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# Motion with variable mass

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A spherical raindrop of radius a, falls from rest under gravity. It falls through a stationary cloud so that, because of condensation, its radius increases with time at a constant rate k. Find the distance fallen by the raindrop after time t.

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A spherical raindrop of radius a, falls from rest under gravity. It falls through a stationary cloud so that, because of condensation, its radius increases with time at a constant rate k. Find the distance fallen by the raindrop after time t

Let the radius of the drop at time t be r

The rate of increase of the radius is given by

As the initial radius is a and it increases in time t by Kt the radius r at time t will be

r = a + K*t
The volume of the drop at time t will be
And hence its mass will be m =
Here is the density of water.

Now according to the second law of motion the rate change of momentum is equal to the force applied and the only force acting on the drop is the gravity (neglecting air resistance).

As the mass of the drop is variable, hence taking downward direction positive, we can write

Where v is the velocity of the drop at time t (initial velocity is zero)

Substituting the value of mass from above we get

Or

Integrating this equation wrt t we get

Here C is the constant of integration.
As at t = 0; v = 0 substituting in the equation we have

Substituting this in above equation we have

Or

Or ------------------ (2)
Now if the distance fallen in time t is x then the velocity is given by v = dx/dt and hence

Integrating wrt to t we have

Where C' is constant of integration.
Or

Now as initially at t = 0; x = 0 gives

Substituting in equation above we get

Or
Or
Or
Or

This is the required distance.

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!