The professor very generally created a very simple conceptual problem as a basis for harder ones, but I don't understand how to answer it.
A piece of clay, with mass (m) and speed (v) collides with a motionless stick of length L (with uniform mass density and total mass M). The stick is aligned along the Z-axis with its center at z=0. The clay is initially moving in the x-direction. The clay collides at the stick's end at z = L/2. Assume there is no gravity or external forces.
What is the momentum in the y direction after the collision? (the clay completely sticks to the stick upon collision).
e. mL/M (mv)
What is the angular momentum of the system after collision?
I'm unsure. I know from my textbook that L = r x p, though we really haven't done cross products so I'm unsure of what that means (it says it's mrvsin(θ), right hand rule direction)
I also know about conservation of momentum, which I assume is mv=(m+M)vf. Also I know of kinetic energy conservation and torque, which causes angular momentum to change.. but I've read my textbook 5 times and I can't understand the steps to solve this very simple problem.
The Attempt at a Solution
At first I thought the initial answer to question 1 was zero, because if a particle moving in x collides at with an object in z, it would just rotate about the xz plane and not have any y momentum. But then I remembered that there's a right hand rule (which I don't understand much either) and I doubt the professor would put a question that didn't require calculations..
In regards to question 2, since we're given the equation L = mrvsinx, and I'm assuming that it's just mrv since the angle is 90 (plus the answers don't have any sin's). But I don't understand how that would work in an inelastic collision (why would it be a ratio of M/m or m/M?)
So my guesses are 1. D and 2. B. (because there's an M and a L/2 involved)
I'm not just interested in just the answer, if someone could try to explain to me the general concept of how it works in a simple language I would really appreciate that. Thanks for your time!