Since the initial linear momentum of the system is 0, the linear momentum of the gun-arm system must be equal in magnitude and opposite in direction to the linear momentum of the bullet.
Find the linear momentum of the bullet:
![\begin{gathered} p_{\text{bullet}}=m_{\text{bullet}}v_{\text{bullet}} \\ =(0.038kg)(385(m)/(s)) \\ =14.63\operatorname{kg}\cdot(m)/(s) \end{gathered}]()
If the gun-arm system recoils with a speed v, the total linear momentum of the gun-arm system is:
![(3.3\operatorname{kg}+15.0\operatorname{kg})v]()
Since the linear momentum of the gun-arm system must have the same magnitude as the linear momentum of the bullet, then:
![(3.3\operatorname{kg}+15.0\operatorname{kg})v=14.63\operatorname{kg}\cdot(m)/(s)]()
Isolate v and find its value:
![\begin{gathered} v=\frac{14.63\operatorname{kg}\cdot(m)/(s)}{3.3\operatorname{kg}+15.0\operatorname{kg}} \\ =0.79945\ldots(m)/(s) \\ \approx0.8(m)/(s) \end{gathered}]()
Therefore, the recoil speed of the shotgun and arm-shoulder combination is 0.8 meters per second.