Cannon shoots a bullet. What if it's not fixed to the platform?

[Ariano AnnaG]

Homework Statement

A cannon is initially attached to a fixed platform (ignore the recoil). Its mass is 5800 kg.
The cannon shoots a bullet(mass=85 kg), and its speed is 551 m/s.
If the cannon is not fixed to the platform anymore, what is the speed of the bullet now?


Is the attempt at the solution shown in the attached photo well organized and correct? Should I add other explanation or diagrams to improve it?

Relevant Equations

K=K' (kinetic energy)
q=q' (momentum)

 

[kuruman]

First it's not easy to read your solution with all the squares interfering. Please use plain white paper if you post more photos of your work.

Your solution is not easy to follow but it is correct. You may consider using sentences here and there to explain what you are doing. For example, you might say something like "The same amount of energy that is imparted to only the bullet in the first case is divided between the bullet and the cannon in the second case."

I congratulate you on deriving the expression in terms of symbols before substituting the numbers. That's as it should be.

 

[haruspex]

I congratulate you on deriving the expression in terms of symbols before substituting the numbers. That's as it should be.

I second that, but I think you have $v_{B_2}$ on the right instead of $v_{B_1}$, and you could simplify to $\frac{v_{B_1}}{\sqrt{1+\frac{m_B}{m_C}}}$.

 

[Ariano AnnaG]

Thank you so much for your replies kuruman and haruspex despite the wrong organization of the post, I'm so sorry about that, I'm new to the forum but I promise I will learn from my mystakes.