Collisions in two dimensions

Not every collision happens along a single straight line. A carrom striker glances off a coin at an angle and both go skidding away in different directions; two hockey pucks meet off-centre on the ice and scatter apart like a small firework; a snooker cue ball clips a red ball and the two split, one veering left, one veering right, the table suddenly full of new velocity vectors. None of this fits the tidy head-on picture of one-dimensional collisions, where everything moves along a single axis both before and after. The moment the impact is even slightly off-centre — which is the ordinary case in real life, not the exception — the outgoing velocities point off in two different directions, and a purely scalar bookkeeping of momentum stops being enough. This is the problem of the two-dimensional, or oblique, collision. The physics you already trust — momentum is always conserved in a collision, and kinetic energy is conserved only if the collision is elastic — does not change one bit. What changes is the bookkeeping: momentum is a vector, so "conserved" now has to be enforced separately along two perpendicular directions at once, turning one collision problem into a small simultaneous-equations problem in $x$ and $y$ . This article builds that machinery from the one-dimensional collisions you already know, shows where the elegant "right-angle" result for equal masses comes from, and walks through two full worked problems — a glancing elastic collision between two identical balls, and a genuinely 2-D perfectly inelastic collision between two cars meeting at a right-angle intersection — of exactly the kind that shows up in board and competitive papers.

Key takeaways

A two-dimensional (oblique) collision is one where the outgoing bodies do not move along the same line as the incoming body — the everyday, glancing kind of collision.
Momentum is a vector, so conservation of momentum must be applied separately along the x-direction and the y-direction — two independent scalar equations from one vector law.
For an elastic 2-D collision, kinetic energy conservation supplies a third, scalar equation, giving three equations in total to pin down unknown speeds and angles.
For an inelastic 2-D collision, kinetic energy is not conserved, so only the two momentum-component equations are available — extra given information (like both angles, or that the bodies stick together) is needed.
A famous special case: an elastic collision between equal masses, one initially at rest, always sends the two bodies off at right angles to each other after impact, $\theta_1 + \theta_2 = 90°$ .
A perfectly inelastic 2-D collision (bodies stick together) is solved most simply by adding momentum components first and dividing by the total mass at the end, rather than tracking two separate velocity vectors.
Key idea: in an oblique collision, momentum conservation is applied component-by-component along perpendicular x and y axes, and — only if the collision is elastic — the scalar kinetic-energy equation supplies a third relation, together enough to solve for unknown final speeds and directions.

The core insight

Because momentum is a vector, a two-dimensional collision is solved by writing separate x- and y-momentum conservation equations, adding a kinetic-energy equation only when the collision is elastic.

Where it fits in the syllabus

This topic sits inside the CBSE Class 11 unit Work, Energy and Power, directly after one-dimensional elastic and inelastic collisions, and it is explicitly named in the prescribed scope: "elastic and inelastic collisions in one and two dimensions" with "momentum conservation in collisions." It leans on vector resolution from Motion in a Plane and on the definitions of kinetic energy and elastic/inelastic collisions you have already met in this same unit. One honest scope note: the board syllabus keeps the treatment at the level of writing and solving the momentum and (for elastic collisions) energy equations for given angles or masses; a formal, derived "coefficient of restitution" law is outside the prescribed scope here and is mentioned only in passing, qualitatively, never as a formula to be derived or applied numerically.

The physical picture

Picture a body of mass $m_1$ moving with speed $u_1$ along a straight line, about to strike a second body of mass $m_2$ that is initially at rest — the standard Class 11 setup, and the one used in both worked examples below. If the collision were perfectly head-on, both bodies would continue along that same original line. An oblique collision is different only in where the impact lands: the moment the point of contact is even a little off-centre, the impulse each body feels has a component perpendicular to the original line of motion, and the outgoing bodies peel off at angles $\theta_1$ and $\theta_2$ , measured from the original direction, on opposite sides of it. Because momentum $\vec p = m\vec v$ is a vector quantity, the law "total momentum before equals total momentum after" is really two independent scalar statements once you resolve everything along a chosen x-axis (usually the original direction of motion) and a perpendicular y-axis:

\sum p_{x,\text{before}} = \sum p_{x,\text{after}}, \qquad \sum p_{y,\text{before}} = \sum p_{y,\text{after}}

where: $p_x$ and $p_y$ are the x- and y-components of the momentum of each body, and the sums run over all bodies in the system. The physical reading is simple — momentum sideways cannot appear from nowhere, and momentum along the original line cannot vanish into nothing; each direction keeps its own separate books. For the standard setup — $m_1$ moving at $u_1$ along the x-axis, striking stationary $m_2$ , with $m_1$ leaving at speed $v_1$ and angle $\theta_1$ and $m_2$ leaving at speed $v_2$ and angle $\theta_2$ on the opposite side of the x-axis — the two component equations become:

m_1u_1 = m_1v_1\cos\theta_1 + m_2v_2\cos\theta_2

0 = m_1v_1\sin\theta_1 - m_2v_2\sin\theta_2

where: the first line balances momentum along the original direction of motion (the x-axis), and the second balances momentum perpendicular to it (the y-axis) — it reads zero on the left because neither body had any sideways momentum before the collision, so the two sideways components after must exactly cancel. If the collision is elastic, a third, purely scalar equation joins these two:

\frac{1}{2}m_1u_1^2 = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2

where: this states that the total kinetic energy is unchanged by the collision — no energy is spent deforming the bodies or generated as heat and sound. Three equations, up to three unknowns: given enough of $m_1, m_2, u_1, \theta_1, \theta_2, v_1, v_2$ , the rest can be solved algebraically, with no calculus required anywhere in this topic.

Figure

Two circles before and after an oblique collision, one moving along a horizontal line striking a stationary one, then splitting into two velocity vectors at angles theta one and theta two above and below the original line

The right-angle rule — derivation

Derivation AWhy equal-mass elastic collisions separate at ninety degrees

The goal is to prove, using only the two momentum-component equations and the kinetic-energy equation already stated above, that when a moving body strikes an equal-mass stationary body in a perfectly elastic collision, the two bodies always fly apart at exactly $90°$ to each other afterward — a clean geometric fact that follows from nothing more than algebra, with no calculus and no extra assumption beyond equal masses and elasticity. This result is worth deriving once carefully because it turns a two-unknown-angle problem into a one-unknown problem, which is exactly what makes Worked Example 1 below solvable so quickly.

Step 1. Set $m_1 = m_2 = m$ in the general momentum equations for a body of mass $m$ , speed $u$ , striking an equal stationary mass, with outgoing speeds $v_1, v_2$ at angles $\theta_1, \theta_2$ on opposite sides of the original line. The mass $m$ cancels out of every term in both the x- and y-momentum equations:

u = v_1\cos\theta_1 + v_2\cos\theta_2 \qquad \text{and} \qquad 0 = v_1\sin\theta_1 - v_2\sin\theta_2

Step 2. Square both equations and add them. Squaring the first gives $u^2 = v_1^2\cos^2\theta_1 + v_2^2\cos^2\theta_2 + 2v_1v_2\cos\theta_1\cos\theta_2$ . Squaring the second gives $0 = v_1^2\sin^2\theta_1 + v_2^2\sin^2\theta_2 - 2v_1v_2\sin\theta_1\sin\theta_2$ . Adding these two results and using $\sin^2\theta + \cos^2\theta = 1$ for each body collapses the squared terms neatly:

u^2 = v_1^2 + v_2^2 + 2v_1v_2\big(\cos\theta_1\cos\theta_2 - \sin\theta_1\sin\theta_2\big)

Step 3. Recognise the compound-angle identity. The bracket $\cos\theta_1\cos\theta_2 - \sin\theta_1\sin\theta_2$ is exactly $\cos(\theta_1+\theta_2)$ , so:

u^2 = v_1^2 + v_2^2 + 2v_1v_2\cos(\theta_1+\theta_2)

This single equation is just the vector law of cosines applied to the triangle formed by $\vec u$ , $\vec v_1$ , $\vec v_2$ — a purely geometric restatement of the two momentum equations combined.

Step 4. Bring in the elastic-collision condition. With equal masses, kinetic-energy conservation $\tfrac12 mu^2 = \tfrac12 mv_1^2 + \tfrac12 mv_2^2$ simplifies to $u^2 = v_1^2 + v_2^2$ once $m$ cancels. Substitute this directly into the result of Step 3:

v_1^2 + v_2^2 = v_1^2 + v_2^2 + 2v_1v_2\cos(\theta_1+\theta_2)

Step 5. Solve for the angle. Subtracting $v_1^2+v_2^2$ from both sides leaves $0 = 2v_1v_2\cos(\theta_1+\theta_2)$ . Since neither ball is at rest after a genuine collision, $v_1$ and $v_2$ are both nonzero, so the only way the product can vanish is:

\cos(\theta_1+\theta_2) = 0 \quad\Longrightarrow\quad \theta_1+\theta_2 = 90°

This is the right-angle rule: for an elastic collision between equal masses, one initially at rest, the two outgoing velocity vectors are always perpendicular, regardless of the actual value of $\theta_1$ or of $u$ . It is a geometric consequence of combining vector-momentum conservation with scalar kinetic-energy conservation for equal masses — change either the mass ratio or the elasticity and the right angle no longer holds.

Worked examples

Board level - equal-mass elastic collision at an angle

A ball of mass $2\ \text{kg}$ moving at $10\ \text{m/s}$ along the x-axis strikes an identical stationary ball of mass $2\ \text{kg}$ , and the collision is elastic. After the collision, the first ball moves off at $30°$ above the x-axis. Find the direction of the second ball's motion and the final speed of each ball.

Step 1. Apply the right-angle rule proved above. Because the masses are equal and the collision is elastic with one ball initially at rest, $\theta_1 + \theta_2 = 90°$ . With $\theta_1 = 30°$ :

\theta_2 = 90° - 30° = 60°

on the opposite side of the x-axis from ball 1, since the y-momentum equation requires the two sideways components to cancel.

Step 2. Write momentum conservation along x and y with the given numbers, $m_1=m_2=2\ \text{kg}$ , $u_1=10\ \text{m/s}$ :

\text{x:} \quad 2(10) = 2v_1\cos30° + 2v_2\cos60°

\text{y:} \quad 0 = 2v_1\sin30° - 2v_2\sin60°

Dividing both by the common mass factor of $2$ :

10 = 0.866\,v_1 + 0.5\,v_2 \quad \text{...(i)} \qquad\qquad 0 = 0.5\,v_1 - 0.866\,v_2 \quad \text{...(ii)}

Step 3. Solve the pair of equations. From (ii), $v_1 = 1.732\,v_2$ . Substituting into (i):

10 = 0.866(1.732\,v_2) + 0.5\,v_2 = 1.5\,v_2 + 0.5\,v_2 = 2\,v_2

\boxed{v_2 = 5\ \text{m/s}}, \qquad \boxed{v_1 = 1.732(5) \approx 8.66\ \text{m/s}}

Step 4. Check with kinetic energy, as a consistency test (never a substitute for the momentum equations). Before: $\tfrac12(2)(10)^2 = 100\ \text{J}$ . After: $\tfrac12(2)(8.66)^2 + \tfrac12(2)(5)^2 = 75 + 25 = 100\ \text{J}$ . The two match, confirming the answer. Ball 1 leaves at about $8.66\ \text{m/s}$ at $30°$ , ball 2 at $5\ \text{m/s}$ at $60°$ , on opposite sides of the original line, separated by exactly $90°$ .

Competitive level - perfectly inelastic 2-D collision at an intersection

A car of mass $1000\ \text{kg}$ travelling east at $15\ \text{m/s}$ collides at a right-angle intersection with a second car of mass $1500\ \text{kg}$ travelling north at $10\ \text{m/s}$ . The two cars lock together on impact (a perfectly inelastic 2-D collision). Find the speed and direction of the combined wreckage immediately after the collision.

Step 1. Take east as the x-axis and north as the y-axis, and find the x- and y-momentum components before the collision separately — this is the key move for a 2-D inelastic problem, since there is only one final velocity to solve for once the masses combine.

p_x = 1000(15) + 1500(0) = 15{,}000\ \text{kg}\,\cdot\,\text{m/s}, \qquad p_y = 1000(0) + 1500(10) = 15{,}000\ \text{kg}\,\cdot\,\text{m/s}

Step 2. Find the combined mass after the two bodies stick together.

M = 1000+1500 = 2500\ \text{kg}

Step 3. Divide each momentum component by the combined mass to get the velocity components of the wreckage — this is exactly the perfectly-inelastic-collision idea from one dimension, applied once along each axis.

v_x = \frac{p_x}{M} = \frac{15{,}000}{2500} = 6\ \text{m/s}, \qquad v_y = \frac{p_y}{M} = \frac{15{,}000}{2500} = 6\ \text{m/s}

Step 4. Combine the components into a single speed and direction using the Pythagorean theorem and the tangent ratio.

v = \sqrt{v_x^2+v_y^2} = \sqrt{36+36} = \sqrt{72} \approx 8.49\ \text{m/s}, \qquad \tan\phi = \frac{v_y}{v_x} = \frac{6}{6}=1 \Rightarrow \phi = 45°

The wreckage moves off at about $8.49\ \text{m/s}$ , at $45°$ north of east — that is, straight along the diagonal, which makes sense since both cars carried equal momentum magnitude ( $15{,}000\ \text{kg}\,\cdot\,\text{m/s}$ each) into the crash. Timing note: in an exam, resist the temptation to hunt for a fancy 2-D shortcut here — a perfectly inelastic collision, in one dimension or two, is always fastest by summing momentum components first and dividing by the total mass only at the very end. Budget about 3 minutes: one minute to get $p_x$ and $p_y$ , one minute for $v_x$ and $v_y$ , one minute for the Pythagorean combination and the angle — do not compute two separate "velocity vectors" and try to add them, which wastes time and invites arithmetic slips.

Common traps

Trying to conserve momentum as a single scalar equation in a 2-D collision — momentum is a vector, and each perpendicular component must be conserved on its own; adding speeds directly (ignoring direction) is not conservation of momentum at all.
Using the kinetic-energy conservation equation for a collision that is stated (or implied) to be inelastic — in 2-D inelastic collisions only the two momentum-component equations are valid, and pulling in an energy equation anyway will give a wrong, overdetermined answer.
Assuming the right-angle separation result applies generally — it is a special result, proved above, that holds only for an elastic collision between equal masses with one initially at rest; it fails for unequal masses and it fails for any inelastic collision.
Forgetting to assign consistent positive x- and y-directions and correct signs to angles measured on opposite sides of the initial line of motion — a sign slip in one component equation throws off the entire solution, since the two equations are solved simultaneously.
Quoting or attempting to derive a numeric "coefficient of restitution" for a 2-D collision — this is outside the current CBSE scope for this topic; restitution is mentioned only as a qualitative idea (how "bouncy" a collision is), never as a formula to compute with here.

Common trapMomentum conservation in 2-D is two separate equations, not one

The single biggest mark-loser in this topic is treating "momentum is conserved" as though it meant one scalar sentence that can be written down once and solved directly for a final speed — the way it works in one dimension. In two dimensions that instinct is wrong: momentum is a vector, so conservation genuinely means two independent scalar equations, one for the x-direction and one for the y-direction, that must be solved together (simultaneously, not one after the other in isolation). Skipping the y-equation, or silently assuming the sideways components must be zero without checking, is the single most common way marks are lost on this topic — always write both component equations explicitly before attempting to solve, even when the answer looks "obvious" from the geometry.

You should be able to…

Write momentum conservation as two separate scalar equations, one along x and one along y, for an oblique (2-D) collision.
State when a third, kinetic-energy equation may be added (only for elastic collisions) and when it must not be (inelastic collisions).
Derive, or at least confidently apply, the right-angle separation result for an elastic collision between equal masses with one initially at rest.
Solve a perfectly inelastic 2-D collision by adding momentum components first, then dividing by the total mass to get the final velocity.
Combine velocity components into a resultant speed and direction using the Pythagorean theorem and the tangent ratio.
Check a 2-D elastic-collision answer for consistency using kinetic-energy conservation as an independent numerical check.

Ask the tutor: Want to see how the right-angle rule changes (or breaks down) when the two masses are unequal — ask the AI tutor to redo the derivation with $m_1 \ne m_2$ and show exactly where the clean cancellation stops working.

Why does the right-angle rule need equal masses AND an elastic collision — can it work with just one of those?

No, both conditions are needed together, and the derivation above shows exactly why. The ninety-degree result came from combining the vector law of cosines (built purely from the two momentum equations, valid for any masses) with the elastic-collision condition $u^2 = v_1^2+v_2^2$ , which itself only simplified to that clean form because the masses were equal (the mass cancelled out of the kinetic-energy equation). If the masses are unequal, kinetic-energy conservation no longer simplifies that way, and $\theta_1+\theta_2$ works out to something other than $90°$ . If the collision is inelastic, there is no kinetic-energy equation to substitute at all, so the derivation cannot even get started. Both ingredients are load-bearing.

In the board-level worked example, how did we know ball two goes on the opposite side of the x-axis from ball one?

It comes straight out of the y-momentum equation. Before the collision there is zero momentum in the y-direction (both balls move purely along x), so after the collision the y-components of the two balls' momenta must add to zero — meaning they must be equal in size and opposite in sign. If ball one's y-component is positive (angled above the x-axis), ball two's y-component must be negative (angled below it), otherwise their sum could never be zero. This is a direct, checkable consequence of the second conservation equation, not an assumption pulled out of a diagram.

Do I need to memorise the right-angle rule, or should I be able to re-derive it in an exam?

For CBSE purposes, knowing and correctly applying the right-angle rule (equal masses, elastic, one at rest, gives $\theta_1+\theta_2=90°$ ) is what boards actually test, and that is enough to solve problems like Worked Example 1 quickly. But walking through the four-step derivation above once, carefully, is worth doing anyway — it is short, uses only algebra you already know (squaring, adding, one trig identity), and it is exactly the kind of "prove this special result" question that occasionally appears in competitive papers. Understanding why the rule holds also protects you from misapplying it to unequal masses or inelastic collisions, which is the trap examiners most often set.

What actually happens physically during the tiny instant of a 2-D collision — why do the bodies get deflected sideways at all?

During the brief contact, each body exerts a force on the other along the line joining their centres at the instant of impact (the "line of impact"), not necessarily along the original direction of motion. If the impact is off-centre, that line of impact is tilted relative to the incoming velocity, so the impulse each body receives has a component perpendicular to its original path — that is the sideways push that deflects it. You are not expected to compute this contact-force geometry directly in Class 11; the point of the momentum and energy conservation equations used throughout this article is that they let you skip past the messy details of the contact force entirely and jump straight to the "before versus after" relationship, which is all that is examinable here.