# A falling object of decreasing mass.

The problem is:

A snowball is dropped from the top of a building, evaporation causes the mass m of the snowball to decrease at a rate proportional to the snowball's speed as the snowball falls, so that

M' = k * abs(y')
where y denotes the vertical distance from the ground and k is a negative constant.

Derive the differential equation governing y(t), neglecting all physical phenomena except the pull of Earth's gravity on the snowball and the variation in the snowball's mass

This is what I have been able to work out:
Position = 0i + y(t)j
v = d/dt(position) = y'(t)
Using newtons Second law​

Net force = (d/dt)[(mass)(velocity)]
-m(t)g = (d/dt)(m(t)y'(t))
-m(t)g = m'(t)y'(t)+m(t)y''(t)

And I'm not able to simplify this expression so I must be missing something in the setup...Any help would be great.

## Answers and Replies

Hootenanny
Staff Emeritus
Science Advisor
Gold Member
Welcome to Physics Forums.

Since the velocity is never positive, we can replace the absolute value sign with a negative sign

$$m^\prime\left(t\right) = -ky^\prime\left(t\right)$$

You can simply substitute this directly into the first term on the RHS. However, for the remaining two terms we need to determine m(t). This can be done by integrating both sides of the above expression with respect to t.

Try LaTex, it makes things easier to read :P

On a sidenote: If you don't account for friction the mass does not matter at all. Just neglect the loss (unless you care about energy).

Hootenanny
Staff Emeritus
Science Advisor
Gold Member
On a sidenote: If you don't account for friction the mass does not matter at all. Just neglect the loss (unless you care about energy).
I'm afraid that you're wrong there. If the mass were constant, then yes you could safely ignore the mass of the snowball. However, since the mass is not constant you need to take this into account.

--edit: I am oviously wrong, sorry for the bs