Question about Conservation of Momentum I have what I believe is a simple question about the conservation of momentum, but I'm not sure what the answer is.
I'm wondering when a stationary ballistic pendulum collides perfectly inelastically with a bullet at an angle (where the bullet's velocity isn't purely horizontal), is all of the bullet's momentum conserved or only its horizontal momentum.
My understanding of conservation of momentum comes from Newton's second law, where $\vec{F}=\frac{d\vec{p}}{dt}$ meaning $\vec{F}=0 \Rightarrow \vec{p}=\vec{k}$ for some constant vector k, and because this is a vector equation, I'm assuming horizontal and vertical components can be considered separately, and only horizontal momentum is conserved. This also makes intuitive sense because the pendulum's swing is basically horizontal after it's hit.
Is this reasoning correct? If not, what's wrong with it?
 A: It's a fact that in reality momentum is conserved in all directions even during an inelastic collision, if you could track and measure the momentum of every particle and object involved. However, these sorts of textbook problems frequently have some conventional assumptions that, IMO, aren't well stated and are misleading to students. In this case there are generally two major assumptions:

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*The pivot point of the pendulum will not break. It's attached to a wall, or the ceiling, and thus indirectly to the Earth. Otherwise, the pendulum might not actually swing if pushed horizontally (and instead just dragging the floating pivot point behind it)

*The connection between the pivot point and the pendulum body is rigid against deformation in one or both directions. I.e., if the pivot point is a rope, it cannot stretch. If it is a rigid rod, as it has been clarified is the case for this particular problem, it also cannot be compressed.

Together, these unstated assumptions mean that all components of momentum may not necessarily be conserved within the conventions of the problem, because the model doesn't permit figuring out how the pendulum's rope or rigid rod, or the pivot point, or the ceiling, or the walls, or the Earth will move in response to an arbitrary impulse.
A: There are different possible scenarios here that can give slightly different results. Both the horizontal and vertical momentum is conserved in all scenarios
One is that the pendulum is hanging with a rigid material, connected to a pivot in the ceiling. In that case the horizontal momentum will be transferred to the pendulum, and the vertical momentum will be transferred to the earth. This is because the pendulum is connected to the ceiling, which in turn is connected to a house or building, which is connected to the earth, and the rigid materials cannot bend or be deformed.
The second is that the the pendulum is connected using a material like a rope, that can bend but not stretch(even it does slightly it is negligible). In that case two things can happen depending on the angle of the bullet. If the bullet is angled downwards, the horizontal momentum will be transferred to the pendulum, and the downward component of the momentum will be transferred to the earth. This is because the rope cannot stretch. If it is angled slightly upwards, both the horizontal and upward momentum will be transferred to the pendulum. This will happen because the rope will bend upwards creating a 'compression' because the distance in a straight line between the joint at the top and the pendulum will decrease. So it will move back, and also have a slightly higher upward momentum. But the path of the pendulum will no longer be purely radial to the pivot, but instead a combination of radial and vertical
A: In short, the momentum must be conserved in all directions.
Let $m$ and $M$ be the masses of the bullet and the pendulum respectively, and let the bullet approach and hit the pendulum at velocity $\mathbf{v_i}$. The collision is perfectly inelastic, so the two masses will stick together and move -- instantaneously after colliding -- with a combined velocity $\mathbf{v_f}$. From the conservation of linear momentum,
$$m \mathbf{v_i} = (m+M) \mathbf{v_f}.$$ Assuming the collision lies in a plane, this is now a vector problem in 2 dimensions. Accordingly, we can it split into two equations:
$$m \mathrm{v_{i,x}} = (m+M) \mathrm{v_{f, x}},$$ and
$$m \mathrm{v_{i, y}} = (m+M) \mathrm{v_{f,y}}.$$ If you know the bullet's angle of impact, say an angle $\theta$ with respect to the horizontal (assuming it is moving toward the right initially and inclined upward: we can define these components of its velocity as being positive in the $x$ as well as the $y$ direction), you can substitute $\mathrm{v_{i,x}} = v_i \cos{\theta},$ and  $\mathrm{v_{i,x}} = v_i \sin{\theta}$ in the above and solve for $\mathrm{v_{f, x}}$ and $\mathrm{v_{f, y}}$.
