Elastic collisions — 2D - angle relationship

1.4.6 · Physics › Momentum & Collisions

HUM KYA DESCRIBE KAR RAHE HAIN?

Mass $m_{1}$ ka ek particle velocity $u_{1}$ se move karta hua mass $m_{2}$ ke particle se rest par takraata hai. Ek elastic collision (kinetic energy conserved) ke baad dono alag-alag directions mein ek plane mein chale jaate hain.

Figure — Elastic collisions — 2D - angle relationship

KAISE: sab kuch first principles se derive karo

Hamare paas do conservation laws hain. 2D mein, momentum do scalar equations deta hai (x aur y) aur energy ek deti hai. Yani total 3 equations hain.

Step 1 — Conservation of momentum (vector)

$m_{1} u_{1} = m_{1} v_{1} + m_{2} v_{2}$

Yeh step kyun? Momentum conserved hota hai kyunki brief impact ke dauran koi external force nahi lagta. Yeh ek vector equation hai — direction matter karta hai.

Components mein split karo (x incoming direction ke along, y perpendicular):

$x: m_{1} u_{1} = m_{1} v_{1} cos θ_{1} + m_{2} v_{2} cos θ_{2}$ $y: 0 = m_{1} v_{1} sin θ_{1} - m_{2} v_{2} sin θ_{2}$

Minus sign kyun? Dono balls line ke opposite sides par jaati hain, isliye unke y-momenta cancel ho jaate hain (initial y-momentum zero tha).

Step 2 — Conservation of kinetic energy (scalar)

$\frac{1}{2} m_{1} u_{1}^{2} = \frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} v_{2}^{2}$

Yeh step kyun? "Elastic" ka definition hi hai KE conserved hona. Energy ek scalar hai — koi direction nahi.

Step 3 — Vector ke zariye clean general result

Momentum ko vector rakhte hue use square karo (dot with itself). Step 1 se:

$m_{1} u_{1} - m_{1} v_{1} = m_{2} v_{2}$

Dono sides square karo ( $∣ a ∣^{2} = a \cdot a$ ):

$m_{1}^{2} (u_{1}^{2} - 2 u_{1} \cdot v_{1} + v_{1}^{2}) = m_{2}^{2} v_{2}^{2} (⋆)$

Square kyun kiya? Yeh awkward vector subtraction ko magnitudes aur ek single dot product $u_{1} \cdot v_{1} = u_{1} v_{1} cos θ_{1}$ mein convert kar deta hai — exactly woh angle jo hum chahte hain.

Step 4 — Special, famous case: $m_{1} = m_{2} = m$

Ab energy (Step 2) simplify ho jaati hai: $u_{1}^{2} = v_{1}^{2} + v_{2}^{2} (1)$

Aur momentum as a vector (Step 1, masses cancel): $u_{1} = v_{1} + v_{2}$

Square karo: $u_{1}^{2} = v_{1}^{2} + 2 v_{1} \cdot v_{2} + v_{2}^{2} (2)$

(2) mein se (1) subtract karo: $0 = 2 v_{1} \cdot v_{2}$

$v_{1} \cdot v_{2} = 0$

Forecast-then-Verify

Step 5 — Actual speeds nikalna (equal mass)

Perpendicularity use karo. $u_{1}$ ko hypotenuse rakho, $v_{1}$ aur $v_{2}$ ko legs of a right triangle, jahan $u_{1}$ aur $v_{1}$ ke beech angle $θ_{1}$ hai:

$v_{1} = u_{1} cos θ_{1}, v_{2} = u_{1} sin θ_{1}$

Yeh kyun kaam karta hai? Right-triangle trig: $θ_{1}$ ke adjacent leg $u cos θ_{1}$ hai; opposite leg $u sin θ_{1}$ hai. Check: $v_{1}^{2} + v_{2}^{2} = u_{1}^{2} (cos^{2} + sin^{2}) = u_{1}^{2}$ ✓ energy se match karta hai.

Common Mistakes (Steel-manned)

Active Recall

Recall Click karke khud test karo

2D elastic collision mein exactly 3 scalar equations kyun hote hain?
Kin do equations se 90° rule nikalta hai, aur unhe kaun si operation link karti hai?
Agar masses unequal hon toh kya break hota hai?
$u$ aur $θ_{1}$ ke terms mein dono outgoing speeds ka formula kya hai?

Recall Feynman: ek 12-saal ke bacche ko samjhao

Socho tum ek marble ko doosre same size ke marble par throw kar rahe ho jo still hai. Takraane ke baad, dono ek "V" shape mein roll karte hain. Cool secret yeh hai: woh V hamesha ek perfect right-angle corner hota hai, jaise kisi square ka corner. Kyun? Kyunki energy rule kehta hai speeds ek Pythagoras triangle banati hain (jaise $3, 4, 5$ ), aur momentum rule kehta hai original throw us triangle ki lambi side hai. Ek lambi side jo do chhoti sides se bani ho square corner par — yahi right angle hai! Yeh tabhi kaam karta hai jab dono marbles ka weight same ho.

Flashcards

For elastic equal-mass 2D collision (target at rest, both balls move), what is the angle between outgoing velocities?

Exactly

9 0^{\circ}

(

v_{1} \cdot 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} \cdot v_{2}

for equal masses, one at rest, elastic?

0

— the velocities are perpendicular.

Outgoing speeds in terms of

u

and cue-ball angle

θ_{1}

(equal mass)?

v_{1} = u cos θ_{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

); with one velocity zero there's no defined angle between them.

m_{1} > m_{2}

(heavy hits light), is the opening angle <, =, or > 90°?

Less than

9 0^{\circ}

For inelastic equal-mass collision, is the opening angle 90°?

No — less than

9 0^{\circ}

; the proof needs energy conservation, which inelastic collisions violate.

In the right-triangle picture, what does the incoming velocity

u

correspond to?

The hypotenuse, with

v_{1}

and

v_{2}

as the perpendicular legs.

Connections

Conservation of Momentum — woh vector law jo yahan aadha kaam karta hai.
Elastic collisions — 1D — head-on limiting case (velocity exchange).
Kinetic Energy — woh scalar constraint jo perpendicularity force karta hai.
Dot Product — $v_{1} \cdot v_{2} = 0 \Leftrightarrow$ perpendicular.
Inelastic collisions — contrast: opening angle $< 90°$ .
Center of Mass Frame — CM frame mein dono balls seedha reverse ho jaati hain aur lab-frame 90° sabse cleanly dikhta hai.