Coefficient of restitution e = (v₂ − v₁) - (u₁ − u₂): Question bank

1.4.8 · D5 · Physics › Momentum & Collisions › Coefficient of restitution e = (v₂ − v₁) - (u₁ − u₂)

Yeh ek question bank hai un chhupi hui galat-fehmiyon ko pakadne ke liye jo coefficient of restitution ke andar chhup jaati hain. Yahan koi bhaari arithmetic nahi hai — har item samajhne ke baare mein hai. Jawab ko cover karo, ek reason pe commit karo, phir reveal karo.

Neeche ka saara content parent note Coefficient of Restitution pe build hota hai. Agar koi term unfamiliar lagti hai, pehle woh note dobara padho — yahan hum assume karte hain ki tumhe already pata hai:

Recall Woh paanch facts jinhe yeh traps attack karte hain

$e = \frac{v _{2} - v _{1}}{u _{1} - u _{2}} = \frac{separation speed}{approach speed}$ , aur $0 \leq e \leq 1$ .
Momentum har $e$ ke liye conserved hoti hai; kinetic energy sirf tab conserved hoti hai jab $e = 1$ .
$e = 1$ = perfect bounce (Elastic Collisions); $e = 0$ = saath chipak jaana (Perfectly Inelastic Collisions).
$e$ materials pe depend karta hai, speeds ya masses pe nahi.
Ball on a fixed floor: $e = h_{2} / h_{1}$ , isliye $h_{n} = e^{2 n} h_{0}$ .

True or false — justify karo

Har ek ek statement hai. True/false decide karo aur reason do — sirf verdict ka koi score nahi hoga.

Agar

e < 1

hai, toh collision ke dauran momentum conserved nahi hoti.

False. Momentum sab

e

ke liye conserved hoti hai kyunki pair pe koi external impulse nahi hota; jab

e < 1

hota hai toh kinetic energy girti hai.

e = 1

wali collision kinetic energy conserve karti hai.

True. Separation speed approach speed ke barabar hoti hai, aur kyunki

KE \propto

speed

^{2}

hai aur momentum fixed hai, koi bhi kinetic energy nahi khoti — yahi Elastic Collisions ki exact definition hai.

e

ki value is baat pe depend karti hai ki impact se pehle dono bodies kitni tez chal rahi thin.

Zyaadatar false hai ek ideal law ke roop mein —

e

ko ek material constant maana jaata hai. (Real materials mein thodi si speed-dependence hoti hai, lekin is level pe

e

is baat pe depend karta hai ki objects kis cheez se bane hain, na ki unki speed pe.)

Dono bodies ke beech collision hone ke liye zaruri hai ki

u_{1} > u_{2}

ho.

True (body 1 ko chaser maante hue). Body 1 ko body 2 ke paas aana chahiye, isliye line ke saath uski velocity body 2 se zyaada honi chahiye, jisse approach speed

u_{1} - u_{2} > 0

hoti hai.

Collision ke baad separation speed approach speed se zyaada ho sakti hai.

False ordinary collisions ke liye — uske liye

e > 1

chahiye hoga, matlab energy create hui. Yeh sirf "super-elastic" events mein hota hai jahan stored chemical/spring energy release hoti hai, un mechanical collisions mein nahi jo hum padhte hain.

Agar dono balls baad mein ek hi direction mein move kar rahi hain, toh collision inelastic rahi hogi.

False. Motion ki direction baad mein masses aur speeds pe depend karti hai,

e

pe nahi; ek perfectly elastic (

e = 1

) collision mein bhi dono bodies aage move kar sakti hain.

Ek perfectly inelastic collision (

e = 0

) saari kinetic energy kho deti hai.

False. Woh momentum conservation ke saath consistent maximum possible KE khoti hai, lekin combined mass chalta rehta hai, isliye common-velocity motion ki KE retain rehti hai.

Drop height ko double karne se ek fixed ball aur floor ke liye rebound height bhi double ho jaati hai.

True.

h_{2} = e^{2} h_{1}

h_{1}

mein linear hai, isliye

e

fixed hone par

h_{1}

ko double karne se

h_{2}

bhi double ho jaati hai.

Coefficient of restitution ki units metres per second hain.

False. Yeh do speeds ka ratio hai, isliye units cancel ho jaate hain —

e

dimensionless hota hai.

Floor pe bounce karne wali ball ke liye

e = h_{2} / h_{1}

hai.

False.

e

speeds ka ratio hai aur height

\propto

speed

^{2}

hoti hai, isliye

e = h_{2} / h_{1}

; plain height ratio

e^{2}

ke barabar hota hai.

Error dhundho

Har line mein ek flawed statement ya step hai. Galti ka naam batao aur use correct karo.

"Separation speed

= v_{1} - v_{2}

taaki

u_{1} - u_{2}

ke order se match kare."

Indices jaanbujhkar swap kiye gaye hain. Kyunki body 2 baad mein aage hoti hai (

v_{2} > v_{1}

), separation

v_{2} - v_{1}

hai;

v_{1} - v_{2}

likhne se negative, meaningless

e

milta hai.

"Kyunki

e < 1

hone par energy khooti hai, momentum bhi zaroor kam hogi."

Energy aur momentum alag-alag conservation questions hain. Momentum ek vector total hai jo collision ke dauran Newton's third law se protected hai; energy loss chahe kuch bhi ho, woh fixed rehti hai.

"Ek clay lump jo wall se takraata hai ka

e = 0

hai, isliye uski final speed zero hai."

e = 0

ka matlab hai koi separation nahi (koi bounce nahi), isliye woh wall ke relative ruk jaata hai — lekin sirf isliye kyunki wall fixed hai. Ek free movable body ke against,

e = 0

ka matlab hai ki woh saath mein ek common velocity se move karte hain, yeh nahi ki sab kuch ruk jaata hai.

e = J_{r} / J_{d}

hai, aur kyunki restoration hamesha deformation se zyaada push return karta hai,

e > 1

hai."

Restoration zyaada se zyaada utna hi impulse return karta hai jitna deformation ne store kiya (

J_{r} \leq J_{d}

), kyunki kuch energy heat/sound ke roop mein nikal jaati hai. Isliye

e \leq 1

hoga, kabhi upar nahi.

"Kaafi saare bounces ke baad

h_{n} = e^{n} h_{0}

hai."

Har bounce speed ko

e

se multiply karta hai, aur height speed

^{2}

ke hisaab se jaati hai, isliye har bounce height ko

e^{2}

se multiply karta hai.

n

bounces ke baad

h_{n} = e^{2 n} h_{0}

e

tumhe batata hai ki collision mein kinetic energy ka kitna fraction bacha rehta hai."

e

relative speed ka fraction hai jo bacha rehta hai, energy ka nahi. Relative motion se judi KE ka fraction

e^{2}

ke roop mein scale hota hai, aur total KE-loss law mein

(1 - e^{2})

involve hota hai — dekho Kinetic Energy Loss in Collisions.

"Kyunki floor nahi hilti, ek bounce mein momentum conserved nahi hoti."

Momentum conserved hoti hai — Earth (floor) equal aur opposite momentum leta hai. Woh mass mein itni badi hai ki uski velocity change undetectable hoti hai; isliye hum

u_{2} = v_{2} = 0

model karte hain.

Why wale questions

Underlying reason ke saath jawab do, sirf restatement nahi.

Humein

e

ko alag equation ki zarurat kyun hai?

Conservation of Linear Momentum ek equation deta hai lekin collision mein do unknown final velocities hoti hain, isliye ek doosra relation chahiye, aur energy generally conserved nahi hoti — Newton's restitution law woh supply karta hai.

Indices swap karne se (

v_{2} - v_{1}

over

u_{1} - u_{2}

)

e

positive kyun rehta hai?

Kyunki physically chaser pehle faster hota hai (

u_{1} > u_{2}

) aur chased baad mein faster hota hai (

v_{2} > v_{1}

), isliye dono differences positive hain aur unka ratio automatically

\geq 0

hai.

Common velocity

v_{c}

final formula se kyun gayab ho jaati hai?

Do impulse ratios pe apply ki gayi "add numerators, add denominators" trick har

v_{c}

term ko cancel kar deti hai, isliye hum

e = \frac{v _{2} - v _{1}}{u _{1} - u _{2}}

recover karte hain bina

v_{c}

jaane.

Bounce-height relation straight ratio kyun nahi balki square root kyun hai?

Kyunki ball ki speed har surface pe

v = 2 g h

obey karti hai, isliye height speed squared pe depend karti hai; height ratio se speed ratio

e

nikaalte waqt isliye square root chahiye (Projectile Motion

v

–

h

link provide karta hai).

e

materials pe depend karta hai, masses pe kyun nahi?

e

measure karta hai ki deformation kitni completely spring back karta hai (

J_{r} / J_{d}

), jo material ki elasticity ki property hai; masses final velocities ko affect karte hain lekin bounce-back fraction ko nahi.

Normal collision mein

e > 1

kyun nahi ho sakta?

Uska matlab hoga ki separation speed approach speed se zyaada hai, yaani mechanical energy kuch nahi se create hui, jo ek passive collision ke liye energy conservation violate karta hai.

e = 0

maximum kinetic energy loss ka case kyun hai?

Zero separation speed ke saath dono bodies ek velocity share karti hain, jo (fixed momentum diya hua) ek hi outcome hai jo sabse kam KE chodta hai — koi bhi bounce kuch relative motion aur isliye zyaada KE return karta.

Edge cases

Woh scenarios jinhe log bhool jaate hain. Har boundary ke through reason karo.

Jab ball drop ki jaaye aur kabhi bounce na kare (floor pe rehe), toh

e

kya hoga?

e = 0

: rebound height

h_{2} = 0

deta hai

e = 0/ h_{1} = 0

, ek perfectly inelastic contact.

e = 1

bouncing ball ke liye kya predict karta hai?

Ball hamesha ke liye apni original drop height pe return aati hai, kyunki

h_{2} = e^{2} h_{1} = h_{1}

— ek idealised perpetual bounce bina kisi energy loss ke.

Equal mass ki do bodies,

e = 1

, ek rest mein — kya hoga?

Woh velocities exchange karti hain: moving wali ruk jaati hai aur stationary wali incoming speed se move karti hai, classic elastic equal-mass swap.

Agar dono bodies ek hi velocity se start karein (

u_{1} = u_{2}

Approach speed zero hai, isliye koi collision hi nahi hai —

e

undefined hai (division by zero) kyunki kuch bhi close nahi aa raha.

Impulse picture mein

J_{r} = 0

kya correspond karta hai?

Bilkul koi restoration push nahi, isliye material kabhi spring back nahi karta:

e = J_{r} / J_{d} = 0

, ek perfectly plastic (splat) collision.

Agar hum poore problem ke liye chosen positive direction reverse kar dein toh

e

ka kya hoga?

Kuch nahi — saari velocities saath mein sign flip karti hain, isliye numerator aur denominator dono flip hote hain aur ratio

e

unchanged rehta hai;

e

convention-independent hai.

Agar "floor" actually large lekin finite mass ka ek free block hai, toh kya

e = h_{2} / h_{1}

exact hai?

Exactly nahi — woh formula assume karta hai ki floor ki mass infinite hai isliye woh still rehti hai. Finite mass ke saath block recoil karta hai, kuch speed leta hai, isliye rebound height true material

e

ko thoda under-read karti hai.

Connections

Conservation of Linear Momentum — woh always-true equation jo yeh traps baar baar test karte hain.
Elastic Collisions — upar explore ki gayi $e = 1$ boundary.
Perfectly Inelastic Collisions — $e = 0$ boundary.
Impulse and Momentum — $e$ ka $J_{r} / J_{d}$ view ka source.
Kinetic Energy Loss in Collisions — kaafi saare traps ke peechhe $(1 - e^{2})$ energy ki kahaani.
Projectile Motion — $v = 2 g h$ link jis par height traps rely karte hain.