## Momentum when object is thrown at angle

1. The problem statement, all variables and given/known data

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?

2. Relevant equations

3. 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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 Recognitions: Homework Help 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) ?

Recognitions:
Homework Help

## Momentum when object is thrown at angle

 Quote by jorcrobe mt(vb)cos(30)=mc(v) ?
That looks promising, but where did the "30" come from?

 Quote by gneill That looks promising, but where did the "30" come from?
Good question. deg 22*

Thanks a bunch!