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The water is falling horizontally on the masses with speed u, and a constant rate of k kg/sec. When the rain hits the block it drips to the sides.

My question is how to deal with the rain dripping to the sides.

Let's take the top block for example:


$$P_x(t+dt)= m(v+dv)$$

If I solve this I'd get $$\frac{dP_x}{dt} = ma$$ as if the water had no effect. but I'm not sure how to incorporate the rain into the equations. Is $$P_x(t+dt)=(m+dm)(v+dv)~?$$


closed as off-topic by John Rennie, ZeroTheHero, Jon Custer, Kyle Kanos, stafusa Feb 12 at 21:45

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  • $\begingroup$ You'd have momentum transfer to the second block moving vertically, though. Somewhat like $( u - v(t)) k$ $\endgroup$ – mikuszefski Feb 11 at 10:50
  • 1
    $\begingroup$ What is the level of approximations? Basically, the right box should get the momentum from the top, but less with the increasing velocity. The left box should get momentum from the right - but zero at zero velocity and more with increasing velocity = increasing angle of the rain $\alpha$. For each droplet $m(w \frac{v}{u} \cos \alpha)$ may be the transferred momentum, velocity $w^2$ is the $u^2+v^2$. I think ? I just made a sketch for me... $\endgroup$ – jaromrax Feb 11 at 15:19