Calculating Momentum of Gun-Fired Shell

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SUMMARY

The discussion focuses on calculating the momentum of a gun-firing shell using the conservation of momentum principle. The equation derived is v = v(1 + m)M, where M represents the mass of the gun, m is the mass of the shell, and v is the velocity of the shell. The initial momentum of the system is zero, leading to the conclusion that the final momentum must also equal zero, expressed as 0 = Mbullet * vbullet - Mgun * vgun. This establishes a clear relationship between the masses and velocities involved in the system.

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Oblio
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Using the conservation of momentum I need to show that a gun firing a shell will equal:

v=v(1+m)M

Where the gun of mass M shoots a shell with mass m and a velocity of v.
There are no external forces, ie the gun is free to recoil.
And its the velocity with respect to the ground...

This is what I've gotten to

v(m +M) = Mv(s) - mv(g)

v=Mv(s) - mv(g) / (m + M)
I'v done more, but I'm not if need to equal something to zero somewhere, since technically v (initial) is zero...

help?
 
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What exactly is the question asking... if the initial velocity of the system is 0, then the initial momentum of the system is 0.

So: initial momentum = final momentum

0 = Mbullet*vbullet - Mgun*vgun
 

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