(1) Conservation of momentum (WHY: no external horizontal force ⇒ total p constant):
m1u1+m2u2=m1v1+m2v2
(2) Conservation of kinetic energy (WHY: elastic ⇒ no KE lost):
21m1u12+21m2u22=21m1v12+21m2v22
Now the clever rearrangement. Group masses together.
From (1):
m1(u1−v1)=m2(v2−u2)(A)
From (2), cancel the 21 and group:
m1(u12−v12)=m2(v22−u22)
Factor each side as a difference of squares:
m1(u1−v1)(u1+v1)=m2(v2−u2)(v2+u2)(B)
Why this step? Both (A) and (B) now contain the factors m1(u1−v1) and m2(v2−u2). Divide (B) by (A):
u1+v1=v2+u2
Rearrange:
u1−u2=−(v1−v2)
Why is the relative-velocity rule easier than energy conservation?
It's linear (no squares), so combined with momentum it gives two linear equations.
Formula for v1 in a 1D elastic collision?
v1=m1+m2m1−m2u1+m1+m22m2u2.
Equal masses, target at rest — what happens?
They exchange velocities: v1=0,v2=u1.
Heavy ball (m1≫m2) hits light ball at rest — light ball's speed?
About 2u1 (double the incoming speed).
Light ball hits a heavy wall at rest — outcome?
Light ball bounces back at ≈−u1, wall stays ~still.
How many equations and unknowns set up the problem?
Two equations (momentum, KE) and two unknowns (v1,v2).
What sign convention must you keep?
Choose a positive direction; velocities are signed (left = negative if right is positive).
Recall Feynman: explain to a 12-year-old
Imagine two carts on a smooth track. They crash but don't crumple — perfect bouncy carts. Two rules never break: the total push (momentum) stays the same, and the total bounciness (kinetic energy) stays the same. Cool trick: the speed they rush toward each other before the crash equals the speed they fly apart after. If they're the same weight and one was sitting still, the moving one stops and passes ALL its motion to the other — like in a Newton's cradle. With those two rules you can always figure out exactly how fast each cart goes afterward.
Elastic collision ka matlab hai do cheezein takraati hain par koi bhi kinetic energy waste nahi hoti — na heat, na sound, na dent. Is wajah se humare paas do conservation laws hote hain: momentum bhi conserve, aur kinetic energy bhi conserve. Do equations, aur do unknowns (dono final velocities v1 aur v2) — isliye exact solve ho jaata hai.
Sabse kaam ka shortcut yaad rakho: approach speed = separation speed. Yaani takraane se pehle dono jis speed se ek dusre ke paas aa rahe the (u1−u2), takraane ke baad usi speed se door jaayenge (v2−v1). Yeh actually energy conservation ka hi simple roop hai, par ismein square nahi karna padta, sirf linear equation banti hai. Isko momentum equation ke saath mila do, do linear equations solve karo, answer mil gaya.
Kuch ready-made cases jo exam mein 80% baar aate hain: agar dono mass barabar hain aur ek rest pe hai, to velocities swap ho jaati hain (moving wala ruk jaata hai, doosra full speed se nikal jaata hai — Newton's cradle wala scene). Agar bhaari ball halki rest wali ko maare, to halki ball lagbhag 2u1 speed se udti hai. Agar halki ball bhaari deewar se takraye, to wahi speed se ulta bounce ho jaati hai.
Sabse common galti: signs bhool jaana. Left direction ki velocity ko negative likho, warna pura answer galat ho jaayega. Aur "speeds swap" sirf equal mass mein hota hai — general case mein formula use karo. Hamesha end mein energy check karke confirm kar lo (Forecast-then-Verify).
