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Homework Statement
The non-moving 10-kg block is held frictionlessly inclined at 30 degrees by a stop at point A . A 0.010-kg bullet is moving at 300 m/s hits the block and embeds into it.
http://img386.imageshack.us/img386/9305/68906392wf8.jpg
I'm unsure which is the horizontal velocity.
Homework Equations
m1v1=m2v2
The Attempt at a Solution
Let m=0.01 kg, M=10 kg, v=300 m/s
[tex]mv=(m+M)v_{f} [/tex]
[tex]v_{f}=(mv)/(M+m) = 0.3 m/s
[/tex]
I thought the final velocity would be the initial horizontal velocity of the two combined masses, however, the solution says the horizontal velocity is vf*cos30=0.26 m/s and implies the velocity up the incline plane as v=0.3 m/s. Don't we need to be consistent with the direction of velocity in the momentum equation?
It kind of makes sense to me because if vx=0.3 m/s, then the inclined velocity would be greater than 0.3 m/s. But equation wise, I can't figure out how it happened.
Thanks
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