Cannon Velocity after firing Cannonball

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    Cannon Velocity
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Homework Help Overview

The problem involves a stationary cannon firing a cannonball horizontally, requiring the calculation of the cannon's velocity after the cannonball is fired. The context is centered around the principles of momentum conservation in physics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the conservation of momentum, with some attempting to set up equations based on initial and final momentum. Questions arise regarding the calculations, particularly the use of the total mass in the momentum equation.

Discussion Status

There is an ongoing exploration of different interpretations of the momentum conservation principle. Some participants provide alternative formulations of the problem, while others express uncertainty about their understanding of the concepts involved.

Contextual Notes

Participants note the assumption of negligible friction and question the setup of the momentum equations, indicating a need for clarity on the definitions and relationships involved in the problem.

HarleyM
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Homework Statement


A stationary 2000 Kg cannon fires a 25 kg cannonball horizontally at 250 m/s. Find the velocity of the cannon after firing the cannonball . You may assume friction is negligible.


Homework Equations



M1V1+M2V2=(M1+M2)V'



The Attempt at a Solution



(25)(250)+ (2000)(0)= (2000+25)V'
V'=(25)(250)/(2025)
V'=3.086 m/s

I just need confirmation as I am not 100% positive with my physics... thanks !
 
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Why 2025?
I would write:
momentum before shot = momentum after shot
0 = mv + MV
0 = 25*250 + 2000*V
 
The total momentum of the situation before the firing is zero. This means that after the firing, the total must remain at zero. Find the momentum of the cannonball once it has been fired. The cannon must have equal momentum in the opposite direction for momentum to be conserved at zero kgm/s

i.e:

Before: (2000*0)+(25*0) = 0

After: (2000*V)-(25*250) = 0, or 2000V = 6250

Taking the cannons velocity to be positive, and the motion of the cannonball in the other direction to be negative. Should be trivial to solve from there (3.125 m/s).
 
Delphi51 said:
Why 2025?
I would write:
momentum before shot = momentum after shot
0 = mv + MV
0 = 25*250 + 2000*V

LOL because my understanding of momentum is limited.. this makes much more sense and gives a velocity backwards which is consistent.

Thanks!
 
Most welcome! It pays to write the general principle down, perhaps highly abbreviated, to orient yourself and your readers. I used to be too brief but had to change when I became a school teacher. Students couldn't understand unless I wrote the whole story.
 

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