Visual walkthrough — Elastic collisions — 2D - angle relationship

1.4.6 · D2 · Physics › Momentum & Collisions › Elastic collisions — 2D - angle relationship

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Step 1 — "Before" wali picture banao

KYA. Ek ball move kar rahi hai; ek doosri identical ball wahan ruki hui hai. Hum woh instant freeze karte hain jab woh touch hone se pehle ki situation hai.

KYUN. Har collision problem mein sabse pehle yeh naam rakhte hain ki kya move kar raha hai aur kya nahi. Agar hum ek picture aur ek direction fix nahi karte, toh "left" aur "right" ka koi matlab nahi hoga aur koi equation nahi likhi ja sakti.

PICTURE. Arrow dekho. Moving ball (magenta) ek velocity carry karti hai jise hum ${\vec{u}}_1$ bolte hain. Letter ke upar chhota arrow ka matlab hai "yeh ek vector hai" — iske paas ek size (kitni fast) aur ek direction (kis taraf) dono hain. Struck ball (violet) ke paas koi arrow nahi hai: woh rest mein hai.

Hum ${\vec{u}}_1$ ko ek horizontal line ke seedha along point karte hain aur us line ko x-direction kehte hain. Jo bhi us se perpendicular hai (upar/neeche) woh y-direction hai. Yeh choice kuch bhi kharab nahi karti aur algebra clean ho jaata hai.

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Step 2 — "After" wali picture banao

KYA. Bump ke baad balls alag ho jaati hain. Ball 1 line ke upar ki taraf jaati hai, ball 2 line ke neeche ki taraf.

KYUN. 2D collision ka matlab hai balls line chhhod sakti hain. Humein do naye directions allow karne hote hain, isliye unhe describe karne ke liye do naye angles chahiye.

PICTURE. Magenta arrow ${\vec{v}}_1$ purani line se ${\theta }_1$ angle upar ki taraf jaata hai; violet arrow ${\vec{v}}_2$ line se ${\theta }_2$ angle neeche ki taraf. Angle ${\theta }_1$ dashed purani line se magenta arrow tak measure kiya jaata hai — iska yahi matlab hai kisi velocity ka "angle": woh reference line se kitna tilt hua hai.

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Step 3 — Momentum law, ek vector picture ke roop mein

KYA. Contact ke chhote se instant mein, koi cheez bahar se pair ko nahi dhakelta. Toh total momentum — mass times velocity, sab add karke — pehle aur baad mein same rehta hai.

YEH TOOL KYUN. Hum Conservation of Momentum use karte hain kyunki jab bhi koi bahari force nahi lagti, tab yeh hamesha sach hota hai, collision ho ya na ho. Aur momentum ek vector hai, isliye yeh ek law secretly direction information carry karta hai — exactly wahi jo ek angle problem ko chahiye.

Kyunki dono masses $m$ ke barabar hain, aur $m$ har term mein multiply hota hai, hum ise cancel kar sakte hain. Jo bachta hai woh sirf velocity arrows ke baare mein ek statement hai:

$\underset{\text{pehle, ek arrow}}{\underbrace{{\vec{u}}_1}}=\underset{\text{ball 1 baad mein}}{\underbrace{{\vec{v}}_1}}+\underset{\text{ball 2 baad mein}}{\underbrace{{\vec{v}}_2}}$

PICTURE. Ise tip-to-tail triangle ki tarah padho: magenta ${\vec{v}}_1$ arrow ki tip par violet ${\vec{v}}_2$ arrow shuru karo, aur bilkul shuruwat se bilkul ant tak seedha arrow ${\vec{u}}_1$ hai. Incoming arrow literally woh teesri side hai jo do outgoing arrows se bane triangle ko close karti hai.

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Step 4 — Energy law, ek length statement ke roop mein

KYA. Ek elastic collision ek extra fact se define hoti hai: total kinetic energy $\frac{1}{2}m{v}^2$ bhi unchanged rehti hai.

YEH TOOL KYUN. Momentum akela angle ko pin down nahi kar sakta — bahut saare triangles ka ek hi closing side hota hai. Energy woh missing constraint add karti hai. Aur energy ek scalar hai (koi direction nahi, sirf size), isliye woh arrows ki lengths ke baare mein baat karegi.

Ise likho, common $\frac{1}{2}m$ cancel karo:

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

Yahan $u_1,v_1,v_2$ bina arrows ke arrows ki lengths (speeds) ka matlab rakhte hain.

PICTURE. Equation $(1)$ Pythagoras relation $c^2=a^2+b^2$ hai jo Step 3 ke usi triangle par draw ki gayi hai: squared long side baaki do squared legs ke sum ke barabar hai.

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Step 5 — Momentum arrow ko square karo (key move)

KYA. Vector equation ${\vec{u}}_1={\vec{v}}_1+{\vec{v}}_2$ lo aur har side ko khud se dot product use karke multiply karo.

YEH TOOL KYUN. Dot product "arrow times itself" ko length squared mein badal deta hai, aur "arrow times doosra arrow" ko $|\vec{a}||\vec{b}|\cos (\text{unke beech ka angle})$ mein — yeh exactly woh angle term banata hai jo chahiye. Square karne ka poora reason yahi hai.

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

Right side ko $(a+b{)}^2$ ki tarah expand karo:

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

PICTURE. Green cross-term $2{\vec{v}}_1\cdot {\vec{v}}_2$ energy line ke comparison mein ek hi naya ingredient hai. Baaki sab equation $(1)$ se term by term match karta hai.

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Step 6 — Subtract karo, aur angle nikal aata hai

KYA. Dono equations align karo. Momentum-squared $(2)$ aur energy $(1)$ teen identical pieces share karte hain: $u_1^2$, $v_1^2$, $v_2^2$. $(2)$ mein se $(1)$ subtract karo.

KYUN. Kyunki woh teen pieces exactly cancel ho jaate hain, sirf cross-term bachta hai — aur woh zero ke barabar set ho jaata hai. Subtraction us ek term ko isolate karne ka sabse clean tarika hai jo angle carry karta hai.

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

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

PICTURE. ${\vec{v}}_1\cdot {\vec{v}}_2=v_1{v}_2\cos \varphi =0$ ke saath, aur dono speeds non-zero hain, toh ek hi rasta hai: $\cos \varphi =0$, yani $\varphi =90^{\circ }$. V-shape ek right angle hai.

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Step 7 — Degenerate case: head-on hit

KYA. Dead-centre aim karo. Incoming ball ruk jaati hai; struck ball poori velocity le kar chali jaati hai.

KYUN yeh alag hai. Hamara proof speeds multiply karta tha: $v_1{v}_2\cos \varphi =0$. Agar $v_1=0$ ho, toh yeh equation kisi bhi $\varphi$ ke liye satisfy ho jaati hai — lekin angle measure karne ke liye koi doosra arrow hi nahi hai. Toh 90° statement ke liye dono speeds non-zero honi chahiye, aur head-on case ek hi exception hai.

PICTURE. Triangle line par collapse ho jaata hai: ${\vec{v}}_2={\vec{u}}_1$ aur ${\vec{v}}_1$ gayab ho jaata hai. Yeh sirf 1D elastic result hai — equal masses ke beech full velocity transfer.

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Step 8 — Doosra broken case: unequal masses / inelastic

KYA. Ek time mein ek assumption badlo aur dekho 90° toot jaata hai.

KYUN. Step 6 mein clean cancellation equal masses par nirbhar thi (taaki $m$ cancel ho) aur energy conservation par (taaki $(1)$ hold kare). Koi bhi ek hatao aur cross-term zero hone par majboor nahi rahega.

PICTURE. Do altered triangles: heavy-hits-light V ko band kar deta hai (opening $<90^{\circ }$); light-hits-heavy ise khol deta hai (opening $>90^{\circ }$, backscatter bhi ho sakta hai). Ek inelastic equal-mass hit energy kho deti hai, isliye line $(1)$ fail ho jaati hai aur opening bhi $90^{\circ }$ se kam ho jaati hai.

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Ek-picture summary

Upar ki sab cheez ek single right triangle mein compress ho jaati hai. Ise clockwise padho: incoming = hypotenuse, do outgoing = legs, energy = Pythagoras, momentum = closing side, conclusion = right-angle corner.

Recall Feynman: poora walkthrough kisi dost ko batao

Ek ball ko right ki taraf roll karte hua draw karo, jo ek identical still ball se takraati hai. Bump ke baad woh "V" mein nikal jaati hain. Ab ek trick khelo: momentum kehta hai jo original arrow tumne pheka tha woh do naye arrows ke tip-to-tail add karne ke barabar hai — toh purana arrow un do naye arrows se bane triangle ki closing side hai. Energy (kyunki yeh elastic hai) kehta hai purane arrow ki length squared dono naye lengths ke squares ke sum ke barabar hai — yeh Pythagoras hai, right triangles ka rule. Ek closing side jo Pythagoras follow kare woh sirf ek right triangle hi close kar sakti hai, isliye do naye arrows ke beech ka corner ek perfect square corner hai: $90^{\circ }$. Ek algebra move jo ise prove karta hai: momentum arrow ko khud se multiply karo (dot product), jisse $u^2=v_1^2+2{\vec{v}}_1\cdot {\vec{v}}_2+v_2^2$ milta hai; energy line subtract karo aur sirf ${\vec{v}}_1\cdot {\vec{v}}_2=0$ bachta hai — perpendicular. Yeh tabhi kaam karta hai jab balls ka weight barabar ho aur koi energy nahi khoyi ho, aur agar dono actually move kar rahi hon (ek dead-centre hit pehli ball ko bas roka deta hai).

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Active Recall

Recall Khud ko test karo

Momentum triangle mein incoming velocity kaunsi side hai? ::: Closing side (hypotenuse). Kaunsa operation vector momentum law ko angle equation mein convert karta hai? ::: Dot product se ise square karna. Squared-momentum mein se energy subtract karne ke baad kaunsa ek term bachta hai? ::: $2{\vec{v}}_1\cdot {\vec{v}}_2$, jo zero hone par majboor hai. Head-on hit 90° rule se kyun bach jaata hai? ::: Ball 1 ruk jaati hai, toh sirf ek arrow hai — do velocities ke beech koi angle nahi. Kaun si do assumptions $m$ aur extra terms ko cancel karti hain? ::: Equal masses aur energy conservation (elastic).