Collisions in two dimensions

A carom striker glances off a coin, or two cars meet at an intersection at an angle — not every collision happens along one straight line. These oblique, two-dimensional collisions still obey momentum conservation, but momentum is a vector, so the bookkeeping must be done separately along two perpendicular directions rather than along one. At this level the treatment stays qualitative: you learn to set up x- and y-component equations, to use them when enough information is given, and to recognise a famous special case — two equal masses, one initially at rest, colliding elastically — where the two bodies always fly apart at exactly $90°$ to each other. The same idea that solved 1-D collisions now solves 2-D ones, just twice over.

The core insight

In any 2-D collision, momentum is conserved independently along the x-axis and the y-axis, giving two scalar equations instead of one.

The core idea

An oblique collision looks messy — bodies come in and leave at odd angles — but the underlying rule is the same conservation law used for head-on collisions, applied component by component.

Vector momentum splits into two scalar laws. Because momentum $\vec{p} = m\vec{v}$ is a vector, "total momentum is conserved" really means its x-component and its y-component are each separately conserved, since no external force acts on the isolated system during the brief collision:

\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 components of each body's momentum along the chosen x- and y-axes, and the sums run over all colliding bodies. This gives two independent scalar equations in place of the single equation used for 1-D collisions.

Why a full solution usually needs extra data. A general oblique collision has more unknowns — two final speeds and two final directions — than the two momentum equations (plus, for an elastic collision, one energy equation) can pin down. Something else must be given, such as the angle at which one body scatters, which in turn depends on the impact parameter (how off-centre the collision is). That is why ISC Class 11 treats the general 2-D case only qualitatively, while still expecting component-wise calculations when the missing pieces are supplied.

The equal-mass perpendicular-paths result. For one particular, well-known case — an elastic collision between two equal masses where the target is initially at rest, and the collision is not head-on — the two bodies always move off at $90°$ to each other after the collision. This single qualitative fact (seen constantly in carom and billiards) follows from combining momentum conservation with kinetic energy conservation for equal masses, without needing the full oblique-elastic-collision algebra.

Elastic versus inelastic in 2-D. Kinetic energy is conserved only in elastic 2-D collisions, exactly as in one dimension. Inelastic 2-D collisions — including the "stick together" case below — still conserve momentum component-wise, but kinetic energy is lost to heat, sound, and deformation.

Figure

Two balls approach along perpendicular directions and merge into one ball moving off at an angle between them

Derivation AComponent-wise momentum conservation for a 2-D collision

The goal is to show, from the vector statement of momentum conservation, exactly how the x- and y-momentum totals before a collision fix the final velocity components after it, so that a combined-mass or general final velocity can be read off directly once each axis is treated on its own.

Step 1. Write total momentum before the collision as a vector sum, $\vec{P}_i = \sum m_k \vec{u}_k$ , where $\vec{u}_k$ is the initial velocity of body $k$ . Resolve every velocity into x- and y-components using the chosen axes, so $\vec{P}_i$ has components $P_{x,i} = \sum m_k u_{x,k}$ and $P_{y,i} = \sum m_k u_{y,k}$ .

Step 2. Since no external force acts during the collision (the mutual contact forces are internal and cancel in pairs by Newton's third law), the total vector momentum is unchanged: $\vec{P}_i = \vec{P}_f$ . A vector equation is equivalent to its two component equations, so this single statement splits into $P_{x,i} = P_{x,f}$ and $P_{y,i} = P_{y,f}$ , which is exactly the pair of scalar laws quoted above.

Step 3. If the bodies stick together (perfectly inelastic) with combined mass $M = \sum m_k$ moving at velocity $(v_x, v_y)$ , the final-momentum components are simply $Mv_x$ and $Mv_y$ , so $v_x = P_{x,i}/M$ and $v_y = P_{y,i}/M$ . The resultant speed and direction then follow from ordinary vector-to-polar conversion: $v = \sqrt{v_x^2 + v_y^2}$ and $\theta = \tan^{-1}(v_y/v_x)$ measured from the x-axis.

Step 4. This is the entire method for any component-wise 2-D collision problem at this level: resolve, apply $P_{x,i} = P_{x,f}$ and $P_{y,i} = P_{y,f}$ separately, then recombine into a magnitude and direction if the bodies move off together, or use the two equations (with any extra given angle) to pin down separate final velocities if they do not.

Worked example

A ball A of mass $2\ \text{kg}$ moving at $3\ \text{m/s}$ along the positive x-direction collides with a ball B of mass $2\ \text{kg}$ moving at $4\ \text{m/s}$ along the positive y-direction, and the two balls stick together. Find the velocity (magnitude and direction) of the combined mass just after the collision.

Step 1. Find the initial momentum along each axis separately. Only ball A has x-momentum, and only ball B has y-momentum:

p_{x,i} = m_Au_A = (2)(3) = 6\ \text{kg}\,\cdot\,\text{m/s}, \qquad p_{y,i} = m_Bu_B = (2)(4) = 8\ \text{kg}\,\cdot\,\text{m/s}

Step 2. Since no external force acts, these two totals are unchanged after the collision. With the combined mass $m_A + m_B = 4\ \text{kg}$ moving with components $v_x, v_y$ :

v_x = \frac{p_{x,i}}{4} = \frac{6}{4} = 1.5\ \text{m/s}, \qquad v_y = \frac{p_{y,i}}{4} = \frac{8}{4} = 2\ \text{m/s}

Step 3. Combine the components into a resultant speed and direction:

v = \sqrt{v_x^2+v_y^2} = \sqrt{1.5^2+2^2} = \sqrt{6.25} = 2.5\ \text{m/s}, \qquad \theta = \tan^{-1}\left(\frac{v_y}{v_x}\right) = \tan^{-1}\left(\frac{2}{1.5}\right) \approx 53.1°

The combined mass moves off at $2.5\ \text{m/s}$ , at about $53.1°$ to ball A's original direction — found entirely by applying momentum conservation independently along the two perpendicular axes and then recombining.

Competitive stretch

A moving ball of mass $m$ collides elastically and obliquely (not head-on) with an identical stationary ball of mass $m$ . One ball is observed to move off at $35°$ to the original direction. Using the perpendicular-paths property of equal-mass elastic collisions, find the direction of the other ball.

Step 1. For an elastic collision between equal masses with the target initially at rest, the two final paths are always perpendicular to each other, so $\theta_1 + \theta_2 = 90°$ .

Step 2. With $\theta_1 = 35°$ , the second ball moves off at $\theta_2 = 90° - 35° = 55°$ on the other side of the original line of motion — obtained without solving the full momentum-and-energy system.

Common traps

Treating 2-D momentum conservation as one scalar equation — it must be applied separately for x and y; adding momenta as plain numbers regardless of direction gives a wrong resultant.
Assuming the equal-mass perpendicular-paths result applies to any 2-D elastic collision — it needs equal masses and an initially stationary target; other mass ratios give a different, non- $90°$ angle.
Forgetting that if the initial momentum along an axis is zero, the final momentum along that axis must also be zero — a fast consistency check.
Trying to solve a general oblique collision fully (all final speeds and angles) from momentum conservation alone — without an extra given piece of information the system is under-determined.

Ask the tutor: Want to work through a few more carom or billiards-style oblique collision problems together?

Why can't we just add the two balls' speeds directly in a 2-D collision?

Because momentum is a vector, and speeds are not added like plain numbers unless the velocities point the same way. In an oblique collision the two initial velocities usually point in different directions, so you must resolve each one into x- and y-components first, conserve each component separately, and only then combine the final $v_x$ and $v_y$ back into a resultant speed and direction using $v = \sqrt{v_x^2+v_y^2}$ . Adding raw speeds ignores direction entirely and gives a wrong, usually too-large, answer.

Does the 90-degree rule for equal masses apply to inelastic collisions too?

No. The perpendicular-paths result depends on both momentum conservation and kinetic energy conservation holding simultaneously for equal masses with one body initially at rest — that combination is unique to elastic collisions. In an inelastic 2-D collision, kinetic energy is lost, so the energy equation that forces the $90°$ result no longer applies, and the two bodies (or the combined body, if they stick together) can move off at any angle consistent with momentum conservation alone.

Do I need to derive the general formula for oblique elastic collisions for my exam?

No — at the ISC Class 11 level, the general oblique (2-D) collision is treated qualitatively, not through a derived general formula. You are expected to know that momentum conservation splits into independent x- and y-equations, to solve numerical problems where enough information (like a given scattering angle) is supplied, and to state and use the equal-mass perpendicular-paths result. A full quantitative treatment of arbitrary oblique elastic collisions, involving the impact parameter, goes beyond this syllabus.

Collisions in two dimensions

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads