Wrong Answer in Textbook for Collision Velocity Calculation

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The discussion centers on a collision velocity calculation problem from a physics textbook, where the user successfully solved parts (a) and (b) but found a discrepancy in part (c) regarding the angle. The user's calculated angle is -48.4°, indicating it is in the fourth quadrant, while the textbook states the angle as +48.4°, which is in the first quadrant. Participants confirm that the textbook's answer is misleading since it does not specify the quadrant. The consensus is that the user's answer is correct, and the textbook should clarify the angle's direction. This highlights the importance of precise language in educational materials.
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Homework Statement


(Problem about a collision) (...). What then is their velocity in (a) unit-vector notation and as (b) magnitude and (c) angle.

I will not post the problem due to copyright issues and that it is absolutely pointless now.

The Attempt at a Solution


I was able to solve (a) and (b).
However, my answer for (c) differs from that of the textbook.
\text{a)} \;(2.67 \;\text{m/s})\hat{i}+(-3.00 \;\text{m/s})\hat{j}
\text{b)} \;4.01 \;\text{m/s}
\text{c)} \;\theta{} = \tan^{-1}\left(-\frac{3.00}{2.67}\right)=-0.845397=-48.4^{\circ{}}

While the book provides +48.4° as the angle.

Am I correct? You can solve (c) from (a) and (b) [my answers are identical to those of the book].

It is from Fundamentals of Physics (Extended) - 9th ed.
 
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Well, without thinking too much about it, the vector given in (a) is in the fourth quadrant. The angle 48.4° is in the first quadrant. Your answer, however, is definitely in the fourth quadrant.
 
You are correct. The text should not have stated the answer as simply 48.4o.
If they had stated it as 48.4o below the positive x direction, then that would have been a good answer. But they didn't.
 
Yes, that's what I thought. I was pretty sure I was right, but as this book has presented very few mistakes already I thought I could be missing something.

Thank you guys for looking at it.
 
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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