How many bricks should Johhny attach to the body?

Question

Don Guzzilani called his leutenant into the office and told him:

'That truck in parking lot has the body of Fat Toni in the trunk.
We finally got him yesterday, Johnny. Drive to Hudson river and dispose of the body. Concrete bricks are in the bed of the truck, 20lb each. Those are good bricks, too, made of high density concrete, twice as dense as water.

And keep in mind that the body will swell in the water and its volume will double. Do not fail me, Johnny, this time. And you may keep Fat Tony's wallet.'

Jonny walked to the car and checked the wallet. Inside he found 20 crispy $100 bills and driver's licence, which read:
Name: Antonio Fattini
DOB: 01/02/60
Weight: 300lb
Height: 5'8''
Expires 01/02/08

Answer 1

G = W / (W-F)

of the bricks, G=2.
2 = 20/(20-F), then F=10.

so each brick only counts as 10lb in water, being submerge.

before volume double, fat toni sinks in the water.
after volume double, fat toni floats.

so johnny needs the bricks' masses to cover for the "doubled" volume. so he needs to double the mass too. that's another 300lb, by the bricks.

how many bricks?
= 300/10
= 30

add another 1 or 2, or as many as he wants too, for safety measure.

Answer 2

Hmmm. What kinda truck has a trunk? Just curious; might come in handy for some social work.

It’s conceivable that Fat Toni is so fat that he has a very large buoyancy when placed in water. Thus, he could displace 300 pounds of water when totally submerged. If his volume then doubled, he would displace 600 pounds of water. At 20 pounds per concrete brick, it would require 30 bricks to keep Toni submerged, but I would probably double that amount, or use whatever number of bricks were on the truck as long as there were at least 30 bricks.

Remember, Johnny doesn’t want to “disappoint” Don Guzzilani this time.

Answer 3

His body swells because it absorbs water. So the volume doubles but that extra volume is filled with water mass, so it can be neglected from the analysis.

W_body = 300 lbs
ρ_body =~ 0.985ρ_water
V_body = 300 / (0.985ρ_water)

W_bricks = 20*n
V_bricks = 20n/(2ρ_water)

W_total = 300 + 20n
V_total = 300 / (0.985ρ_water) + 20n/(2ρ_water)

To have a combined density equal to water
300/0.985 + 20n/2 = 300 + 20n
n = 0.46

So 1 brick will do.

If however you assume that his body swells with air (utterly ridiculous), then you'd get n = 30.91 in which case 31 bricks will be the answer.

Answer 4

15 bricks

Answer 5

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