In one dimension everything happens along a single line, so you just add momenta with plus and minus signs. The moment a collision spreads out into a plane — a snooker break, two cars meeting at a junction, a firework bursting — you need a smarter trick. Here it is: split every velocity into an across (x) part and an up-down (y) part, and treat those two directions as two completely separate 1D problems. Momentum is conserved in each direction on its own.
📘 What you need to know
In 2D, momentum is conserved separately in the x (horizontal) and y (vertical) directions
Resolve each velocity into components using cos for the horizontal part and sin for the vertical part
Total x-momentum before = total x-momentum after; the same holds for y
This gives you two equations — enough to solve for two unknowns (like a missing speed and angle)
An explosion starts with zero momentum, so all the fragments’ momenta must add (as vectors) back to zero
Momentum is always conserved; kinetic energy is only conserved in an elastic collision
Why one dimension isn’t enough
Think back to a head-on collision. Both objects move along the same straight line, so you pick a positive direction and every momentum is either + or −. Easy. But when a moving ball glances off another, the two fly apart at angles. Their velocities now point in genuinely different directions, and you can’t capture that with a single number line.
The fix is to notice that a vector can be broken into two pieces at right angles: how much it points sideways, and how much it points up or down. Because those two directions are independent, momentum bookkeeping in the x direction never “leaks” into the y direction. So one messy 2D problem becomes two tidy 1D problems that you already know how to solve.
Here’s the mental image I use: shine a torch straight down on the moving balls and watch only their shadows sliding left-right — that’s the x problem. Then shine it from the side and watch the shadows moving up-down — that’s the y problem. Each shadow obeys conservation of momentum by itself. Solve both, and you’ve solved the whole thing.
Resolving a velocity into components
Every velocity at an angle θ to the horizontal splits into two parts. The part running along the horizontal is the cosine part; the part running up (or down) is the sine part.
Components of a velocityvx = v cos θvy = v sin θ
Any velocity at angle θ splits into a horizontal part (v cos θ) and a vertical part (v sin θ). The two components and the velocity form a right-angled triangle.
Collisions in two dimensions
Picture a moving ball A striking a stationary ball B and glancing off. After the hit, A heads off at some angle above the original line and B shoots off below it. Both now have horizontal and vertical velocity components.
A glancing collision: A comes in along the dashed line, then A and B leave at angles above and below it. Each has both an x and a y velocity component.
Now write conservation of momentum twice — once for each direction. Taking rightwards and upwards as positive, and remembering ball B starts at rest:
Horizontal (x) momentummAuA = mAvA cos θA + mBvB cos θB
Vertical (y) momentum
0 = mAvA sin θA − mBvB sin θB
The vertical total starts at zero because nothing was moving up or down before the hit. The minus sign appears because A goes up while B goes down — they carry opposite vertical momentum, and those must cancel.
A lovely shortcut worth remembering: when two objects of equal mass have a perfectly elastic collision and one was at rest, they always fly apart at exactly 90° to each other. So θA + θB = 90°. If a question quietly tells you the collision is elastic and the masses match, you already know the angle between them.
🛠️ Solving a 2D collision
Draw before and after — mark every mass, speed and angle you’re given.
Pick your positive directions — usually right for x, up for y.
Write the x equation: total momentum before = total after, using cos for each angled velocity.
Write the y equation: same again, using sin. Watch the signs — up is +, down is −.
Solve the pair for your two unknowns. The y equation is often the simplest one to start with.
WE 1
A puck A of mass 0.20 kg slides right at 3.0 m s−1 and strikes a stationary puck B of the same mass. After the collision, A moves off at 30° above the original line and B at 60° below it. Find the speed of each puck after the collision.
Step 1 — vertical momentum (starts at zero)
0 = mAvA sin30 − mBvB sin60
masses are equal, so they cancel: vA sin30 = vB sin60Step 2 — horizontal momentum(0.20)(3.0) = 0.20 vA cos30 + 0.20 vB cos60
3.0 = vA cos30 + vB cos60
Step 3 — solve the pairvB = 1.5 m s⁻¹, then vA = 1.5 × sin60 ÷ sin30vA = 2.6 m s⁻¹, vB = 1.5 m s⁻¹The angles add to 90° and the masses match — a nice check that this is an elastic collision.
Explosions in two dimensions
An explosion is a collision run backwards. One object sits still, then bursts into pieces that scatter in different directions. Because nothing was moving beforehand, the total momentum before is zero — so the momenta of all the fragments must add up (as vectors) to zero afterwards.
In practice that means the x-momenta of the fragments cancel among themselves, and so do the y-momenta. If two fragments fly one way, a third must carry exactly the right momentum to balance them out.
Fragments 1 and 2 carry equal momentum east and north. The third fragment must carry momentum to the south-west so that all three add to zero — the same total the object had before it burst.
WE 2
A stationary shell bursts into three fragments. A 2.0 kg piece flies east at 6.0 m s−1, and a 2.0 kg piece flies north at 6.0 m s−1. The third piece has mass 3.0 kg. Find its speed and direction.
Step 1 — total momentum before = 0, so fragments must sum to 0
p₁ = 2.0 × 6.0 = 12 kg m s⁻¹ east
p₂ = 2.0 × 6.0 = 12 kg m s⁻¹ north
Step 2 — the third piece balances bothp₃ must be 12 west and 12 south
p₃ = √(12² + 12²) = √288 = 17.0 kg m s⁻¹
Step 3 — divide by its mass for speedv₃ = 17.0 ÷ 3.0v₃ = 5.7 m s⁻¹, 45° south of westEqual east and south momentum → the direction sits exactly halfway between, at 45°.
💡 Top tips
Two directions, two equations. Always write momentum conservation once for x and once for y — never try to do it in one line.
cos for along, sin for across. If the angle is measured from the horizontal, the horizontal part uses cos and the vertical part uses sin.
Start with the zero equation. The direction where total momentum is zero (often the vertical) usually gives the cleanest first step.
Check your calculator is in degrees if the angles are given in degrees — a classic silent error.
Quick recap: Split every velocity into x and y parts (cos and sin), then conserve momentum in each direction separately. That turns a 2D collision or explosion into two ordinary 1D problems. Explosions start from zero momentum, so all the fragments’ momenta must vector-add back to zero.
⚠ Common mistakes
Trying to add momenta at angles directly instead of resolving them into components first
Swapping sin and cos — check whether your angle is measured from the horizontal or the vertical
Forgetting the sign on a vertical component when one object goes up and the other goes down
Assuming kinetic energy is conserved in every collision — only elastic ones conserve it, though momentum always is
Leaving a final answer as a component only, and forgetting to combine the x and y parts back into a single speed and direction
That completes the “Forces & Momentum” collision story — you can now handle momentum in a straight line, at an angle, and in a full plane. Next we swing away from straight-line motion entirely and start going round in circles, beginning with Angular Velocity: how we measure motion around a circle using radians instead of metres.
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