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

Before anything, one reminder of the symbols so nobody is lost from line one:

The scenario matrix

Every CM problem falls into one of these cells. We will hit all of them.

Cell	What makes it different	Example that covers it
A. $F_{e x t} = 0$ , both signs	No external force → CM frozen; velocities have opposite signs	Ex 1 (skaters)
B. $F_{e x t} = 0$ , position bookkeeping	Displacements, one object shifts back	Ex 2 (man on boat)
C. $F_{e x t} \neq = 0$ constant	CM follows a real trajectory (gravity)	Ex 3 (exploding shell)
D. Zero / degenerate input	A mass or velocity is 0, or masses equal	Ex 4 (equal masses, dropped piece)
E. Limiting case	One mass $\to \infty$ or $\to 0$	Ex 5 (heavy wall / light fleck)
F. 2-D vector case	Motion not on a line; components	Ex 6 (L-shaped scatter)
G. Real-world word problem	Translate messy words → equation	Ex 7 (person climbs a ladder-cart)
H. Exam twist	Looks like it needs more; CM shortcut wins	Ex 8 (two blocks + spring, find where they meet)

Let's clear every cell.

Ex 1 — Cell A: opposite signs, frozen CM

Worked example Two skaters push apart

Skaters $m_{1} = 40$ kg and $m_{2} = 60$ kg stand at rest on frictionless ice and shove each other. Afterwards $m_{1}$ glides right at $v_{1} = 3$ m/s. Find $v_{2}$ and the CM velocity.

Forecast: Guess — does the CM drift toward the lighter skater, the heavier one, or stay put?

Step 1. Identify external forces. Ice is frictionless, gravity is vertical and cancelled by the ice pushing up. So horizontal $F_{e x t} = 0$ . Why this step? If $F_{e x t} = 0$ then $V_{C M}$ is constant (parent corollary). It was $0$ at the start, so it stays $0$ .

Step 2. Write momentum conservation. Total momentum before $= 0$ (both at rest): $0 = m_{1} v_{1} + m_{2} v_{2} .$ Why this step? $P = M V_{C M}$ ; since $V_{C M} = 0$ , total momentum is zero at all times.

Step 3. Solve for $v_{2}$ : $v_{2} = - \frac{m _{1} v _{1}}{m _{2}} = - \frac{40 \cdot 3}{60} = - 2 m/s .$ The minus sign says $m_{2}$ goes left — opposite to $m_{1}$ . Both signs appear, as promised.

Verify: Plug back. $P = 40 (+ 3) + 60 (- 2) = 120 - 120 = 0$ ✓. Units: $kg \cdot m/s$ on both terms ✓. Heavier skater is slower ( $2 < 3$ ) — correct, since $m_{2} v_{2}$ must match $m_{1} v_{1}$ .

Answer to the forecast: the CM stays put. It never had a reason to move.

Ex 2 — Cell B: position bookkeeping

Worked example Man walks on a boat

A man $m = 60$ kg walks $L = 2$ m forward along a boat $M = 120$ kg floating on still water (ignore water drag). How far does the boat move, and which way?

Forecast: The boat moves backward — but more or less than 1 m?

Step 1. External horizontal force $= 0$ (no drag) → $X_{C M}$ is fixed in the water frame. Why this step? No external force means the balance point cannot slide; whatever the man gains forward, the boat must give back.

Step 2. Let the boat slide back by $d$ (so boat displacement $= - d$ ). The man walks $L$ along the boat, so relative to the ground he moves $L - d$ forward. Why this step? " $2$ m forward" is measured along the boat, not the ground. Ground displacement subtracts the boat's own retreat.

Step 3. Fix the CM: mass-weighted displacements sum to zero: $0 = m (L - d) + M (- d) .$

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

Verify: Man's ground move $= L - d = 2 - 0.667 = 1.333$ m forward. Check CM: $60 (1.333) + 120 (- 0.667) = 80 - 80 = 0$ ✓. Units: kg·m both sides ✓. Boat is heavier, so it moves less than the man — sensible.

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

Worked example Exploding shell

A shell of mass $M = 10$ kg is fired and at the top of its arc moves horizontally at $u = 20$ m/s. It bursts into two equal $5$ kg pieces. One piece falls straight down (horizontal velocity $0$ ) right after the burst. Find the horizontal velocity of the other piece.

Forecast: Faster than $20$ , exactly $20$ , or slower?

Step 1. External force during the tiny burst: only gravity (vertical). Horizontal $F_{e x t} = 0$ , so horizontal momentum is conserved through the explosion. Why this step? Explosion forces are internal → they cancel and can't change horizontal $P$ .

Step 2. Horizontal momentum before $=$ after: $M u = m_{1} v_{1} + m_{2} v_{2}, 10 (20) = 5 (0) + 5 v_{2} .$ Why this step? One piece has $v_{1} = 0$ horizontally (falls straight down).

Step 3. Solve: $v_{2} = \frac{200}{5} = 40 m/s .$ Faster than $20$ — it grabs the missing momentum of the dropped piece.

Step 4 (the CM point). The CM keeps moving horizontally at $V_{C M} = 20$ m/s and keeps falling under gravity — the same parabola the whole shell would have followed. Why this step? $F_{e x t} = M g$ still, so $a_{C M} = g$ downward regardless of the burst.

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

Answer to forecast: faster — $40$ m/s.

Ex 4 — Cell D: zero / degenerate inputs

Worked example Equal masses, one dropped from rest

Two identical blocks each $m = 3$ kg sit on frictionless ice. Block 1 is at rest; block 2 is nudged to $v_{2} = 4$ m/s. Later a third, massless tag ( $m_{3} = 0$ ) is attached to block 1. Find $V_{C M}$ of the whole system and confirm the massless tag changes nothing.

Forecast: Does the $0$ kg tag drag $V_{C M}$ toward block 1?

Step 1. Total mass: $M = 3 + 3 + 0 = 6$ kg. Why this step? A zero mass adds nothing to $M$ — that's the degenerate check.

Step 2. CM velocity: $V_{C M} = \frac{m _{1} v _{1} + m _{2} v _{2} + m _{3} v _{3}}{M} = \frac{3 ( 0 ) + 3 ( 4 ) + 0 \cdot v _{3}}{6} = \frac{12}{6} = 2 m/s .$ Why this step? The $m_{3} v_{3}$ term is $0 \cdot (anything) = 0$ , so the tag's motion is irrelevant.

Step 3. Because the masses are equal, notice $V_{C M}$ is just the plain average: $\frac{0 + 4}{2} = 2$ 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 contributes $0$ ✓.

Answer to forecast: no — a $0$ kg object never moves the CM.

Ex 5 — Cell E: limiting case (one mass huge)

Worked example Ball bounces off a wall

A ball $m = 0.5$ kg hits a wall of mass $M$ head-on at $v = 6$ m/s and bounces back at $6$ m/s. Take the limit $M \to \infty$ (an immovable wall). Find the wall's velocity change and the CM velocity.

Forecast: As $M \to \infty$ , does the wall's speed go to something finite or to $0$ ?

Step 1. Momentum conservation for the ball–wall system (external floor/foundation forces ignored for the collision instant): $m v + M (0) = m (- v) + M V_{w} .$ Why this step? The ball's momentum flips sign; the wall must absorb the difference.

Step 2. Solve the wall speed: $V_{w} = \frac{m v - m ( - v )}{M} = \frac{2 m v}{M} = \frac{2 ( 0.5 ) ( 6 )}{M} = \frac{6}{M} .$ Why this step? Isolate $V_{w}$ to see how it shrinks with $M$ .

Step 3. Take the limit: $lim_{M \to \infty} V_{w} = lim_{M \to \infty} \frac{6}{M} = 0.$ Why this step? This is the whole point of a limiting case — an infinitely heavy wall doesn't budge, yet it still reverses the ball's momentum. The wall's momentum change is $M V_{w} = 6$ kg·m/s (finite!), even though its speed $\to 0$ .

Verify: At $M = 1 0^{6}$ kg, $V_{w} = 6/1 0^{6} = 6 \times 1 0^{- 6}$ m/s ≈ $0$ ✓. Momentum given to wall $= M V_{w} = 6$ kg·m/s $= 2 m v = 2 (0.5) (6)$ ✓. Units m/s and kg·m/s consistent ✓.

Answer to forecast: speed $\to 0$ , but the momentum it receives stays a finite $6$ kg·m/s.

Ex 6 — Cell F: 2-D vector case

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

Worked example L-shaped scatter

A stationary object of mass $M = 4$ kg explodes on frictionless ice into two pieces. Piece 1 ( $m_{1} = 1$ kg) flies east (positive $x$ ) at $12$ m/s. Piece 2 ( $m_{2} = 3$ kg) — find its velocity vector.

Forecast: Which direction must piece 2 go, and roughly how fast compared to piece 1?

Step 1. Initial momentum is zero (at rest), and $F_{e x t} = 0$ , so total momentum stays $0$ in both $x$ and $y$ . Why this step? In 2-D, momentum is conserved component by component — two equations, not one.

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: $0 = m_{1} v_{1 y} + m_{2} v_{2 y} = 1 (0) + 3 v_{2 y}$ . $v_{2 y} = 0.$ Why this step? Piece 1 had no north–south motion, so piece 2 can't either — otherwise $P_{y} \neq = 0$ . Look at the red arrow in the figure: it points straight back along the line of piece 1, no sideways tilt.

Step 4. So $v_{2} = (- 4, 0)$ m/s — due west, slower than piece 1 because it is heavier.

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

Ex 7 — Cell G: real-world word problem

Worked example Person climbs a wheeled cart

A cart of mass $M = 50$ kg carries a $B = 30$ kg box and a person $m = 70$ kg, all on frictionless rails, initially at rest. The person walks $1.5$ m toward the box (which sits fixed on the cart). By how much does the cart-plus-box slide, and which way?

Forecast: Bigger or smaller shift than the man-on-boat case (0.667 m for a lighter walker)?

Step 1. Group masses: the "walker" is the person $m = 70$ kg; everything he walks on is the cart+box $= M + B = 80$ kg. Why this step? The box doesn't move relative to the cart, so cart and box move together — treat them as one mass.

Step 2. No external horizontal force → $X_{C M}$ fixed. Person walks $L = 1.5$ m along the cart, cart slides back by $d$ ; person's ground move $= L - d$ . Why this step? Same bookkeeping as the boat: walk distance is measured relative to the cart.

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

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's ground displacement $= 0.8$ m; cart back $0.7$ m; walker heavier than the surface here so shift is larger than the boat case ✓. Units m ✓.

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

Worked example Two blocks meet via a spring — where?

Blocks $m_{1} = 2$ kg and $m_{2} = 6$ kg sit on frictionless ice, $m_{1}$ at $x = 0$ and $m_{2}$ at $x = 8$ m, connected by a compressed spring, held at rest. Released, they rush together. Where do they collide? (Ignore block sizes.)

Forecast: Nearer $x = 0$ , the middle ( $x = 4$ ), or nearer $x = 8$ ?

Step 1. Spring forces are internal; $F_{e x t} = 0$ ; system starts at rest. So $X_{C M}$ never moves. Why this step? The trap is to solve spring dynamics. But the meeting point must be the fixed CM — no differential equations needed.

Step 2. Compute $X_{C M}$ : $X_{C M} = \frac{m _{1} x _{1} + m _{2} x _{2}}{m _{1} + m _{2}} = \frac{2 ( 0 ) + 6 ( 8 )}{8} = \frac{48}{8} = 6 m.$ Why this step? They can only collide at the one point that stays put — the CM.

Step 3. So they meet at $x = 6$ m — nearer the heavier $m_{2}$ , exactly as mass-weighting predicts.

Verify: $m_{1}$ travels $6$ m right, $m_{2}$ travels $2$ m left. Mass-weighted displacement: $2 (+ 6) + 6 (- 2) = 12 - 12 = 0$ ✓ (CM unmoved). Meeting point sits $3 \times$ closer to the heavy block since $m_{2} = 3 m_{1}$ ✓. Units m ✓.

Answer to forecast: nearer $x = 8$ — the heavier block barely moves.

Active Recall

Connections

Parent topic ↩ — the law these examples exercise.
Conservation of Linear Momentum — every $F_{e x t} = 0$ cell above.
Newton's Third Law — why internal spring/explosion forces cancel.
Collisions (Elastic & Inelastic) — Ex 3, 5, 6 are collision/explosion cases.
Rocket Propulsion — the momentum-swap idea of Ex 5 pushed further.
Rigid Body Dynamics — cart+box treated as one mass in Ex 7.