# Momentum and energy

1. Nov 23, 2008

### 2RIP

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.

I'm unsure which is the horizontal velocity.

2. Relevant equations
m1v1=m2v2

3. The attempt at a solution
Let m=0.01 kg, M=10 kg, v=300 m/s

$$mv=(m+M)v_{f}$$
$$v_{f}=(mv)/(M+m) = 0.3 m/s$$

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

2. Nov 23, 2008

### Nolgosphere

The horizontal velocity of the combined mass immediately before it interacts with the inclined plane is the final velocity you found by momentum conservation. But it's likely you must find the horizontal velocity due to the plane.

To do this you must decompose the final velocity into its components parallel and perpendicular to the inclined plane. If you think about where the perpendicular component is directed, you will see that you're just left with the parallel part. This is the initial velocity of the combined mass along the plane that you must decompose further into its horizontal, and vertical, component.

Having said all that though, my calculations disagree with the solution you have but I don't see what I'm doing wrongly.