Conservation of momentum (relative speed)

[weirdlycool]

Homework Statement

Consider a gun of mass M (when unloaded) that fires a shell of mass m with muzzle speed v. (That is, the shell's speed relative to the gun is v.) Assuming that the gun is completely free to recoil (no external forces on gun or shell), use conservation of momentum to show that the shell's speed relative to the ground is $\frac{v}{(1+ \frac{m}{M})}$

Homework Equations

$p=mv$
$m_1 v_1 = m_2 v_2$
$M =$ mass of the gun
$m =$ mass of the shell
$v =$ speed of the shell relative to the gun's muzzle
$v_g =$ shell's speed relative to the ground

The Attempt at a Solution

$p_{initial} = Mv$
$p_{final} = ( M + m ) v_g$
I'm not sure if the initial and final momentum is correct
$Mv = ( M + m ) v_g$
$v = (1 + \frac{m}{M}) v_g$
$v_g = \frac{v}{(1+ \frac{m}{M})}$

 

[Whitehole]

Your equation is wrong.

 

[Figaro]

In conservation of momentum, you have to always identify first what is the initial state and the final state then you will know how to construct the initial and final momentum. Now, the initial state is when you are NOT yet firing the gun then the final state is when you already fired the gun with the bullet having a speed v. So, what is your initial momentum given that you haven't fired the gun yet? Also after firing, is the bullet and gun together? You should also construct the relationship of the velocities in vector form.

v(gun rel ground) + v(bullet rel gun) = v(bullet rel ground)

 

[weirdlycool]

In conservation of momentum, you have to always identify first what is the initial state and the final state then you will know how to construct the initial and final momentum. Now, the initial state is when you are NOT yet firing the gun then the final state is when you already fired the gun with the bullet having a speed v. So, what is your initial momentum given that you haven't fired the gun yet? Also after firing, is the bullet and gun together? You should also construct the relationship of the velocities in vector form.

v(gun rel ground) + v(bullet rel gun) = v(bullet rel ground)

Thanks, I think I got it already. The initial momentum is 0
$p_{initial} =$ 0
$p_{final} =$ M v(gun rel ground) + m v(bullet rel ground) therefore,
0 = M [ v(bullet rel ground) - v(bullet rel gun) ] + m v(bullet rel ground)
m v(bullet rel ground) = - M v(bullet rel ground) + M v(bullet rel gun)
( m + M ) v(bullet rel ground) = M v(bullet rel gun)

$v(bullet \ rel \ ground) = \frac{v(bullet \ rel \ gun)}{(\frac{m}{M}+1)}$

 

[Figaro]

That's right.