Elastic collisions — 1D - solve for final velocities: Question bank

1.4.5 · D5 · Physics › Momentum & Collisions › Elastic collisions — 1D - solve for final velocities

Intuition Is page ko kaise use karein

Neeche har line ek question ::: answer reveal hai. Ye questions bilkul wahi traps spring karne ke liye banaye gaye hain jo 1D elastic collision mein aate hain — sign confusion, equal-mass swap ko over-generalise karna, momentum aur energy ko mix up karna, aur weird boundary cases (zero mass, infinite mass, target already moving). Answer ko cover karo, pehle apni reasoning zor se bolo, phir reveal karo. Agar tumhari reasoning sahi thi lekin galat wajah se, to woh bhi miss hi count hota hai — "kyun" ka jawab dena hi asli point hai.

Yahan koi bhari-bhaarkam calculation nahi hai. Sab kuch parent note ke do rules se answer ho jaata hai:

Momentum conserved: $m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$ .
Relative-velocity rule (= energy, but linear): approach speed $u_{1} - u_{2}$ equals separation speed $v_{2} - v_{1}$ .

Jahan symbols ka matlab hai: $m$ = mass (kitna "stuff" hai, hamesha positive), $u$ = velocity pehle sign ke saath (right = $+$ , left = $-$ ), $v$ = velocity baad mein sign ke saath. Puri derivation ke liye Elastic collisions — 1D - solve for final velocities (index 1.4.5) dekho jis par ye sab based hain.

True or false — justify karo

Elastic collision ke baad pehli ball hamesha slow down ho jaati hai.

False. Agar woh kisi bhari cheez se takraaye to woh ulti bounce ho sakti hai (velocity reverse ho jaati hai, toh sirf slow down nahi hoti), aur agar target itni tezi se uski taraf aa rahi ho toh pehli ball aur tez bhi ho sakti hai. "Slow down" sirf special mass/velocity combos ke liye hi hota hai.

Elastic collision mein dono balls ki speeds simply swap ho jaati hain.

False in general. Velocities tab swap hoti hain jab masses equal hon. Unequal masses ke liye poore formulas use karne padte hain; speed-swapping

m_{1} \neq = m_{2}

ke liye momentum conservation violate karta hai.

Elastic collision mein momentum conserve hota hai lekin kinetic energy nahi.

False. "Elastic" ki definition hi yahi hai ki kinetic energy bhi conserve hoti hai. Dono quantities conserve hoti hain — isliye exactly hume do unknowns ke liye do equations milti hain.

Agar dono balls end mein right ki taraf move kar rahi hain, toh momentum conserve nahi hua.

False. Momentum signed sum

m_{1} v_{1} + m_{2} v_{2}

ke baare mein hai, directions opposite hone ke baare mein nahi. Dono ka right move karna bilkul theek hai jab tak woh sum initial total momentum ke barabar ho.

Relative-velocity rule ek alag physical law hai jo momentum aur energy ke upar add hota hai.

False. Yeh algebraically equivalent hai energy conservation ke (yeh wohi hai jo aapko energy equation ko factor karke momentum equation se divide karne par milta hai). Yeh messy squared equation ki jagah ek clean linear equation se replace kar deta hai.

Ek stationary ball elastically hit hone ke baad incoming ball se tez kabhi move nahi kar sakti.

False. Jab ek heavy ball ek light ball ko rest mein hit karti hai, to light ball

2 u_{1}

tak ki speed se nikal sakti hai — incoming speed se lagbhag double. Yeh cap

2 u_{1}

tab approach hoti hai jab

m_{1} ≫ m_{2}

Agar target initially at rest ho, toh incoming ball kabhi dead stop nahi le sakti.

False. Equal masses aur target at rest — yahi exact case hai jahan incoming ball dead stop le leti hai (

v_{1} = 0

) aur apni saari velocity transfer kar deti hai. Yahi Newton's-cradle result hai — Newton's Cradle dekho.

Collision se pehle total momentum zero hona dono final velocities ko zero hone par majboor karta hai.

False. Zero total momentum ka matlab hai ki dono contributions cancel ho jaati hain (

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

), na ki motion ruk jaata hai. Elastically, Center of Mass Frame mein har object simply reverse ho jaata hai; energy abhi bhi hai, toh woh alag ho jaate hain, freeze nahi hote.

Error dhundo

"Elastic collision mein speeds conserve hoti hain, toh main har speed rakh ke direction swap kar doonga."

Galat: individual speeds conserve nahi hoti jab tak masses equal na hon. Jo conserve hota hai woh total momentum aur total KE hai. Sirf directions reverse karna aksar momentum tod deta hai.

u_{2} = - 1

m/s matlab target ki speed

1

hai, toh main

+ 1

plug karoonga."

Sign drop ho gaya. Velocity

- 1

hai (woh left move kar rahi hai).

+ 1

plug karne se approach speed

u_{1} - u_{2}

change ho jaati hai aur poora answer flip ho jaata hai. Hamesha signed velocities use karo.

"Heavy hits light, toh

v_{1} = u_{1}

exactly — heavy ball apni speed maintain karti hai."

Yeh sirf approximation hai, valid tab jab

m_{1} ≫ m_{2}

ki limit mein ho. Kisi bhi finite mass ratio ke liye heavy ball thodi speed lose karti hai; jab numbers diye gaye hon to momentum properly conserve karo.

"Energy conservation deta hai

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

Nahi — energy quadratic hai. Sahi linear consequence relative-velocity rule hai

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

(note karo minus aur right side par swapped order).

"Maine sirf momentum use kiya, toh

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

v_{1}

aur

v_{2}

poori tarah determine ho jaate hain."

Ek equation, do unknowns — infinitely many solutions. Answer pin down karne ke liye tumhe doosri equation bhi chahiye (energy / relative-velocity rule).

"Formula

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

tab bhi kaam karta hai jab target move kar raha ho."

Sirf tab jab

u_{2} = 0

ho. Poore formula mein ek doosra term bhi hai

\frac{2 m _{2}}{m _{1} + m _{2}} u_{2}

; jab target move kar raha ho to ise drop karne se galat answer aata hai.

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

mein minus sign ek typo hai, yeh

+

hona chahiye."

Minus essential hai. Approach (

u_{1} - u_{2}

) aur separation (

v_{2} - v_{1}

) ki magnitude same hai lekin pair reverse ho jaata hai ki kaun catch up kar raha hai — sign flip usi reversal ko encode karta hai.

Why questions

Hume exactly do conservation laws kyun chahiye, na zyada na kam?

Exactly do unknowns hain (

v_{1}, v_{2}

). Do independent equations unique solution deti hain — ek se undetermined rehta hai, teen se over-constrain ho jaata (aur clash karta agar collision truly elastic na hoti).

Energy ke liye squaring ki jagah relative-velocity rule kyun prefer kiya jaata hai?

Squaring se ek quadratic equation milti hai jo solve karna painful hai. Relative-velocity rule linear hai, toh momentum ke saath pair karke do aasaan linear equations solve hoti hain — same physics, bahut kam algebra.

Ek light ball heavy wall se seedha kyun bounce karti hai (lagbhag) same speed pe?

Heavy wall barely move karti hai, toh uske frame mein ball

u_{1}

se approach karti hai aur

u_{1}

pe separate karni chahiye (elastic) — iska matlab lab mein

- u_{1}

pe reverse hona hai. Wall limit

m_{1} ≪ m_{2}

dekho.

Heavy ball ke hit karne par light ball double speed pe kyun jaati hai?

Heavy ball ke frame mein light ball ek wall hai jo

u_{1}

se approach kar rahi hai; woh heavy ball ke relative

u_{1}

pe bounce back karti hai, aur heavy ball ki apni

u_{1}

add karne par lab frame mein

u_{1} + u_{1} = 2 u_{1}

milta hai.

Labels

1 \leftrightarrow 2

swap karne par

v_{1}

formula

v_{2}

formula kyun ban jaata hai?

Physics ko koi fark nahi padta ki tum kaun si ball ko "1" kahte ho. Equations relabelling ke under symmetric hain, toh solution bhi hona chahiye — yeh symmetry formulas par ek free correctness check hai.

Real steel-ball collision perfectly elastic kyun nahi ho sakti?

KE hamesha thodi sound, heat, aur tiny deformation mein leak hoti hai. Perfect elasticity ek idealisation hai (

e = 1

in Coefficient of Restitution); real collisions mein

e < 1

hota hai aur thodi energy lose hoti hai.

Elastic collisions ke baare mein sochne ka slick tarika Center of Mass Frame kyun hai?

Us frame mein total momentum zero hai, aur ek elastic collision simply har object ki velocity reverse kar deti hai. Us frame mein transform karo, signs flip karo, wapas transform karo — koi quadratic kabhi appear nahi hoti.

Edge cases

Kya hota hai agar

m_{2} \to 0

(target ka practically koi mass nahi)?

Incoming ball unaffected rehti hai (

v_{1} \to u_{1}

) kyunki kuch push back nahi karta, aur tiny target

v_{2} \to 2 u_{1} - u_{2}

pe phek diya jaata hai — woh momentum ya energy carry away nahi kar sakta, toh simply closing speed ke double tak sweep ho jaata hai.

Kya hota hai agar

m_{2} \to \infty

(immovable wall)?

Wall wahi rehti hai (

v_{2} \to u_{2}

, usually

0

) aur incoming ball reflect karti hai:

v_{1} \to 2 u_{2} - u_{1}

, yani rest mein wall ke liye

- u_{1}

. Energy aur momentum tab bhi balance karte hain kyunki "infinite" mass vanishing velocity absorb karti hai.

Agar dono balls same velocity se start karein (

u_{1} = u_{2}

) toh kya hoga?

Approach speed

u_{1} - u_{2} = 0

hai, toh collision hoti hi nahi — woh kabhi close in nahi karti. Formulas correctly

v_{1} = u_{1}

v_{2} = u_{2}

return karte hain: kuch nahi badalta.

Agar

m_{1} = m_{2}

ho lekin dono balls move kar rahi hon?

Woh phir bhi simply velocities exchange karti hain:

v_{1} = u_{2}

v_{2} = u_{1}

. Equal-mass swap ke liye target ka rest mein hona zaroori nahi — woh sirf sabse zyada quoted special case hai.

Kya elastic collision ke baad dono objects rest mein ho sakte hain (pehle nonzero motion ke saath)?

Nahi. Iska matlab zero final KE hoga, lekin KE conserve hai aur pehle positive thi. Zyada se zyada ek object baad mein rest mein ho sakta hai (jaise equal-mass swap mein incoming ball).

Agar

u_{2}

aisa chosen ho ki total momentum zero ho — kya woh ruk jaate hain?

Woh nahi rukte. Zero total momentum ka matlab hai final momenta bhi cancel karte hain, lekin energy conserve hai, toh dono direction reverse karte hain (CM frame mein har speed simply flip hoti hai) aur alag ho jaate hain. Zero momentum ≠ zero energy.

Kya ek head-on collision possible hai jahan incoming ball tez ho jaaye?

Haan — agar target uski taraf move kar raha ho, toh target incoming ball mein momentum transfer kar sakta hai. Yeh Inelastic Collisions — 1D intuition se alag hai; yahan energy preserved hai toh fast approach pehli ball ko shuru se tez chhodh sakti hai.

Recall Aage badhne se pehle ek-line self-test

Ise cover karo aur recite karo: do conserved quantities, relative-velocity rule ki exact form (minus sign ke saath), aur woh ek condition jisme velocities simply swap hoti hain. Answer ::: Momentum aur kinetic energy conserved hote hain; $u_{1} - u_{2} = v_{2} - v_{1}$ ; velocities tab swap hoti hain jab sirf $m_{1} = m_{2}$ ho.

Connections

Elastic collisions — 1D - solve for final velocities (index 1.4.5) — parent derivation jis par yahan ke har trap ka test hota hai.
Conservation of Momentum — hamesha-true law; ise energy se alag samjho.
Kinetic Energy — extra elastic condition jo speeds pin down karti hai.
Coefficient of Restitution — elastic $e = 1$ boundary hai; edge cases $e < 1$ ki taraf blur hote hain.
Center of Mass Frame — "har speed reverse karo" wala viewpoint jo kaafi answers ke peeche hai.
Newton's Cradle — equal-mass swap physical form mein.