Reference frames — Galilean transformations: Question bank

1.1.22 · D5 · Physics › Measurement, Vectors & Kinematics › Reference frames — Galilean transformations

Shuru karne se pehle, symbols ka ek-line reminder taaki kuch bhi mystery na rahe:

$S$ = ground frame, $S^{'}$ = frame jo constant velocity $V$ se $S$ ke relative move kar raha hai.
$r$ position, $u$ velocity, $a$ acceleration; primed = $S^{'}$ mein measure kiya gaya.
Dictionary: $r^{'} = r - V t$ , $t^{'} = t$ , $u^{'} = u - V$ , $a^{'} = a$ .

True or false — justify

Ek ball jo smoothly chalti train ke andar drop ki gayi woh thrower ke paon ke paas girti hai, isliye train zaroor rest mein hogi.

False. Apne paon ke paas girna kisi bhi inertial frame mein hota hai — ball poore time train ki horizontal velocity share karti hai; andar koi experiment aapko constant speed nahi bata sakta.

Do alag inertial frames mein baithe observers hamesha kisi object ki velocity par disagree karte hain.

Mostly true, ek exception hai. Dono

V

se differ karte hain, isliye agree tabhi karte hain jab special case mein

V = 0

ho (same frame) ya jab object ki velocity sirf us direction mein measure ho jo purely... nahi — har component mein

V

ke along differ karte hain, isliye jab bhi

V \neq = 0

ho, genuine disagreement hoti hai.

Sabhi inertial observers kisi bhi object ke acceleration par agree karte hain.

True. Kyunki

V

constant hai,

d V / d t = 0

, isliye

a^{'} = a

; yahi invariance exactly wajah hai ki

F = m a

har inertial frame mein hold karta hai.

Galilean physics mein do events jo

S

mein simultaneous hain,

S^{'}

mein bhi simultaneous hoti hain.

True. Kyunki

t^{'} = t

(universal time) hai, agar

S

mein

t_{1} = t_{2}

hai toh

S^{'}

mein

t_{1}^{'} = t_{2}^{'}

— simultaneity absolute hai, special relativity ke ulat.

Do events ke beech ka time interval sabhi Galilean observers ke liye same hota hai.

True.

t^{'} = t

se, koi bhi difference

Δ t^{'} = Δ t

; Galilean duniya mein koi time dilation nahi hota — woh sirf Special relativity — Lorentz transformation mein appear karta hai.

Do simultaneous events ke beech ki spatial distance sabhi Galilean observers ke liye same hoti hai.

True.

r^{'} = r - V t

; same

t

wale do events ke liye,

V t

shift identical hai aur difference mein cancel ho jaata hai, isliye

Δ r^{'} = Δ r

Ek akele moving particle dwara do alag times ke beech tay ki gayi distance frame-independent hoti hai.

False. Yahan do events ke alag

t

hain, isliye

V t

terms cancel nahi hote; path length aur displacement dono frames ke beech differ karte hain (platform ek curve dekhta hai, train ek seedha drop dekhti hai).

Constant speed par turning car se attached frame inertial hota hai.

False. Constant speed ke saath direction badalna matlab

V

constant nahi hai (

A \neq = 0

), isliye pseudo-forces appear hoti hain — Inertial and non-inertial frames dekhein.

Kisi object ki kinetic energy sabhi inertial frames mein same hoti hai.

False. KE

∣ u ∣^{2}

par depend karta hai, aur

u

V

se shift hota hai; ek ball train mein zero KE rakh sakti hai phir bhi platform se large KE dikha sakti hai.

Edge cases

Galilean transform

V = 0

par kya deta hai?

Yeh identity tak reduce ho jaata hai:

r^{'} = r

t^{'} = t

u^{'} = u

— do frames coincide karte hain aur sabhi measurements agree karti hain.

Agar koi object

S

mein exactly

V

par move kare toh

S^{'}

mein uski velocity kya hogi?

u^{'} = u - V = V - V = 0

; woh

S^{'}

mein rest mein hai — yeh precisely ek passenger hai jo train mein still baitha hai.

t = 0

par dono origins coincide karte hain; tab position transform kya kehta hai?

r^{'} = r - V \cdot 0 = r

; positions us instant par agree karti hain jab frames overlap karte hain, aur baad mein time ke saath linearly diverge karte hain.

Agar "object" moving frame ka khud origin

O^{'}

ho, toh velocity rule kya deta hai?

S

mein woh

V

par move karta hai; apne frame mein

u^{'} = V - V = 0

, jo consistent hai

O^{'}

S^{'}

mein fixed hone ke saath.

Agar do cars same direction mein equal speed par move karein toh closing-speed formula kya hoga?

u_{A / B} = 30 - 30 = 0

; woh mutually rest mein hain, jo us intuition se match karta hai ki same-velocity travellers ek-doosre ko frozen dekhte hain.

Recall Ek-sentence self-test

Agar aap explain kar sako ki "position $V t$ carry karta hai, velocity $V$ carry karti hai, acceleration kuch nahi carry karta" sirf yeh phrase use karke — "har derivative $t$ ki ek power hata deta hai" — toh aapne poora page samajh liya hai.

Question bank — Reference frames — Galilean transformations

True or false — justify

Spot the error

Why questions

Edge cases