How Does a Cannon's Recoil Relate to Momentum Conservation?

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The discussion revolves around the conservation of momentum in the context of a 400kg cannon firing a cannonball at 20 m/s. The initial momentum of the system is zero since both the cannon and the ball are at rest. After firing, the momentum of the cannonball is calculated to be 8000 kg•m/s. The recoil speed of the cannon is derived from the conservation of momentum principle, ensuring the total momentum remains constant. The participant confirms they received assistance from a teacher to clarify their calculations.
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a 400kg cannon fires a cannon ball that exits at a speed of 20 m/s.

A) what was the momentum of the cannon and ball before it was fired ?

B) what is the momentum of the cannon ball alone after it was fired ?

C) what is the speed of the cannon as it recoils ?

Given
m = 400kg
vƒ (ƒ = final) = 20 m/s
vo= 0 m/s
(o = initial, that's the symbol i use for initial :P )
since the cannon ball was sitting in the cannon. so the initial velocity is 0 since
its sitting at rest.

Unknown
p (before)
p (after)
speed

Equations
v = p total / m

Δpƒ = m₁v₂ - m₁v₁ ( is this equation right ? )

also if it recoils meaning the momentum after would be the same as before ?
due to -p = p ?

so it would be like this..

Δp = 400kg(20m/s) - 400kg(0 m/s)
Δp = 8000 kg•m/s

then...

v= 8000 kg•m/s
400kg

v = 20 m/s

i want to make sure i did it right && i think i didn't answer all the questions correctly ; - ;..
 
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Hi TeaWay! Welcome to PF! :smile:

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erm :redface:what is the mass of the ball? :confused:
 
oh uh oops nvm about this question..sorry. i got help on solving this question after school with my teacher today :DD ! but thanks for the warm welcome lol.but thank you for trying to help out (: !
 
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