Intuition The one-line ideaWhen a moving ball hits an identical, stationary ball in a 2D elastic collision, the two balls always fly apart at exactly 90° to each other. This is not a coincidence — it falls straight out of "momentum is a vector, energy is a scalar."
A particle of mass m 1 m_1 m 1 u ⃗ 1 \vec{u}_1 u 1 m 2 m_2 m 2 at rest . After an elastic collision (kinetic energy conserved) they move off in different directions in a plane.
Definition Setup & symbols
Before: u ⃗ 1 \vec{u}_1 u 1 m 2 m_2 m 2
After: v ⃗ 1 \vec{v}_1 v 1 θ 1 \theta_1 θ 1 v ⃗ 2 \vec{v}_2 v 2 θ 2 \theta_2 θ 2 below it.
Elastic ⇒ kinetic energy conserved.Scattering angle = θ 1 + θ 2 \theta_1 + \theta_2 θ 1 + θ 2
We have two conservation laws. In 2D, momentum gives two scalar equations (x and y) and energy gives one . That's 3 equations.
m 1 u ⃗ 1 = m 1 v ⃗ 1 + m 2 v ⃗ 2 m_1\vec{u}_1 = m_1\vec{v}_1 + m_2\vec{v}_2 m 1 u 1 = m 1 v 1 + m 2 v 2
Why this step? Momentum is conserved because no external force acts during the brief impact. It's a vector equation — direction matters.
Split into components (x along incoming direction, y perpendicular):
x: m 1 u 1 = m 1 v 1 cos θ 1 + m 2 v 2 cos θ 2 \text{x:}\quad m_1 u_1 = m_1 v_1\cos\theta_1 + m_2 v_2\cos\theta_2 x: m 1 u 1 = m 1 v 1 cos θ 1 + m 2 v 2 cos θ 2 y: 0 = m 1 v 1 sin θ 1 − m 2 v 2 sin θ 2 \text{y:}\quad 0 = m_1 v_1\sin\theta_1 - m_2 v_2\sin\theta_2 y: 0 = m 1 v 1 sin θ 1 − m 2 v 2 sin θ 2
Why the minus sign? The two balls fly to opposite sides of the line, so their y-momenta cancel (initial y-momentum was zero).
1 2 m 1 u 1 2 = 1 2 m 1 v 1 2 + 1 2 m 2 v 2 2 \tfrac12 m_1 u_1^2 = \tfrac12 m_1 v_1^2 + \tfrac12 m_2 v_2^2 2 1 m 1 u 1 2 = 2 1 m 1 v 1 2 + 2 1 m 2 v 2 2
Why this step? "Elastic" is defined as KE conserved. Energy is a scalar — no direction.
Keep momentum as a vector and square it (dot with itself). From Step 1:
m 1 u ⃗ 1 − m 1 v ⃗ 1 = m 2 v ⃗ 2 m_1\vec{u}_1 - m_1\vec{v}_1 = m_2\vec{v}_2 m 1 u 1 − m 1 v 1 = m 2 v 2
Square both sides (∣ a ⃗ ∣ 2 = a ⃗ ⋅ a ⃗ |\vec{a}|^2 = \vec{a}\cdot\vec{a} ∣ a ∣ 2 = a ⋅ a
m 1 2 ( u 1 2 − 2 u ⃗ 1 ⋅ v ⃗ 1 + v 1 2 ) = m 2 2 v 2 2 ( ⋆ ) m_1^2\left(u_1^2 - 2\vec{u}_1\cdot\vec{v}_1 + v_1^2\right) = m_2^2 v_2^2 \quad (\star) m 1 2 ( u 1 2 − 2 u 1 ⋅ v 1 + v 1 2 ) = m 2 2 v 2 2 ( ⋆ )
Why square? It converts the awkward vector subtraction into magnitudes plus a single dot product u ⃗ 1 ⋅ v ⃗ 1 = u 1 v 1 cos θ 1 \vec{u}_1\cdot\vec{v}_1 = u_1 v_1\cos\theta_1 u 1 ⋅ v 1 = u 1 v 1 cos θ 1
Now energy (Step 2) simplifies to:
u 1 2 = v 1 2 + v 2 2 ( 1 ) u_1^2 = v_1^2 + v_2^2 \quad (1) u 1 2 = v 1 2 + v 2 2 ( 1 )
And momentum as a vector (Step 1, masses cancel):
u ⃗ 1 = v ⃗ 1 + v ⃗ 2 \vec{u}_1 = \vec{v}_1 + \vec{v}_2 u 1 = v 1 + v 2
Square it:
u 1 2 = v 1 2 + 2 v ⃗ 1 ⋅ v ⃗ 2 + v 2 2 ( 2 ) u_1^2 = v_1^2 + 2\vec{v}_1\cdot\vec{v}_2 + v_2^2 \quad (2) u 1 2 = v 1 2 + 2 v 1 ⋅ v 2 + v 2 2 ( 2 )
Subtract (1) from (2):
0 = 2 v ⃗ 1 ⋅ v ⃗ 2 0 = 2\,\vec{v}_1\cdot\vec{v}_2 0 = 2 v 1 ⋅ v 2
v ⃗ 1 ⋅ v ⃗ 2 = 0 \boxed{\vec{v}_1\cdot\vec{v}_2 = 0} v 1 ⋅ v 2 = 0
Intuition Why 90° feels inevitable once you see itEnergy says "the lengths satisfy Pythagoras: u 2 = v 1 2 + v 2 2 u^2 = v_1^2 + v_2^2 u 2 = v 1 2 + v 2 2 u ⃗ \vec u u v ⃗ 1 \vec v_1 v 1 v ⃗ 2 \vec v_2 v 2 and whose length-squared equals the sum of their length-squares — that's exactly the hypotenuse of a right triangle . So the two pieces must be perpendicular.
Worked example Predict before you computeA cue ball hits an identical stationary ball. After collision the cue ball moves at 30 ∘ 30^\circ 3 0 ∘ Forecast: what angle is the struck ball below the line, and is the cue ball faster or slower?
Verify: By the 90° rule, struck ball goes at 90 ∘ − 30 ∘ = 60 ∘ 90^\circ - 30^\circ = 60^\circ 9 0 ∘ − 3 0 ∘ = 6 0 ∘ v 1 = u cos θ 1 v_1 = u\cos\theta_1 v 1 = u cos θ 1 v 2 = u sin θ 1 v_2 = u\sin\theta_1 v 2 = u sin θ 1 v 1 = u cos 30 ∘ ≈ 0.87 u v_1 = u\cos30^\circ \approx 0.87u v 1 = u cos 3 0 ∘ ≈ 0.87 u v 2 = u sin 30 ∘ = 0.5 u v_2 = u\sin30^\circ = 0.5u v 2 = u sin 3 0 ∘ = 0.5 u
Use the perpendicularity. Place u ⃗ 1 \vec u_1 u 1 v ⃗ 1 \vec v_1 v 1 v ⃗ 2 \vec v_2 v 2 θ 1 \theta_1 θ 1 u ⃗ 1 \vec u_1 u 1 v ⃗ 1 \vec v_1 v 1
v 1 = u 1 cos θ 1 , v 2 = u 1 sin θ 1 v_1 = u_1\cos\theta_1, \qquad v_2 = u_1\sin\theta_1 v 1 = u 1 cos θ 1 , v 2 = u 1 sin θ 1
Why this works? Right-triangle trig: the leg adjacent to θ 1 \theta_1 θ 1 u cos θ 1 u\cos\theta_1 u cos θ 1 u sin θ 1 u\sin\theta_1 u sin θ 1 v 1 2 + v 2 2 = u 1 2 ( cos 2 + sin 2 ) = u 1 2 v_1^2 + v_2^2 = u_1^2(\cos^2+\sin^2)=u_1^2 v 1 2 + v 2 2 = u 1 2 ( cos 2 + sin 2 ) = u 1 2
Worked example Full worked problemm m m u = 10 m/s u = 10\,\text{m/s} u = 10 m/s m m m θ 2 = 53 ∘ \theta_2 = 53^\circ θ 2 = 5 3 ∘
Step A: Angle of cue ball θ 1 = 90 ∘ − 53 ∘ = 37 ∘ \theta_1 = 90^\circ - 53^\circ = 37^\circ θ 1 = 9 0 ∘ − 5 3 ∘ = 3 7 ∘ Why? The 90° rule.
Step B: v 1 = u cos θ 1 = 10 cos 37 ∘ = 10 ( 0.8 ) = 8 m/s v_1 = u\cos\theta_1 = 10\cos37^\circ = 10(0.8) = 8\,\text{m/s} v 1 = u cos θ 1 = 10 cos 3 7 ∘ = 10 ( 0.8 ) = 8 m/s Why? Cue ball is the leg adjacent to its own angle.
Step C: v 2 = u sin θ 1 = 10 sin 37 ∘ = 10 ( 0.6 ) = 6 m/s v_2 = u\sin\theta_1 = 10\sin37^\circ = 10(0.6) = 6\,\text{m/s} v 2 = u sin θ 1 = 10 sin 3 7 ∘ = 10 ( 0.6 ) = 6 m/s Why? Struck ball is the opposite leg.
Check energy: 8 2 + 6 2 = 64 + 36 = 100 = 10 2 8^2 + 6^2 = 64+36 = 100 = 10^2 8 2 + 6 2 = 64 + 36 = 100 = 1 0 2 Check momentum y: 8 sin 37 − 6 sin 53 = 8 ( 0.6 ) − 6 ( 0.8 ) = 4.8 − 4.8 = 0 8\sin37 - 6\sin53 = 8(0.6) - 6(0.8) = 4.8 - 4.8 = 0 8 sin 37 − 6 sin 53 = 8 ( 0.6 ) − 6 ( 0.8 ) = 4.8 − 4.8 = 0
Common mistake "The 90° rule works for any elastic 2D collision."Why it feels right: You always see two balls scattering, so why wouldn't the angle always be 90°? The fix: It needs equal masses . If m 1 ≠ m 2 m_1 \ne m_2 m 1 = m 2 v ⃗ 1 ⋅ v ⃗ 2 = ( m 2 − m 1 ) 2 m 2 ( … ) ≠ 0 \vec v_1\cdot\vec v_2 = \tfrac{(m_2-m_1)}{2m_2}(\dots)\ne 0 v 1 ⋅ v 2 = 2 m 2 ( m 2 − m 1 ) ( … ) = 0 < 90 ° <90° < 90° > 90 ° >90° > 90°
Common mistake "If it's a perfect head-on hit, the angle is still 90°."Why it feels right: The rule "always says 90°." The fix: In a head-on equal-mass elastic hit, the incoming ball stops (v ⃗ 1 = 0 \vec v_1 = 0 v 1 = 0 v ⃗ 1 ⋅ v ⃗ 2 = 0 \vec v_1\cdot\vec v_2 = 0 v 1 ⋅ v 2 = 0 both balls moving.
Common mistake "Inelastic collisions also give 90°."Why it feels right: Momentum is conserved in all collisions. The fix: The proof used energy conservation (equation 1). Drop that and the v ⃗ 1 ⋅ v ⃗ 2 = 0 \vec v_1\cdot\vec v_2 = 0 v 1 ⋅ v 2 = 0 < 90 ° < 90° < 90°
Recall Click to test yourself
Why are there exactly 3 scalar equations in a 2D elastic collision?
From which two equations does the 90° rule pop out, and what operation links them?
What breaks if masses are unequal?
Give the formula for both outgoing speeds in terms of u u u θ 1 \theta_1 θ 1
Recall Feynman: explain to a 12-year-oldImagine throwing a marble at another marble of the same size sitting still. After they bump, they roll away in a "V" shape. The cool secret: that V is always a perfect right-angle corner , like the corner of a square. Why? Because the energy rule says the speeds make a Pythagoras triangle (like 3 , 4 , 5 3,4,5 3 , 4 , 5
"Equal pals split square." Equal masses, one at rest, split apart at a square (90°) corner. And the proof recipe: "Square the sum, subtract the energy."
For elastic equal-mass 2D collision (target at rest, both balls move), what is the angle between outgoing velocities? Exactly
90 ∘ 90^\circ 9 0 ∘ (
v ⃗ 1 ⋅ v ⃗ 2 = 0 \vec v_1\cdot\vec v_2 = 0 v 1 ⋅ v 2 = 0 ).
Which two conservation laws produce the 90° result? Momentum (squared as a vector) and kinetic energy; you subtract the energy equation from the squared momentum equation.
Why does momentum give 2 equations but energy gives 1 in 2D? Momentum is a vector (x and y components), energy is a scalar.
What is v ⃗ 1 ⋅ v ⃗ 2 \vec v_1\cdot\vec v_2 v 1 ⋅ v 2 0 0 0 — the velocities are perpendicular.
Outgoing speeds in terms of u u u θ 1 \theta_1 θ 1 v 1 = u cos θ 1 v_1 = u\cos\theta_1 v 1 = u cos θ 1 ,
v 2 = u sin θ 1 v_2 = u\sin\theta_1 v 2 = u sin θ 1 .
Why must BOTH balls move for the 90° rule to apply? A head-on hit stops the cue ball (
v 1 = 0 v_1=0 v 1 = 0 ); with one velocity zero there's no defined angle between them.
If m 1 > m 2 m_1 > m_2 m 1 > m 2 Less than
90 ∘ 90^\circ 9 0 ∘ .
For inelastic equal-mass collision, is the opening angle 90°? No — less than
90 ∘ 90^\circ 9 0 ∘ ; the proof needs energy conservation, which inelastic collisions violate.
In the right-triangle picture, what does the incoming velocity u ⃗ \vec u u The hypotenuse, with
v ⃗ 1 \vec v_1 v 1 and
v ⃗ 2 \vec v_2 v 2 as the perpendicular legs.
Conservation of Momentum — the vector law doing half the work here.Elastic collisions — 1D — the head-on limiting case (velocity exchange).Kinetic Energy — the scalar constraint that forces perpendicularity.Dot Product — v ⃗ 1 ⋅ v ⃗ 2 = 0 ⇔ \vec v_1\cdot\vec v_2 = 0 \Leftrightarrow v 1 ⋅ v 2 = 0 ⇔ Inelastic collisions — contrast: opening angle < 90 ° < 90° < 90° Center of Mass Frame — in the CM frame both balls just reverse and the lab-frame 90° is cleanest to see.
Moving ball hits identical stationary ball
u1 squared = v1 sq + v2 sq
Head-on hit is degenerate
Intuition Hinglish mein samjhoDekho, scene ye hai: ek ball move kar rahi hai aur same mass ki doosri ball rest pe rakhi hai. Jab pehli ball usse takraati hai (aur collision elastic hai, yaani kinetic energy bachi rehti hai), to dono balls ke beech ka angle hamesha 90 degree hota hai. Billiards/carrom mein ye cheez tum roz dekhte ho — "V" shape banta hai jo ek perfect square corner jaisa hota hai.
Iska reason simple hai. Do conservation laws lagao. Momentum ek vector hai, isliye 2D mein wo do equations deta hai (x aur y). Energy ek scalar hai, ek hi equation. Ab equal mass case mein: energy bolti hai u 2 = v 1 2 + v 2 2 u^2 = v_1^2 + v_2^2 u 2 = v 1 2 + v 2 2 u ⃗ = v ⃗ 1 + v ⃗ 2 \vec u = \vec v_1 + \vec v_2 u = v 1 + v 2 perpendicular hi honi chahiye. Bas, ho gaya proof: v ⃗ 1 ⋅ v ⃗ 2 = 0 \vec v_1 \cdot \vec v_2 = 0 v 1 ⋅ v 2 = 0
Trick yaad rakhne ke liye: "Square the sum, subtract the energy " — momentum equation ko square karo, usme se energy equation ghatao, dot product zero aa jaata hai. Aur shortcut formula: v 1 = u cos θ 1 v_1 = u\cos\theta_1 v 1 = u cos θ 1 v 2 = u sin θ 1 v_2 = u\sin\theta_1 v 2 = u sin θ 1
Par ek warning — ye 90 degree wala rule sirf tab kaam karta hai jab dono masses equal ho, target rest pe ho, collision elastic ho, aur dono balls move karein (perfectly head-on hit mein pehli ball ruk jaati hai, tab angle define hi nahi hota). Agar masses alag hain ya collision inelastic hai, to angle 90 se kam ya zyada ho sakta hai. Exam mein ye conditions check karna mat bhoolna!
