According to the law of conservation of linear momentum, the total momentum of an isolated system remains constant. Before firing, both the gun and the bullet are at rest, so the initial momentum of the system is zero .

After firing, the final momentum must also be zero: $P_{\text{initial}} = P_{\text{final}}$ $0 = m_b v_b + m_g v_g$

Given: Mass of bullet, $m_b = 200 \text{ g} = 0.2 \text{ kg}$ Velocity of bullet, $v_b = 5 \text{ m/s}$ Mass of gun, $m_g = 1 \text{ kg}$ Recoil velocity of gun $= v_g$

Substituting the values into the equation: $0 = (0.2 \text{ kg} \times 5 \text{ m/s}) + (1 \text{ kg} \times v_g)$ $0 = 1 + v_g$ $v_g = -1 \text{ m/s}$

The negative sign indicates that the gun rebounds in the opposite direction to the bullet. Therefore, the magnitude of the recoil velocity is $1 \text{ m/s}$.