Irodov 1.183: Sand That Joins Steals Momentum – Amit & Deepali Sinha Foundation

Problem Irodov 1.183

A flatcar of initial mass $m_0$ , starting from rest on a horizontal frictionless track, is pushed by a constant horizontal force $\vect F$ . Sand pours straight down onto it from a hopper that is fixed in place above the track, at the constant rate $\mu$ (kilograms per second). Find the flatcar’s velocity $v$ and acceleration $w$ as functions of the elapsed time $t$ since loading and pushing began together.

The car, of instantaneous mass m(t), is driven by the constant force F while sand falls vertically onto it from the fixed, stationary hopper; each grain arrives with zero horizontal velocity and must be brought up to the car's own speed. — The car, of instantaneous mass $m(t)$ , is driven by the constant force $\vect F$ while sand falls vertically onto it from the fixed, stationary hopper; each grain arrives with zero horizontal velocity and must be brought up to the car’s own speed.

Given

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

Find

v(t)

w(t)

Solution

Step 1The mass and the momentum of the loaded car

Sand joins at the constant rate $\mu$ , so the car’s total mass grows linearly from its initial value.

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

The body whose momentum is worth tracking is the car together with all the sand already aboard, because that collection moves as one at the car’s own speed. Its horizontal momentum is therefore the running mass times that speed.

p(t) = m(t)\,v(t) \tag{2}

Step 2Momentum balance for the growing car

Take that collection at $t$ , of mass $m$ and velocity $v$ , and again at $t+dt$ , after a mass $dm=\mu\,dt$ has landed on it with zero horizontal velocity of its own. Writing (2) at the two ends of the interval and expanding the product,

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

Subtracting the first line from the last, and dropping the second order term $dm\,dv$ , leaves the increment of momentum over the interval.

dp = m\,dv + v\,dm \tag{3}

Dividing (3) by $dt$ writes the same statement as a rate, with $dm/dt=\mu$ constant.

\frac{dp}{dt} = m\,\frac{dv}{dt} + v\,\frac{dm}{dt} = m\,\frac{dv}{dt} + \mu v \tag{4}

The sand brings in no horizontal momentum, so the only thing changing the collection’s momentum is the impulse of $\vect F$ , and that rate equals $F$ .

\dfrac{dp}{dt} = \dfrac{d(mv)}{dt} = F

(5)

Step 3The velocity by integrating the momentum

The right side of (5) is a constant, so the momentum simply accumulates $\vect F$ ’s impulse. The car starts from rest with no sand aboard, so $p(0)=0$ .

\begin{align*}\int_0^{p(t)} dp' &= \int_0^{t} F\,dt' \\ p(t) &= Ft\end{align*}

Reading that back through (2) and (1) gives an equation in $v$ alone.

(m_0+\mu t)\,v(t) = Ft

v(t) = \dfrac{Ft}{m_0(1+\mu t/m_0)}

(6)

Step 4The acceleration from the velocity

Differentiating the velocity (6) by the quotient rule, with $m(t)$ from (1), gives the acceleration.

\begin{align*}w(t) = \frac{dv}{dt} &= \frac{d}{dt}\left(\frac{Ft}{m_0+\mu t}\right) \\ &= \frac{F(m_0+\mu t) - Ft\cdot\mu}{(m_0+\mu t)^2} \\ &= \frac{Fm_0}{(m_0+\mu t)^2}\end{align*}

w(t) = \dfrac{F}{m_0(1+\mu t/m_0)^2}

(7)

Check

At $t=0$ , $w(0)=F/m_0$ and $v(0)=0$ , consistent with the car starting from rest under the whole applied force before any sand has joined. Comparing to the naive guess $F/m(t)=F/(m_0+\mu t)$ , the true acceleration is smaller by the factor $m_0/(m_0+\mu t)<1$ : part of $\vect F$ ’s effort each instant goes into bringing the newly arrived, momentarily stationary sand up to the car’s own speed, which is the term $\mu v$ carried by (4).

Remark

Contrast this with a cart that instead loses sand at the same rate through a floor hole (problem 1.182): there the spilled grains keep the cart’s own velocity as they leave, so $u=v$ and the loss supplies no reaction at all, giving the larger acceleration $F/m(t)$ . Here the incoming grains arrive at $u=0\neq v$ , and it is exactly that mismatch, not the changing mass by itself, that produces the extra retarding term $v\,dm/dt$ in the equation of motion.