Project 14 Solution: Part A: Showdown at boar tiger corral

In order to determine the speed of the center of mass, we have to use momentum conservation. Accordingly, we must have \begin{eqnarray*}p_{i}&=&p_{f}\\mv&=&(m+M)v_{\rm cm},\end{eqnarray*} where $m$ is the mass of the cannon ball, $v$ is the velocity of the cannon ball, $M$ is the mass of the block. Thus, we can solve for the center of mass speed $$v_{\rm cm}=\frac{mv}{m+M}.$$

Now, in order to determine the speed of the end of the block with respect to the center of mass motion as it spins, we have to use angular momentum conservation. If we assume that the cannon ball strikes the rectangular block at its end, we must have \begin{eqnarray*}L_{i}&=&L_{f}\\mvr&=&\tilde{I}\omega\\mv(L/2)&=&\big[m(L/2)^{2}+I\big]\frac{v_{\rm end}}{L/2},\end{eqnarray*} where $$I=\frac{1}{12}*M\big(L^{2}+W^{2})$$ is the moment of inertia of a block about its geometric center along the height axis with length $L$ and width $W$ (they should Google this), and we have used the fact that $v_{\rm end}=\omega(L/2)$.

Thus, we can solve for the speed of the end of the block with respect to the center of mass $$v_{\rm end}=\frac{mL^{2}v}{4I+mL^{2}}$$ Now, assuming that the contraption will strike the boar tigers at its corner, they will see a different velocity depending on their location with respect to the center of mass: $$\vec{v}_{\rm tot}=\vec{v}_{\rm cm}+\vec{v}_{\rm end}$$

Tutor Questions:

Question: What system have you chosen?
Expected Answer: The cannon ball and the block.

Question: Why does conservation of linear momentum work here?
Expected Answer: Since the cannon ball and the block are in the system, we have no external forces to cause a change in momentum.

Question: Where is the axis of rotation of the cannon ball and block system?
Expected Answer: We assume that the center of mass of the cannon ball and block system is not significantly different from that of the block alone. Thus, the cannon ball and block system rotates about the geometric center of the block.

Tutor Question: Why does conservation of angular momentum work here?
Expected Answer: Again, considering the cannon ball and block in the system, we have no external torque to cause a change in angular momentum.

Tutor Question: Does energy conservation work here?
Expected Answer: No, we can't use energy conservation because of the inelastic collision of the cannon ball and block.

Tutor Question: What motion will the end of a block trace out (draw it on the white board) as it moves towards the boar tigers?
Expected Answer: …

Tutor Question: How does the speed of the center of mass and the speed of the end of the block depend on where you shoot the cannon ball at the block? What would happen if you shoot the cannon ball at the center of the block?
Expected Answer: The further out you shoot the cannon ball, the faster the rotation. If you shoot the cannon ball at the center of the block, there will be no torque and so no rotation. The center of mass speed is independent of the location of the collision.

Tutor Question: How does the speed of each point on the block due to rotation compare to its neighbor?
Expected Answer: Each point on the block is moving faster (going through a larger distance in the same amount of time) as you move away from the axis of rotation.

Tutor Question: What is the maximum velocity with which the boar tigers can be struck?
Expected Answer: This occurs when the length axis is perpendicular to the velocity of the center of mass, where we have the simple sum (at least on one side of the block) $v_{\rm max}=v_{\rm cm}+v_{\rm end}$.

QR code for conceptual questions.