Question

and open railroad car of mass M is moving with initial velocity field on a straight horizontal frictionless track it suddenly starts raining at a time T = 0 the raindrops fall vertically with velocity v and add a mass of Mu kg per second of water Determine the velocity of v of car after t seconds​

Answer 1

Given :

Mass of open railroad car  = M

The 1st raindrop falls at time , T = 0

The raindrops are falling vertically with velocity =  v

To Find :

The velocity of car after t seconds = ?

Solution :

Here by applying conservation of momentum we have :

Initial momentum  = final momentum

For 1st drop :

$Mu + mv = (M+m)v' - (1)$

Here  , u = initial velocity of car , m = mass of drop , v = velocity of drop and v' = final velocity of car

For t sec duration :

$mu + Mu.t.v = (M + Mu.t)v'$     - (2)

So, we have :

$v' = \frac{(Mu + Mutv)}{(M + Mut)} \frac{m}{s}$

So, the final velocity of car after t seconds is $v' = \frac{(Mu + Mutv)}{(M + Mut)} \frac{m}{s}$