System with external forces — conditions for conservation: Exercises

The master result you are applying throughout (built from scratch in the parent note): $\frac{d P}{d t} = F_{e x t}^{n e t}, F_{e x t}^{n e t} = 0 ⟺ P = constant$

Here $P = \sum_{i} m_{i} v_{i}$ is the total momentum (add up mass×velocity of every tracked object), and $F_{e x t}^{n e t}$ is the net external force (the sum of only the pushes coming from outside your chosen system).

Level 1 — Recognition

Goal: just decide "is $P$ conserved, and on which axis?" — no arithmetic yet.

Recall Solution L1·Q1

System = all 30 fragments. The explosion pushes are between fragments → internal → cancel in pairs (Newton's 3rd law), so they cannot change $P$ . The only external force is gravity, pointing straight down.

Horizontal axis: no external force → $P_{x}$ conserved. ✅
Vertical axis: gravity acts → $P_{y}$ not conserved. ❌ Newton's law is component-wise, so you win on $x$ even while losing on $y$ .

Recall Solution L1·Q2

System = ball only. External forces: gravity (down, from Earth which is outside) and the table's normal force (up, table is outside). Here they are equal and opposite, so $F_{e x t}^{n e t} = 0$ at this instant → $P$ (which is zero, ball at rest) stays zero. Conserved, but trivially — the ball just isn't moving. The teaching point: gravity and normal are external here only because Earth and table are outside the boundary you drew.

Level 2 — Application

Goal: plug into $P_{i} = P_{f}$ on the safe axis and solve.

Recall Solution L2·Q1

System = cannon + ball. The firing force is now internal. External forces (gravity, normal) are vertical, so horizontal $P_{x}$ is conserved. Before firing, everything is at rest: $P_{x, i} = 0$ . After: ball goes $+ v$ , cannon recoils at $- V$ : $0 = m v + M (- V) \Rightarrow V = \frac{m v}{M} = \frac{5 ( 180 )}{500} = 1.8 m/s$ The cannon rolls back at $1.8 m/s$ .

Recall Solution L2·Q2

System = A + B. Rail is frictionless, gravity/normal vertical → horizontal $P_{x}$ conserved. $P_{i} = 3 (4) + 2 (0) = 12 kg\cdotm/s$ Stuck together they share mass $5 kg$ at speed $u$ : $12 = 5 u \Rightarrow u = 2.4 m/s$ Common final speed $2.4 m/s$ (this is a perfectly inelastic collision — see Elastic vs Inelastic Collisions).

Level 3 — Analysis

Goal: reason about which relaxation (component / impulsive / average) is doing the work.

Recall Solution L3·Q1

The correct test compares impulses ( $=$ force $\times$ time), not forces (parent, Impulse–Momentum Theorem). Bat impulse = ball's momentum change: $J_{ba t} = m Δ v = 0.15 (50) = 7.5 N\cdots$ Gravity impulse during contact: $J_{g} = m g Δ t = 0.15 (9.8) (0.0012) \approx 1.76 \times 1 0^{- 3} N\cdots$ Ratio $J_{g} / J_{ba t} \approx 2.4 \times 1 0^{- 4}$ , i.e. gravity contributes about $0.024%$ . Since $J_{g} ≪ J_{ba t}$ , treat $P$ as conserved through the impact, then re-add gravity for the flight. Impulsive approximation justified. ✅

Figure — System with external forces — conditions for conservation

Recall Solution L3·Q2

System = both disks. Ice frictionless, gravity/normal vertical → both horizontal axes ( $x$ and $y$ ) conserved. So we write $P_{i} = P_{f}$ separately on each. Disk 1 after: $v_{1 x} = 33 cos 3 0^{\circ} = 33 \cdot \frac{3}{2} = 4.5$ , $v_{1 y} = 33 sin 3 0^{\circ} = 33 \cdot \frac{1}{2} = \frac{3 3}{2} \approx 2.598$ . x-axis: $2 (6) + 2 (0) = 2 (4.5) + 2 (v_{2 x})$ $12 = 9 + 2 v_{2 x} \Rightarrow v_{2 x} = 1.5 m/s$ y-axis: $0 = 2 (2.598) + 2 (v_{2 y})$ $0 = 5.196 + 2 v_{2 y} \Rightarrow v_{2 y} = - 2.598 m/s$ Disk 2 moves at $(1.5, - 2.598) m/s$ , i.e. speed $1. 5^{2} + 2.59 8^{2} = 3 m/s$ at $3 0^{\circ}$ below the $x$ -axis. The two disks fly apart symmetrically — that's the geometry of equal masses.

Level 4 — Synthesis

Goal: combine phases — conservation during impact, then dynamics before/after.

Recall Solution L4·Q1

Two phases with different valid laws. Phase 1 — embedding (impulsive): the collision is fast, so gravity's impulse is negligible → momentum conserved horizontally. $m v_{0} = (m + M) V, m = 0.010, M = 2.00$ Phase 2 — swing up (energy): after impact, no non-conservative force does net work along the swing, so KE converts to gravitational PE. $\frac{1}{2} (m + M) V^{2} = (m + M) g h \Rightarrow V = 2 g h = 2 (9.8) (0.20) = 1.9799 m/s$ Back-substitute: $v_{0} = \frac{m + M}{m} V = \frac{2.010}{0.010} (1.9799) = 398.0 m/s$ Bullet speed $v_{0} \approx 398 m/s$ . Why not one law for both phases? Momentum is conserved in phase 1 but not KE (embedding = inelastic, heat lost). Energy is conserved in phase 2 but not horizontal momentum (strings pull inward). Each phase uses the law whose condition it satisfies.

Recall Solution L4·Q2

System = person + raft. No horizontal external force → $P_{x} = 0$ stays $0$ → the center of mass stays fixed in the water (this is Center of Mass Motion: $F_{e x t} = 0 \Rightarrow a_{c m} = 0$ , and it started at rest). Let the raft shift by $x$ (in the direction opposite the walk). The person walks $4 m$ relative to the raft, so relative to water the person moves $(4 - x)$ backward-adjusted; enforce fixed CM: $m_{p} d_{p} = m_{r} d_{r}, d_{p} + d_{r} = 4$ With $m_{p} = 60$ , $m_{r} = 120$ : the raft's displacement is $x = \frac{m _{p}}{m _{p} + m _{r}} (4) = \frac{60}{180} (4) = 1.333 m$ The raft slides $1.33 m$ opposite to the person's walking direction (person moves $2.67 m$ over the water). CM never budged.

Level 5 — Mastery

Goal: boundary-choice subtleties, degenerate/limiting cases, sign bookkeeping.

Recall Solution L5·Q1

Take up as positive. (a) Ball alone: $Δ p = m v_{f} - m v_{i} = 1 (+ 4) - 1 (- 5) = + 9 kg\cdotm/s$ . Nonzero because gravity and the floor's normal are external to a one-ball system — the floor delivers a real upward impulse. $P$ of the ball is not conserved. (b) Ball + Earth: now the floor's push is internal (floor is part of Earth). No external horizontal/vertical force → total $P$ conserved. Earth must absorb $- 9 kg\cdotm/s$ : $M_{\oplus} V_{\oplus} = 9 \Rightarrow V_{\oplus} = \frac{9}{5.97 \times 1 0 ^{24}} \approx 1.5 \times 1 0^{- 24} m/s$ Utterly imperceptible, but nonzero — same event, opposite conservation verdict, purely from where the boundary is drawn.

Recall Solution L5·Q2

System = both pieces, no external force → $P = 0$ before and after (explosion forces internal). $0 = m u + (K m) (- w) \Rightarrow w = \frac{u}{K}$ (b) Limits:

$K \to \infty$ : $w = u / K \to 0$ — a hugely heavy chunk barely moves, like a rifle recoiling almost none against a light bullet. ✅
$K \to 0$ : $w = u / K \to \infty$ — a vanishingly light second piece is flung arbitrarily fast, so all the "kick" goes into the tiny fragment. Both limits respect $P = 0$ exactly: heavy·slow always balances light·fast.

Recall One-line self-test after all levels

The single question to ask on every momentum problem ::: "Where is my boundary, and is the net external force zero on the axis I care about?"

Connections

Conservation of Linear Momentum — the $F_{e x t} = 0$ special case powering L1–L2, L5.
Impulse–Momentum Theorem — the impulse comparison in L3·Q1.
Center of Mass Motion — fixed-CM reasoning in L4·Q2.
Elastic vs Inelastic Collisions — why L2/L4 use only momentum, not energy.
Newton's Third Law — why explosion/collision forces cancel internally.
Parent topic