Mass flow and rolling carts (kleppner 3.11)

1. Dec 13, 2014

geoffrey159

1. The problem statement, all variables and given/known data
Material is blown into cart A from cart B at a rate b kilograms per
second. The material leaves the chute vertically down-
ward, so that it has the same horizontal velocity u as cart B. At the
moment of interest, cart A has mass M and velocity v. Find dv/dt,
the instantaneous acceleration of A.

2. Relevant equations
Momentum

3. The attempt at a solution

System studied: cart A and the chute.

Let's put that $M_A$ is the mass of cart A (unloaded), $m(t)$ is the material's mass in cart A at time $t$, and $\triangle m = m(t+\triangle t) - m(t)$ is a small amount of material falling from the chute into cart A in a $\triangle t$ seconds.

At a given time $t$, the horizontal momentum will be:
$P(t) = (M_A + m(t)) v(t) + \triangle m \ u(t)$
$P(t+\triangle t ) = (M_A + m(t) + \triangle m) v(t+\triangle t)$

So that in time $t$:
$\frac{dP}{dt} = (M_A + m) \frac{dv}{dt} + \frac{dm}{dt} (v - u)$

In this system, only friction from the wheels of cart A contribute to horizontal external force :
$f_{ext} (t) = -\mu g (M_A +m(t))$

Since at time of interest $t_i$, we are given:
$M_a + m(t_i) = M$,
$v(t_i) = v$,
$u(t_i) = u$,
$\frac{dm}{dt} = b$,

the acceleration should be:
$\frac{dv}{dt}(t_i) = \frac{1}{M_A + m(t_i)}(f_{ext}(t_i) + \frac{dm}{dt} (u - v) = -\mu g +\frac{b}{M} (u-v)$

Do you agree with this solution ?

2. Dec 13, 2014

haruspex

Yes, but you did not need to separate MA from m. Since friction is not mentioned, I would have assumed it zero.

3. Dec 13, 2014

geoffrey159

Thank you haruspex!
I did separate because it wasn't so clear at first, but now I see it would work just the same ! Thanks