Velocity of an object with varying mass

[Nasa123123]

Homework Statement

I am working on a problem where a box is moving on a frictionless surface with velocity v0, and is gaining mass by the rate of dm/dt = a. There are no external forces working on the box. The problem is to find an expression for the velocity as a function of time as the mass increases. The rate of change in mass is just given as «a» and I interpret this as it is not supposed to be a variable in the differential equation.

Relevant Equations

F = ma = m dP/dt = m dv/dt

I have tried several things but I am a little uncertain if I’m thinking right so a little hint goes a long way. I think I have to use the law of conservation of momentum as the collision between the raindrops and the box is inelastic. But I am unsure how to set up the equation.

 

[scottdave]

Instead of m * (dv / dt) try this:
d(m*v) / dt.

When mass is constant, they are equivalent. What does your problem say about external forces?

 

[Nasa123123]

Instead of m * (dv / dt) try this:
d(m*v) / dt.

When mass is constant, they are equivalent. What does your problem say about external forces?

There are no external forces working on the box.

 

[haruspex]

d(m*v) / dt

Ummm... I wage a personal war against using that to encompass a varying mass (and I'm not alone).
In the real world, the mass of a closed system does not change. If the mass is increasing it is coming from somewhere else, and it does so with its own momentum. Using d(mv)/dt happens to work if that momentum is zero in the reference frame.
In the present case, it seems these are raindrops, and I would guess they arrive with a velocity normal to that of the box (@Nasa123123 , please confirm). So you would get away with it in this case.

 

[Nasa123123]

Ummm... I wage a personal war against using that to encompass a varying mass (and I'm not alone).
In the real world, the mass of a closed system does not change. If the mass is increasing it is coming from somewhere else, and it does so with its own momentum. Using d(mv)/dt happens to work if that momentum is zero in the reference frame.
In the present case, it seems these are raindrops, and I would guess they arrive with a velocity normal to that of the box (@Nasa123123 , please confirm). So you would get away with it in this case.

Yes, the raindrops are falling vertically from the sky with no impact from wind or other forces.

 

[haruspex]

Yes, the raindrops are falling vertically from the sky with no impact from wind or other forces.

As I wrote, you can use @scottdave's equation because the raindrops bring no horizontal momentum with them.
A more rigorous approach is to consider the horizontal momentum of the box+rain system. This is conserved, so the rate at which the rain gains momentum equals the rate at which the (dry) box loses it.

 

[Orodruin]

Ummm... I wage a personal war against using that to encompass a varying mass (and I'm not alone).
In the real world, the mass of a closed system does not change. If the mass is increasing it is coming from somewhere else, and it does so with its own momentum. Using d(mv)/dt happens to work if that momentum is zero in the reference frame.

I support your cause! If one used d(mv)/dt=0 blindly one would build very poor rockets.