Elastic collisions — 1D - solve for final velocities

1.4.5 · Physics › Momentum & Collisions

Derivation scratch se

Humare paas do facts hain. Inhe likhte hain aur solve karte hain.

(1) Conservation of momentum (KYU: koi external horizontal force nahi ⇒ total $p$ constant): $m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$

(2) Conservation of kinetic energy (KYU: elastic ⇒ koi KE nahi khooti): $\frac{1}{2} m_{1} u_{1}^{2} + \frac{1}{2} m_{2} u_{2}^{2} = \frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} v_{2}^{2}$

Ab clever rearrangement. Masses ko group karo.

(1) se: $m_{1} (u_{1} - v_{1}) = m_{2} (v_{2} - u_{2}) (A)$

(2) se, $\frac{1}{2}$ cancel karo aur group karo: $m_{1} (u_{1}^{2} - v_{1}^{2}) = m_{2} (v_{2}^{2} - u_{2}^{2})$ Har side ko difference of squares ki tarah factor karo: $m_{1} (u_{1} - v_{1}) (u_{1} + v_{1}) = m_{2} (v_{2} - u_{2}) (v_{2} + u_{2}) (B)$

Yeh step kyun? Ab dono (A) aur (B) mein factors $m_{1} (u_{1} - v_{1})$ aur $m_{2} (v_{2} - u_{2})$ hain. (B) ko (A) se divide karo: $u_{1} + v_{1} = v_{2} + u_{2}$ Rearrange karo: $u_{1} - u_{2} = - (v_{1} - v_{2})$

Ab do clean equations solve karo

Hum simple linear pair lete hain: $m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2} (1)$ $v_{2} = v_{1} + (u_{1} - u_{2}) (relative-velocity rule se)$

Doosre ko (1) mein substitute karo aur $v_{1}$ ke liye solve karo (algebra sirf brevity ke liye omit ki hai — har step ek substitution hai):

Special cases (80/20 — yeh zyatatar exam questions cover karte hain)

Worked example (full numbers)

Worked example

$m_{1} = 2$ kg at $u_{1} = 3$ m/s, $m_{2} = 1$ kg at $u_{2} = - 1$ m/s se takraata hai (left move kar raha hai) Step 1 — rule likho. Approach speed $u_{1} - u_{2} = 3 - (- 1) = 4$ . Toh separation $v_{2} - v_{1} = 4$ . Kyun? Elastic 1D relative-velocity law.

Step 2 — momentum. $m_{1} u_{1} + m_{2} u_{2} = 2 (3) + 1 (- 1) = 5$ . Kyun? Total $p$ conserved hai.

Step 3 — solve. $2 v_{1} + v_{2} = 5$ aur $v_{2} = v_{1} + 4$ . Substitute karo: $2 v_{1} + (v_{1} + 4) = 5 \Rightarrow 3 v_{1} = 1 \Rightarrow v_{1} = \frac{1}{3}$ m/s. Phir $v_{2} = \frac{1}{3} + 4 = \frac{13}{3}$ m/s. Yeh step kyun? Do linear equations, substitution ke baad ek unknown.

Step 4 — energy verify karo. Pehle: $\frac{1}{2} (2) (9) + \frac{1}{2} (1) (1) = 9.5$ J. Baad mein: $\frac{1}{2} (2) (\frac{1}{9}) + \frac{1}{2} (1) (\frac{169}{9}) = \frac{1}{9} + \frac{169}{18} = \frac{2 + 169}{18} = 9.5$ J. ✓ Verify kyun? Forecast-then-Verify — agar energy match nahi karti, toh koi arithmetic slip hui hai.

Common mistakes

Flashcards

1D elastic collision mein kaun se do quantities conserved hoti hain?

Total momentum AUR total kinetic energy.

1D elastic collisions ke liye relative-velocity rule batao.

Approach speed = separation speed:

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

Relative-velocity rule, energy conservation se aasaan kyun hai?

Yeh linear hai (koi squares nahi), toh momentum ke saath milke do linear equations deta hai.

1D elastic collision mein

v_{1}

ka formula?

v_{1} = \frac{m _{1} - m _{2}}{m _{1} + m _{2}} u_{1} + \frac{2 m _{2}}{m _{1} + m _{2}} u_{2}

Equal masses, target at rest — kya hota hai?

Yeh velocities exchange karte hain:

v_{1} = 0, v_{2} = u_{1}

Heavy ball (

m_{1} ≫ m_{2}

) light ball ko at rest maarta hai — light ball ki speed?

Lagbhag

2 u_{1}

(incoming speed ka double).

Light ball heavy wall ko at rest maarta hai — outcome?

Light ball

\approx - u_{1}

par bounce back karta hai, wall ~still rehti hai.

Problem setup mein kitni equations aur kitne unknowns hain?

Do equations (momentum, KE) aur do unknowns (

v_{1}, v_{2}

Kaun sa sign convention maintain karna hai?

Ek positive direction choose karo; velocities signed hain (left = negative agar right positive hai).

Recall Feynman: 12-year-old ko explain karo

Socho do carts ek smooth track par hain. Yeh crash karte hain lekin crumple nahi hote — perfect bouncy carts. Do rules kabhi nahi tootte: total push (momentum) same rehta hai, aur total bounciness (kinetic energy) same rehti hai. Cool trick: crash se pehle jis speed se yeh ek dusre ki taraf bhaagte hain, crash ke baad usi speed se alag hote hain. Agar dono same weight ke hain aur ek still tha, toh chalta hua ruk jaata hai aur SAARI apni motion doosre ko de deta hai — jaise Newton's cradle mein. Un do rules se tum hamesha pata kar sakte ho ki har cart baad mein kitni fast jaayegi.

Connections

Conservation of Momentum — yahan do pillars mein se pehla.
Kinetic Energy — elastic condition jo doosri equation add karti hai.
Inelastic Collisions — 1D — KE conserved nahi; coefficient of restitution $e < 1$ use karta hai.
Coefficient of Restitution — elastic, $u_{1} - u_{2} = - e (v_{1} - v_{2})$ ka $e = 1$ special case hai... wait, $= - (v_{1} - v_{2})$ .
Center of Mass Frame — us frame mein har speed simply reverse ho jaati hai, inhe re-derive karne ka ek slick tarika.
Newton's Cradle — equal-mass swap in action.