Cannon Muzzle Velocity Question ( Solved )

[ChiHawksFan]

Homework Statement

A cannon that weighs 500 lbs shoots a cannonball horizontally that weighs 20 lbs at 400 m/s while mounted in place so that it does not move. If the same cannon and cannonball are shot in the same way with the cannon mounted on wheels allowing it to recoil, what is the new muzzle velocity?

Homework Equations

p=mv

p₁=p₂

KE= (1/2)mv²

m_cv_ci + m_pv_pi = m_cv_cf + m_pv_pf
Where:
m_c= Mass of the cannon, m_p= Mass of the projectile, v_ci= Initial velocity of the cannon, v_pi= Initial velocity of the projectile, v_cf= Final velocity of the cannon, v_pf= Final velocity of the projectile.

The Attempt at a Solution

I am really quite lost here. I have tried a bunch of stuff, but can't seem to get the answer. I tried finding the velocity of the cannon's recoil and using it to figure out the new muzzle velocity, but I got an answer that was too low. I have also tried using KE₁=KE₂, but I don't think that that formula is applicable because I don't think that energy is conserved. Thanks for any help!

 

[ChiHawksFan]

Actually, I just solved this problem, but I will show my solution incase someone else might find this helpful.

 

[ChiHawksFan]

Mass of Cannon = 500 lbs = 226.8 kg, Mass of Projectile = 20 lbs = 9.1 kg, v_pi (Initial muzzle velocity) = 400 m/s

First, I calculated the KE of the projectile in the first situation in which the cannon remains stationary.

KE = (1/2)mv² --------> KE = (1/2)(9.1)(400)² -------> KE = 728,000 J

Then, I used the equation KE₁ = KE₂. This is possible because the same amount of energy is used in both situations. However, in the second situation the cannon has recoil, so it moves in addition to the bullet. Due to this, the right side of the above equation will look like this.

KE = (1/2)(m_c)(V_c)² + (1/2)(m_p)(v_p)²
(m_c = Mass of the cannon, m_p = Mass of the projectile, v_c = Velocity of the cannon, v_p = Velocity of the projectile)

Now, we can set our kinetic energy from the first situation equal to the above equation to get: 728,000 = (1/2)(m_c)(V_c)² + (1/2)(m_p)(v_p)²

After plugging in all of our known values, we are left with: 728,000 = 113.4v_c² + 4.55v_p²

The next step is to put v_c in terms of v_p in order to figure out the muzzle velocity in the second situation.

To do this, I used the equation m_cv_ci + m_pv_pi = m_cv_cf + m_pv_pf

Since nothing is moving before the canon is fired, the initial momentum is 0. This leaves us with 0 = m_cv_cf + m_pv_pf

Next, we will plug in our known values to get 0 = 226.8v_c + 9.1v_p

If we solve for v_c, we are left with v_c = .04 v_p

Now that we have v_c in terms of v_p, we can plug it into our previous equation 728,000 = 113.4v_c² + 4.55v_p²

After we do this, we are left with: 728,000 = .181v_p² + 4.55v_p² --------> 728,000 = 4.731v_p²

If we solve for v_p, we get that v_p = 392.3 m/s, which is the correct answer.

I hope that this helps!