Motion of centre of mass — external force determines a_CM: Exercises

1.4.11 · D4 · Physics › Momentum & Collisions › Motion of centre of mass — external force determines a_CM

Is page par saara kaam do master facts karte hain. Chalte hain inhe plain words mein dobara state karte hain taaki kuch bhi mystery na rahe:

Level 1 — Recognition

L1·Q1 — Do blocks ka balance point kahan hai?

Do blocks ek line par rakhe hain: $m_{1} = 2 kg$ at $x_{1} = 0 m$ aur $m_{2} = 6 kg$ at $x_{2} = 4 m$ . $x_{C M}$ nikalo. Kya yeh light block ke paas hai ya heavy block ke?

Recall Solution

Ek dimension mein definition apply karo (positions bas line par numbers hain): $x_{C M} = \frac{m _{1} x _{1} + m _{2} x _{2}}{m _{1} + m _{2}} = \frac{( 2 ) ( 0 ) + ( 6 ) ( 4 )}{2 + 6} = \frac{24}{8} = 3 m .$ DIKHTA KYA HAI: balance point $3$ m par hai — heavy $6$ kg block ki taraf teen-chauthai raaste tak (neeche figure dekho). KYU: heavier mass average ko apni taraf "kheenchta" hai, bilkul waise jaise see-saw heavy bache ke paas balance karta hai.

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

Figure s01 — Ek line par do blocks. Light teal $2$ kg block $x_{1} = 0$ par hai, bada orange $6$ kg block $x_{2} = 4$ par. Plum triangle mass-weighted balance point $x_{C M} = 3$ m mark karta hai, jo heavy block ki taraf teen-chauthai raaste par hai.

L1·Q2 — Inme se kaun $V_{C M}$ change karta hai?

Ek band box deep space mein float kar raha hai (na gravity, na contact). Andar, ek fan hawa chhod raha hai, ek spring ek ball ko wall ke against launch karti hai, do magnets snap karke saath aate hain. Kya inme se koi bhi box ke centre of mass ki velocity badal deta hai?

Recall Solution

Nahi — inme se koi nahi. Fan thrust, spring force, magnet attraction — sab internal hain (pusher aur pushed dono box ke andar hain). Newton ke 3rd law se har ek ke saath ek equal-and-opposite partner aata hai, isliye $\sum f_{in t er na l} = 0$ . $F_{e x t} = 0$ hone par $a_{C M} = 0$ milta hai, isliye $V_{C M}$ constant rehta hai. YEH KYUN MATTER KARTA HAI: box ke balance point ko hilane ke liye kuch bahar se push karna padega — koi wall jise yeh touch kare, gravity, ya gas ka jet jo box ke bahar phenka gaya ho.

Figure s02 — Floating box ke andar har internal push (fan, spring, magnets) ke saath ek equal-and-opposite reaction arrow aata hai. Teal aur orange arrows pairwise cancel ho jaate hain, isliye plum CM dot exactly wahin rehta hai jahan tha — koi bahar ka force nahi, koi CM motion nahi.

Level 2 — Application

L2·Q1 — Do skaters, unequal push

Skaters $m_{1} = 50 kg$ aur $m_{2} = 75 kg$ frictionless ice par rest mein khade hain aur push karke alag ho jaate hain. Skater 1 $v_{1} = 3 m/s$ se chala jaata hai. $v_{2}$ aur CM ki speed nikalo.

Recall Solution

Step 1 — KYU momentum conserved hai: ice frictionless hai, isliye koi horizontal external force nahi lagg raha. Initial total momentum $= 0$ . Step 2 — Likhte hain (skater 1 ki direction positive maano): $0 = m_{1} v_{1} + m_{2} v_{2} \Rightarrow v_{2} = - \frac{m _{1} v _{1}}{m _{2}} = - \frac{( 50 ) ( 3 )}{75} = - 2 m/s .$ Minus sign ka matlab hai skater 2 opposite direction mein $2 m/s$ se glide karta hai. Step 3 — CM speed aur conceptual link: kyunki total momentum conserved hai aur zero se shuru hua tha, yeh zero hi rehta hai. CM velocity total momentum divided by total mass hai, isliye $V_{C M} = \frac{p _{t o t a l}}{M} = \frac{m _{1} v _{1} + m _{2} v _{2}}{M} = \frac{0}{125} = 0$ . YEH KYUN POORI BAAT HAI: conserved zero momentum ka matlab $M V_{C M} = 0$ hai, isliye balance point frozen hai — push, internal hone ki wajah se, CM ko kabhi motion mein nahi daal sakti chahe skaters kisi bhi taraf jaayein.

Figure s03 — Do skaters frozen plum CM line par shuru karte hain. Push ke baad, light skater (orange, $3$ m/s) right jaata hai aur heavy skater (teal, $2$ m/s) left jaata hai. Arrow lengths dikhate hain ki lighter skater faster move karta hai; plum CM dot apni line kabhi nahi chhodta.

L2·Q2 — Exploding shell, doosre piece ki position

$M = 4 kg$ ka ek shell right mein $V_{C M} = 10 m/s$ se move kar raha hai (gravity external hai, lekin burst ke instant ke baare mein poochh rahe hain isliye fall ignore karo) aur do equal pieces mein $2 kg$ each explode hota hai. Ek piece backward $5 m/s$ se jaata hai. Doosre piece ki velocity nikalo.

Recall Solution

KYU burst ke through momentum conserved hai: explosion internal hai; burst ke tiny instant mein gravity ka impulse negligible hai, isliye total momentum unchanged rehta hai. $M V_{C M} = m_{a} v_{a} + m_{b} v_{b} .$ $(4) (10) = (2) (- 5) + (2) v_{b} \Rightarrow 40 = - 10 + 2 v_{b} \Rightarrow v_{b} = \frac{50}{2} = 25 m/s .$ Doosra piece forward $25 m/s$ se jaata hai. CM check: $\frac{( 2 ) ( - 5 ) + ( 2 ) ( 25 )}{4} = \frac{40}{4} = 10 m/s$ — unchanged, exactly jaisa $F_{e x t} = M a_{C M}$ demand karta hai.

Figure s04 — Pehle: ek shell (plum) right mein $10$ m/s se move kar raha hai. Burst ke baad: piece A (teal) left mein $5$ m/s se recoil karta hai, piece B (orange) right mein $25$ m/s se rocket karta hai. Dono ke neeche plum CM arrow apni original $10$ m/s ki length rakhta hai — burst momentum ko reshuffle karta hai total change kiye bina.

Level 3 — Analysis

L3·Q1 — Man boat par chalta hai, boat recoil karti hai

Ek man $m = 60 kg$ stern se bow tak $L = 3 m$ ki distance chalta hai (boat ke deck ke along measure kiya gaya). Boat $M = 180 kg$ frictionless water par float kar rahi hai. Boat water ke relative kitni door move karti hai, aur kis direction mein?

Recall Solution

Step 1 — Golden shortcut: koi external horizontal force nahi → $X_{C M}$ frozen rehta hai. Step 2 — Bookkeeping (LETTERS KA MATLAB): maano boat $d$ distance back slide karti hai (water ke relative). Man $L$ forward deck ke along chalta hai, lekin deck khud $d$ back slide hua, isliye man ka displacement water ke relative $(L - d)$ forward hai. Step 3 — CM freeze karo: total mass-weighted displacement zero hona chahiye. $m (L - d) + M (- d) = 0.$ $m L - m d - M d = 0 \Rightarrow d = \frac{m L}{m + M} = \frac{( 60 ) ( 3 )}{240} = 0.75 m .$ Boat $0.75 m$ backward move karti hai (man ke walk ke opposite).

Figure s05 — Upar: man stern par, boat rest mein. Neeche: walk ke baad boat $d = 0.75$ m back slide kar chuki hai (plum arrow) jabki man bow tak pahunch gaya. Dashed plum line frozen $X_{C M}$ mark karti hai jo dono configurations share karte hain.

L3·Q2 — Door wala skater kahan pahunchta hai?

L2 wale skaters ( $m_{1} = 50 kg$ , $m_{2} = 75 kg$ , rest mein, phir push apart) par wapas aao. $t = 2 s$ gliding ke baad, woh kitne apart hain, aur har ek start line se kitna door gaya hai?

Recall Solution

L2 se speeds: $v_{1} = 3 m/s$ , $v_{2} = - 2 m/s$ . Distances: $d_{1} = 3 \times 2 = 6 m$ ; $d_{2} = 2 \times 2 = 4 m$ . Separation $= 6 + 4 = 10 m$ . KYU yeh differ karte hain: CM fixed rehta hai, isliye displacements $m_{1} d_{1} = m_{2} d_{2}$ follow karte hain: check karo $50 \times 6 = 300 = 75 \times 4$ . ✓ Lighter skater farther jaata hai, jaise frozen balance point par hinge ho.

Level 4 — Synthesis

L4·Q1 — Projectile split ho jaata hai; doosra fragment kahan girega

Mass $M$ ka ek firework launch hota hai. Yeh apne arc ke top par do equal halves mein burst hota hai. Ek half straight down girta hai aur $x_{1} = 40 m$ par land karta hai (launch point se measure kiya). Agar unexploded shell $x_{f u l l} = 100 m$ par land karta, toh doosra half kahan land karega?

Recall Solution

Key idea: explosion internal hai, isliye CM original parabola follow karta rehta hai aur wahin land karta hai jahan poora shell karta — $x_{C M} = 100 m$ par. CM landing point likhte hain do landing spots ka mass-weighted average ke roop mein (equal masses $M /2$ each): $x_{C M} = \frac{\frac{M}{2} x _{1} + \frac{M}{2} x _{2}}{M} = \frac{x _{1} + x _{2}}{2} .$ $100 = \frac{40 + x _{2}}{2} \Rightarrow x_{2} = 200 - 40 = 160 m .$ Doosra half $160 m$ par land karta hai — original target se kaafi aage, kyunki ise CM ka "missing" ground carry karna hai.

Figure s06 — Dashed plum parabola CM path hai, $x_{f u l l} = 100$ m par land karti hai. Burst par, ek half (teal) straight down $x_{1} = 40$ m par girta hai; doosra half (orange) $x_{2} = 160$ m tak aage phirak jaata hai taaki do landing points ka mass-weighted average exactly CM target par ho.

L4·Q2 — 2D mein momentum se CM velocity

Frictionless ice par do pucks: $m_{1} = 3 kg$ jo $v_{1} = (4, 0) m/s$ se move kar raha hai aur $m_{2} = 1 kg$ jo $v_{2} = (0, 8) m/s$ se move kar raha hai. $V_{C M}$ aur uski speed nikalo.

Recall Solution

KYU $P = M V_{C M}$ use karo: yeh messy individual motions ko ek clean CM velocity mein convert karta hai. $V_{C M} = \frac{m _{1} v _{1} + m _{2} v _{2}}{M} = \frac{( 3 ) ( 4 , 0 ) + ( 1 ) ( 0 , 8 )}{4} = \frac{( 12 , 8 )}{4} = (3, 2) m/s .$ Speed: $∣ V_{C M} ∣ = 3^{2} + 2^{2} = 13 \approx 3.606 m/s$ .

Figure s07 — Puck 1 (teal) right drive karta hai, puck 2 (orange) upar drive karta hai. Plum CM velocity $(3, 2)$ mass-weighted average hai, heavier puck 1 ki taraf tila hua; uski length $13 \approx 3.61$ m/s hai.

Level 5 — Mastery

L5·Q1 — Man chalta hai aur boat drift karti hai

Ek man $m = 80 kg$ ek boat $M = 320 kg$ ke stern par khada hai, dono frictionless water par rest mein. Man bow tak chalta hai, deck distance $L = 5 m$ , phir ruk jaata hai. (a) Boat kitni door gayi? (b) Jis instant par man water ke relative $2 m/s$ se chal raha hai, boat ki velocity kya hai? (c) Poore time $V_{C M}$ kya hai?

Recall Solution

(a) Boat displacement: external horizontal force $= 0$ , isliye $X_{C M}$ freeze karo. Maano boat $d$ back slide kare; man ka ground displacement tab $(L - d)$ hai aur boat ka $- d$ hai: $m (L - d) - M d = 0 \Rightarrow d = \frac{m L}{m + M} = \frac{( 80 ) ( 5 )}{400} = 1.0 m backward .$ (b) Man ke chalte waqt boat velocity: $v_{man}$ man ki water ke relative velocity ko denote kare, jo $+ 2 m/s$ di gayi hai. Kyunki system rest se shuru hua aur koi external horizontal force nahi hai, total momentum zero rehta hai: $0 = m v_{man} + M v_{b o a t} \Rightarrow v_{b o a t} = - \frac{m v _{man}}{M} = - \frac{( 80 ) ( 2 )}{320} = - 0.5 m/s .$ Boat $0.5 m/s$ se back drift karti hai jab man chal raha hota hai. (c) Poore time CM velocity: external horizontal force $= 0$ aur system rest se shuru hua, isliye sab waqt $V_{C M} = 0$ — yahi woh anchor hai jisne (a) aur (b) dono kaam karaye.

L5·Q2 — Teen-fragment explosion, teesri velocity nikalo

Mass $M = 9 kg$ ka ek stationary bomb teen pieces mein explode hota hai. Piece A ( $2 kg$ ) $v_{A} = (6, 0) m/s$ par jaata hai; piece B ( $3 kg$ ) $v_{B} = (0, 4) m/s$ par jaata hai. Piece C ( $4 kg$ ) ki velocity nikalo. Confirm karo ki CM rest mein rehta hai.

Recall Solution

KYU momentum conserved hai: bomb rest se shuru hota hai aur burst internal hai, isliye $\sum m_{i} v_{i} = 0$ . $m_{A} v_{A} + m_{B} v_{B} + m_{C} v_{C} = 0.$ $(2) (6, 0) + (3) (0, 4) + (4) v_{C} = (0, 0) .$ $(12, 0) + (0, 12) + 4 v_{C} = (0, 0) \Rightarrow 4 v_{C} = (- 12, - 12) \Rightarrow v_{C} = (- 3, - 3) m/s .$ CM check: $\frac{( 12 , 0 ) + ( 0 , 12 ) + ( - 12 , - 12 )}{9} = \frac{( 0 , 0 )}{9} = (0, 0)$ . ✓ Balance point kabhi nahi hila — shelf par rest mein rakha bomb apna CM exactly wahin chhodta hai jahan tha.

Figure s08 — Teen momentum vectors tip-to-tail draw kiye gaye hain aur loop origin par close hona chahiye (total momentum zero). Piece A (teal, right) aur piece B (orange, up) piece C (plum, down-left at $(- 3, - 3)$ ) se balance hote hain, jiska momentum $(- 12, - 12)$ baaki dono ko exactly cancel karta hai.

L5·Q3 — Degenerate case: zero aur collinear masses

(a) Agar do particles mein se ek ka mass $0$ ho, toh CM kahan hai? (b) Agar dono particles same point par hain, toh CM kahan hai? (c) Agar $m_{1} = m_{2}$ ho, dikhao ki $x_{C M}$ midpoint hai.

Recall Solution

(a) $m_{2} = 0$ ke saath: $x_{C M} = \frac{m _{1} x _{1} + 0}{m _{1} + 0} = x_{1}$ . Ek massless particle average mein koi weight nahi carry karta — CM poori tarah real mass par hota hai. (Yahi limiting behaviour hai jo "midpoint" thinking ko rokta hai.) (b) Agar $x_{1} = x_{2} = x$ ho: $x_{C M} = \frac{m _{1} x + m _{2} x}{m _{1} + m _{2}} = x$ . Dono ek point par → CM us point par, koi bhi masses ho. (c) Agar $m_{1} = m_{2} = m$ ho: $x_{C M} = \frac{m x _{1} + m x _{2}}{2 m} = \frac{x _{1} + x _{2}}{2}$ — plain midpoint. Equal masses ek hi case hai jahan "geometric centre" correct hai, aur yahi wajah hai ki midpoint ki aadat andar ghus jaati hai.

Active Recall

Recall Poore page ke one-line answers

Boat recoil formula? ::: $d = \frac{m L}{m + M}$ , boat man ke opposite direction mein move karti hai. Frictionless water par $X_{C M}$ kyun freeze hota hai? ::: Koi external horizontal force nahi, isliye $a_{C M} = 0$ aur (rest se) $X_{C M}$ constant. Explosion: burst ke through kaun si quantity conserved hoti hai? ::: Total momentum $\sum m_{i} v_{i}$ ; internal forces Newton's Third Law se cancel ho jaate hain. Lighter skater vs heavier skater — kaun farther jaata hai? ::: Lighter wala, kyunki $m_{1} d_{1} = m_{2} d_{2}$ . Do-body system ki CM velocity? ::: $V_{C M} = \frac{m _{1} v _{1} + m _{2} v _{2}}{m _{1} + m _{2}}$ , phir speed ke liye magnitude lo.

Connections

Conservation of Linear Momentum — yahaan har $F_{e x t} = 0$ problem is law ka disguise hai.
Newton's Third Law — kyun explosion aur push forces cancel hote hain.
Collisions (Elastic & Inelastic) — agla step: kya hota hai jab pieces saath aate hain.
Rocket Propulsion — CM idea with mass thrown out over time.
Rigid Body Dynamics — extended bodies jahan $F = M a_{C M}$ abhi bhi rule karta hai.