Motion of centre of mass — external force determines a_CM: Worked examples

1.4.11 · D3 · Physics › Momentum & Collisions › Motion of centre of mass — external force determines a_CM

Kuch bhi shuru karne se pehle, symbols ki ek reminder taaki koi pehli line se hi confused na ho:

Scenario matrix

Har CM problem inhi cells mein se kisi ek mein aati hai. Hum sab ko cover karenge.

Cell	Kya cheez ise alag banati hai	Example jo ise cover karta hai
A. $F_{e x t} = 0$ , dono signs	Koi external force nahi → CM frozen; velocities ke opposite signs	Ex 1 (skaters)
B. $F_{e x t} = 0$ , position bookkeeping	Displacements, ek object peeche shift hota hai	Ex 2 (man on boat)
C. $F_{e x t} \neq = 0$ constant	CM ek real trajectory follow karta hai (gravity)	Ex 3 (exploding shell)
D. Zero / degenerate input	Koi mass ya velocity 0 hai, ya masses equal hain	Ex 4 (equal masses, dropped piece)
E. Limiting case	Ek mass $\to \infty$ ya $\to 0$	Ex 5 (heavy wall / light fleck)
F. 2-D vector case	Motion ek line par nahi; components	Ex 6 (L-shaped scatter)
G. Real-world word problem	Ulajhe hue words ko equation mein translate karo	Ex 7 (person climbs a ladder-cart)
H. Exam twist	Lagta hai aur kuch chahiye; CM shortcut jeetta hai	Ex 8 (two blocks + spring, find where they meet)

Chalo har cell clear karte hain.

Ex 1 — Cell A: opposite signs, frozen CM

Worked example Do skaters alag push karte hain

Skaters $m_{1} = 40$ kg aur $m_{2} = 60$ kg frictionless ice par rest mein khade hain aur ek doosre ko shove karte hain. Baad mein $m_{1}$ right taraf $v_{1} = 3$ m/s par glide karta hai. $v_{2}$ aur CM velocity nikalo.

Forecast: Andazo lagao — kya CM lighter skater ki taraf drift karega, heavier wale ki taraf, ya wahi ruk jaega?

Step 1. External forces identify karo. Ice frictionless hai, gravity vertical hai aur ice ke upar push se cancel ho jaati hai. Toh horizontal $F_{e x t} = 0$ . Yeh step kyun? Agar $F_{e x t} = 0$ toh $V_{C M}$ constant hai (parent corollary). Shuru mein $0$ tha, toh $0$ hi rehega.

Step 2. Momentum conservation likho. Pehle total momentum $= 0$ (dono rest mein): $0 = m_{1} v_{1} + m_{2} v_{2} .$ Yeh step kyun? $P = M V_{C M}$ ; kyunki $V_{C M} = 0$ hai, total momentum har waqt zero hai.

Step 3. $v_{2}$ solve karo: $v_{2} = - \frac{m _{1} v _{1}}{m _{2}} = - \frac{40 \cdot 3}{60} = - 2 m/s .$ Minus sign kehta hai $m_{2}$ left jaata hai — $m_{1}$ ke opposite. Dono signs appear hote hain, jaise bataya tha.

Verify: Wapas plug karo. $P = 40 (+ 3) + 60 (- 2) = 120 - 120 = 0$ ✓. Units: dono terms par $kg \cdot m/s$ ✓. Heavier skater slower hai ( $2 < 3$ ) — sahi hai, kyunki $m_{2} v_{2}$ ka $m_{1} v_{1}$ se match karna zaroori hai.

Forecast ka jawab: CM wahi ruka raha. Uske paas move karne ki koi wajah hi nahi thi.

Ex 2 — Cell B: position bookkeeping

Worked example Aadmi naav par chalta hai

Ek aadmi $m = 60$ kg, still water par floating ek naav $M = 120$ kg par $L = 2$ m aage chalta hai (water drag ignore karo). Naav kitni door khisak jaati hai, aur kis direction mein?

Forecast: Naav peeche jaati hai — lekin 1 m se zyada ya kam?

Step 1. External horizontal force $= 0$ (koi drag nahi) → $X_{C M}$ water frame mein fixed hai. Yeh step kyun? Koi external force nahi matlab balance point slide nahi kar sakta; jo aadmi aage gain karta hai, naav ko woh peeche de dena padta hai.

Step 2. Maano naav $d$ peeche khisak jaati hai (toh naav ka displacement $= - d$ ). Aadmi naav par $L$ chalta hai, toh zameen ke relative woh $L - d$ aage move karta hai. Yeh step kyun? " $2$ m aage" naav ke along measure hota hai, zameen se nahi. Ground displacement se naav ka apna retreat minus ho jaata hai.

Step 3. CM fix karo: mass-weighted displacements ka sum zero ho: $0 = m (L - d) + M (- d) .$

Step 4. Solve karo: $m L = (m + M) d \Rightarrow d = \frac{m L}{m + M} = \frac{60 \cdot 2}{180} = 0.667 m peeche.$

Verify: Aadmi ka ground move $= L - d = 2 - 0.667 = 1.333$ m aage. CM check: $60 (1.333) + 120 (- 0.667) = 80 - 80 = 0$ ✓. Units: dono taraf kg·m ✓. Naav heavier hai, toh aadmi se kam move karti hai — sensible hai.

Ex 3 — Cell C: constant external force (gravity)

Worked example Exploding shell

$M = 10$ kg mass ka ek shell fire kiya jaata hai aur apne arc ke top par horizontally $u = 20$ m/s par move karta hai. Yeh do equal $5$ kg pieces mein burst ho jaata hai. Ek piece burst ke turant baad seedha neeche girta hai (horizontal velocity $0$ ). Doosre piece ki horizontal velocity nikalo.

Forecast: $20$ se tez, bilkul $20$ , ya slow?

Step 1. Tiny burst ke dauran external force: sirf gravity (vertical). Horizontal $F_{e x t} = 0$ , toh horizontal momentum explosion ke through conserve hota hai. Yeh step kyun? Explosion forces internal hain → yeh cancel ho jaate hain aur horizontal $P$ change nahi kar sakte.

Step 2. Pehle horizontal momentum $=$ baad mein: $M u = m_{1} v_{1} + m_{2} v_{2}, 10 (20) = 5 (0) + 5 v_{2} .$ Yeh step kyun? Ek piece ka horizontally $v_{1} = 0$ hai (seedha neeche girta hai).

Step 3. Solve karo: $v_{2} = \frac{200}{5} = 40 m/s.$ $20$ se tez — yeh gire hue piece ka missing momentum le leta hai.

Step 4 (CM point). CM horizontally $V_{C M} = 20$ m/s par chalti rehti hai aur gravity ke under girti rehti hai — wahi parabola jo poora shell follow karta. Yeh step kyun? $F_{e x t} = M g$ abhi bhi hai, toh burst chahe ho ya na ho, $a_{C M} = g$ neeche ki taraf rehta hai.

Verify: Baad mein CM horizontal speed $= \frac{5 ( 0 ) + 5 ( 40 )}{10} = 20$ m/s ✓ — unchanged. Units m/s ✓.

Forecast ka jawab: tez — $40$ m/s.

Ex 4 — Cell D: zero / degenerate inputs

Worked example Equal masses, ek rest se drop kiya gaya

Do identical blocks, har ek $m = 3$ kg, frictionless ice par baithe hain. Block 1 rest mein hai; block 2 ko $v_{2} = 4$ m/s par nudge kiya gaya hai. Baad mein ek teesra, massless tag ( $m_{3} = 0$ ) block 1 se attach kiya jaata hai. Poore system ka $V_{C M}$ nikalo aur confirm karo ki massless tag kuch nahi badalta.

Forecast: Kya $0$ kg ka tag $V_{C M}$ ko block 1 ki taraf kheench leta hai?

Step 1. Total mass: $M = 3 + 3 + 0 = 6$ kg. Yeh step kyun? Zero mass $M$ mein kuch add nahi karta — yahi degenerate check hai.

Step 2. CM velocity: $V_{C M} = \frac{m _{1} v _{1} + m _{2} v _{2} + m _{3} v _{3}}{M} = \frac{3 ( 0 ) + 3 ( 4 ) + 0 \cdot v _{3}}{6} = \frac{12}{6} = 2 m/s.$ Yeh step kyun? $m_{3} v_{3}$ term $0 \cdot (kuch bhi) = 0$ hai, toh tag ki motion irrelevant hai.

Step 3. Kyunki masses equal hain, notice karo ki $V_{C M}$ bas plain average hai: $\frac{0 + 4}{2} = 2$ m/s. Equal masses $\Rightarrow$ mass-weighted average $=$ ordinary average.

Verify: Momentum $P = 3 (0) + 3 (4) + 0 = 12$ kg·m/s; $M V_{C M} = 6 (2) = 12$ ✓. Massless tag $0$ contribute karta hai ✓.

Forecast ka jawab: nahi — $0$ kg wala object CM ko kabhi move nahi karta.

Ex 5 — Cell E: limiting case (ek mass bahut bada)

Worked example Ball wall se bounce karti hai

Ek ball $m = 0.5$ kg, $M$ mass ki ek wall se head-on $v = 6$ m/s par takraati hai aur $6$ m/s par wapas bounce karti hai. Limit $M \to \infty$ lo (ek immovable wall). Wall ki velocity change aur CM velocity nikalo.

Forecast: $M \to \infty$ hone par, kya wall ki speed kisi finite value par jaati hai ya $0$ par?

Step 1. Ball–wall system ke liye momentum conservation (collision instant ke liye external floor/foundation forces ignore kiye gaye): $m v + M (0) = m (- v) + M V_{w} .$ Yeh step kyun? Ball ka momentum sign flip karta hai; wall ko difference absorb karna hoga.

Step 2. Wall speed solve karo: $V_{w} = \frac{m v - m ( - v )}{M} = \frac{2 m v}{M} = \frac{2 ( 0.5 ) ( 6 )}{M} = \frac{6}{M} .$ Yeh step kyun? $V_{w}$ isolate karo taaki dekh sako yeh $M$ ke saath kaise shrink karta hai.

Step 3. Limit lo: $lim_{M \to \infty} V_{w} = lim_{M \to \infty} \frac{6}{M} = 0.$ Yeh step kyun? Yahi limiting case ka poora point hai — infinitely heavy wall hilti nahi, phir bhi ball ka momentum reverse kar deti hai. Wall ka momentum change $M V_{w} = 6$ kg·m/s (finite!) hai, chahe uski speed $\to 0$ ho jaaye.

Verify: $M = 1 0^{6}$ kg par, $V_{w} = 6/1 0^{6} = 6 \times 1 0^{- 6}$ m/s ≈ $0$ ✓. Wall ko mila momentum $= M V_{w} = 6$ kg·m/s $= 2 m v = 2 (0.5) (6)$ ✓. Units m/s aur kg·m/s consistent ✓.

Forecast ka jawab: speed $\to 0$ , lekin wall ko mila momentum finite $6$ kg·m/s rehta hai.

Ex 6 — Cell F: 2-D vector case

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

Worked example L-shaped scatter

$M = 4$ kg mass ka ek stationary object frictionless ice par do pieces mein explode ho jaata hai. Piece 1 ( $m_{1} = 1$ kg) $12$ m/s par east (positive $x$ ) mein uda jaata hai. Piece 2 ( $m_{2} = 3$ kg) — uska velocity vector nikalo.

Forecast: Piece 2 ko kis direction mein jaana chahiye, aur piece 1 se roughly kitna fast?

Step 1. Initial momentum zero hai (rest mein), aur $F_{e x t} = 0$ , toh total momentum dono $x$ aur $y$ mein $0$ rehta hai. Yeh step kyun? 2-D mein, momentum component by component conserve hota hai — do equations, ek nahi.

Step 2. $x$ -direction: $0 = m_{1} v_{1 x} + m_{2} v_{2 x} = 1 (12) + 3 v_{2 x}$ . $v_{2 x} = - \frac{12}{3} = - 4 m/s (west).$

Step 3. $y$ -direction: $0 = m_{1} v_{1 y} + m_{2} v_{2 y} = 1 (0) + 3 v_{2 y}$ . $v_{2 y} = 0.$ Yeh step kyun? Piece 1 mein koi north–south motion nahi thi, toh piece 2 mein bhi nahi ho sakti — warna $P_{y} \neq = 0$ ho jaayega. Figure mein red arrow dekho: yeh piece 1 ki line ke seedha wapas point karta hai, koi sideways tilt nahi.

Step 4. Toh $v_{2} = (- 4, 0)$ m/s — seedha west, piece 1 se slower kyunki yeh heavier hai.

Verify: $P = 1 (12, 0) + 3 (- 4, 0) = (12 - 12, 0) = (0, 0)$ ✓. CM rest mein rehta hai ✓. Units m/s ✓.

Ex 7 — Cell G: real-world word problem

Worked example Aadmi ek wheeled cart par chadta hai

Ek cart jiska mass $M = 50$ kg hai, ek $B = 30$ kg box aur ek aadmi $m = 70$ kg ko carry kar rahi hai, sab frictionless rails par, shuruaat mein rest mein. Aadmi box ki taraf $1.5$ m chalta hai (jo cart par fixed baitha hai). Cart-plus-box kitna slide karega, aur kis direction mein?

Forecast: Man-on-boat case (0.667 m ek lighter walker ke liye) se bada ya chota shift?

Step 1. Masses group karo: "walker" person $m = 70$ kg hai; jo kuch bhi woh chal raha hai woh cart+box $= M + B = 80$ kg hai. Yeh step kyun? Box cart ke relative nahi hilta, toh cart aur box saath move karte hain — unhe ek mass treat karo.

Step 2. Koi external horizontal force nahi → $X_{C M}$ fixed. Person $L = 1.5$ m cart par chalta hai, cart $d$ peeche slide karta hai; person ka ground move $= L - d$ aage. Yeh step kyun? Naav wali bookkeeping: walk distance cart ke relative measure hoti hai.

Step 3. CM fix karo: $0 = m (L - d) - (M + B) d \Rightarrow d = \frac{m L}{m + M + B} = \frac{70 ( 1.5 )}{150} = 0.7 m.$

Verify: Total mass $70 + 80 = 150$ kg. CM check: $70 (1.5 - 0.7) + 80 (- 0.7) = 70 (0.8) - 56 = 56 - 56 = 0$ ✓. Person ka ground displacement $= 0.8$ m; cart $0.7$ m peeche; walker yahaan surface se heavier hai toh shift boat case se bada hai ✓. Units m ✓.

Ex 8 — Cell H: exam twist (spring, CM shortcut)

Worked example Do blocks spring ke zariye milte hain — kahan?

Blocks $m_{1} = 2$ kg aur $m_{2} = 6$ kg frictionless ice par baithe hain, $m_{1}$ $x = 0$ par aur $m_{2}$ $x = 8$ m par, ek compressed spring se connected, rest mein rakhe hue. Release hone par, yeh saath rush karte hain. Yeh kahan collide karte hain? (Block sizes ignore karo.)

Forecast: $x = 0$ ke paas, middle ( $x = 4$ ) par, ya $x = 8$ ke paas?

Step 1. Spring forces internal hain; $F_{e x t} = 0$ ; system rest se start karta hai. Toh $X_{C M}$ kabhi nahi hilta. Yeh step kyun? Trap yeh hai ki spring dynamics solve karo. Lekin meeting point wahi hoga jahan fixed CM hai — koi differential equations nahi chahiye.

Step 2. $X_{C M}$ compute karo: $X_{C M} = \frac{m _{1} x _{1} + m _{2} x _{2}}{m _{1} + m _{2}} = \frac{2 ( 0 ) + 6 ( 8 )}{8} = \frac{48}{8} = 6 m.$ Yeh step kyun? Yeh sirf us ek point par collide kar sakte hain jo wahi ruka hua hai — CM.

Step 3. Toh yeh $x = 6$ m par milte hain — heavier $m_{2}$ ke paas, bilkul jaisa mass-weighting predict karta hai.

Verify: $m_{1}$ $6$ m right move karta hai, $m_{2}$ $2$ m left. Mass-weighted displacement: $2 (+ 6) + 6 (- 2) = 12 - 12 = 0$ ✓ (CM nahi hila). Meeting point heavy block ke $3 \times$ paas baitha hai kyunki $m_{2} = 3 m_{1}$ ✓. Units m ✓.

Forecast ka jawab: $x = 8$ ke paas — heavier block bahut kam hila.

Active Recall

Connections

Parent topic ↩ — woh law jise yeh examples exercise karte hain.
Conservation of Linear Momentum — upar har $F_{e x t} = 0$ cell.
Newton's Third Law — internal spring/explosion forces cancel kyun hote hain.
Collisions (Elastic & Inelastic) — Ex 3, 5, 6 collision/explosion cases hain.
Rocket Propulsion — Ex 5 ka momentum-swap idea aur aage.
Rigid Body Dynamics — Ex 7 mein cart+box ko ek mass treat kiya gaya.