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 What's the difference between snowmen and snowwomen? (Posted on 2004-06-14)
Sometime this past winter, I was in a snowball fight... and I had two left when we had to call it quits. As it happened, they were both spheres and one had exactly twice the diameter as the other.

I left the two on the ground, when we quit... and the weather started to get warmer. The snowballs started to melt. The melting only occurred at their surface, so the speed at which the balls melted was proportional to only the surface of the (remainder of the) snowballs.

How much (volume) was left of the small snowball when half the volume of the larger had melted?

 No Solution Yet Submitted by SilverKnight Rating: 4.0000 (2 votes)

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 solution | Comment 6 of 16 |

The radius of each snowball is decreasing at a constant rate, the same for each.  When the volume of the large ball has reached 1/2 its original volume, its radius is down to (1/2)^(1/3), or about .7937 of its original radius. It had lost 1-.7937 of that larger radius, or .2063.  So the smaller ball had lost this fraction of the larger ball's radius also, or .4126 of its own, leaving .5874 of its original radius. Cubing this we get .2027 of its original volume remaining. That's   (1-2*(1-(1/2)^(1/3)))^3 = .20267685653536.

As calculus backup for the statement that "each snowball is decreasing at a constant rate, the same for each":

dV/dt=V^(2/3)

dt/dV=V^(-2/3)

t=3(V^(1/3)) + c

Let c=0 for time measured before disappearance. Volumes/times are arbitrary and we're only interested in the proportion, so time is proportional to the cube root of the volume, that is, to the radius.

A non-calculus justification is that melting volume proportional to the area, leaves a quotient that is the (delta) radius.

 Posted by Charlie on 2004-06-14 13:39:44

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