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

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The discussion focuses on the principles of momentum conservation as applied to a 400 kg cannon firing a cannonball at 20 m/s. The initial momentum of the system is zero, and after firing, the momentum of the cannonball is calculated to be 8000 kg•m/s. The recoil speed of the cannon is derived using the conservation of momentum principle, confirming that the momentum before and after firing remains equal, thus validating the calculations presented.

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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 ; - ;..
 
Last edited:
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Hi TeaWay! Welcome to PF! :smile:

(try using the X2 icon just above the Reply box :wink:)

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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