Classical-Mechanics · Unit 13 · Video 6 · Interactive Practice

Pushed by What It Catches: The Boat and the Fire Hose

IKey Formulas

Formula	Name	What it takes
$0 = m_b\dfrac{dv}{dt} + \dfrac{dm_b}{dt}(v - u)$	Equation of motion	System $=$ boat $+$ the element $\Delta m_w$ about to land; no external $x$ -force
$v(t) = u\left(1 - \dfrac{m_0}{m_b(t)}\right)$	Speed as a function of mass	Separation of variables from $v = 0$ , $m_b = m_0$
$m_b v = m_w u$	Momentum-transfer check	The same result read as one conservation statement
$\dfrac{dm_b}{dt} = \alpha\left(1 - \dfrac{v}{u}\right)$ and $m_b(t) = m_0\sqrt{1 + \dfrac{2\alpha t}{m_0}}$	Intake rate, and mass in time	Jet density $\lambda = \alpha/u$ ; only the closing length lands

Key Insight: The driving term $\dfrac{dm_b}{dt}(v - u)$ is alive only while $v \lt u$ — the jet can push the boat only while it can still catch it. And $\alpha$ is a setting on the hose, never the boat's intake rate: the boat fills at $\alpha\left(1 - v/u\right)$ , because filling is a chase.

IIThe Chase: Why the Boat Fills Slower Than the Hose Sprays

The nozzle ejects at $\alpha$ , but the deck is running away from the jet.

Δt elapsed 0.0 s boat speed v 5.0 m/s

IIIOne Line of Conservation: $m_b v = m_w u$

The boat ends up owning exactly the momentum its captured water flew in with.

Water aboard Your guess: v 6.0 m/s

💡 The one-line statement is exact because every splash is internal to the system "boat $+$ all the water that ends up aboard", and water that lands never leaves it again.

IVMass and Speed as Functions of Time

Time enters through the mass alone: the heavier the boat, the slower it fills.

Step 1

Step 1 — Two relations on the table

\frac{dm_b}{dt} = \alpha\left(1 - \frac{v}{u}\right), \qquad v = u\left(1 - \frac{m_0}{m_b}\right)

Step 2 — Substitute the speed into the rate

1 - \frac{v}{u} = \frac{m_0}{m_b} \quad\Longrightarrow\quad \frac{dm_b}{dt} = \frac{\alpha m_0}{m_b}

Step 3 — Separate and integrate

\int_{m_0}^{m_b(t)} m_b\,dm_b = \alpha m_0 \int_0^t dt \quad\Longrightarrow\quad \tfrac{1}{2}\left(m_b(t)^2 - m_0^2\right) = \alpha m_0 t

Step 4 — Solve for the mass

m_b(t) = m_0\sqrt{1 + \frac{2\alpha t}{m_0}}

Step 5 — Feed the mass back into the speed

v(t) = u\left(1 - \frac{1}{\sqrt{1 + 2\alpha t/m_0}}\right)

VQuiz Questions

Problem 1 · Speed from Mass

Given: a boat that started from rest has taken on water until $m_b = 3m_0$ , with jet speed $u = 20\ \text{m/s}$ — find its speed $v$ .

✅ Correct!

v = u\left(1 - \tfrac{m_0}{3m_0}\right) = \tfrac{2}{3}u

, and the check agrees:

m_w = 2m_0

, so

v = m_w u/m_b = 2u/3

❌ That is the fraction you subtract.

m_0/m_b = \tfrac{1}{3}

is the shortfall; the speed is

u

times

1 - m_0/m_b

❌ Not quite.

v = u

is the ceiling the boat approaches only as

m_b \to \infty

— at

v = u

the jet would never land at all.

❌ Not quite.

v

can never exceed

u

: the water pushes the boat only while it can still catch it.

Show solution

Part (b) of the example gives speed directly from mass:

v = u\left(1 - \frac{m_0}{m_b}\right) = 20\left(1 - \frac{m_0}{3m_0}\right) = 20\left(1 - \frac{1}{3}\right) = \frac{40}{3} \approx 13.3\ \text{m/s}

Check with the one-line statement. The water aboard is $m_w = m_b - m_0 = 2m_0$ , so

m_b v = m_w u \;\Longrightarrow\; v = \frac{m_w u}{m_b} = \frac{2m_0 (20)}{3m_0} = \frac{40}{3}\ \text{m/s}\ \checkmark

Tripling the mass buys two thirds of the jet speed — and no amount of water ever buys all of it.

Problem 2 · The Rate Trap

Given: the hose ejects water at $\alpha = 8.0\ \text{kg/s}$ with $u = 20\ \text{m/s}$ . At the instant the boat is moving at $v = 5.0\ \text{m/s}$ — find $dm_b/dt$ .

✅ Correct!

\alpha(1 - v/u) = 8.0\left(1 - \tfrac{5}{20}\right) = 8.0(0.75) = 6.0\ \text{kg/s}

— three quarters of the stream still lands.

❌ This is the chapter's trap.

\alpha

is the rate water leaves the nozzle. The stern is running away at

v

, so only the closing length

(u-v)\Delta t

ever reaches the deck.

❌ You used the wrong factor.

\alpha\,v/u

is the part of the stream that never catches up; the part that lands carries the factor

1 - v/u

❌ Not quite. The intake rate must lie between

\alpha

(boat at rest) and

0

(boat at

v = u

Show solution

Step 1 — the jet's mass per unit length. Through a fixed cross-section, a length $u\,\Delta t$ passes in time $\Delta t$ , so $\alpha = \lambda u$ and

\lambda = \frac{\alpha}{u} = \frac{8.0}{20} = 0.40\ \text{kg/m}

Step 2 — move the surface onto the stern. In $\Delta t$ the water advances $u\,\Delta t$ while the stern advances $v\,\Delta t$ , so the jet gains only $(u - v)\Delta t$ on the boat:

\Delta m_w = \lambda (u - v)\Delta t = \frac{\alpha}{u}(u - v)\Delta t

Step 3 — divide and take the limit. Since $m_b = m_0 + m_w$ , we have $dm_b/dt = dm_w/dt$ :

\frac{dm_b}{dt} = \alpha\left(1 - \frac{v}{u}\right) = 8.0\left(1 - \frac{5.0}{20}\right) = 6.0\ \text{kg/s}

At $v = 0$ this returns $\alpha$ ; at $v = u$ it returns $0$ . The intake rate is not a setting on the hose — it is the outcome of a chase.

Problem 3 · Mass and Speed at a Given Instant

Given: $m_0 = 400\ \text{kg}$ at rest, $\alpha = 8.0\ \text{kg/s}$ , $u = 20\ \text{m/s}$ — find the boat's mass and speed at $t = 75\ \text{s}$ .

What is $m_b(75\ \text{s})$ ?

And the speed $v(75\ \text{s})$ ?

✅ Correct! The mass has exactly doubled, so the speed is exactly

u/2

— and

m_b v = 800(10) = 8000 = 400(20) = m_w u\ \checkmark

❌ That is $m_0 + \alpha t$ . It assumes the boat fills at the full hose rate for all

75\ \text{s}

, but it is moving for almost all of that time.

❌ You dropped the square root.

1 + 2\alpha t/m_0 = 4

, and

m_b = m_0\sqrt{4} = 2m_0

, not

4m_0

❌ Check the mass. Use

m_b = m_0\sqrt{1 + 2\alpha t/m_0}

with

2\alpha t/m_0 = 2(8)(75)/400 = 3

❌ Check the speed. Put your mass into

v = u(1 - m_0/m_b)

— a doubled mass gives half the jet speed.

Show solution

Step 1 — the mass. Integrating $m_b\,dm_b = \alpha m_0\,dt$ gives $\tfrac{1}{2}(m_b^2 - m_0^2) = \alpha m_0 t$ , so

m_b(t) = m_0\sqrt{1 + \frac{2\alpha t}{m_0}} = 400\sqrt{1 + \frac{2(8.0)(75)}{400}} = 400\sqrt{4} = 800\ \text{kg}

Step 2 — the speed.

v = u\left(1 - \frac{m_0}{m_b}\right) = 20\left(1 - \frac{400}{800}\right) = 10\ \text{m/s}

Step 3 — the pocket check. The water aboard is $m_w = 800 - 400 = 400\ \text{kg}$ , and

m_b v = 800(10) = 8000 = 400(20) = m_w u\ \checkmark

Note how far the naive answer is off: at the full hose rate the boat would have swallowed $\alpha t = 600\ \text{kg}$ in $75\ \text{s}$ . It swallowed only $400\ \text{kg}$ , because it spent that time running away from its own supply.

Problem 4 · How Long to Reach Two Thirds of the Jet Speed

Given: the same boat ( $m_0 = 400\ \text{kg}$ at rest, $\alpha = 8.0\ \text{kg/s}$ , $u = 20\ \text{m/s}$ ) — find the time at which $v = \tfrac{2}{3}u$ .

✅ Correct!

v = \tfrac{2}{3}u

needs

m_b = 3m_0 = 1200\ \text{kg}

, and

\tfrac{1}{2}(1200^2 - 400^2) = \alpha m_0 t

gives

t = 200\ \text{s}

❌ That is $m_w/\alpha$ . The boat does need

800\ \text{kg}

of water, but it only collects at

\alpha(1 - v/u)

— so it takes twice as long as the hose alone would suggest.

❌ You forgot to square. From

m_b = 3m_0

you need

1 + 2\alpha t/m_0 = 3^2 = 9

, not

3

❌ You dropped the factor $\tfrac{1}{2}$ . The integral of

m_b\,dm_b

\tfrac{1}{2}m_b^2

, which halves your time.

❌ Not quite. Two steps: turn the target speed into a target mass, then invert

m_b(t) = m_0\sqrt{1 + 2\alpha t/m_0}

Show solution

Step 1 — speed target becomes a mass target.

u\left(1 - \frac{m_0}{m_b}\right) = \frac{2}{3}u \;\Longrightarrow\; \frac{m_0}{m_b} = \frac{1}{3} \;\Longrightarrow\; m_b = 3m_0 = 1200\ \text{kg}

Step 2 — invert the mass-time law. From $\tfrac{1}{2}\left(m_b^2 - m_0^2\right) = \alpha m_0 t$ :

t = \frac{m_b^2 - m_0^2}{2\alpha m_0} = \frac{1200^2 - 400^2}{2(8.0)(400)} = \frac{1{,}280{,}000}{6400} = 200\ \text{s}

Why not $100\ \text{s}$ ? The boat must swallow $m_w = 800\ \text{kg}$ , which at the full hose rate would take $800/8.0 = 100\ \text{s}$ . But by then the boat is already moving fast, and its intake has fallen to

\frac{dm_b}{dt} = \alpha\left(1 - \frac{v}{u}\right) = 8.0\left(1 - \frac{2}{3}\right) = 2.7\ \text{kg/s}

The chase doubles the time. This is also why $v \to u$ only like $1 - 1/\sqrt{t}$ : the intake that drives the boat is choked by the boat's own escape.

Solved: 0 / 4

Pushed by What It Catches: The Boat and the Fire Hose

IKey Formulas

IIThe Chase: Why the Boat Fills Slower Than the Hose Sprays

IIIOne Line of Conservation: mbv=mwum_b v = m_w umb​v=mw​u

IVMass and Speed as Functions of Time

VQuiz Questions

IIIOne Line of Conservation: $m_b v = m_w u$