System with external forces — conditions for conservation: Question bank

1.4.4 · D5 · Physics › Momentum & Collisions › System with external forces — conditions for conservation

Symbols jo tumhe chahiye (ek baar yahan define kiye gaye)

Taaki is page par koi bhi symbol aisa na ho jo tumne pehle na dekha ho:

Master equation — picture ke saath

Shuru karne se pehle, wo ek fact yaad karo jis par sab kuch tika hai: $\frac{d P}{d t} = F_{e x t}^{n e t}$ Ise padho: "system ka total momentum sirf utni rate se badalta hai jitni net bahari push tay kare." Internal pushes yahan kabhi appear nahi karte — equal-and-opposite pairs mein cancel ho jaate hain (Newton's Third Law).

Figure 1 ise visual banata hai: internal arrows pairs mein aate hain jo cancel ho jaate hain, isliye sirf bacha hua bahar wala arrow group ke momentum ko move kar sakta hai.

Figure — System with external forces — conditions for conservation

Teen relaxations — define kiye aur picture ke saath

Parent note teen tarike batata hai jab momentum phir bhi (partly) conserved ho sakta hai chahe external force ho. Yahan har ek ka asli matlab hai, taaki neeche ka Q&A inhe freely use kar sake:

True or false — justify karo

Total momentum conserved hota hai jab bhi system ke kisi bhi part par koi force act na kare.

False hai jaise likha hai — jo zaroori hai wo ye hai ki net external force zero ho; internal forces bahut badi ho sakti hain (ek explosion) phir bhi total

P

untouched rehta hai kyunki wo pairs mein cancel ho jaate hain.

Agar ek nonzero external force act kare, toh momentum kisi bhi direction mein conserved nahi ho sakta.

False — conservation component-wise hoti hai (relaxation D). Agar

F_{e x t, x}^{n e t} = 0

hai, tab

P_{x}

constant rehta hai chahe

P_{y}

badal raha ho; ek cannon horizontally fire kiya gaya gravity ke neeche act karne ke bawajood horizontal momentum conserve karta hai.

Momentum aur kinetic energy hamesha saath conserve hoti hain.

False — ye dono alag conditions follow karte hain.

P

F_{e x t}^{n e t} = 0

chahiye; kinetic energy ko collision elastic hona chahiye. Dekho Elastic vs Inelastic Collisions.

Perfectly inelastic collision mein objects ek saath chipak jaate hain, isliye momentum lost ho jaata hai.

False — "chipakna" kinetic energy lose karta hai, momentum nahi. Jab tak external impulse negligible hai, total

P

puri tarah conserved rehta hai; gaya hua KE heat/deformation ban jaata hai.

Gravity hamesha ek external force hoti hai.

False — "external" us boundary par depend karta hai jo tum draw karo. Gravity external hai agar system sirf ball hai, lekin internal hai agar system ball + Earth hai. Same force, alag label.

Earth par bat–ball collision ke dauran, momentum exactly conserved hota hai.

False agar tum exactly maango; relaxation I ke zariye approximately true hai. Gravity act karti hai, lekin millisecond contact ke dauran uska impulse

\int F_{g r a v} d t

negligible hai, isliye hum

P

ko impact ke dauraan conserved maante hain.

Agar total momentum constant hai, toh koi external force act nahi kar rahi.

Is sense mein True ki net external force zero honi chahiye — lekin individual external forces exist kar sakti hain aur simply cancel ho sakti hain (jaise gravity aur horizontal table par normal force balance karte hain).

A system with zero total momentum can never have its momentum change.

False — zero sirf ek value hai. Agar ek net external force act kare, toh

d P / d t \neq = 0

aur momentum zero se door chala jaata hai; starting value ise protect nahi karti.

Ek system jisme

F_{e x t}^{n e t} = 0

ho, uska center of mass constant velocity par straight line mein move karta hai.

True — kyunki

F_{e x t}^{n e t} = M a_{c m}

(jisme

M

total mass hai,

a_{c m}

center-of-mass acceleration hai), zero net external force ka matlab zero CM acceleration hai, isliye

v_{c m}

constant rehti hai — center of mass internal chaos ke bawajood uniformly drift karta rehta hai.

Error dhundho

"Deep space mein ek rocket speed up karta hai, isliye uska momentum increase hota hai — rocket ke liye momentum conserved nahi hai."

Error hai galat system choose karna. Rocket + expelled exhaust lo: koi external force nahi, isliye total

P

conserved hai. Akela rocket ek closed system nahi hai — ejected gas balancing momentum carry karta hai.

"Ek explosion mein internal forces huge hoti hain, isliye unhe total momentum bahut zyada change karna chahiye."

Galat — magnitude irrelevant hai. Newton ke 3rd law ke zariye har internal force

f_{ij}

ke saath

f_{j i} = - f_{ij}

paired hoti hai; sab pairs ka vector sum exactly

0

deta hai, isliye koi bhi internal force (chahe kitni bhi badi ho) total

P

nahi badal sakti.

"Ek ball zameen se bounce karti hai aur direction reverse karti hai, isliye momentum kuch nahi se create hua."

Error ye hai ki akeli ball ko closed system maana. Floor ek external normal impulse exert karta hai jo ball ka momentum reverse karta hai; ball + Earth include karo toh total

P

conserved hai (Earth imperceptibly recoil karta hai).

"Kyunki

F = \dot{p}

aur ek force present hai,

p

zaroor change ho raha hai — isliye total momentum conserved nahi ho sakta."

Trap hai small-p/big-P ki galti:

\dot{p}

ek particle ka momentum hai, jise force zaroor change karta hai. Lekin total momentum

P = \sum_{i} p_{i}

obey karta hai

d P / d t = F_{e x t}^{n e t}

; internal forces cancel ho jaate hain aur balanced external forces zero net deti hain, isliye

P

constant reh sakta hai chahe individual

p_{i}

change ho rahe hon.

"Do cars crash hoti hain, energy crumpling mein lost hoti hai, isliye momentum bhi utni hi amount mein lost hota hai."

Momentum aur energy alag ledgers hain. Lost KE deformation/heat mein jaati hai, lekin momentum ka aisa koi sink nahi — brief crash ke dauran external impulse negligible hota hai, isliye total

P

conserved rehta hai chahe KE girta rahe.

"Rough table par collision mein friction external hai, isliye momentum kabhi conserved nahi hota."

Short collision time ke dauran friction ka impulse bade contact impulse ke comparison mein tiny hota hai (relaxation I), isliye

P

impact ke dauraan approximately conserved hota hai. Friction sirf baad ke longer sliding phase mein matter karta hai.

"Ek insaan zameen se upar jump karta hai; unka momentum increase hota hai, jo conservation violate karta hai."

Akele insaan ke liye ground ki normal force external hai, isliye unka momentum legitimately change hota hai. Insaan + Earth lo toh total

P

conserved hai — Earth equal downward momentum acquire karta hai jo tum perceive nahi kar sakte.

Why questions

Hum momentum conservation apply karne se pehle boundary check karne ki koshish kyun karte hain?

Kyunki koi force internal hai ya external — aur isliye

P

conserved hai ya nahi — ye puri tarah depend karta hai ki system ke andar kaun se objects hain. Physics nahi badlti, lekin sahi bookkeeping badal jaati hai.

Impulsive approximation ke liye sahi test forces ka comparison kyun nahi hai, balki impulses ka kyun hai?

Kyunki jo momentum change karta hai wo hai

Δ P = \int F d t

(relaxation A) — force times time, sirf force nahi. Lambe time tak ek choti force ek tiny time ke liye ek badi force ko beat kar sakti hai.

Internal forces total momentum kyun kabhi nahi badal sakti, chahe wo kisi bhi arrangement mein hon?

Har internal force

f_{ij}

ke saath Newton's Third Law ke zariye

f_{j i} = - f_{ij}

paired hoti hai; sab pairs ka sum karo toh exactly

0

milta hai, isliye ye

d P / d t

se drop out ho jaate hain — exactly wahi cancellation jo Figure 1 mein draw ki gayi hai.

Level ground par fire kiye gaye cannon ke liye horizontal momentum conserved kyun hota hai, lekin vertical momentum nahi?

External forces (gravity neeche, normal upar) sirf vertically act karti hain, isliye

F_{e x t, x}^{n e t} = 0

P_{x}

ko protect karta hai, jabki nonzero vertical net force

P_{y}

ko change hone deti hai — Newton ka law har component par alag se hold karta hai (relaxation D).

F_{e x t}^{n e t} = 0 \Leftrightarrow P = constant

" mein "if and only if" kyun use hota hai?

Kyunki implication dono taraf jaati hai: zero net external force

P

ko constant karne par majboor karti hai, aur constant

P

(

d P / d t = 0

) net external force ko zero hone par majboor karta hai. Ye logically equivalent hain.

F_{e x t}^{n e t} = M a_{c m}

"same equation ko alag tarah se bolna" kyun hai?

Kyunki

P = M v_{c m}

(total mass times center-of-mass velocity), isliye differentiate karne par milta hai

d P / d t = M a_{c m}

. Master equation aur CM equation ek hi identity ke do readings hain — dekho Center of Mass Motion.

Ek collision mein momentum conserved kyun ho sakta hai chahe bodies ke beech violent force ho?

Wo violent force do-body system ke liye internal hai, isliye ye pairs mein cancel ho jaati hai; sirf bahut kamzor external forces (gravity, friction)

P

badal sakti hain, aur short contact ke dauran unka impulse negligible hota hai.

Edge cases

Agar net external force nonzero ho lekin motion ke perpendicular ho, toh momentum ka kya hota hai?

Motion ke along momentum component conserved hota hai, jabki perpendicular component badal jaata hai — conservation axis by axis decide hoti hai, sab-ya-kuch nahi (relaxation D).

Ek system ki net external force sirf ek instant ke liye zero hai. Kya momentum conserved hai?

Sirf instantaneously — us instant

d P / d t = 0

ka matlab

P

momentarily stationary hai, lekin agar force wapas aaye,

P

phir se change hone lagta hai. Ek interval mein conservation ke liye net force puri tarah zero rehni chahiye.

Maan lo ek single free particle hai jis par koi force nahi. Kya uska momentum "conserved" hai?

Haan, trivially —

F_{e x t}^{n e t} = 0

ke saath uska momentum constant hai. Ek one-body system Conservation of Linear Momentum ka sabse simple case hai.

Do external forces ek system par act karti hain lekin exactly cancel ho jaati hain. Kya momentum conserved hai?

Haan — sirf net external force

d P / d t

mein enter karti hai. Do canceling externals (jaise gravity aur table ki normal force)

F_{e x t}^{n e t} = 0

chhodti hain, isliye

P

conserved hai.

Collision ke liye

Δ t \to 0

ki limit mein, ek badi finite external force kyun matter karna band kar deti hai?

Kyunki uska impulse

F_{e x t} Δ t \to 0

jab

Δ t \to 0

; ek finite force ek vanishing time se multiply hone par vanishing momentum change deti hai, isliye ye collision analysis se drop out ho jaati hai (relaxation I).

Ek ball drop ki jaati hai aur tumhare haath se pakdi jaati hai — momentum clearly change hota hai. Kahan gaya?

Tumhare aur Earth mein tumhare pairon ke zariye: ball-alone system ko haath se ek external upward impulse milta hai, lekin ball + haath + Earth total

P

conserve karta hai, balancing momentum planet mein flow karta hai.

Agar external forces poore time present hain, kya koi aisa window hai jahan momentum conservation phir bhi apply hoti hai?

Haan — ek sufficiently short sub-interval (ek collision) ke dauran, external impulse negligible hota hai, isliye

P

us window ke across conserved hota hai chahe poori motion mein conserved na ho.

Connections

Newton's Third Law — reason kyun internal-force pairs cancel hoti hain.
Impulse–Momentum Theorem — impulse, force nahi, impulsive approximation ke liye sahi yardstick hai.
Center of Mass Motion — $F_{e x t}^{n e t} = M a_{c m}$ , same law alag tarah se pehna hua.
Elastic vs Inelastic Collisions — kinetic energy ke liye alag condition.
Conservation of Linear Momentum — clean special case $F_{e x t}^{n e t} = 0$ .