees.

This result is specific to equal masses and elastic collisions. If the masses are unequal, the energy equation becomes $m_1 v_0^2 = m_1 v_1^2 + m_2 v_2^2$, which is not a Pythagorean relation between the speeds, and the right-angle conclusion breaks down.

Connection to Prior Ideas

Component independence (Section 3.4). In projectile motion, you treated $x$ and $y$ as separate problems: constant velocity horizontally, constant acceleration vertically. Here the same principle returns in a new context. The $x$- and $y$-components of momentum are conserved independently. Each direction has its own equation, its own accounting. The principle of component independence is not specific to kinematics --- it is a deep structural feature of how vectors work in physics.

Collision types (Section 8.4). Everything you learned about elastic versus inelastic collisions in one dimension carries over. Elastic collisions conserve kinetic energy; inelastic ones do not. The classification matters just as much in two dimensions. What changes is the geometry: the outgoing objects can now scatter in a plane, and the directional information is essential.

Vector decomposition (Chapter 3). The skill of breaking vectors into components and working with each component separately is the foundation of this entire section. If you are comfortable with $v_x = v\cos\theta$ and $v_y = v\sin\theta$, the physics here is the same conservation principle you already know --- just applied in two directions.

Spaced Retrieval

Before moving to practice, check your recall of earlier material.

Recall prompt 1: What is the relationship between impulse and momentum change? State it as an equation. (Section 8.1)

Recall prompt 2: Under what conditions is the total momentum of a system conserved? (Section 8.3)

Recall prompt 3: In a perfectly inelastic collision, what quantity is conserved and what quantity is not? (Section 8.4)

Practice Layers

Layer 1: Concrete --- Solve a 2D Collision

Problem 1. A 2.0 kg ball moving east at 5.0 m/s strikes a 3.0 kg ball initially at rest. After the collision, the 2.0 kg ball moves at 3.0 m/s in a direction 40 degrees north of east.

(a) Find the $x$- and $y$-components of the 3.0 kg ball's velocity after the collision.

(b) Find the speed and direction of the 3.0 kg ball after the collision.

(c) Is this collision elastic, inelastic, or perfectly inelastic? Justify with a calculation.

Check your answer **Setup.** Choose east as $+x$ and north as $+y$. Before the collision: ball 1 has $\vec{v}_{1i} = (5.0, \, 0)$ m/s and ball 2 has $\vec{v}_{2i} = (0, \, 0)$ m/s. After the collision: ball 1 has $v_1 = 3.0$ m/s at $\theta_1 = 40°$ north of east, so $v_{1x,f} = 3.0\cos 40° = 2.30$ m/s and $v_{1y,f} = 3.0\sin 40° = 1.93$ m/s. **(a) $x$-component:** $$m_1 v_{1x,i} + m_2 v_{2x,i} = m_1 v_{1x,f} + m_2 v_{2x,f}$$ $$(2.0)(5.0) + 0 = (2.0)(2.30) + (3.0)v_{2x,f}$$ $$10.0 = 4.60 + 3.0\,v_{2x,f}$$ $$v_{2x,f} = \frac{5.40}{3.0} = 1.80 \text{ m/s}$$ **$y$-component:** $$0 = m_1 v_{1y,f} + m_2 v_{2y,f}$$ $$0 = (2.0)(1.93) + (3.0)v_{2y,f}$$ $$v_{2y,f} = \frac{-3.86}{3.0} = -1.29 \text{ m/s}$$ The negative sign means ball 2 moves south (downward), as expected --- ball 1 went north, so ball 2 must go south to conserve the zero initial $y$-momentum. **(b)** Speed: $v_2 = \sqrt{(1.80)^2 + (-1.29)^2} = \sqrt{3.24 + 1.66} = \sqrt{4.90} = 2.21$ m/s. Direction: $\phi = \arctan\left(\frac{1.29}{1.80}\right) = 35.6°$ south of east. **(c)** Total kinetic energy before: $KE_i = \frac{1}{2}(2.0)(5.0)^2 = 25.0$ J. Total kinetic energy after: $KE_f = \frac{1}{2}(2.0)(3.0)^2 + \frac{1}{2}(3.0)(2.21)^2 = 9.0 + 7.33 = 16.3$ J. Since $KE_f < KE_i$ (and the balls do not stick together), this is an **inelastic** collision. About 8.7 J of kinetic energy was lost to deformation, sound, or heat.

Layer 2: Pattern --- The 90-Degree Rule

Problem 2. Two identical billiard balls (same mass $m$) collide elastically. Ball A has initial speed $v_0$ and ball B is at rest.

(a) After the collision, ball A is deflected 25 degrees from its original direction. What angle does ball B make with the original direction?

(b) Now suppose ball A is deflected 10 degrees. What angle does ball B make?

(c) In a nearly head-on elastic collision ($\theta_1 \approx 0°$), ball A nearly stops and ball B moves nearly straight ahead. Check that the 90-degree rule is still satisfied in this limiting case.

(d) Use the three conservation equations ($p_x$, $p_y$, and $KE$) to verify that $v_1^2 + v_2^2 = v_0^2$ implies $\theta_1 + \theta_2 = 90°$.

Check your answer **(a)** By the 90-degree rule: $\theta_1 + \theta_2 = 90°$, so $\theta_2 = 90° - 25° = 65°$. Ball B moves at 65 degrees from the original direction (on the opposite side). **(b)** $\theta_2 = 90° - 10° = 80°$. A barely deflected ball A means ball B moves nearly perpendicular to the original direction. **(c)** In the head-on limit, $\theta_1 \to 0°$. Ball A stops ($v_1 \to 0$) and ball B moves with speed $v_0$ along the original direction ($\theta_2 \to 0°$, but more precisely, $\theta_2$ is undefined since ball A isn't really deflected). The 90-degree rule still holds in the sense that the velocity triangle degenerates: $v_1 = 0$ means the "triangle" collapses to a line, and the right angle is between a zero-length leg and the full-length leg. Mathematically, $v_0^2 = 0^2 + v_0^2$ still satisfies the Pythagorean relation, so the structure is consistent. **(d)** The momentum equations give $\vec{v}_0 = \vec{v}_1 + \vec{v}_2$ (since the masses cancel). This means $\vec{v}_0$, $\vec{v}_1$, and $\vec{v}_2$ form a closed triangle. The energy equation gives $v_0^2 = v_1^2 + v_2^2$, which is the Pythagorean theorem with $v_0$ as the hypotenuse. By the converse of the Pythagorean theorem, the angle between the two legs ($\vec{v}_1$ and $\vec{v}_2$) must be 90 degrees. Therefore $\theta_1 + \theta_2 = 90°$.

Layer 3: Structure --- Why 2D Needs More Information

Problem 3. A student claims: "In 1D, I can solve an elastic collision with just the masses and initial velocities. Why can't I do the same in 2D?"

(a) In a 1D elastic collision, how many unknowns are there after the collision? How many independent equations do conservation of momentum and conservation of kinetic energy provide?

(b) In a 2D elastic collision (one object initially at rest), how many unknowns are there after the collision? How many independent equations do the conservation laws provide?

(c) What additional information do you typically need to fully solve a 2D collision? Give an example.

(d) Why does the dimensionality of the problem increase the number of unknowns faster than the number of equations?

Check your answer **(a)** In 1D, the unknowns after the collision are $v_{1f}$ and $v_{2f}$ --- two unknowns. Conservation of momentum provides one equation. Conservation of kinetic energy provides a second equation. Two equations, two unknowns: the system is fully determined. **(b)** In 2D, the unknowns after the collision are $v_1$, $\theta_1$, $v_2$, $\theta_2$ --- four unknowns (two speeds and two angles). Momentum conservation gives two equations ($x$ and $y$). Energy conservation gives one equation. That is three equations for four unknowns: the system is underdetermined. You cannot solve it without additional information. **(c)** You typically need one of the following: a measured outgoing angle (from experiment), the impact parameter (which determines the geometry of the contact), or some other constraint like "perfectly inelastic" (which removes two unknowns by requiring the objects to move together). In billiards, the impact parameter determines everything: the offset between the ball centers at contact fixes the geometry of the collision. **(d)** Each new dimension adds one new velocity component per object (one new unknown per object). With two objects, that is two new unknowns per added dimension. But each new dimension adds only one new momentum conservation equation. So the unknowns grow faster than the equations. In 1D: 2 unknowns, 2 equations. In 2D: 4 unknowns, 3 equations. In 3D: 6 unknowns, 4 equations. The gap widens with each dimension. The physics (conservation laws) does not fully constrain the outcome --- the *details of the interaction* (impact geometry, forces during contact) determine the rest.

Layer 4: Debug --- Scalar Momentum Is Not Enough

Problem 4. A student attempts to solve a 2D collision problem by conserving the total magnitude of momentum rather than its components. They write:

$$|\vec{p}{\text{total, before}}| = |\vec{p}|$$}

and use this as their conservation equation.

(a) Is this equation always true? Sometimes true? Never true?

(b) Consider a specific example: ball 1 ($m = 1$ kg) moves east at 4 m/s; ball 2 ($m = 1$ kg) moves north at 3 m/s. They collide and stick together. The student computes the "total momentum before" as $4 + 3 = 7$ kg m/s. What is the actual total momentum magnitude before the collision? Where did the student go wrong?

(c) Why does conserving components work when conserving magnitudes does not?

Check your answer **(a)** The equation $|\vec{p}_{\text{total, before}}| = |\vec{p}_{\text{total, after}}|$ is **always true** --- the magnitude of the total momentum is indeed conserved, because the total momentum *vector* is conserved, and the magnitude of a conserved vector is conserved. The equation is correct but *useless* in a different way than the student thinks. The problem is not with this equation itself; it is with how the student *computes* the magnitude. See part (b). **(b)** The student added the magnitudes of the individual momenta: $|p_1| + |p_2| = (1)(4) + (1)(3) = 7$ kg m/s. But the *total momentum* is a vector sum: $$\vec{p}_{\text{total}} = (4, \, 0) + (0, \, 3) = (4, \, 3) \text{ kg m/s}$$ $$|\vec{p}_{\text{total}}| = \sqrt{4^2 + 3^2} = \sqrt{25} = 5 \text{ kg m/s}$$ The magnitude of the total momentum is 5 kg m/s, not 7 kg m/s. The student confused the sum of magnitudes with the magnitude of the sum. These are only equal when all momenta point in the same direction. In general, $|p_1| + |p_2| \geq |\vec{p}_1 + \vec{p}_2|$ (the triangle inequality). After the collision (sticking together): combined mass is 2 kg, and the combined momentum is $(4, \, 3)$ kg m/s, so $v = 5/2 = 2.5$ m/s at $\arctan(3/4) = 36.9°$ north of east. The magnitude of the total momentum is still 5 kg m/s. **(c)** Conserving components works because momentum conservation is a *vector* equation: $\vec{p}_{\text{before}} = \vec{p}_{\text{after}}$. Two vectors are equal if and only if their corresponding components are equal. Components respect the directional structure of momentum. Simply adding magnitudes discards direction entirely --- it treats momentum as if balls moving east and balls moving north contribute to the same "total" in the same way, which they do not. Momentum in the $x$-direction and momentum in the $y$-direction are separate conserved quantities. They must be tracked separately.

Reflection

What does 2D collision analysis add beyond the 1D case?

In one dimension, direction was binary: left or right, positive or negative. Momentum conservation gave one equation, and that was often enough. In two dimensions, direction becomes continuous --- objects can scatter at any angle in a plane --- and momentum conservation gives two equations that must be satisfied simultaneously. The vector structure of momentum, which was always present but easy to overlook in 1D, becomes impossible to ignore.

The deeper lesson is that conservation of momentum is fundamentally a vector statement. Each component has its own independent conservation equation. You do not conserve "total momentum" as a single number; you conserve each direction's contribution separately. This is both more powerful (more equations to work with) and more demanding (more unknowns to manage, more geometry to navigate).

The 90-degree rule for equal-mass elastic collisions is a striking example of what emerges when you take the vector structure seriously. It is not an additional assumption --- it is a geometric consequence of momentum and energy conservation applied together. Results like this are the reward for keeping the vector framework intact.

Looking Ahead

You have now extended collision analysis from a line to a plane. The component-by-component approach --- treating each direction's momentum as its own conserved quantity --- is the essential technique.

But there is one more perspective on momentum that ties everything together. When two objects interact, each one does something complicated. One speeds up, the other slows down. One goes left, the other goes right. Tracking each object individually can be messy. In Section 8.6, you will discover that there is a single point --- the center of mass --- that moves as if none of the internal complexity exists. The center of mass of a system responds only to external forces, gliding along its predictable path while the individual pieces scatter and collide around it. It is the ultimate system-level simplification, and it will change how you think about every multi-body problem from here on.