Visual walkthrough — Elastic collisions — 2D - angle relationship

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Step 1 — Draw the "before" picture

WHAT. One ball is moving; a second identical ball sits still. We freeze the instant before they touch.

WHY. Every collision problem starts by naming what's moving and what isn't. If we don't fix a picture and a direction, "left" and "right" have no meaning and no equation can be written.

PICTURE. Look at the arrow. The moving ball (magenta) carries a velocity we call ${\vec{u}}_1$. The little arrow over the letter means "this is a vector" — it has both a size (how fast) and a direction (which way). The struck ball (violet) has no arrow: it is at rest.

We point ${\vec{u}}_1$ straight along a horizontal line and call that line the x-direction. Everything perpendicular to it (up/down) is the y-direction. That choice costs nothing and makes the algebra clean.

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Step 2 — Draw the "after" picture

WHAT. After the bump the balls separate. Ball 1 flies off above the line, ball 2 flies off below it.

WHY. A 2D collision means the balls can leave the line. We must allow two new directions, so we need two new angles to describe them.

PICTURE. The magenta arrow ${\vec{v}}_1$ leaves at angle ${\theta }_1$ above the old line; the violet arrow ${\vec{v}}_2$ leaves at angle ${\theta }_2$ below it. The angle ${\theta }_1$ is measured from the dashed old line up to the magenta arrow — that is what an angle "of a velocity" means: how far it has tilted from the reference line.

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Step 3 — The momentum law, as one vector picture

WHAT. During the tiny instant of contact, nothing pushes the pair from outside. So the total momentum — mass times velocity, added up — is the same before and after.

WHY this tool. We use Conservation of Momentum because it is always true when no outside force acts, collision or not. And momentum is a vector, so this single law secretly carries direction information — exactly what an angle problem needs.

Because both masses equal $m$, and $m$ multiplies every term, we can cancel it. What's left is a statement purely about the velocity arrows:

$\underset{\text{before, one arrow}}{\underbrace{{\vec{u}}_1}}=\underset{\text{ball 1 after}}{\underbrace{{\vec{v}}_1}}+\underset{\text{ball 2 after}}{\underbrace{{\vec{v}}_2}}$

PICTURE. Read this as a tip-to-tail triangle: start the violet ${\vec{v}}_2$ arrow at the tip of the magenta ${\vec{v}}_1$ arrow, and the straight arrow from the very start to the very end is ${\vec{u}}_1$. The incoming arrow is literally the third side that closes the triangle made by the two outgoing arrows.

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Step 4 — The energy law, as a length statement

WHAT. An elastic collision is defined by one extra fact: the total kinetic energy $\frac{1}{2}m{v}^2$ is also unchanged.

WHY this tool. Momentum alone can't pin the angle down — many triangles have the same closing side. Energy adds the missing constraint. And energy is a scalar (no direction, just size), so it will speak about the lengths of the arrows.

Write it, cancel the common $\frac{1}{2}m$:

$\frac{1}{2}m{u}_1^2=\frac{1}{2}m{v}_1^2+\frac{1}{2}m{v}_2^2⟹\underset{{\text{(long side)}}^2}{\underbrace{u_1^2}}=\underset{{\text{(leg 1)}}^2}{\underbrace{v_1^2}}+\underset{{\text{(leg 2)}}^2}{\underbrace{v_2^2}}(1)$

Here $u_1,v_1,v_2$ without arrows mean the lengths (speeds) of the arrows.

PICTURE. Equation $(1)$ is the Pythagoras relation $c^2=a^2+b^2$ drawn onto the very triangle from Step 3: the squared long side equals the sum of the two squared legs.

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Step 5 — Square the momentum arrow (the key move)

WHAT. Take the vector equation ${\vec{u}}_1={\vec{v}}_1+{\vec{v}}_2$ and multiply each side by itself using the dot product.

WHY this tool. The dot product turns "arrow times itself" into a length squared, and "arrow times a different arrow" into $|\vec{a}||\vec{b}|\cos (\text{angle between them})$ — it manufactures exactly the angle term we want. That is the whole reason to square.

${\vec{u}}_1\cdot {\vec{u}}_1=({\vec{v}}_1+{\vec{v}}_2)\cdot ({\vec{v}}_1+{\vec{v}}_2)$

Expand the right side like $(a+b{)}^2$:

$\underset{{\vec{u}}_1\cdot {\vec{u}}_1}{\underbrace{u_1^2}}=\underset{{\vec{v}}_1\cdot {\vec{v}}_1}{\underbrace{v_1^2}}+\underset{\text{the cross term}}{\underbrace{2{\vec{v}}_1\cdot {\vec{v}}_2}}+\underset{{\vec{v}}_2\cdot {\vec{v}}_2}{\underbrace{v_2^2}}(2)$

PICTURE. The green cross-term $2{\vec{v}}_1\cdot {\vec{v}}_2$ is the only new ingredient compared to the energy line. Everything else matches equation $(1)$ term for term.

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Step 6 — Subtract, and the angle falls out

WHAT. Line up the two equations. Momentum-squared $(2)$ and energy $(1)$ share three identical pieces: $u_1^2$, $v_1^2$, $v_2^2$. Subtract $(1)$ from $(2)$.

WHY. Because those three pieces cancel exactly, only the cross-term survives — and it is set equal to zero. Subtraction is the cleanest way to isolate the one term that carries the angle.

$\left(u_1^2\right)-\left(u_1^2\right)=(v_1^2+2{\vec{v}}_1\cdot {\vec{v}}_2+v_2^2)-(v_1^2+v_2^2)$

$0=2{\vec{v}}_1\cdot {\vec{v}}_2⟹\boxed{{\vec{v}}_1\cdot {\vec{v}}_2=0}$

PICTURE. With ${\vec{v}}_1\cdot {\vec{v}}_2=v_1{v}_2\cos \varphi =0$, and both speeds non-zero, the only escape is $\cos \varphi =0$, i.e. $\varphi =90^{\circ }$. The V-shape is a right angle.

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Step 7 — The degenerate case: the head-on hit

WHAT. Aim dead-centre. The incoming ball stops; the struck ball leaves with the full velocity.

WHY it's separate. Our proof multiplied speeds: $v_1{v}_2\cos \varphi =0$. If $v_1=0$, this equation is satisfied for any $\varphi$ — but there's no second arrow to measure an angle from. So the 90° statement needs both speeds non-zero, and the head-on case is the one exception.

PICTURE. The triangle collapses onto the line: ${\vec{v}}_2={\vec{u}}_1$ and ${\vec{v}}_1$ vanishes. This is just the 1D elastic result — full velocity transfer between equal masses.

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Step 8 — The other broken case: unequal masses / inelastic

WHAT. Change one assumption at a time and watch 90° break.

WHY. The clean cancellation in Step 6 relied on equal masses (so $m$ cancels) and energy conservation (so $(1)$ holds). Remove either and the cross-term is no longer forced to zero.

PICTURE. Two altered triangles: heavy-hits-light bends the V shut (opening $<90^{\circ }$); light-hits-heavy opens it wide (opening $>90^{\circ }$, can even backscatter). An inelastic equal-mass hit loses energy, so line $(1)$ fails and the opening also shrinks below $90^{\circ }$.

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The one-picture summary

Everything above compresses into a single right triangle. Read it clockwise: incoming = hypotenuse, two outgoing = the legs, energy = Pythagoras, momentum = the closing side, conclusion = the right-angle corner.

Recall Feynman: tell the whole walkthrough to a friend

Draw a ball rolling right, hitting an identical still ball. After the bump they leave in a "V". Now play a trick: momentum says the original arrow you threw equals the two new arrows added tip-to-tail — so the old arrow is the closing side of a triangle made from the two new arrows. Energy (because it's elastic) says the old arrow's length squared equals the two new lengths squared added up — that's Pythagoras, the rule for right triangles. A closing side that obeys Pythagoras can only close a right triangle, so the corner between the two new arrows is a perfect square corner: $90^{\circ }$. The one algebra move that proves it: multiply the momentum arrow by itself (dot product), which spits out $u^2=v_1^2+2{\vec{v}}_1\cdot {\vec{v}}_2+v_2^2$; subtract the energy line and you're left with ${\vec{v}}_1\cdot {\vec{v}}_2=0$ — perpendicular. It only works if the balls weigh the same and no energy is lost, and if both actually move (a dead-centre hit just stops the first ball).

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Active Recall

Recall Test yourself

Which side of the momentum triangle is the incoming velocity? ::: The closing side (hypotenuse). What operation converts the vector momentum law into an angle equation? ::: Squaring it with the dot product. After subtracting energy from squared-momentum, what single term survives? ::: $2{\vec{v}}_1\cdot {\vec{v}}_2$, forced to zero. Why does a head-on hit escape the 90° rule? ::: Ball 1 stops, so there's only one arrow — no angle between two velocities. Which two assumptions make $m$ and the extra terms cancel? ::: Equal masses and energy conservation (elastic).