Question bank — Elastic collisions — 2D - angle relationship

False — iske liye equal masses chahiye aur target rest par honi chahiye. Unequal masses se nonzero ${\vec{v}}_1\cdot {\vec{v}}_2$ banta hai, isliye opening angle $90^{\circ }$ se drift ho jaata hai.

True — momentum conservation ko sirf impact ke dauran zero external force chahiye. Yeh energy ke baare mein kuch nahi kehta, isliye inelastic hits mein bhi survive karta hai.

False — proof energy equation subtract karta hai, jo inelastic collisions violate karti hain; opening angle $90^{\circ }$ se kam aata hai.

False side ke baare mein: uska angle magnitude $60^{\circ }$ hai lekin opposite ($-y$) side par point karta hai, yaani uska signed direction $-60^{\circ }$ hai. Unke y-momenta ($+v_1\sin {\theta }_1$ aur $-v_2\sin {\theta }_2$) phir original zero par cancel ho jaate hain.

True — yahi exactly kinetic-energy equation hai masses cancel hone ke baad; yeh right triangle ka Pythagoras statement hai. (Dekho Kinetic Energy ki energy ek single scalar kyun hai, vector nahi.)

False — cue ball ruk jaati hai, isliye koi doosri direction nahi hai; ${\vec{v}}_1\cdot {\vec{v}}_2=0$ trivially true hai lekin opening angle undefined hai (upar head-on definition dekho).

False — heavy hits light mein opening angle $90^{\circ }$ se kam hota hai; heavy ball muskil se deflect hoti hai. Light-hits-heavy wala hai jo $90^{\circ }$ se zyada hota hai.

False — $u$ completely cancel ho jaata hai. Perpendicularity purely geometric hai; koi bhi speed same right-angle split deta hai.

Error: 2D mein momentum ke sirf do components hain (x aur y) → 2 equations. Energy ke saath yeh 3 equations total hain, 4 nahi.

Error: vectors ke magnitudes add nahi kar sakte. Sirf jab woh parallel hoon tab yeh hold karta hai; yahan woh ek angle par milte hain, isliye $u_1^2=v_1^2+2{\vec{v}}_1\cdot {\vec{v}}_2+v_2^2$.

Error: squaring se cross term $2{\vec{v}}_1\cdot {\vec{v}}_2$ milta hai. Yeh sirf energy equation subtract karne ke baad vanish hota hai — cancellation hi poora proof hai, assumption nahi. (Dekho Dot Product ki $\vec{a}\cdot \vec{b}=0$ ka matlab perpendicular kyun hota hai.)

Error: woh opposite ($-y$) side par jaati hai, isliye uska y-momentum $-m{v}_2\sin {\theta }_2$ hai. Hamare fixed y-axis ke saath, opposite sides ka matlab opposite signs hai — aur tabhi total y-momentum zero rehta hai.

Error: KE ki koi direction nahi hoti — yeh ek scalar number hai. Yeh exactly ek equation contribute karta hai, kabhi components mein split nahi hota.

Error: incoming ball ka backscatter tab hota hai jab woh heavier target se takraati hai (light-hits-heavy), opening angle ko ek full split ki taraf push karta hai lekin pahuncha nahi; ek heavy incomer kabhi backscatter nahi karta. (Dekho Elastic collisions — 1D bounce-back ke peeche mass-ratio rules ke liye.)

Squaring (khud se dotting) vector subtraction ko magnitudes plus ek single ${\vec{v}}_1\cdot {\vec{v}}_2$ term mein badal deta hai — precisely woh quantity jiska vanish hona perpendicular ka matlab hai.

Momentum equation ($m\vec{u}=m{\vec{v}}_1+m{\vec{v}}_2$) aur energy equation ($\frac{1}{2}m{u}^2=\dots$) dono har term par common $m$ carry karte hain jo divide out ho jaata hai, ek purely kinematic relation chodke.

Humne y-axis ek baar fix kiya. Cue ball $+y$ ki taraf scatter hoti hai, target $-y$ ki taraf; $-y$ ki taraf point karne wali velocity ka by definition negative y-component hota hai, isliye uska contribution $-v_2\sin {\theta }_2$ hai. Dono signs opposite hone chahiye taaki total initial zero par rahe.

Angle do directions ke beech define hota hai. Agar ek velocity zero hai (head-on stop), toh koi doosri direction nahi hai, isliye "unke beech $90^{\circ }$" khaali hai.

Elastic ka matlab $\frac{1}{2}m{u}^2=\frac{1}{2}m{v}_1^2+\frac{1}{2}m{v}_2^2$ hai, yaani $u^2=v_1^2+v_2^2$ — bilkul hypotenuse-legs relation ka form.

Lost kinetic energy outgoing legs ko incoming hypotenuse ke relative mein short kar deta hai; $\vec{u}$ mein (vectors ki tarah) add hone ke liye, legs ko ek doosre ki taraf lean karna padta hai, V ko $90^{\circ }$ se neeche narrow karta hai.

Impact parameter (hit ka sideways offset) sirf yeh set karta hai ki $90^{\circ }$ kaise share ho (${\theta }_1$ vs ${\theta }_2$); conservation laws unka sum $90^{\circ }$ par fix karte hain regardless.

Momentum akela $\vec{u}$ ko do vectors mein koi bhi split allow karta hai; sirf energy add karne par (length constraint) cross term zero par pin hota hai. (Dekho Conservation of Momentum ki vector sum kyun fixed hai.)

Cue ball muskil se deflect hoti hai (${\theta }_1\to 0$), isliye struck ball almost perpendicular jaati hai (${\theta }_2\to 90^{\circ }$) aur slowly move karti hai — V ek right angle tak open hota hai ek bahut chhote leg ke saath.

Undefined — cue ball ruk jaati hai aur target saari velocity le leta hai; sirf ek moving direction hai, isliye koi angle exist nahi karta (yeh ek single degenerate case hai jo upar define kiya gaya hai, $0^{\circ }$ ya $90^{\circ }$ value nahi).

Nahi — yeh extreme light-hits-heavy case hai; incoming ball reflect hoti hai aur target essentially move nahi karta, isliye koi equal-mass right angle exist nahi karta.

CM frame mein equal-mass balls hamesha bilkul opposite ($180^{\circ }$) emerge karte hain; neat $90^{\circ }$ lab frame ki feature hai jahan target rest se start karta hai. (Dekho Center of Mass Frame frames switch karne se picture kaise badlti hai.)

$90^{\circ }$ rule fail ho jaata hai — yeh assume karke derive kiya gaya tha ki $m_2$ ka initial momentum zero hai. Dono moving hone par full component equations par wapas jaana padta hai.

Nahi — equal masses, target at rest, aur dono balls moving ke liye, ${\vec{v}}_1\cdot {\vec{v}}_2=0$ exact hai; exceptions sirf degenerate head-on aur grazing limits hain, koi intermediate angle nahi. (Sub-$90^{\circ }$ opening angles Inelastic collisions mein rehte hain, jahan energy lost hoti hai.)