Linear momentum p = mv: Question bank — Hinglish deep dive

1.4.1 · D5 · Physics › Momentum & Collisions › Linear momentum p = mv

Woh ek symbol jis par yeh poora page tika hai

Kisi bhi trap se pehle, is page ka star character yahan pin down karte hain taaki kuch bhi borrowed on trust na rahe.

Do words jo throughout kaam aate hain — unhe side by side neeche ki picture mein dekho:

Ek vector = size aur direction (ek arrow). Yeh 1D mein negative ho sakta hai ya 2D mein $x$ / $y$ parts mein split ho sakta hai.
Ek scalar = sirf size, koi direction nahi (ek plain number jaise $10$ kilograms ya $9$ joules).

Figure dekho: same object $m$ ko longer arrow milta hai jab $v$ longer hoti hai, aur flipped arrow milta hai jab $v$ reverse hoti hai — lekin thickness (mass) kabhi sign change nahi karti. Yeh single picture is page ke roughly aadhe traps ke peeche hai.

True or false — justify karo

TRUE or FALSE: "Momentum ka answer complete hai jab tum uska size bata do, jaise

10 kg\cdotm/s

— direction optional hai."

FALSE. Momentum

p = m v

ek vector hai, isliye uski direction answer ka hissa hai; ek bare number se woh half-finished rehta hai (upar figure mein dono arrows compare karo).

Ek heavy truck rest par badi momentum rakhti hai kyunki uski mass badi hai.

False.

p = m v

; agar

v = 0

toh

p = 0

chahe

m

kitna bhi bada ho. Mass akela momentum nahi hai — woh move karna chahiye.

Same speed wale do objects ki momentum hamesha same hoti hai.

False. Same speed matlab same velocity nahi (direction alag ho sakta hai) aur same mass bhi nahi. Momentum ke liye matching mass aur matching velocity vector dono chahiye.

Agar do carts ki momentum equal hai, toh unki kinetic energy bhi equal hogi.

False.

K E = p^{2} /2 m

, isliye same

p

ke liye lighter cart mein zyada

K E

hogi. Equal momentum ke saath alag mass alag energies deti hai (baad mein

K E

-vs-

m

curve dekho).

Kisi object ki speed double karne se uski momentum aur kinetic energy dono double ho jaati hain.

Energy ke liye False.

p = m v

double hoti hai (

v

mein linear hai), lekin

K E = \frac{1}{2} m v^{2}

quadruple hoti hai (

v^{2}

ke saath jaati hai). Sirf momentum double hoti hai.

Ek dimension mein, momentum ek negative number ho sakta hai.

True. 1D mein velocity direction ke liye sign carry karti hai, isliye leftward motion ke liye

p = m v

negative hota hai — minus sign physical hai, koi error nahi.

Formula

p = m v

mein mass negative ho sakti hai.

False. Mass hamesha ek positive scalar hai; momentum ka sign poori tarah velocity ke direction se aata hai, kabhi mass se nahi.

F = ma

aur

F = \frac{d p}{d t}

do bilkul alag laws hain.

False.

F = d p / d t

general law hai;

F = ma

woh special case hai jo tab milta hai jab mass

m

constant ho,

m

ko derivative se bahar nikalte hue.

Momentum ki SI unit,

kg\cdotm/s

, impulse ki unit ke same hai.

True. Impulse hai

F d t

, aur

N\cdots = kg\cdotm/s^{2} \cdot s = kg\cdotm/s

— impulse aur momentum units share karte hain kyunki impulse hai hi momentum mein change (neeche force–time area figure aur Impulse–Momentum Theorem dekho).

Momentum har collision mein conserve hoti hai, lekin kinetic energy nahi.

True. Momentum tab conserve hoti hai jab koi external force nahi act karti (Newton's third law internal pushes ko cancel karta hai);

K E

sirf elastic collisions mein conserve hoti hai.

Algebra links, step by step dikhaye

Neeche ke do traps formula-links par turn karte hain jo students often memorise karte hain samjhte nahi. Yahan se aate hain woh, ek ek honest step ke saath.

Error dhundo

"Ball ki momentum

5

joules hai." — galti dhundo.

Unit galat hai: joules energy measure karte hain (ek scalar). Momentum

kg\cdotm/s

mein hoti hai aur direction chahiye. Student ne

p

K E

se confuse kiya.

"Cart A (

+ 8 kg\cdotm/s

) cart B (

- 6 kg\cdotm/s

) se milti hai; total momentum

= 8 + 6 = 14 kg\cdotm/s

Signs drop kar diye. Vector sum ke roop mein yeh hai

+ 8 + (- 6) = + 2 kg\cdotm/s

; opposite-direction arrows partially cancel hote hain (neeche head-to-tail picture dekho), add nahi hote.

"Ek rocket fuel jalata hai, isliye main current mass ke saath

F = ma

use karunga."

Galat law. Rocket ki mass change hoti hai jaise fuel nikalta hai, isliye

m

ko derivative se bahar nahi nikal sakte — tumhe

F = \frac{d p}{d t}

use karna hoga, jo changing mass ko andar rakhta hai.

"Dono pucks mein

∣ p ∣ = 1.0 kg\cdotm/s

hai, isliye unki velocities equal hain."

Magnitude poora vector nahi hoti. Equal

∣ p ∣

kehta hai arrows same length ke hain, same direction ya same mass nahi — bahut saare alag

(m, v)

combinations same magnitude dete hain.

"2D mein total momentum paane ke liye, main magnitudes add karta hoon:

∣ p_{1} ∣ + ∣ p_{2} ∣

Magnitudes add nahi kar sakte. Components alag alag add karo,

p_{x} = p_{1 x} + p_{2 x}

aur

p_{y} = p_{1 y} + p_{2 y}

, phir

p_{x}^{2} + p_{y}^{2}

lo. Magnitudes tab hi add hoti hain jab vectors same direction mein point karein — neeche parallelogram figure dikhata hai kyun

∣ (3, 0) ∣ + ∣ (0, 4) ∣ = 7

lekin true sum

5

hai.

"Kinetic energy equals

p \times v

hai, kyunki

K E = \frac{1}{2} m v^{2}

aur

p = m v

Ek factor se off hai.

p v = (m v) v = m v^{2} = 2 \cdot K E

(proof bilkul upar). Correct link hai

K E = \frac{1}{2} p v = \frac{p ^{2}}{2 m}

Why questions

m v

combination kyun special hai,

m v^{2}

m + v

kyun nahi?

Kyunki Newton's law kehta hai

F = \frac{d p}{d t}

— momentum exactly woh quantity hai jiska rate of change force ke equal hota hai. Neeche force–time area figure yeh literally dikhata hai: ek force curve ke neeche ka area wahi momentum hai jo hand over hua, jo

m

aur

v

ka koi aur combination nahi deta.

Momentum conserve kyun ho sakti hai jabki kinetic energy loss ho rahi ho?

Internal forces equal-opposite pairs mein aate hain (Newton's Third Law), isliye woh vector total

p

mein cancel ho jaate hain (equal arrows opposite directions mein point karte hue zero sum dete hain) chahe koi bhi energy heat, sound, ya deformation mein turn ho jaaye. Energy ka aisa koi cancellation guarantee nahi hai.

Momentum ko vector treat karna collisions ko easier kyun banata hai?

+ / -

signs (1D) ya components (2D) cancelling aur adding automatically karte hain, isliye "before = after" signs wale numbers par clean arithmetic ban jaata hai, direction ke baare mein case-by-case reasoning ki jagah.

Ek fast light object aur ek slow heavy object equal momentum ke saath alag kyun hurt karte hain?

Equal momentum matlab equal "stopping difficulty," lekin light-fast wala zyada kinetic energy carry karta hai (

K E = p^{2} /2 m

— curve ka steep small-

m

end dekho), isliye woh rukne par zyada work kar sakta hai (zyada damage).

Ek poore system ke liye total momentum

M v_{c m}

kyun likha jaata hai?

\sum m_{i} v_{i}

ko sum karna total mass times centre-of-mass velocity mein factor ho jaata hai, isliye messy multi-particle motion ek single point mein collapse ho jaata hai jo total momentum ke saath move karta hai.

F = ma

ek car ke liye perfectly kyun kaam karta hai, lekin rocket ke liye fail kyun karta hai?

Ek car ki mass essentially constant hai, isliye

m

derivative se cleanly bahar aa jaata hai. Ek rocket har second mass shed karti hai, isliye

\frac{d m}{d t}

part ignore nahi kiya ja sakta — sirf

F = d p / d t

use karke yeh capture hota hai.

Edge cases

Ek object bilkul still baitha hai. Uski momentum kya hai, aur uski direction?

p = m v = m \cdot 0 = 0

. Zero momentum ki koi direction nahi — yeh woh ek case hai jahan "vector" kuch bhi nahi collapse ho jaata hai.

Ek ball right direction mein

3 m/s

se move kar rahi hai aur ek identical ball left direction mein

3 m/s

se: total momentum?

m (+ 3) + m (- 3) = 0

. System ki total momentum zero hai chahe har ball clearly move kar rahi ho — arrows exactly cancel kar dete hain.

Kya ek system ki total momentum zero ho sakti hai lekin total kinetic energy nonzero?

Haan. Upar ke do opposite-moving balls

p = 0

mein sum ho jaate hain, lekin har ek positive

K E = \frac{1}{2} m v^{2} > 0

contribute karta hai, isliye energy total strictly positive hai.

K E = \frac{p ^{2}}{2 m}

mein kya hota hai jab object ki mass fixed momentum par zero ki taraf shrink hoti hai?

K E

infinity ki taraf blow up hota hai — same momentum par ek lighter object ko ever faster move karna padta hai, aur

v^{2}

dominate karta hai.

K E

-vs-

m

curve (figure s02) apni left edge par shoot up karta hai — yeh exactly wahi behaviour hai.

Ek single free particle jis par koi force nahi: kya uski momentum time ke saath change hoti hai?

Nahi.

F = 0

ke saath,

\frac{d p}{d t} = 0

, isliye

p

constant rehta hai — yeh Newton's first law hai jo momentum form mein chhupa hua hai (dekho Conservation of Linear Momentum).

Do particles aise move karte hain ki unka centre of mass rest par hai. System ki total momentum kya hai?

Exactly zero, kyunki total momentum

= M v_{c m}

aur

v_{c m} = 0

. Har particle move kar sakta hai, lekin unki momenta equal-and-opposite hain.

Agar mass truly constant hai, toh

F = ma

aur

F = d p / d t

use karne mein koi physical difference hai?

Answer mein koi difference nahi — dono identical results dete hain. Distinction sirf tab appear hota hai jab

m

change ho; constant mass ke liye woh literally same equation rearranged hai.

Connections

Impulse–Momentum Theorem — impulse hai area-under-force-time (figure s04) aur momentum mein change ke equal hai.
Conservation of Linear Momentum — yahan ke "no external force" edge cases.
Newton's Second Law — $F = ma$ vs $F = d p / d t$ traps.
Newton's Third Law — kyun internal collision forces cancel hote hain.
Elastic and Inelastic Collisions — momentum hamesha, energy kabhi kabhi.
Kinetic Energy — $K E = p^{2} /2 m$ contrast jo traps ke core mein hai.
Centre of Mass — total momentum $= M v_{c m}$ .