A 1. 2-kg spring-activated toy bomb slides on a smooth surface along the x-axis with a speed of 0. 50 m/s. At the origin 0, the bomb explodes into two fragments. Fragment 1 has a mass of 0. 40 kg and a speed of 0. 90 m/s along the negative y-axis. In the figure, the angle θ, made by the velocity vector of fragment 2 and the x-axis, is closest to.

The velocity vector of fragment 2 80.54° and the x-axis is 0.075.Identify:m = 1.2 kg (mass of the spring)v = 0.05 m/s (speed of the spring)m₁ = 40 kg (mass of the first fragment)u₁ = 0.9 m/s (initial speed of first fragment)m₂ = 0.8 kg (mass of the second fragment)Applying the theory of linear momentum conservation, determine the velocity of the second segment:- Find the x-axis:mucos θ = m₁ v₁ cos θ + m₂ v₂ cos θ1.2 x 0.05 x cos (0) = 0 + 0.8 v₂ cos θv₂ cos θ = [tex]\frac{0.06}{0.8}[/tex]v₂ cos θ = 0.075- Find the y-axis:1.2 (0) = -0.4 (0.9) + 0.8 (v₂ sin θ)v₂ sin θ = [tex]\frac{0.36}{0.8}[/tex]v₂ sin θ = 0.45- Solve the  equation:[tex]\frac{V_{2} sin0}{V_{2} cos0}[/tex] = [tex]\frac{0.45}{0.075}[/tex]tan θ = 6θ = [tex]tan^{-1}(6)[/tex]θ = 80.54°- The value of velocity:v₂ sin θ = 0.45V₂ = [tex]\frac{0.45}{sin0}[/tex]V₂ = [tex]\frac{0.45}{sin (80.54)}[/tex]V₂ = 0.46 m/sLearn more about velocity fragment here:https://brainly.com/question/19979064#SPJ4