Simple collision between rocks in outer space. Webassign

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Two rocks in outer space collide, with one rock having a mass of 14 kg and a velocity of ‹4450, −2950, 2250› m/s, and the other rock having a mass of 7 kg and a velocity of ‹−600, 1900, 3600› m/s. After a 1 kg chunk breaks off from the first rock and sticks to the second, the new velocity of the 13 kg rock is ‹1450, 300, 1900› m/s. The discussion focuses on applying the law of conservation of momentum to find the final velocity of the now 8 kg second rock. Initially, the calculations were incorrect due to an error in managing zeros, but after correcting this mistake, the final velocity was successfully determined. The resolution highlights the importance of attention to detail in physics problems.
najatau
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Homework Statement


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Two rocks collide in outer space. Before the collision, one rock had mass 14 kg and velocity ‹4450, −2950, 2250› m/s. The other rock had mass 7 kg and velocity ‹−600, 1900,3600› m/s. A 1 kg chunk of the first rock breaks off and sticks to the second rock. After the collision the 13 kg rock has velocity ‹1450, 300, 1900› m/s. After the collision, what is the velocity of the other rock, whose mass is now 8 kg?

Homework Equations


This is an introductory college level course. We were talking about the law of conservation of momentum, but I keep getting the problem wrong when I try to set the two total momentums equal to each other. I keep getting the sample Webassign problem wrong too. I've been working on this for an hour with no luck.

The Attempt at a Solution



14*<4450,-2950,2250>+7*<-600,1900,3600>=8*final velocity of other rock+13*<1450,300,1900>

<62300,-41300,31500>+<-4200,13300,25200>=8*final velocity of other rock+<18850,3900,24700>

<58100,-2800,56700>-<18850,3900,24700>=8*final velocity of other rock

<39250,-6700,32000>/8=final velocity of other rock

<4906.25,-837.5,4000> m/s =final velocity of other rock
[/B]
This isn't the correct answer. :( I tried rounding to 3 significant figures as well, and it's still not the right answer.
 
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najatau said:
<58100,-2800,56700>-<18850,3900,24700>=8*final velocity of other rock
Check the number of zeros.
 
Oh, thank you! *cries for joy and does a little dance* I don't know why I kept missing that! I got my problem right. It's a bigger relief to know that I actually do understand the concept. :)
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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