Collisions in Two Dimensions

In carrom, you rarely hit a coin dead-centre. Most shots are glancing — the striker comes in at an angle, clips the edge of the coin, and both pieces scatter in different directions. The striker deflects one way, the coin flies off another way. If you watch closely, you notice something: the two paths after the collision look roughly perpendicular. The striker goes one way, the coin goes the other, and the angle between them is close to a right angle.

That observation is not an accident. It is a theorem — and it takes only three lines of algebra to prove. But first, you need the machinery to handle collisions that are not head-on: momentum conservation in two dimensions.

Momentum is a vector — three equations, four unknowns

In a head-on elastic collision, momentum conservation gives you one equation and energy conservation gives you another — two equations for two unknowns (v_1 and v_2), cleanly solvable. The collision is fully determined by the masses and the initial velocities.

In two dimensions, the problem opens up. A ball moving along the x-axis hits a stationary ball, and both scatter at angles. The final state has four unknowns: two speeds (v_1, v_2) and two angles (\theta_1, \theta_2) measured from the original direction of motion.

Conservation of momentum is now a vector equation:

This single vector equation splits into two scalar equations — one for each component:

Why: the initial momentum is entirely along x (ball 2 is at rest), so the y-component of total momentum was zero before and must be zero after. The opposite signs in equation (2) account for ball 1 scattering above the x-axis and ball 2 scattering below it.

Conservation of kinetic energy adds one more equation:

Three equations, four unknowns. Unlike the 1D case, the system is underdetermined. You cannot solve it without one additional piece of information — either a measured scattering angle or the geometry of how the two objects met. That missing input comes from the line of impact.

The line-of-impact decomposition

When two smooth spheres collide, the contact force between them acts along the line connecting their centres at the instant of contact. No other direction — Newton's third law guarantees that the force one sphere exerts on the other is directed along the line joining their centres. This line is the line of impact.

The insight that makes 2D collisions tractable: decompose the problem into two independent directions. Along the line of impact, the collision obeys the same 1D elastic formulas you already know. Perpendicular to it, nothing happens at all.

Suppose ball A approaches ball B (at rest), and the line of impact makes angle \alpha with ball A's velocity direction. Decompose ball A's velocity into two components:

• Along the line of impact (n-direction): \;u_{1n} = u_1 \cos\alpha
• Perpendicular to the line of impact (t-direction): \;u_{1t} = u_1 \sin\alpha

Ball B is at rest, so both its components are zero: u_{2n} = 0, u_{2t} = 0.

The contact force has no component in the t-direction. It acts purely along the line connecting the two centres. So the impulse during the collision is entirely along the n-direction, and the t-component of each ball's velocity is unchanged:

Why: no force means no impulse in the t-direction, so no change in t-momentum for either ball. Ball A keeps its tangential speed intact. Ball B, which had no tangential speed, still has none after the collision.

Along the line of impact, you have a standard one-dimensional elastic collision — mass m_1 arriving at u_{1n} hitting mass m_2 at rest. The 1D elastic collision formulas apply directly:

Why: along the line of impact, the physics is identical to a head-on collision. The formulas are the same — you just replace u_1 with the component u_1\cos\alpha that is aimed along the collision axis.

That is the entire method. Decompose. Solve the 1D problem along the line of impact. Keep the perpendicular component. Reconstruct the final velocity vectors. Ball B always moves purely along the line of impact (since v_{2t} = 0), at angle \alpha from the original direction. Ball A's final velocity combines its unchanged tangential component with its (possibly reversed) normal component — the worked examples below show the reconstruction step by step.

Equal masses — the 90° separation theorem

When m_1 = m_2, the oblique elastic collision produces a beautiful geometric result. The mass cancels from every equation:

Momentum: \;\vec{u}_1 = \vec{v}_1 + \vec{v}_2

Kinetic energy: \;u_1^2 = v_1^2 + v_2^2

The momentum equation says the three velocity vectors form a closed triangle — \vec{u}_1 is the vector sum of \vec{v}_1 and \vec{v}_2. The kinetic energy equation says the square of one side of that triangle equals the sum of the squares of the other two sides.

That is the Pythagorean theorem — in reverse. A triangle where c^2 = a^2 + b^2 is a right triangle, with the right angle between the sides of length a and b. So the angle between \vec{v}_1 and \vec{v}_2 must be 90°.

Here is the formal proof in three steps:

Step 1. Square the momentum equation.

Why: the identity |\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + 2\vec{a}\cdot\vec{b} + |\vec{b}|^2 expands the magnitude of a vector sum. The cross term 2\vec{v}_1\cdot\vec{v}_2 is the piece that will vanish.

Step 2. Subtract the kinetic energy equation from Step 1.

Subtracting:

Why: the v_1^2 and v_2^2 terms appear identically in both equations and cancel, leaving only the dot product.

Step 3. Conclude.

Why: two non-zero vectors whose dot product is zero are perpendicular. The final velocities make a 90° angle with each other — regardless of the impact angle \alpha.

This is the 90° separation theorem: in an elastic oblique collision between equal masses (with one initially at rest), the two final velocity vectors are always perpendicular. Watch it happen:

This is why billiard balls and carrom pieces scatter at roughly right angles. Every time you see a striker clip a coin off-centre, the two paths after the hit open up to nearly 90°. The result holds for any impact angle \alpha (as long as 0 < \alpha < 90°, i.e., an actual oblique collision occurs). At \alpha = 0 (head-on), ball A stops completely and v_1 = 0 — the angle is undefined. At \alpha = 90° (a grazing touch), ball A keeps all its speed and ball B barely moves.

When masses are unequal, the 90° result breaks. For a heavier target (m_2 > m_1), the projectile deflects more sharply and the separation angle exceeds 90°. For a lighter target (m_2 < m_1), the projectile barely deflects and the separation angle is less than 90°. The equal-mass case sits exactly on the boundary. Experimentally, measuring a separation angle tells you something about the mass ratio — an idea that becomes central in nuclear scattering.

Speeds and angles for equal masses

With m_1 = m_2, the 1D elastic formulas give v_{1n} = 0 and v_{2n} = u_{1n} along the line of impact — a complete transfer, just like the head-on case. The results are clean:

• Ball A keeps only its tangential component: v_1 = u_1 \sin\alpha, deflected at angle \theta_1 = 90° - \alpha from the original direction.
• Ball B moves along the line of impact: v_2 = u_1 \cos\alpha, at angle \theta_2 = \alpha from the original direction.

Check: \theta_1 + \theta_2 = (90° - \alpha) + \alpha = 90°. Always perpendicular. ✓

Check: v_1^2 + v_2^2 = u_1^2\sin^2\alpha + u_1^2\cos^2\alpha = u_1^2. Kinetic energy conserved. ✓

The total kinetic energy is always shared between the two balls. The impact angle \alpha determines the split: at \alpha = 45°, each ball gets exactly half the kinetic energy. At small \alpha (nearly head-on), most energy goes to ball B. At large \alpha (grazing), ball A keeps most of it.

Explore the impact angle

Drag the red point to see how the two final speeds vary with the impact angle for equal masses. The red curve is ball A's speed after the collision; the dark curve is ball B's.

Notice that \left(\frac{v_1}{u_1}\right)^2 + \left(\frac{v_2}{u_1}\right)^2 = \sin^2\alpha + \cos^2\alpha = 1 — kinetic energy conservation is encoded directly in the Pythagorean identity. The energy is always fully accounted for; the impact angle just determines how it is divided.

Worked examples

Common confusions

• "Two-dimensional collisions need different formulas." They do not. The elastic collision formulas v_{1n} = \frac{m_1 - m_2}{m_1 + m_2}\,u_{1n} and v_{2n} = \frac{2m_1}{m_1 + m_2}\,u_{1n} are the same 1D formulas from the previous article. The only new step is decomposing the velocity into components along and perpendicular to the line of impact. There is no new physics — only a new coordinate system.

• "The 90° theorem works for all elastic collisions." Only for equal masses with one at rest. If the masses differ, the separation angle differs: greater than 90° when the target is heavier, less than 90° when the target is lighter. If both objects are initially moving, the 90° result holds in the centre-of-mass frame but not in the lab frame. And for inelastic collisions, kinetic energy is lost, so the Pythagorean argument fails entirely.

• "A glancing collision barely transfers any energy." True for equal masses at large impact angles — ball A keeps most of its speed. But look at the energy split: ball A gets \sin^2\alpha of the total kinetic energy and ball B gets \cos^2\alpha. At \alpha = 45°, each gets exactly half. The key insight is that sin and cos are complementary — whatever ball A keeps, ball B gets the rest.

• "The scattering angle is the same as the impact angle." No — these are different quantities. The impact angle \alpha is the angle between the incoming velocity and the line of impact (a geometric property of how the balls met). The scattering angle \theta_1 is the angle between the final velocity of ball A and its original direction (an observable). For equal masses, \theta_1 = 90° - \alpha, so they are complementary, not equal.

• "The line of impact is always horizontal." It depends on where the target ball sits relative to the projectile's path. If the target is directly ahead, the line of impact is along the incoming velocity (\alpha = 0, head-on). If the target is offset above or below, the line of impact tilts by the corresponding angle. The line of impact is determined by the geometry of the collision, not by any preferred direction.

Where this leads next

• Elastic Collisions — the 1D foundation: head-on elastic collisions, the velocity-exchange formulas, and the three special cases.
• Inelastic Collisions — what happens when kinetic energy is not conserved, including the 2D inelastic case where objects stick together at angles.
• Centre of Mass — Definition and Calculation — the reference frame that simplifies all collision problems, especially in 2D and 3D.
• Conservation of Linear Momentum — the foundational law: total momentum is always conserved, regardless of how complex the collision geometry is.
• Variable Mass Systems — Rockets — extending momentum conservation to systems where mass changes during the interaction.