System with external forces — conditions for conservation: Worked examples

Before anything else, three symbols we will lean on. If any feels shaky, this page re-earns it as we go.

The parent's master equation ties them together: $\frac{d P}{d t} = F_{e x t}^{n e t} ⟺ Δ P = J_{e x t}$

Read it as: "the only thing that can change the group's total momentum is an outside impulse."

The scenario matrix

Every problem this topic throws is one (or a mix) of these cells. The worked examples below are tagged with the cell they cover.

Cell	What makes it distinct	The trap it hides
A. Zero total, opposite signs	System starts at rest; parts fly apart with opposite-sign velocities	Forgetting the minus sign on the recoiling piece
B. One-axis conservation	External force acts along one axis only (gravity in a launch)	Claiming all momentum is conserved when only $P_{x}$ is
C. Boundary choice flips the answer	Same physics, momentum conserved or not depending on where you draw the system	Thinking "conserved" is a property of nature, not of your choice
D. Impulsive approximation	Collision + a background force (gravity, friction) acting during contact	Testing forces instead of impulses
E. Degenerate / limiting mass	One mass $\to \infty$ (a wall) or $\to 0$	Assuming momentum "vanishes" when a partner is a wall
F. 2-D angled collision	Velocities point at angles; must split into $x$ and $y$	Adding speeds like scalars instead of vector components
G. Momentum yes, energy no	Inelastic collision — check both laws separately	Assuming KE is conserved because $P$ is
H. Exam twist: continuous mass — stream hitting a target, and a rocket losing mass	Mass, not velocity, is what changes	Using $F = ma$ when mass is changing

We now walk all eight cells (Cell H gets two examples: an inflow stream and a mass-losing rocket).

Example 1 — Cell A: zero total, opposite signs (explosion)

Worked example A firecracker at rest bursts into two pieces

A stationary firecracker of mass $5$ kg bursts into two pieces on a frictionless table. A $2$ kg piece flies right at $30$ m/s. Find the velocity of the $3$ kg piece.

Forecast: The other piece must go left. Faster or slower than $30$ ? (It's heavier, so guess slower.)

Step 1 — Choose the system and check external force. System = both pieces. The bursting force is between the pieces ⇒ internal. Table is frictionless; gravity and normal are vertical and cancel. So horizontal $F_{e x t} = 0$ . Why this step? Only after confirming $F_{e x t, x} = 0$ are we allowed to write $P_{x}$ = constant.

Step 2 — Write initial and final $P_{x}$ . Initially at rest: $P_{x, i} = 0$ . Take right as $+$ . $0 = (2) (+ 30) + (3) (v)$ Why this step? Conservation means "before = after"; we set them equal.

Step 3 — Solve. $v = \frac{- 60}{3} = - 20 m/s \Rightarrow 20 m/s to the left .$ Why this step? The minus sign is the physics — it tells us the direction, matching the forecast.

Verify: Total after $= (2) (30) + (3) (- 20) = 60 - 60 = 0 =$ total before. ✓ Units: kg·m/s throughout. Heavier piece slower (20 < 30), as forecast. ✓

Example 2 — Cell B: conservation on one axis only

Worked example Cannon on the ground (revisited, both axes checked)

A cannon ( $M = 400$ kg) fires a ball ( $m = 4$ kg) horizontally at $200$ m/s. Find the recoil speed, and explicitly show which axis is conserved and which is not.

Forecast: Horizontal is safe (nothing pushes sideways from outside). Vertical is not (gravity, normal). Guess recoil $\approx 2$ m/s.

Step 1 — Split forces by axis. Firing force = internal. External: gravity (down) and ground normal (up) — both vertical. So $F_{e x t, x} = 0$ but $F_{e x t, y} \neq = 0$ in general. Why this step? Newton's law is one equation per axis; we treat $x$ and $y$ separately.

Step 2 — Conserve $P_{x}$ only. Right $= +$ , initially rest: $0 = M (- V) + m v \Rightarrow V = \frac{m v}{M} = \frac{4 \cdot 200}{400} = 2 m/s .$ Why this step? We are forbidden from writing a $P_{y}$ conservation line — the ground supplies vertical impulse. We simply don't need $y$ .

Verify: $P_{x, after} = 400 (- 2) + 4 (200) = - 800 + 800 = 0$ . ✓ In the figure, the pink (ball) and blue (cannon) horizontal arrows are equal-and-opposite; the green vertical ground arrow shows why $P_{y}$ is not conserved. Units: kg·m/s. ✓

Example 3 — Cell C: the boundary choice flips the answer

Worked example A ball bounces off the floor

A $0.5$ kg ball hits the floor moving down at $6$ m/s and rebounds up at $4$ m/s. (a) Is its momentum conserved? (b) What impulse did the floor deliver? (c) For which system is momentum conserved?

Forecast: For the ball alone, momentum clearly changed (it reversed) — so not conserved. The floor must have supplied the impulse.

Step 1 — System = ball alone. Take up $= +$ . Before: $p_{i} = 0.5 (- 6) = - 3$ kg·m/s. After: $p_{f} = 0.5 (+ 4) = + 2$ kg·m/s. Why this step? With the ball alone, gravity and the floor's normal are external — so we expect $P$ to change, and we compute by how much.

Step 2 — Impulse from the change. $J = Δ p = p_{f} - p_{i} = 2 - (- 3) = + 5 N\cdots (upward) .$ Why this step? $Δ P = J_{e x t}$ — the change is the impulse. This is the Impulse–Momentum Theorem doing the work.

Step 3 — Enlarge the system to ball + Earth. Now the floor's push and gravity are internal (Earth is inside). By Newton's 3rd law they cancel ⇒ total $P$ (ball + Earth) is conserved. Earth recoils by an unmeasurably tiny amount. Why this step? Shows the punchline: "conserved" is a statement about your chosen boundary, not about nature.

Verify: $J = + 5$ N·s is positive (upward), consistent with a floor that pushes up. Magnitude sanity: $∣ J ∣ = 5 > ∣ p_{i} ∣ = 3$ because the ball not only stopped ( $+ 3$ ) but was thrown back up ( $+ 2$ ). ✓

Example 4 — Cell D: impulsive approximation (collision under gravity)

Worked example Two pucks collide in mid-air

Two pucks (combined mass $2$ kg) collide during a contact lasting $Δ t = 2$ ms. A typical contact impulse in this hit is about $5$ N·s. Show that gravity can be ignored during the collision, and quantify the tiny error.

Forecast: Gravity acts for only $2$ ms during contact — its impulse should be minuscule next to $5$ N·s.

Step 1 — Compute gravity's impulse during contact. $J_{g} = m g Δ t = 2 \times 9.8 \times 0.002 = 0.0392 N\cdots .$ Why this step? The correct test compares impulses, not forces — impulse is what changes $P$ .

Step 2 — Form the ratio. $\frac{J _{g}}{J _{c}} = \frac{0.0392}{5} \approx 0.0078 = 0.78%.$ Why this step? A ratio under $1%$ means treating $P$ as conserved through the impact introduces under a percent of error.

Step 3 — Two-phase strategy. Conserve $P$ through the collision (ignore gravity), then re-include gravity for the free flight before and after. Why this step? This is how the approximation is actually used: it isolates the collision as an instantaneous event.

Verify: $J_{g} = 0.0392$ N·s and ratio $= 0.00784 < 0.01$ . ✓ Units: (kg)(m/s²)(s) = kg·m/s = N·s. ✓

Example 5 — Cell E: degenerate mass (bouncing off a wall)

Worked example A ball bounces elastically off a fixed wall

A $0.2$ kg ball hits a rigid wall at $10$ m/s and rebounds at $10$ m/s (elastic). Is the ball's momentum conserved? Is momentum conserved for ball + wall? What is the limiting picture as the wall's mass $\to \infty$ ?

Forecast: The ball's momentum flips sign — not conserved for the ball alone. The wall (attached to Earth) absorbs the impulse.

Step 1 — Ball alone. Take incoming direction $= +$ . Before $p_{i} = 0.2 (+ 10) = + 2$ ; after $p_{f} = 0.2 (- 10) = - 2$ . $Δ p = - 2 - 2 = - 4 N\cdots .$ Why this step? The wall's normal force is external to the ball, so a nonzero $Δ p$ is expected.

Step 2 — The limit "wall = infinite mass". Momentum conservation of ball+wall says the wall gains $+ 4$ N·s. Its velocity change is $Δ V = J / M_{w a l l} = 4/ M_{w a l l}$ . As $M_{w a l l} \to \infty$ , $Δ V \to 0$ : the wall stays still yet carries the momentum. Why this step? This is the degenerate case — the "wall" is really an object so massive its velocity change is negligible, exactly like the Earth in Example 3.

Verify: $lim_{M \to \infty} 4/ M = 0$ . ✓ Ball's KE before $= \frac{1}{2} (0.2) (1 0^{2}) = 10$ J, after $= 10$ J — elastic, KE conserved, consistent with speed unchanged. ✓

Example 6 — Cell F: 2-D angled collision (must resolve components)

Worked example Glancing collision on a frictionless table

Ball A ( $1$ kg) moves right at $4$ m/s and strikes stationary ball B ( $1$ kg). After impact, A moves off at $3 0^{\circ}$ above the original line at $23$ m/s. Find B's velocity vector.

Forecast: B must carry the "missing" sideways momentum downward to keep $P_{y} = 0$ , and share the forward momentum.

Step 1 — Check external force, pick axes. Frictionless table ⇒ no horizontal external force ⇒ both $P_{x}$ and $P_{y}$ conserved. Let $x$ = A's original direction. Why this step? On a horizontal frictionless table the whole momentum vector is conserved — both axes are fair game.

Step 2 — Conserve $P_{x}$ . Before: $(1) (4) + 0 = 4$ . $4 = (1) (23 cos 3 0^{\circ}) + (1) v_{B x} = 23 \cdot \frac{3}{2} + v_{B x} = 3 + v_{B x} \Rightarrow v_{B x} = 1.$ Why this step? $x$ -components add like ordinary numbers once resolved.

Step 3 — Conserve $P_{y}$ . Before: $0$ . $0 = (1) (23 sin 3 0^{\circ}) + (1) v_{B y} = 23 \cdot \frac{1}{2} + v_{B y} = 3 + v_{B y} \Rightarrow v_{B y} = - 3 .$ Why this step? The initial sideways momentum was zero, so the two final sideways parts must cancel — B goes down.

Step 4 — Assemble. $v_{B} = (1, - 3)$ , speed $= 1 + 3 = 2$ m/s at angle $arctan (3 /1) = 6 0^{\circ}$ below the line. Why this step? Convert components back to a magnitude-and-direction answer.

Verify: $P_{x}$ : $3 + 1 = 4$ ✓. $P_{y}$ : $3 - 3 = 0$ ✓. In the figure the pink (A') and blue (B') arrows tip-to-tail rebuild the original yellow arrow — that's momentum conservation drawn as a vector triangle. Note A' at $3 0^{\circ}$ and B' at $6 0^{\circ}$ are perpendicular — the signature of an equal-mass elastic hit. ✓

Example 7 — Cell G: momentum yes, kinetic energy no (perfectly inelastic)

Worked example Two clay lumps stick together

A $3$ kg lump moving right at $4$ m/s hits a $1$ kg lump moving left at $4$ m/s. They stick. Find the common velocity, then check kinetic energy.

Forecast: The heavier lump wins ⇒ they move right, but slower than $4$ . KE will drop (clay deforms).

Step 1 — Conserve $P_{x}$ (frictionless, no external horizontal force). Right $= +$ : $3 (4) + 1 (- 4) = (3 + 1) v_{f} \Rightarrow 12 - 4 = 4 v_{f} \Rightarrow v_{f} = 2 m/s (right) .$ Why this step? Sticking is an internal force, so $P$ is conserved regardless of the mess.

Step 2 — Compare kinetic energy before/after. $K E_{i} = \frac{1}{2} (3) (4^{2}) + \frac{1}{2} (1) (4^{2}) = 24 + 8 = 32 J, K E_{f} = \frac{1}{2} (4) (2^{2}) = 8 J .$ Why this step? Momentum and energy have different conditions — we must test energy separately (see Elastic vs Inelastic Collisions).

Step 3 — Interpret the loss. $Δ K E = 8 - 32 = - 24$ J, lost to heat and deformation. $P$ still conserved; KE not. Why this step? Directly refutes the parent's steel-manned mistake "momentum and energy conserved together or not at all."

Verify: $P_{i} = 12 - 4 = 8 = P_{f} = 4 (2) = 8$ ✓. $K E_{f} = 8 < 32 = K E_{i}$ ✓. Fraction of KE lost $= 24/32 = 75%$ . ✓

Example 8 — Cell H exam twist (part 1): continuous inflow (a stream hits a plate)

Before this example, one new symbol.

Worked example A stream of water hits a plate

Water leaves a hose at $20$ m/s and strikes a wall horizontally at a rate of $3$ kg/s, then falls straight down (no rebound). Find the horizontal force on the wall.

Forecast: Force = rate of momentum destroyed. Guess around $3 \times 20 = 60$ N.

Step 1 — Momentum arriving per second. In one second, $\overset{m}{˙} = 3$ kg of water carrying horizontal velocity $20$ m/s arrives: $\frac{Δ p _{x}}{Δ t} = \overset{m}{˙} v = 3 \times 20 = 60 kg\cdotm/s per s .$ Why this step? When mass flows, the right tool is $F = \frac{d p}{d t} = \overset{m}{˙} v$ , not $F = ma$ (mass isn't fixed).

Step 2 — All horizontal momentum is destroyed. Water leaves with zero horizontal velocity (it just slides down), so its horizontal momentum drops from $\overset{m}{˙} v$ to $0$ each second. $F_{on water} = \frac{Δ p _{x}}{Δ t} = 0 - 60 = - 60 N .$ Why this step? The question asks for a horizontal force, so we track only the horizontal momentum; the wall must supply exactly the impulse per second that kills the incoming horizontal momentum. The minus sign says the wall pushes the water backward (against its motion).

Step 3 — Newton's 3rd law: force on the wall. The wall pushes the water with $- 60$ N; by Newton's Third Law the water pushes the wall with the opposite force: $F_{on wall} = + 60 N (in the direction of the stream) .$ Why this step? The question asked for the force on the wall, which is the reaction to the force on the water — so we flip the sign.

Verify: $∣ F_{on wall} ∣ = \overset{m}{˙} v = 3 (20) = 60$ N, matching the forecast of "around 60 N." The sign is positive — the wall is pushed the way the water was going, exactly as a hose visibly shoves a plate. Units: (kg/s)(m/s) = kg·m/s² = N. ✓

Example 9 — Cell H exam twist (part 2): continuous outflow (a rocket loses mass)

The stream example added mass to a target. The mirror case is a rocket: it throws mass away to push itself forward. Same tool ( $F = \overset{m}{˙} v$ ), opposite bookkeeping.

Worked example A hovering rocket

A rocket burns fuel at $\overset{m}{˙}_{ex} = 4$ kg/s, ejecting exhaust at $u = 500$ m/s relative to itself. (a) Find the thrust. (b) The rocket + unburnt fuel has mass $180$ kg at this instant. Can it lift off ( $g = 9.8$ m/s²)? (c) What is its instantaneous upward acceleration?

Forecast: Thrust $= 4 \times 500 = 2000$ N. Weight $\approx 180 \times 9.8 \approx 1764$ N. Thrust beats weight, so yes — it lifts, with a modest acceleration.

Step 1 — Thrust from the outflow rule. $F_{thrust} = u \overset{m}{˙}_{ex} = 500 \times 4 = 2000 N (upward) .$ Why this step? Momentum is being poured into the exhaust at rate $u \overset{m}{˙}_{ex}$ ; by Newton's 3rd law the reaction of that same size drives the rocket. We use $F = \overset{m}{˙} v$ , not $F = ma$ , because the rocket's mass is dropping.

Step 2 — Compare thrust with weight (the external force here). $W = M g = 180 \times 9.8 = 1764 N (down) .$ Net upward force on the rocket body: $F_{net} = 2000 - 1764 = 236 N$ . Why this step? Gravity is the genuine external force on the rocket system; thrust is the internal ejection reaction. Lift-off needs thrust > weight.

Step 3 — Instantaneous acceleration. Apply $F_{net} = M a$ at this instant (mass is $180$ kg right now): $a = \frac{F _{net}}{M} = \frac{236}{180} \approx 1.31 m/s^{2} upward .$ Why this step? Once we have the net force, ordinary Newton's 2nd law gives the acceleration for that instant — the mass-change is already accounted for inside the thrust term.

Verify: Thrust $2000$ N $>$ weight $1764$ N ⇒ lifts off. ✓ $a = 236/180 = 1.3 \overline{1}$ m/s² $> 0$ , consistent with rising. Units: thrust (m/s)(kg/s) = kg·m/s² = N ✓; $a$ in N/kg = m/s² ✓. Sanity: exhaust ( $500$ m/s) far exceeds any rocket speed here, so treating $u$ as the ejection speed is valid.

Recall Quick self-test across the matrix

Explosion from rest — sign of the second piece's velocity? ::: Opposite to the first (total stays zero). Cannon fires horizontally — which momentum component is conserved? ::: Horizontal only; vertical is broken by gravity and normal. Ball bounces off floor — conserved for ball alone? ::: No; yes for ball + Earth. Correct test for ignoring gravity in a collision? ::: Compare impulses, not forces. Ball off a fixed wall as $M_{w a l l} \to \infty$ — wall's velocity change? ::: Zero, yet it carries the momentum. Sticking (inelastic) collision — is KE conserved? ::: No; only momentum is. Water stream hitting a wall — which formula for force? ::: $F = \overset{m}{˙} v$ , not $F = ma$ . Rocket thrust from ejecting exhaust — formula? ::: $F_{thrust} = u \overset{m}{˙}_{ex}$ (exhaust speed × burn rate).

Connections

Parent topic — the rule these examples exercise.
Conservation of Linear Momentum — the $F_{e x t} = 0$ special case (Examples 1, 6, 7).
Impulse–Momentum Theorem — powers Examples 3, 4, 8.
Newton's Third Law — why internal/stream/exhaust forces pair up (Examples 1, 8, 9).
Elastic vs Inelastic Collisions — the separate KE test in Example 7.
Center of Mass Motion — the same master equation, restated.