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 energy released by the explosion is closest to.

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smecloudbird18 smecloudbird18 · Answer 1 · 2022-11-14T07:15:32+01:00

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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