An open topped rail road car of mass M has an initial velocity $v_0$  along a straight horizontal frictionless track. It suddenly starts raising at time t = 0. The rain drops fall vertically with velocity u and add a mass m kg/sec of water. The velocity of car after t second will be (assuming that it is not completely filled with water)

The rain drops falling vertically with velocity u do not affect the momentum along the horizontal track. A vector has no component in a perpendicular direction

Rain drops add to the mass of the car

Mass added in t sec = (mt)kg

Momentum is conserved along horizontal track.

Initial mass of car = M

Initial velocity of car = $v_0$

Final velocity of (car + water) = v

Mass of (car + water) after time t = (M + m)

$\therefore$  Final momentum = Initial momentum

$\left(M+mt\right)v=M{v}_0$

$\therefore v=\frac{M{v}_0}{\left(M+mt\right)}$