Momentum when object is thrown at angle

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SUMMARY

The discussion focuses on calculating the velocity of a student on a skateboard after throwing a textbook at an angle. The student and skateboard have a combined mass of 104 kg, while the textbook has a mass of 1.05 kg and is thrown at a velocity of 2.25 m/s at an angle of 22 degrees. The initial momentum equation set by the student is mt(vb) = mc(v), where the vertical component of the throw is acknowledged but deemed irrelevant for the horizontal momentum calculation. The correct approach involves using the cosine of the angle to resolve the horizontal component of the throw.

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Homework Statement



A student standing on a stationary skateboard tosses a textbook, mt = 1.05 kg, to a friend standing in front of him. The student and the skateboard have a combined mass of mc = 104 kg and the book leaves his hand at a velocity of 2.25 m/s at and angle of 22 degrees with respect to the horizontal.

Randomized Variables
mt = 1.05 kg
mc = 104 kg
Vb = 2.25 m/s
θ = 22 degrees.

what is an expression for the magnitude of the velocity the student has, Vs, after throwing the book?

Homework Equations





The Attempt at a Solution



I thought that the momentum would be equal and opposite, so I set:

mt(vb)=mc(v)

solved for v, and got .023

I feel as if I am working this terribly wrong.
 
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Momentum is a vector. It has components (vertical, horizontal). The book is thrown at an angle to the horizontal. The vertical component won't do much since it's directed into the Earth. What does that leave you with?
 
mt(vb)cos(30)=mc(v)

?
 
jorcrobe said:
mt(vb)cos(30)=mc(v)

?

That looks promising, but where did the "30" come from?
 
gneill said:
That looks promising, but where did the "30" come from?

Good question. deg 22*


Thanks a bunch!
 
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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