Classical-Mechanics · Unit 13 · Video 5 · Interactive Practice

Why Leaking Sand Changes Nothing and Blown-In Grain Changes Everything

IKey Formulas

Formula	Name	What it takes
$F_{\text{ext}} = m(t)\dfrac{dv}{dt} + \dfrac{dm}{dt}\,(v - u)$	Equation of motion with mass flow	$u$ = velocity of the transferred matter in the ground frame
$v_c(t) = \dfrac{F}{b}\ln\!\left[\dfrac{m_c + m_s}{m_c + m_s - bt}\right]$	Emptying car — sand out at $u = v_c$	Constant $F$ , leak rate $b$ , released from rest
$v_A(t) = \dfrac{bt}{m_{A,0} + bt}\,u$	Filling car — grain in at $u$	No external force along the track, starts from rest
$\left(m_{A,0} + bt\right)v_A(t) = bt\,u$	Momentum ledger of the filling car	Every kilogram of grain hands over the momentum it arrived with

Key Insight: $p = mv$ was defined for a fixed collection of matter, so differentiating it with the product rule on a system whose mass is changing is illegal — the stray $v\,dm/dt$ even changes value from one reference frame to another. Follow the same matter through $\Delta t$ instead, and the only survivor is $\dfrac{dm}{dt}(v - u)$ : mass rate times relative velocity.

IISand Out: The Leak the Equation Never Feels

Sand drops with the car's own horizontal velocity, so only a shrinking $m_c(t)$ survives in $F = m_c(t)\,dv_c/dt$ .

Time t 0.0 s Applied force

IIIGrain In: How Fast Once the Mass Has Doubled?

Grain arrives carrying $u$ ; how fast is car A once that grain has doubled its mass?

Guess: v_A ÷ u 0.80

💡 Integrating in time instead of in mass gives the identical curve: $\dfrac{dv_A}{u - v_A} = \dfrac{b\,dt}{m_{A,0} + bt}$ , where the antiderivative's factor $1/b$ cancels the $b$ upstairs.

IVOne Term Decides

Every mass-flow problem turns on one term: the mass rate times the relative velocity $u - v$ .

Time t 8.0 s

VQuiz Questions

Problem 1 · Emptying the Freight Car

Given: a car of mass $m_c = 3000\ \text{kg}$ carrying $m_s = 2000\ \text{kg}$ of sand is released from rest under a constant force $F = 2000\ \text{N}$ , while sand streams straight down through a floor port at $b = 100\ \text{kg/s}$ — find the car's speed at the instant the last sand leaves.

✅ Correct! The same

2000\ \text{N}

buys more acceleration every second as the mass falls from

5000\ \text{kg}

3000\ \text{kg}

(F/b)\ln(5/3) = 20(0.511) = 10.2\ \text{m/s}

❌ That is the constant-mass answer.

Ft/(m_c+m_s) = 2000(20)/5000

would hold only if the car never got lighter; the real car ends up faster than

8.0\ \text{m/s}

❌ That treats the car as empty the whole way.

Ft/m_c = 2000(20)/3000

is an upper bound — the true answer must lie between the full-mass and empty-mass estimates.

❌ Check what the denominator collapses to. At

t = m_s/b

the remaining mass is

m_c + m_s - bt = m_c = 3000\ \text{kg}

, not

m_s = 2000\ \text{kg}

❌ Not quite. Evaluate

v_c = (F/b)\ln[(m_c+m_s)/(m_c+m_s-bt)]

t = m_s/b = 20\ \text{s}

Show solution

The sand leaves through the floor with the car's own horizontal velocity, so no $dm/dt$ term appears and Newton's second law keeps its familiar shape with a time-dependent mass:

F = m_c(t)\,\frac{dv_c}{dt},\qquad m_c(t) = m_c + m_s - bt = 5000 - 100t

Separate the variables and integrate from rest:

\int_0^{v_c(t)} dv_c' = \int_0^t \frac{F\,dt'}{m_c + m_s - bt'} \;\Longrightarrow\; v_c(t) = \frac{F}{b}\ln\!\left[\frac{m_c+m_s}{m_c+m_s-bt}\right]

The sand runs out when $bt = m_s$ , i.e. at $t = m_s/b = 20\ \text{s}$ , where the denominator collapses to the empty car $m_c$ :

v_c = \frac{2000}{100}\ln\!\left[\frac{5000}{3000}\right] = 20\ln\!\left(\frac{5}{3}\right) = 20(0.5108) = 10.2\ \text{m/s}

Sanity check. The full car ( $5000\ \text{kg}$ ) would have reached $8.0\ \text{m/s}$ and the empty car ( $3000\ \text{kg}$ ) would have reached $13.3\ \text{m/s}$ ; $10.2\ \text{m/s}$ sits between them, exactly where a mass that shrinks from one to the other belongs.

Problem 2 · The Product-Rule Trap

Given: at one instant that same leaking car has mass $m_c(t) = 4000\ \text{kg}$ and speed $v_c = 5.0\ \text{m/s}$ , with $F = 2000\ \text{N}$ still applied and $dm_c/dt = -100\ \text{kg/s}$ — find its acceleration.

✅ Correct! The sand separates at the car's own horizontal velocity, so

v_c - u = 0

, the mass-flow term dies, and

F = m_c(t)\,dv_c/dt

gives

2000/4000

❌ That is the product-rule shortcut. Writing

F = m\,dv/dt + v\,dm/dt

and solving gives

(F - v\,dm/dt)/m = (2000 + 500)/4000

. But

p = mv

was defined for a fixed collection of matter, and that leftover term is no thrust — it even changes value when you change reference frames.

❌ The same shortcut with the opposite sign:

(F + v\,dm/dt)/m = 1500/4000

. The term should be absent altogether, not merely re-signed.

❌ That is $F/m_c$ with the empty-car mass $3000\ \text{kg}$ . At this instant the car still carries

1000\ \text{kg}

of sand, so the mass in the denominator is

4000\ \text{kg}

❌ Not quite. With

u = v_c

the equation of motion is simply

F = m_c(t)\,dv_c/dt

Show solution

Follow a fixed collection of matter — the car plus the sand still inside at time $t$ . The element that drops out separates while moving horizontally with the car, so at $t + \Delta t$ both pieces share the velocity $v_c + \Delta v_c$ and the mass changes cancel:

\Delta p_{\text{sys}} = \big(\Delta m_s + m_c + \Delta m_c\big)(v_c + \Delta v_c) - m_c v_c = m_c(t)\,\Delta v_c \quad (\Delta m_c = -\Delta m_s)

F = m_c(t)\frac{dv_c}{dt} = \frac{2000}{4000} = 0.50\ \text{m/s}^2

Why the shortcut fails. The general mass-flow term is $\dfrac{dm}{dt}(v - u)$ , the mass rate times the relative velocity. Sand falling through the floor has $u = v_c$ , so the term is exactly zero. The product rule instead keeps $v\,dm/dt = (5.0)(-100) = -500\ \text{N}$ , an invented force of $500\ \text{N}$ — and its value would change if you measured $v_c$ from a moving platform, a sure sign it is not a law.

Corollary. Set $F = 0$ and the result is $dv_c/dt = 0$ : a rolling, leaking car holds its speed exactly, however much sand it sheds.

Problem 3 · Filling the Freight Car

Given: car A starts from rest with $m_{A,0} = 2000\ \text{kg}$ while grain blown from car B arrives at $b = 50\ \text{kg/s}$ carrying car B's horizontal velocity $u = 12\ \text{m/s}$ ; no external force acts along the track.

What is car A's speed at $t = 20\ \text{s}$ ?

How much grain must be aboard for car A to reach $v_A = 9.0\ \text{m/s}$ ?

✅ Correct!

v_A/u

is the fraction of car A's current mass that arrived as grain:

1000/3000

20\ \text{s}

, and reaching

0.75u

demands

6000\ \text{kg}

of grain inside an

8000\ \text{kg}

car.

❌ $u$ is the limit, not the value.

v_A = [bt/(m_{A,0}+bt)]u

climbs toward car B's speed but reaches it only as

t \to \infty

❌ The fraction is upside down.

v_A/u = bt/(m_{A,0}+bt) = 1000/3000

— the grain's share of the mass, not the car's share

m_{A,0}/m_A = 2000/3000

❌ You divided by the initial mass.

bt\,u/m_{A,0} = 12000/2000

hands all the momentum to the original car; the grain aboard shares it, so divide by

m_A(20) = 3000\ \text{kg}

❌ Check the ledger.

m_A(t)\,v_A(t) = bt\,u

with

m_A(20) = 2000 + 1000 = 3000\ \text{kg}

and

bt\,u = 12000\ \text{kg}\cdot\text{m/s}

❌ That is car A's total mass at that moment. The grain aboard is

8000 - 2000 = 6000\ \text{kg}

❌ Not enough grain. With

3000\ \text{kg}

aboard the mass is

5000\ \text{kg}

and

v_A = (3000/5000)(12) = 7.2\ \text{m/s}

❌ That is $0.75\,m_{A,0}$ . The ratio

0.75

is a fraction of the current mass, which the grain has itself inflated.

❌ Not quite. Solve

bt/(m_{A,0}+bt) = 9.0/12 = 0.75

for the grain mass

bt

Show solution

The system is car A, the grain already aboard, and the element $\Delta m_g$ about to arrive — the same matter at both instants. With no external force along the track,

0 = m_A(t)\frac{dv_A}{dt} + \frac{dm_A}{dt}\big(v_A - u\big)

Separating against mass rather than time and integrating gives $\ln[u/(u - v_A)] = \ln[m_A/m_{A,0}]$ , hence

v_A = \frac{m_A - m_{A,0}}{m_A}\,u = \frac{bt}{m_{A,0} + bt}\,u

Part 1. At $t = 20\ \text{s}$ the grain aboard is $bt = 1000\ \text{kg}$ and $m_A = 3000\ \text{kg}$ :

v_A = \frac{1000}{3000}(12) = 4.0\ \text{m/s}

Ledger check: $(3000)(4.0) = 12000 = (1000)(12)$ — the car carries exactly the momentum the grain brought in.

Part 2. Demand $v_A/u = 9.0/12 = 0.75$ :

\frac{bt}{2000 + bt} = 0.75 \;\Longrightarrow\; 1500 + 0.75\,bt = bt \;\Longrightarrow\; bt = 6000\ \text{kg}

So $m_A = 8000\ \text{kg}$ , reached at $t = 6000/50 = 120\ \text{s}$ . Ledger check: $(8000)(9.0) = 72000 = (6000)(12)$ .

Problem 4 · Sand from a Hopper at Rest

Given: a $2000\ \text{kg}$ car rolls freely at $4.0\ \text{m/s}$ on level track when sand starts pouring in from a hopper hanging at rest above the track at $50\ \text{kg/s}$ , with no applied force — find the car's speed $20\ \text{s}$ later.

✅ Correct! The sand arrives with

u = 0

, so

v - u = v

: the term is alive and brakes the car. Horizontal momentum is conserved,

(2000)(4.0) = (3000)v

❌ That is the leaking car's answer. It worked only because the departing sand left at the car's own velocity; sand arriving at

u = 0

must be dragged up to speed, and only the car's momentum can pay for it.

❌ The mass ratio is upside down.

v = v_0\,m_0/m(t)

, so growing from

2000\ \text{kg}

3000\ \text{kg}

must make the car slower, not faster.

❌ Only $1000\ \text{kg}$ was added, not $2000\ \text{kg}$ . The mass goes

2000 \to 3000\ \text{kg}

, so the speed falls by the factor

2/3

, not

1/2

❌ Not quite. Sand falling from a hopper at rest brings no horizontal momentum, so

m(t)v(t)

stays at its initial

8000\ \text{kg}\cdot\text{m/s}

Show solution

Mass arrives, so the transfer term is present; what matters is the velocity it arrives with. Here $u = 0$ , and with no external force along the track:

0 = m(t)\frac{dv}{dt} + \frac{dm}{dt}\big(v - 0\big) \;\Longrightarrow\; m\frac{dv}{dt} = -bv

That is exactly $d(mv)/dt = 0$ : the horizontal momentum of car-plus-sand is conserved because the incoming sand brings none.

m(t) = 2000 + 50(20) = 3000\ \text{kg}

v = \frac{m_0v_0}{m(t)} = \frac{(2000)(4.0)}{3000} = 2.7\ \text{m/s}

The three cases side by side. Sand out through the floor has $u = v$ , the term vanishes, and the speed holds. Grain blown in from car B has $u > v$ , the term acts like a forward thrust, and the speed climbs toward $u$ . Sand poured from a hopper at rest has $u = 0 < v$ , the term brakes, and the speed decays like $1/m(t)$ . One term, $\dfrac{dm}{dt}(v-u)$ , decides all three.

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