Classical-Mechanics · Unit 13 · Video 4 · Interactive Practice

When Mass Flows: Four Kinds of Problem, One Momentum Principle

IKey Formulas

Formula	Name	What it takes
$\vec{F}^{\,\text{total}}_{\text{ext}} = \dfrac{d\vec{p}_{\text{sys}}}{dt}$	Momentum principle	Survives flowing mass; $F = ma$ does not
$\vec{F}^{\,\text{total}}_{\text{ext}} = \lim\limits_{\Delta t \to 0} \dfrac{\vec{p}_{\text{sys}}(t + \Delta t) - \vec{p}_{\text{sys}}(t)}{\Delta t}$	The $[t,\ t+\Delta t]$ strategy	Identify the transferred $\Delta m$
$p_{\text{sys}} = p_{\text{object}} + p_{\Delta m}$	Constant-mass ledger	$\Delta m$ counted at both instants
$v_f = \dfrac{F t_f}{m_0 + b t_f} = \dfrac{F m_c}{b\,(m_0 + m_c)}$	Coal car (Example 12.1)	$m_c = b\,t_f$ , coal falls vertically

Key Insight: The system boundary follows the mass, not the object. $\Delta m$ belongs to the system at $t$ and at $t + \Delta t$ alike — wherever it happens to be — and only then does $\vec{F}_{\text{ext}} = d\vec{p}/dt$ hold.

IISorting Every Mass-Transfer Problem

Whether $\Delta m$ carries momentum along the motion — not merely whether it arrives or departs — decides the outcome.

Mass flow Momentum along the motion

IIIThe Ledger from $t$ to $t + \Delta t$

The momentum principle needs a system whose mass never changes, so $\Delta m_r$ is counted at both instants.

Step 1

1 · Identify $\Delta m$ , then fix the system

\Delta m_r

is the rain that lands during

\Delta t

. The system is the car together with that rain — at both instants.

2 · Momentum at $t$

p_x(t) = m v + \Delta m_r \cdot 0 = m v

The rain is still falling, so it contributes mass but no

x

-momentum.

3 · Momentum at $t + \Delta t$

p_x(t + \Delta t) = (m + \Delta m_r)(v + \Delta v)

The parcel is aboard now, so it moves with the car.

4 · Divide by $\Delta t$

\frac{\Delta p_x}{\Delta t} = \frac{m\,\Delta v + v\,\Delta m_r + \Delta m_r \Delta v}{\Delta t}

5 · Let $\Delta t \to 0$

The product

\Delta m_r \Delta v

vanishes faster than

\Delta t

, and on frictionless track

F_{\text{ext},x} = 0

m\frac{dv}{dt} + v\frac{dm}{dt} = 0 \quad\Longrightarrow\quad v(t) = \frac{m_0 v_0}{m(t)}

💡 Only the middle term of the ledger changes from category to category: $p_{\Delta m}$ is zero for falling rain, $\Delta m_w u$ for the fire hose, and points backward for rocket exhaust.

IVFilling the Coal Car

Coal falling from a hopper at rest adds mass but no forward momentum, so the same impulse must move more.

t 0.0 s rate b 20 kg/s

💡 As $t$ grows the loaded car approaches $F/b$ — the speed at which the entire applied force is spent accelerating newly arrived coal.

VQuiz Questions

Problem 1 · Classify the Transfer

Given: a railroad car of mass $m$ rolls forward at speed $v$ on frictionless track while grain is blown horizontally forward into it at speed $u > v$ — which category is this?

✅ Correct! Each parcel arrives with

p_x = \Delta m\,u \neq 0

and

u > v

, so the grain both loads the car and drives it forward — the fire-hose situation.

❌ That is the rain car. Rain falls vertically, so its

x

-momentum is zero; this grain is blown forward at

u

, arriving with

p_x = \Delta m\,u

❌ Check the direction of flow. Grain is entering the car, so its mass grows — categories 2 and 4 are the ones where mass leaves.

❌ Not quite. Ask two questions in order: does mass enter or leave, and does the transferred

\Delta m

carry momentum along the direction of motion?

Show solution

The classification needs only two answers.

1 — Which way does the mass flow? The grain lands in the car, so the car's mass increases: this is one of the two mass in categories.

2 — Does $\Delta m$ carry momentum along $x$ ? The grain is blown horizontally at speed $u$ , so a parcel arrives with

p_x(\Delta m) = \Delta m\,u \neq 0

Because $u > v$ , the arriving grain is faster than the car and delivers forward momentum to it on every impact. Mass in with momentum is category 3 — the same physics as the fire hose driving the boat.

Contrast this with vertically falling rain: identical mass flow, but $p_x(\Delta m) = 0$ , which is category 1 and slows the car instead.

Problem 2 · The Skater's Leaking Bag

Given: a skater of total mass $m$ glides at speed $v$ on frictionless ice; sand leaks straight down relative to her, and a mass $\Delta m$ has escaped. No external horizontal force acts — find her speed afterwards.

✅ Correct! The departed sand still travels forward at

v

, so it carries exactly the momentum it always had — the skater's share is untouched and

\Delta v = 0

❌ You dropped the sand from the system. At

t + \Delta t

it is still inside the constant-mass system, contributing

\Delta m\,v

; keep that term and the answer collapses to

\Delta v = 0

❌ Half the bookkeeping. You removed the sand's momentum

\Delta m\,v

but left its mass in the skater — remove both, or neither.

❌ That is the rain-car result. There mass arrives with no momentum; here mass leaves already carrying its own share, so nothing is diluted.

❌ Not quite. Write

p_x

of the whole system — skater plus departed sand — at both instants and set the total change to zero.

Show solution

Choose the constant-mass system: skater, bag, and the sand that will leave. Nothing crosses this boundary, and no external horizontal force acts, so $p_x$ is the same at both instants.

Before: everything moves together,

p_x(t) = m v

After: the skater carries $m - \Delta m$ at her new speed $v + \Delta v$ , while the departed sand — still in the system — keeps the forward speed $v$ it had when it left:

p_x(t + \Delta t) = (m - \Delta m)(v + \Delta v) + \Delta m\, v

Equate:

(m - \Delta m)(v + \Delta v) + \Delta m\,v = m v \quad\Longrightarrow\quad (m - \Delta m)\,\Delta v = 0

Since $m > \Delta m$ , we get $\Delta v = 0$ : the skater's speed is unchanged.

This is the signature of category 2. Relative to the skater the sand simply drops, taking no momentum along the direction of motion — mass leaves, speed does not change.

Problem 3 · Loading the Coal Car

Given: an empty car of mass $m_0 = 300\ \text{kg}$ starts from rest, pulled by a constant force $F = 150\ \text{N}$ , while coal falls vertically from a hopper at rest at the steady rate $b = 25\ \text{kg/s}$ until a total $m_c = 450\ \text{kg}$ has been transferred — find the transfer time $t_f$ and the final speed $v_f$ .

How long does the transfer take?

What is the final speed?

✅ Correct! The impulse

F t_f = 2700\ \text{N}\cdot\text{s}

is shared by the full final mass

750\ \text{kg}

, giving

v_f = 3.6

m/s.

❌ Wrong mass in the numerator. The rate

b

measures the coal arriving, so

t_f = m_c/b

— the car's own

300

kg was never poured from the hopper.

❌ Only the coal falls.

t_f = m_c/b

uses the transferred mass alone;

(m_0 + m_c)/b

would have the hopper delivering the car as well.

❌ Check the units.

F/b

has units of

\text{N}/(\text{kg/s}) = \text{m/s}

— that is a speed, not a time.

❌ Not quite. Coal accumulates as

m(t) = b t

, so the transfer ends when

b t_f = m_c

❌ Right method, wrong time. That is

F(12)/750

; the transfer runs for

t_f = m_c/b = 18

s, not

m_0/b

❌ You forgot the car itself. At

t_f

the moving mass is

m_0 + m_c = 750

kg, not the

450

kg of coal alone.

❌ That is the empty car.

F t_f/m_0

ignores the coal — but the same impulse now has to move

750

kg, so the loaded car is far slower.

❌ Not quite. Use

F t_f = (m_0 + b t_f)\,v_f

: the whole effect of the mass flow sits in that denominator.

Show solution

Step 1 — Classify. The coal falls vertically from a hopper at rest, so it arrives with zero horizontal velocity: category 1, mass in with no momentum in.

Step 2 — Choose a constant-mass system. Take the car plus the entire coal mass $m_c$ . Nothing then enters or leaves; the coal merely moves from hopper to car inside the boundary.

Step 3 — Transfer time. Coal accumulates at the steady rate $b$ :

m_c = b\,t_f \quad\Longrightarrow\quad t_f = \frac{m_c}{b} = \frac{450}{25} = 18\ \text{s}

Step 4 — The two states, along $x$ . Initially the car is at rest and the coal sits in the motionless hopper:

p_x(0) = 0

Finally the loaded car of mass $m_0 + b t_f$ moves at $v_f$ :

p_x(t_f) = (m_0 + b\,t_f)\,v_f = (300 + 450)\,v_f

Step 5 — Momentum principle. The force is constant, so the impulse is simply $F t_f$ :

\int_0^{t_f} F\,dt = F t_f = p_x(t_f) - p_x(0)

150 \times 18 = 2700 = 750\,v_f \quad\Longrightarrow\quad v_f = 3.6\ \text{m/s}

Check with the closed form:

v_f = \frac{F m_c}{b\,(m_0 + m_c)} = \frac{(150)(450)}{(25)(750)} = \frac{67500}{18750} = 3.6\ \text{m/s}\ \checkmark

An empty car pulled for the same $18$ s would reach $2700/300 = 9.0$ m/s. Same impulse, more mass, less speed.

Problem 4 · Rain on a Free-Rolling Car

Given: an open car of mass $m$ rolls at speed $v$ on frictionless, level track with no applied force; rain falls straight down at speed $u$ and a mass $\Delta m$ collects in the car — find the car's new speed.

✅ Correct! With

F_{\text{ext},x} = 0

the horizontal momentum

mv

is fixed, so a larger mass must move more slowly:

v' = mv/(m + \Delta m)

❌ The mass is what changed. The rain adds no

x

-momentum — but it does add mass, and the fixed

p_x = mv

must now be shared by

m + \Delta m

❌ That speeds the car up. You divided the conserved momentum by the wrong mass;

p_x = mv

stays fixed while the mass grows, so the speed must fall.

❌ $u$ is a vertical speed. It contributes no

x

-momentum at all — that is precisely what makes this category 1; the track supplies whatever vertical impulse is needed.

❌ Not quite. Take the car plus the rain as one constant-mass system and set

p_x(t + \Delta t) = p_x(t)

Show solution

The system is the car together with the rain parcel $\Delta m$ — counted at both instants, whether it is still in the air or already aboard.

At $t$ : the car moves at $v$ ; the falling rain moves straight down, so its $x$ -momentum is zero:

p_x(t) = m v + \Delta m \cdot 0 = m v

At $t + \Delta t$ : the rain has landed and rides along at the new speed $v'$ :

p_x(t + \Delta t) = (m + \Delta m)\,v'

No external horizontal force (frictionless, level track), so $p_x$ does not change:

(m + \Delta m)\,v' = m v \quad\Longrightarrow\quad v' = \frac{m v}{m + \Delta m}

The car slows down. Nothing pushed it backwards — the same forward momentum simply has more mass to carry. Repeating this over many parcels gives $m\,dv/dt + v\,dm/dt = 0$ , whose solution is $v(t) = m_0 v_0 / m(t)$ .

The vertical speed $u$ never appears: the rails absorb the vertical impulse, and only the $x$ -component of the ledger matters.

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When Mass Flows: Four Kinds of Problem, One Momentum Principle

IKey Formulas

IISorting Every Mass-Transfer Problem

IIIThe Ledger from ttt to t+Δtt + \Delta tt+Δt

IVFilling the Coal Car

VQuiz Questions

IIIThe Ledger from $t$ to $t + \Delta t$