(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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 [Broken]

I'm unsure which is the horizontal velocity.

2. Relevant equations

m_{1}v_{1}=m_{2}v_{2}

3. 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 v_{f}*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 v_{x}=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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# Homework Help: Momentum and energy

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