Answer :
Fragment 2's velocity vector is 80.54°, and the x-axis is at 0.075 and the velocity of the y- axis is 0.46m/s.
We are given that,
Mass of the spring = m = 1.2kg
Speed of the spring = v = 0.05 m/s
Mass of the first fragment = m₁ = 40 kg
Initial speed of first fragment =u₁ = 0.9 m/s
m₂ = 0.8 kg
To find the second segment's velocity using the idea of linear momentum conservation,
Locate the x-axis,
m u cos θ = m₁ v₁ cos θ + m₂ v₂ cos θ
1.2 x 0.05 x cos (0) = 0 + 0.8 v₂ cos θ
v₂ cosθ = 0.075
Find the y-axis:
1.2 (0) = -0.4 (0.9) + 0.8 (v₂ sin θ)
v₂ sinθ = 0.36/0.8
v₂ sinθ = 0.45
By solving above two equation we can get ,
tanθ = 6.0
θ = 80.54°
Putting the value of θ in y- axis component then we get,
v₂ sinθ = 0.45
v₂ = 0.45/sinθ
v₂ = 0.46m/s
Therefore , fragment 2's velocity vector is 80.54°, and the x-axis is at 0.075 and the velocity of the y- axis is 0.46m/s.
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