# Cannon Muzzle Velocity Question ( Solved )

1. Feb 12, 2015

### ChiHawksFan

1. The problem statement, all variables and given/known data
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?

2. Relevant equations
p=mv

p1=p2

KE= (1/2)mv2

mcvci + mpvpi = mcvcf + mpvpf
Where:
mc= Mass of the cannon, mp= Mass of the projectile, vci= Initial velocity of the cannon, vpi= Initial velocity of the projectile, vcf= Final velocity of the cannon, vpf= Final velocity of the projectile.

3. 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!

2. Feb 12, 2015

### ChiHawksFan

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

Last edited: Feb 12, 2015
3. Feb 12, 2015

### ChiHawksFan

Mass of Cannon = 500 lbs = 226.8 kg, Mass of Projectile = 20 lbs = 9.1 kg, vpi (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)(mc)(Vc)2 + (1/2)(mp)(vp)2
(mc = Mass of the cannon, mp = Mass of the projectile, vc = Velocity of the cannon, vp = 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)(mc)(Vc)2 + (1/2)(mp)(vp)2

After plugging in all of our known values, we are left with: 728,000 = 113.4vc2 + 4.55vp2

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

To do this, I used the equation mcvci + mpvpi = mcvcf + mpvpf

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

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

If we solve for vc, we are left with vc = .04 vp

Now that we have vc in terms of vp, we can plug it into our previous equation 728,000 = 113.4vc2 + 4.55vp2

After we do this, we are left with: 728,000 = .181vp2 + 4.55vp2 --------> 728,000 = 4.731vp2

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

I hope that this helps!

Last edited: Feb 12, 2015