Irodov 1.182: Leaking Sand Never Pushes Back – Amit & Deepali Sinha Foundation

Problem Irodov 1.182

A cart loaded with sand is pushed along a horizontal, frictionless plane by a constant force $\vect F$ directed along the cart’s own velocity at every instant. Sand leaks continuously through a hole in the cart’s floor at a constant rate $\mu$ (kilograms per second). At the initial moment $t=0$ the loaded cart has mass $m_0$ and is at rest. Find the cart’s acceleration $w$ and its velocity $v$ as functions of the elapsed time $t$ .

The cart, of instantaneous mass m(t), is driven by the constant force F; sand leaves through a floor hole at the constant rate μ, dropping straight down and so carrying away no momentum along the direction of motion relative to the cart. — The cart, of instantaneous mass $m(t)$ , is driven by the constant force $\vect F$ ; sand leaves through a floor hole at the constant rate $\mu$ , dropping straight down and so carrying away no momentum along the direction of motion relative to the cart.

Given

$\vect F$
$\mu$
$m_0$
$v(0)=0$

Find

w(t)

v(t)

Solution

Step 1The mass remaining in the cart

Throughout, $dm=\mu\,dt$ is the mass that leaves during $dt$ , a positive quantity, and the cart’s own mass falls by exactly that much. Sand leaves at the constant rate $\mu$ , so the total mass carried falls off linearly from its initial value.

m(t) = m_0 - \mu t \tag{1}

Step 2An equation of motion with no thrust term

Consider the cart-plus-sand system at $t$ , of mass $m$ and velocity $v$ , and again at $t+dt$ , after a mass $dm=\mu\,dt$ has fallen through the hole carrying the horizontal velocity $v$ with it, leaving the cart itself with mass $m-dm$ . Momentum conservation for the whole system, together with the impulse of $\vect F$ , gives

\begin{align*}(m-dm)(v+dv) + dm\,v &= mv + F\,dt \\ mv + m\,dv - dm\,v - dm\,dv + dm\,v &= mv + F\,dt \\ m\,dv &= F\,dt,\end{align*}

the second order term $dm\,dv$ dropped as negligible and the two first order terms in $dm$ canceling exactly, which is the algebra saying that the departing sand takes nothing with it.

m\,\dfrac{dv}{dt} = F

(2)

Step 3The acceleration at time

t

Dividing (2) by the running mass (1) gives the acceleration directly, with no separate integration needed.

w(t) = \frac{dv}{dt} = \frac{F}{m(t)}

w(t) = \dfrac{F}{m_0-\mu t}

(3)

Step 4The velocity by integrating the equation of motion

Separating variables in (2), with $m(t)$ from (1), and integrating from rest at $t=0$ gives the speed at time $t$ .

\begin{align*}\int_0^{v} dv' &= \int_0^{t} \frac{F}{m_0-\mu t'}\,dt' \\ v &= F\left[-\frac{1}{\mu}\ln(m_0-\mu t')\right]_0^{t} \\ v &= \frac{F}{\mu}\Bigl(\ln m_0-\ln(m_0-\mu t)\Bigr)\end{align*}

v(t) = \dfrac{F}{\mu}\ln\dfrac{m_0}{m_0-\mu t}

(4)

Check

At $t=0$ , $w(0)=F/m_0$ , exactly Newton’s second law applied to the whole initial mass, as it must be before any sand has left; and $v(0)=(F/\mu)\ln 1=0$ , matching the given rest condition. As $\mu\to0$ (no leak), $m(t)\to m_0$ and both results reduce to the constant-mass case $w=F/m_0$ , $v=Ft/m_0$ , found by expanding the logarithm to first order in $\mu t/m_0$ .

Remark

Because the departing sand keeps the cart’s own instantaneous velocity, the mass loss supplies no reaction thrust: all of the acceleration’s growth with time comes from $F$ acting on a shrinking mass, never from a push by the escaping sand. A rocket loses mass the opposite way, ejecting it backward relative to the body, and that relative velocity is exactly the extra term this problem’s equation of motion does not have.