Momentum when object is thrown at angle

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In the discussion, a student on a skateboard throws a textbook at an angle of 22 degrees with a velocity of 2.25 m/s, while the combined mass of the student and skateboard is 104 kg. The key focus is on calculating the student's velocity after the throw, using the principle of conservation of momentum. The initial momentum of the book is set equal to the momentum of the student and skateboard, leading to the equation mt(vb) = mc(v). There is some confusion regarding the use of angles in the momentum calculation, specifically the incorrect reference to "30 degrees" instead of the correct 22 degrees. The discussion emphasizes the importance of considering momentum as a vector quantity with horizontal and vertical components.
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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!
 
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