A \(1000 .-\mathrm{kg}\) cannon shoots a \(30.0-\mathrm{kg}\) shell at an angle of \(25.0^{\circ}\) above the horizontal and a speed of \(500 . \mathrm{m} / \mathrm{s}\). What is the recoil velocity of the cannon?

Before firing the shell, both the cannon and the shell are stationary, so their initial momenta are zero.

To find the final momentum of the shell, first find its velocity components in the horizontal and vertical directions using the angle and speed of the projectile. The horizontal and vertical components of the shell's velocity are given by: Horizontal component: \(v_{x_{shell}} = v_{shell} \cos(25.0^{\circ})\) Vertical component: \(v_{y_{shell}} = v_{shell}\sin(25.0^{\circ})\) Substitute the given values for \(v_{shell}\) and angle to find \(v_{x_{shell}}\) and \(v_{y_{shell}}\). Now find the momentum of the shell in each direction by multiplying each velocity component by the mass of the shell: Momentum in x-direction: \(p_{x_{shell}} = m_{shell} v_{x_{shell}}\) Momentum in y-direction: \(p_{y_{shell}} = m_{shell} v_{y_{shell}}\)

Since the total momentum is conserved in both horizontal and vertical directions, the final momentum of the cannon will be equal in magnitude but opposite in direction to the final momentum of the shell. Momentum in x-direction: \(p_{x_{cannon}} = -p_{x_{shell}}\) Momentum in y-direction: \(p_{y_{cannon}} = -p_{y_{shell}}\)

Now we have the momentum of the cannon in both the x and y directions. To find the recoil velocity of the cannon, we can divide each momentum component by the mass of the cannon: Recoil velocity in x-direction: \(v_{x_{cannon}} = \frac{p_{x_{cannon}}}{m_{cannon}}\) Recoil velocity in y-direction: \(v_{y_{cannon}} = \frac{p_{y_{cannon}}}{m_{cannon}}\) Then, we can find the magnitude of the recoil velocity using the Pythagorean theorem: Recoil velocity: \(v_{recoil} = \sqrt{v_{x_{cannon}}^2 + v_{y_{cannon}}^2}\) Calculate and answer with the recoil velocity of the cannon.