Calculating Momentum of Gun-Fired Shell

[Oblio]

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?

 

[learningphysics]

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

 

[m2003]

I can confirm that your calculation is correct. In order to show that the gun firing a shell will result in an equal velocity, we can use the conservation of momentum principle, which states that the total momentum before an event (in this case, the gun firing the shell) is equal to the total momentum after the event.

In this scenario, the initial momentum is zero since the gun and the shell are at rest before the firing. Therefore, we can set the initial momentum equal to the final momentum:

0 = Mv(i) + mv(i) = Mv(f) + mv(f)

Where v(i) is the initial velocity and v(f) is the final velocity.

Since we are interested in the final velocity of the gun and the shell combined, we can rearrange the equation to solve for v(f):

Mv(f) = -mv(f)

And dividing both sides by (m+M), we get:

v(f) = -mv(f) / (m+M)

This is the same as the equation you have derived, except for the negative sign. This negative sign indicates that the gun and the shell will have opposite velocities, as expected from the conservation of momentum principle.

To address your concern about equating something to zero, note that the initial momentum is already equal to zero, so there is no need to set anything equal to zero in this case.

I hope this helps clarify your understanding of the conservation of momentum and its application to this scenario. Keep up the good work!