Motion of centre of mass — external force determines a_CM: Question bank

1.4.11 · D5 · Physics › Momentum & Collisions › Motion of centre of mass — external force determines a_CM

Figure — Motion of centre of mass — external force determines a_CM

True or false — justify

Shell, skaters, boat aur jump traps ke liye neeche diagram:

True or false: Ek exploding shell ka centre of mass apna trajectory us pal badal leta hai jis pal woh explode hota hai.

False — explosion internal hai, isliye woh pairs mein cancel ho jaata hai; sirf gravity (external,

- y

direction mein) act karti hai, toh

a_{C M} = (0, - g)

pehle bhi aur baad mein bhi same rehta hai, aur CM usi parabola par rehta hai jab tak koi fragment land nahi karta.

True or false: Agar total momentum conserved hai, toh CM zaroor rest mein hoga.

False — conserved momentum ka matlab hai

V_{C M}

constant hai, jo koi bhi fixed non-zero value ho sakta hai; woh rest mein tabhi hoga jab woh shuru se rest mein tha.

True or false: Centre of mass object ke material ke andar hi hona chahiye.

False — ek ring ke liye CM hole ke centre mein hota hai (khaali jagah), boomerang ke liye woh arms se bahar hota hai; CM ek computed average point hai, matter ka tukda nahi (neeche ring figure dekho).

True or false: Frictionless ice par push apart karne wale do skaters apna common CM move kar lete hain.

False — push ek internal

+ f, - f

pair hai aur

F_{e x t, x} = 0

hai, isliye

X_{C M}

kabhi nahi badlata chahe kitna bhi zor se dhakka dein; skater 1

- x

mein jaata hai, skater 2

+ x

mein jaata hai, lekin mass-weighted centre fixed rehta hai.

True or false: Ek zyada strong internal force CM ko ek weak force se zyada shift kar sakti hai.

False — magnitude irrelevant hai; har internal force ka ek equal-and-opposite partner hota hai, isliye vector sum exactly zero hota hai chahe size kuch bhi ho.

True or false: Equal masses ke system ke liye, CM geometric midpoint hota hai.

True — equal

m_{i}

ke saath weighting

\frac{\sum m _{i} r _{i}}{\sum m _{i}}

simple positional average

\frac{1}{N} \sum r_{i}

ban jaata hai, jo geometric centre hi hai.

True or false: Ek jumping person sirf muscles se, bina floor ke, apna CM utha sakta hai.

False — muscles internal hain; upar (

+ y

) launch ke liye floor ki external normal reaction chahiye. Ek baar hawa mein aane ke baad,

F_{e x t} = (0, - M g)

aur CM sirf gravity follow karta hai.

True or false: Totally inelastic collision mein CM chalti rehti hai chahe kinetic energy loss ho jaaye.

True — momentum (hence

V_{C M}

) bina kisi external force ke conserved rehta hai, jabki internal deformation kinetic energy drain karta hai; CM bina badlaav ke chalti rehti hai.

Spot the error

Skater / boat / rocket errors ke liye reference schematic:

Flaw dhundo: "Deep space mein ek rocket apna CM accelerate nahi kar sakta kyunki fuel eject karna ek internal force hai."

Fuel system chhod deta hai — ek baar eject hone ke baad woh external ho jaata hai, isliye exhaust baaki bache rocket par ek genuine external force hai; dekho Rocket Propulsion.

Flaw dhundo: "Feather-and-anvil pair ka CM unke midpoint par hota hai."

Galat —

x_{C M} = \frac{m _{1} x _{1} + m _{2} x _{2}}{m _{1} + m _{2}}

mein plug karo: jab anvil ka

m_{2} ≫ m_{1}

ho toh fraction

m_{2} x_{2}

se dominate hota hai, isliye CM almost anvil ke upar hi hota hai; sirf equal masses midpoint dete hain.

Flaw dhundo: "

P = M V_{C M}

se, har particle ki speed double karne par CM speed double hogi sirf tab jab masses equal hoon."

Yeh relation kisi bhi masses ke liye valid hai; saare

v_{i}

double karne par

\sum m_{i} v_{i}

ka har term double ho jaata hai, isliye

V_{C M}

bhi double hoti hai chahe mass distribution kuch bhi ho.

Flaw dhundo: "Ice ki friction two-skater system ke liye internal hai, isliye woh CM ko affect nahi kar sakti."

Friction ice dwara exert hoti hai, jo two-skater system ke bahar hai, isliye yeh external hai aur CM ko accelerate kar sakti hai — hum fixed CM tab maante hain jab hum ise negligible assume kar lete hain.

Flaw dhundo: "Boat par chal raha aadmi CM ko forward move karta hai kyunki woh khud forward move karta hai."

Jab

F_{e x t, x} = 0

ho toh CM move nahi kar sakta;

m (+ Δ x_{man}) + M (- Δ x_{b o a t}) = 0

se boat

- x

mein exactly utna slide karta hai jitna zaroori hai taaki

X_{C M}

fixed rahe jabki aadmi

+ x

mein aage badhta hai.

Flaw dhundo: "

F_{e x t} = M a_{C M}

ke liye zaroori hai ki har particle ko same external force feel ho."

Nahi — iske liye sirf net external force chahiye; individual particles bilkul alag-alag externals feel kar sakte hain, aur sirf unka vector sum CM ko drive karta hai.

Why questions

Internal forces CM equation se kyon gayab ho jaate hain jabki woh clearly individual particles ko move karte hain?

Kisi bhi pair ke liye, Newton ki 3rd law deti hai

f_{ij} = - f_{j i}

, isliye pair sum

0

hai; saare pairs mein sum karne par net internal force zero hai — woh pieces ko rearrange karte hain lekin mass-weighted average ko kabhi shift nahi karte (Newton's Third Law).

CM mass-weighted kyun hota hai instead of simple positional average?

\frac{\sum m _{i} r _{i}}{\sum m _{i}}

mein har position apni mass se scale hoti hai, isliye zyada bhaari particle zyada contribute karta hai aur balance point ko apni taraf kheenchta hai; plain average

\frac{1}{N} \sum r_{i}

is baat ko ignore karta hai aur CM ko light parts ki taraf mislocate kar deta hai.

CM frame mein collisions analyse karna unhe simplify kyun karta hai?

CM frame mein total momentum zero hota hai, isliye do objects hamesha equal-and-opposite momenta

+ p

aur

- p

carry karte hain pehle aur baad mein, jo algebra ko symmetric banata hai; dekho Collisions (Elastic & Inelastic).

Jab hum

F = M a

apply karte hain toh hum ek wobbling, spinning extended body ko ek single point kyun treat kar sakte hain?

Har mass element par Newton ki 2nd law sum karo; saare internal stresses

+ f, - f

pairs mein cancel ho jaate hain, bakee bachta hai

F_{e x t} = M a_{C M}

— CM ek particle ki tarah move karta hai mass

M

ke saath (Rigid Body Dynamics ki neenv).

F_{e x t} = 0

directly conservation of momentum kyun deta hai?

F_{e x t} = 0 \Rightarrow M a_{C M} = 0 \Rightarrow V_{C M}

constant, aur kyunki

P = M V_{C M}

hai aur

M

fixed hai,

P

bhi constant hai; yahi hai Conservation of Linear Momentum.

Edge cases

Ring/boomerang "CM in empty space" edge case:

Edge case: Ek uniform ring — uska CM kahan hai?

Symmetry se har mass element

+ r

par ek doosre se

- r

par match hota hai, isliye integral

\int r d m = 0

aur CM geometric centre par baith jaata hai — hole ke andar, jahan koi material nahi hai.

Edge case: Exploding shell ki flight ke bilkul upar, burst hone se ek pal pehle, CM velocity kya hai?

Apex par

v_{y} = 0

hai (gravity ne upward motion ko abhi cancel kiya hai) jabki

v_{x}

unchanged hai, isliye

V_{C M} = (v_{x}, 0)

— purely horizontal; explosion ise aur uske future evolution ko untouched chhod deta hai.

Edge case: Varying density wala body (ek end par zyada dense) — kya discrete formula phir bhi kaam karta hai?

Continuous form

R_{C M} = \frac{1}{M} \int r ρ d V

use karo; CM denser end ki taraf shift hota hai, exactly jaise bhaare discrete masses balance point ko kheenchte hain.

Edge case: Ek single free particle — uska CM kahan hai aur usse kya govern karta hai?

CM particle khud hi hai, aur

F_{e x t} = M a_{C M}

ordinary

F = m a

ban jaata hai; framework

N = 1

par consistent hai.

Edge case: Ek hi point par do equal masses — kya CM defined hai?

Haan — woh usi shared point par hota hai; formula tab tak valid rehta hai jab tak total mass

M > 0

ho.

Edge case: Jis pal exploded shell ka aakhri fragment ground pe girta hai, CM ka kya hota hai?

Ground ab ek external normal force provide karta hai, isliye CM equation phir bhi valid hai lekin

F_{e x t}

ab sirf

(0, - M g)

nahi raha — woh clean parabola argument wahan khatam ho jaata hai.

Edge case: Ek closed system jisme har particle ki velocity zero hai lekin forces internally act kar rahi hain — kya CM fixed rehta hai?

Haan momentarily aur hamesha ke liye agar

F_{e x t} = 0

ho:

V_{C M} = 0

zero hi rehta hai kyunki internal forces kabhi

a_{C M}

mein contribute nahi karti.

Edge case: Kya

V_{C M}

non-zero ho sakta hai jabki har individual particle momentarily zero velocity pe ho?

Nahi —

P = \sum m_{i} v_{i} = M V_{C M}

, isliye agar saare

v_{i} = 0

hain toh

V_{C M} = 0

exactly us instant par.

Connections

Neeche ka map dikhata hai ki har trap do roots mein se ek tak kaise jaata hai: ya toh definition of mass-weighted CM, ya Newton's 3rd law jo internal forces ko khatam karta hai.

Conservation of Linear Momentum — $F_{e x t} = 0$ special case jo inn adhe traps ke peeche hai.
Newton's Third Law — kyun internal forces hamesha cancel ho jaati hain.
Collisions (Elastic & Inelastic) — upar poochha gaya CM-frame simplification.
Rocket Propulsion — "internal vs ejected" boundary case.
Rigid Body Dynamics — CM as the point jahan $F = M a$ extended bodies ke liye valid hai.