What happens to the momentum of this rainfall? Are these expressions describing the rainfall correct/significant?

Question

An element of mass $Delta m_r$ (mass of rainfall) falls vertically onto a mass $m_c$ (a cart).

The empty cart starts from rest and there is no friction.

A constant horizontal force $F = <F, 0>$ is applied

and the system of the cart and rain inside the cart travel together at velocity $v = <v_x(t), 0>$.

The rainfall transfers rain mass at rate $b ∴ m_r(t) = bt$.

I use $hat{i}$ and $hat{j}$ to analyze the momentum of the system $p(t) = <p_x(t), p_y(t)>$

I analyze $hat{i}$ for the horizontal velocity of the system, which I do in two ways correctly, while I have questions about my analysis of the system in the $hat{j}$ coordinate axis.

$hat{i}$: I described the system as two constituent elements, the cart and the rain in the cart, and the rain that falls into the cart. I equate $Delta p = FDelta t$, and use a Free Body Diagram on the cart and the rain in the cart to build $F = dp/dt = bv_x + (m_c + bt)dv_x/dt$. Both expressions result in the horizontal velocity, correctly. The work is shown here.

$hat{j}$: I considered the momentum of the rainfall $Delta m_r$. The work is also shown in the image above. I use $c = v_t$ (terminal velocity) to add a physical sense to the expression $cDelta m_r$, and derived some expressions using $Delta p = FDelta t$, and a Free Body Diagram, again:$-btv_t = int_0^t F_external = int_0^t (btg)dt = (bt^2g)/2$

and

$F = ma_y = btg$

Specifically to these last two expressions, I don’t know what I’m looking at, if they are right, if they are relevant/what they reveal.

kinematics momentum

ytlu · Answer

Since there is no friction, the forces in the $x$- and the $y$-directions are completely decoupled. The momentum in the $y$-direction is totally balanced by the normal force, brings no effects to the $x$-direction.
 The motion in the $x$-direction is the inverse rocket motion (rocket have a negative mass changing rate, while this problem has a positive mass changing rate.) Lets follow the the analysis in the rocket motion: 
Momentum change from time $t$ to time $t+dt$: assume at time $t$, the mass $m(t)$ and speed $v_x(t)$; while at $t + dt$, the mass $m(t) + Delta m$ and speed $v_x(t) + Delta v_x$.
$$
Delta p_x = p_x(t+dt) - p_x(t) = (m(t) + Delta m) (v_x(t) + Delta v_x) - m(t) v_x(t).
$$
Neglect the second order term:
$$
   Delta p_x = m(t) Delta v_x + v_x(t) Delta m = F Delta t.
$$
We then has the inhomogeneous first order differetial equation:
$$
  m(t) frac{ d v_x}{dt} + v_x(t) frac{d m}{dt} = F.
$$
Using $m(t) = m_c + b t$, and $d m / dt = b$.
$$
    (m_c + b t) frac{ d v_x}{dt} + b v_x = F. 
$$
You have to solve this differential equation with initial condition $v_x(0) = 0$ to obtain the function $v_x(t)$.

What happens to the momentum of this rainfall? Are these expressions describing the rainfall correct/significant?

One Answer

Add your own answers!

Ask a Question