# Hockey stick and puck

1. Nov 9, 2013

1. The problem statement, all variables and given/known data
A hockey stick of mass $m_{s}$ and length $L$ is at rest on the ice (which is assumed to be frictionless). A puck with mass $m_{p}$ hits the stick a distance $D$ from the middle of the stick. Before the collision, the puck was moving with speed $v_{0}$ in a direction perpendicular to the stick, as indicated in the figure. The collision is completely inelastic, and the puck remains attached to the stick after the collision.

Part A
Find the speed $v_{f}$ of the center of mass of the stick+puck combination after the collision.
Express $v_{f}$ in terms of the following quantities: $v_{0}, m_{p}, m_{s}, \text{and } L$.

Part B
After the collision, the stick and puck will rotate about their combined center of mass. How far is this center of mass from the point at which the puck struck? In the figure, this distance is $(D−b)$.

Part C
What is the angular momentum $L_{cm}$ of the system before the collision, with respect to the center of mass of the final system?
Express $L_{cm}$ in terms of the given variables.

Part D
What is the angular velocity $\omega$ of the stick+puck combination after the collision? Assume that the stick is uniform and has a moment of inertia $I_{0}$ about its center.
Your answer for $\omega$ should not contain the variable b.

2. Relevant equations

Part A velocity of centre of mass $u_{cm}=\frac{m_{1}u_{1}+m_{2}u_{2}}{m_{1}+m_{2}}$

Part B equation of CoM $r_{cm}=\frac{m_{1}r_{1}+m_{2}r_{2}}{m_{1}+m_{2}}$

Part C Angular momentum $\vec{L}=\vec{r}\times\vec{p}$

Part D ???

3. The attempt at a solution
Part A substituting in gives $u_{cm}=\frac{0m_{s}+m_{p}v_{0}}{m_{s}+m_{p}}=\frac{m_{p}v_{0}}{m_{s}+m_{p}}=v_{f}$ because the momentum of the CoM remains unchanged.

Part B We start from the point where the ball strikes the stick, as we already know the distance to the combined CoM from there, $D-b$, so $D-b=\frac{Dm_{s}+0m_{p}}{m_{s}+m_{p}}=\frac{Dm_{s}}{m_{s}+m_{p}}$

Part C the perpendicular distance is just $D-b$ so $L=(D-b)m_{p}v_{0}=m_{p}v_{0}\frac{Dm_{s}}{m_{s}+m_{p}}$

Part D I have no idea where to go here, so any help would be greatly appreciated!

2. Nov 9, 2013