I am working on a question in which a cart is moving on frictionless tracks and it begins to rain. The cart has a drainage hole to let the rain out at exactly the same rate at which it enters $k~kg/s$.
My question is, can I simple neglect the fact that it is raining and let the velocity after $\ell$ seconds equal the initial velocity $u$?
If not how would I calculate it?
I have worked out that the momentum of the cart with no drain hole in the rain after $t$ seconds is $$\frac{mu}{m+(k(t))}.$$
Assuming no air resistance and no rolling resistance, the cart's momentum can only be transferred to the rain.
As BowlOfRed says, assume that rain leaving the cart at any one instant does so with the same horizontal velocity component as the cart has at that moment. And assume that before then, its horizontal velocity component was zero.
You'll also need to assume a value for the change in the rain's vertical velocity. Assuming zero makes the calculation simpler, and might be close enough to reality. You could test the sensitivity of that assumption later.
So now, you've got a differential equation: you know the cart's rates of loss of horizontal momentum, expressed in terms of the instantaneous speed of the cart. Integrate that from time zero to time t, to find the cart's total loss of horizontal momentum. And once you've got that, you can subtract it from the original value, to find the cart's residual horizontal momentum at time t. And then convert that into a velocity.
If you want to get fancy, and you have sufficient information, you could calculte how the rain's mass changes the cart's total mass (as the cart's total mass will include some non-zero mass of rain at any t>0, but zero rain at time t=0). It could be a reasonable simplification to assume that the additional mass of the cart due to rain is negligible, unless the question indicates otherwise.
