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

Two master facts do all the heavy lifting on this page. Let us restate them in plain words so nothing is a mystery:

Level 1 — Recognition

L1·Q1 — Where is the balance point of two blocks?

Two blocks sit on a line: $m_{1} = 2 kg$ at $x_{1} = 0 m$ and $m_{2} = 6 kg$ at $x_{2} = 4 m$ . Find $x_{C M}$ . Is it nearer the light block or the heavy one?

Recall Solution

Apply the definition in one dimension (positions are just numbers on a line): $x_{C M} = \frac{m _{1} x _{1} + m _{2} x _{2}}{m _{1} + m _{2}} = \frac{( 2 ) ( 0 ) + ( 6 ) ( 4 )}{2 + 6} = \frac{24}{8} = 3 m .$ WHAT this looks like: the balance point sits at $3$ m — three-quarters of the way toward the heavy $6$ kg block (see figure below). WHY: the heavier mass "pulls" the average toward itself, exactly as a see-saw balances closer to the heavy child.

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

Figure s01 — Two blocks on a line. The light teal $2$ kg block sits at $x_{1} = 0$ , the larger orange $6$ kg block at $x_{2} = 4$ . The plum triangle marks the mass-weighted balance point $x_{C M} = 3$ m, sitting three-quarters of the way toward the heavy block.

L1·Q2 — Which of these changes $V_{C M}$ ?

A closed box floats in deep space (no gravity, no contact). Inside, a fan blows air, a spring launches a ball against a wall, two magnets snap together. Do any of these change the velocity of the box's centre of mass?

Recall Solution

No — none of them. Fan thrust, spring force, magnet attraction are all internal (both the pusher and the pushed are inside the box). By Newton's 3rd law each comes with an equal-and-opposite partner, so $\sum f_{in t er na l} = 0$ . With $F_{e x t} = 0$ we get $a_{C M} = 0$ , hence $V_{C M}$ is constant. WHY it matters: to move the box's balance point you need something outside to push — a wall it touches, gravity, a jet of gas thrown out of the box.

Figure s02 — Inside the floating box every internal push (fan, spring, magnets) comes with an equal-and-opposite reaction arrow. The teal and orange arrows cancel pairwise, so the plum CM dot stays exactly where it was — no outside force, no CM motion.

Level 2 — Application

L2·Q1 — Two skaters, unequal push

Skaters $m_{1} = 50 kg$ and $m_{2} = 75 kg$ stand at rest on frictionless ice and push apart. Skater 1 moves off at $v_{1} = 3 m/s$ . Find $v_{2}$ and the speed of the CM.

Recall Solution

Step 1 — WHY momentum is conserved: ice is frictionless, so no horizontal external force acts. Initial total momentum $= 0$ . Step 2 — Write it out (take skater 1's direction as positive): $0 = m_{1} v_{1} + m_{2} v_{2} \Rightarrow v_{2} = - \frac{m _{1} v _{1}}{m _{2}} = - \frac{( 50 ) ( 3 )}{75} = - 2 m/s .$ The minus sign means skater 2 glides the opposite way at $2 m/s$ . Step 3 — CM speed and the conceptual link: because total momentum is conserved and started at zero, it stays zero. The CM velocity is total momentum divided by total mass, so $V_{C M} = \frac{p _{t o t a l}}{M} = \frac{m _{1} v _{1} + m _{2} v _{2}}{M} = \frac{0}{125} = 0$ . WHY this is the whole point: conserved zero momentum means $M V_{C M} = 0$ , so the balance point is frozen — the push, being internal, can never set the CM in motion no matter how the skaters scatter.

Figure s03 — The two skaters start on the frozen plum CM line. After the push, the light skater (orange, $3$ m/s) glides right and the heavy skater (teal, $2$ m/s) glides left. Arrow lengths show the lighter skater moving faster; the plum CM dot never leaves its line.

L2·Q2 — Exploding shell, position of the second piece

A shell of mass $M = 4 kg$ moving right at $V_{C M} = 10 m/s$ (only gravity external, but ask about the instant of the burst so ignore the fall) explodes into two equal pieces of $2 kg$ . One piece flies backward at $5 m/s$ . Find the velocity of the other piece.

Recall Solution

WHY momentum is conserved through the burst: the explosion is internal; over the tiny burst instant gravity's impulse is negligible, so total momentum is unchanged. $M V_{C M} = m_{a} v_{a} + m_{b} v_{b} .$ $(4) (10) = (2) (- 5) + (2) v_{b} \Rightarrow 40 = - 10 + 2 v_{b} \Rightarrow v_{b} = \frac{50}{2} = 25 m/s .$ The second piece shoots forward at $25 m/s$ . Check the CM: $\frac{( 2 ) ( - 5 ) + ( 2 ) ( 25 )}{4} = \frac{40}{4} = 10 m/s$ — unchanged, exactly as $F_{e x t} = M a_{C M}$ demands.

Figure s04 — Before: one shell (plum) moving right at $10$ m/s. After the burst: piece A (teal) recoils left at $5$ m/s, piece B (orange) rockets right at $25$ m/s. The plum CM arrow beneath both keeps its original $10$ m/s length — the burst reshuffles momentum without changing the total.

Level 3 — Analysis

L3·Q1 — Man walks on a boat, boat recoils

A man $m = 60 kg$ walks a distance $L = 3 m$ (measured along the boat's deck) from stern to bow. The boat $M = 180 kg$ floats on frictionless water. How far does the boat move relative to the water, and in which direction?

Recall Solution

Step 1 — The golden shortcut: no external horizontal force → $X_{C M}$ stays frozen. Step 2 — Bookkeeping (WHAT the letters mean): let the boat slide back a distance $d$ (relative to water). The man walks $L$ forward along the deck, but the deck itself slid back $d$ , so the man's displacement relative to water is $(L - d)$ forward. Step 3 — Freeze the CM: total mass-weighted displacement must be zero. $m (L - d) + M (- d) = 0.$ $m L - m d - M d = 0 \Rightarrow d = \frac{m L}{m + M} = \frac{( 60 ) ( 3 )}{240} = 0.75 m .$ The boat moves $0.75 m$ backward (opposite to the man's walk).

Figure s05 — Top: man at the stern, boat at rest. Bottom: after the walk the boat has slid back $d = 0.75$ m (plum arrow) while the man reached the bow. The dashed plum line marks the frozen $X_{C M}$ that both configurations share.

L3·Q2 — Where does the far skater end up?

Return to the L2 skaters ( $m_{1} = 50 kg$ , $m_{2} = 75 kg$ , at rest, then push apart). After $t = 2 s$ of gliding, how far apart are they, and how far has each moved from the start line?

Recall Solution

Speeds from L2: $v_{1} = 3 m/s$ , $v_{2} = - 2 m/s$ . Distances: $d_{1} = 3 \times 2 = 6 m$ ; $d_{2} = 2 \times 2 = 4 m$ . Separation $= 6 + 4 = 10 m$ . WHY they differ: the CM stays fixed, so displacements obey $m_{1} d_{1} = m_{2} d_{2}$ : check $50 \times 6 = 300 = 75 \times 4$ . ✓ The lighter skater travels farther, as if hinged at the frozen balance point.

Level 4 — Synthesis

L4·Q1 — Projectile splits; find where the second fragment lands

A firework of mass $M$ is launched. It bursts into two equal halves at the top of its arc. One half falls straight down and lands at $x_{1} = 40 m$ (measured from the launch point). If the unexploded shell would have landed at $x_{f u l l} = 100 m$ , where does the second half land?

Recall Solution

Key idea: the explosion is internal, so the CM keeps following the original parabola and lands where the whole shell would have — at $x_{C M} = 100 m$ . Write the CM landing point as the mass-weighted average of the two landing spots (equal masses $M /2$ each): $x_{C M} = \frac{\frac{M}{2} x _{1} + \frac{M}{2} x _{2}}{M} = \frac{x _{1} + x _{2}}{2} .$ $100 = \frac{40 + x _{2}}{2} \Rightarrow x_{2} = 200 - 40 = 160 m .$ The second half lands at $160 m$ — far beyond the original target, because it must carry the CM's "missing" ground.

Figure s06 — The dashed plum parabola is the CM path, landing at $x_{f u l l} = 100$ m. At the burst, one half (teal) drops straight down to $x_{1} = 40$ m; the other half (orange) is flung forward to $x_{2} = 160$ m so the mass-weighted average of the two landing points lands exactly on the CM target.

L4·Q2 — CM velocity from momentum in 2D

Two pucks on frictionless ice: $m_{1} = 3 kg$ moving at $v_{1} = (4, 0) m/s$ and $m_{2} = 1 kg$ moving at $v_{2} = (0, 8) m/s$ . Find $V_{C M}$ and its speed.

Recall Solution

WHY use $P = M V_{C M}$ : it converts messy individual motions into one clean CM velocity. $V_{C M} = \frac{m _{1} v _{1} + m _{2} v _{2}}{M} = \frac{( 3 ) ( 4 , 0 ) + ( 1 ) ( 0 , 8 )}{4} = \frac{( 12 , 8 )}{4} = (3, 2) m/s .$ Speed: $∣ V_{C M} ∣ = 3^{2} + 2^{2} = 13 \approx 3.606 m/s$ .

Figure s07 — Puck 1 (teal) drives right, puck 2 (orange) drives up. The plum CM velocity $(3, 2)$ is the mass-weighted average, tilted toward the heavier puck 1; its length is $13 \approx 3.61$ m/s.

Level 5 — Mastery

L5·Q1 — Man walks and the boat drifts

A man $m = 80 kg$ stands at the stern of a boat $M = 320 kg$ , both at rest on frictionless water. The man walks to the bow, a deck distance $L = 5 m$ , then stops. (a) How far did the boat move? (b) At the instant the man is walking at $2 m/s$ relative to the water, what is the boat's velocity? (c) What is $V_{C M}$ throughout?

Recall Solution

(a) Boat displacement: external horizontal force $= 0$ , so freeze $X_{C M}$ . Let the boat slide back $d$ ; the man's ground displacement is then $(L - d)$ and the boat's is $- d$ : $m (L - d) - M d = 0 \Rightarrow d = \frac{m L}{m + M} = \frac{( 80 ) ( 5 )}{400} = 1.0 m backward .$ (b) Boat velocity while the man moves: let $v_{man}$ denote the man's velocity relative to the water, given as $+ 2 m/s$ . Since the system started at rest with no external horizontal force, total momentum stays zero: $0 = m v_{man} + M v_{b o a t} \Rightarrow v_{b o a t} = - \frac{m v _{man}}{M} = - \frac{( 80 ) ( 2 )}{320} = - 0.5 m/s .$ The boat drifts back at $0.5 m/s$ while the man walks. (c) CM velocity throughout: external horizontal force $= 0$ and the system began at rest, so $V_{C M} = 0$ for all time — this is the anchor that made both (a) and (b) work.

L5·Q2 — Three-fragment explosion, find the third velocity

A stationary bomb of mass $M = 9 kg$ explodes into three pieces. Piece A ( $2 kg$ ) flies at $v_{A} = (6, 0) m/s$ ; piece B ( $3 kg$ ) flies at $v_{B} = (0, 4) m/s$ . Find the velocity of piece C ( $4 kg$ ). Confirm the CM stays at rest.

Recall Solution

WHY momentum is conserved: the bomb starts at rest and the burst is internal, so $\sum m_{i} v_{i} = 0$ . $m_{A} v_{A} + m_{B} v_{B} + m_{C} v_{C} = 0.$ $(2) (6, 0) + (3) (0, 4) + (4) v_{C} = (0, 0) .$ $(12, 0) + (0, 12) + 4 v_{C} = (0, 0) \Rightarrow 4 v_{C} = (- 12, - 12) \Rightarrow v_{C} = (- 3, - 3) m/s .$ CM check: $\frac{( 12 , 0 ) + ( 0 , 12 ) + ( - 12 , - 12 )}{9} = \frac{( 0 , 0 )}{9} = (0, 0)$ . ✓ The balance point never moved — a bomb at rest on the shelf leaves its CM exactly where it lay.

Figure s08 — Three momentum vectors drawn tip-to-tail must close the loop back to the origin (total momentum zero). Piece A (teal, right) and piece B (orange, up) are balanced by piece C (plum, down-left at $(- 3, - 3)$ ), whose momentum $(- 12, - 12)$ exactly cancels the other two.

L5·Q3 — Degenerate case: zero and collinear masses

(a) If one of two particles has mass $0$ , where is the CM? (b) If both particles sit at the same point, where is the CM? (c) If $m_{1} = m_{2}$ , show $x_{C M}$ is the midpoint.

Recall Solution

(a) With $m_{2} = 0$ : $x_{C M} = \frac{m _{1} x _{1} + 0}{m _{1} + 0} = x_{1}$ . A massless particle carries no weight in the average — the CM sits entirely on the real mass. (This is the limiting behaviour that stops "midpoint" thinking.) (b) If $x_{1} = x_{2} = x$ : $x_{C M} = \frac{m _{1} x + m _{2} x}{m _{1} + m _{2}} = x$ . Both at one point → CM at that point, any masses. (c) If $m_{1} = m_{2} = m$ : $x_{C M} = \frac{m x _{1} + m x _{2}}{2 m} = \frac{x _{1} + x _{2}}{2}$ — the plain midpoint. Equal masses is the only case where "geometric centre" is correct, which is exactly why the midpoint habit sneaks in.

Active Recall

Recall One-line answers to the whole page

Boat recoil formula? ::: $d = \frac{m L}{m + M}$ , boat moves opposite to the man. Why does $X_{C M}$ freeze on frictionless water? ::: No external horizontal force, so $a_{C M} = 0$ and (from rest) $X_{C M}$ constant. Explosion: what quantity is conserved through the burst? ::: Total momentum $\sum m_{i} v_{i}$ ; internal forces cancel by Newton's Third Law. Lighter skater vs heavier skater — who travels farther? ::: The lighter one, since $m_{1} d_{1} = m_{2} d_{2}$ . CM velocity of a two-body system? ::: $V_{C M} = \frac{m _{1} v _{1} + m _{2} v _{2}}{m _{1} + m _{2}}$ , then take magnitude for speed.

Connections

Conservation of Linear Momentum — every $F_{e x t} = 0$ problem here is this law in disguise.
Newton's Third Law — why explosion and push forces cancel.
Collisions (Elastic & Inelastic) — next step: what happens when pieces come together.
Rocket Propulsion — the CM idea with mass thrown out over time.
Rigid Body Dynamics — extended bodies where $F = M a_{C M}$ still rules.