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A pressure vessel at rest at the origin of an xy coordinate system explodes into three pieces that remain in the xy plane. Just after the explosion, one piece, of mass m, moves with velocity (−30 �/�)� and a second piece, also of mass m, moves with velocity (−30 �/�)�. The third piece has mass 3m. Just after the explosion, what is the velocity of the third piece? Use unit vector notation.

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jolis1796

Answer:

v₃ = (10) i + (10) j

v₃ = 10√2

Explanation:

Given info

Before the explosion

ux = 0

uy = 0

After the explosion

v₁x = -30

v₁y = 0

v₂x = 0

v₂y = -30

We can use the Principle of Conservation of Momentum as follows

pi = pf

where

pix = M*ux = M*0 = 0

piy = M*uy = M*0 = 0

pfx = p₁x + p₂x + p₃x = (m*v₁x + m*v₂x + 3m*v₃x) = m*(-30) + m*(0) + 3m*v₃x

⇒   pfx = -30m + 3m*v₃x

if

pix = pfx     ⇒    0 = -30m + 3m*v₃x    ⇒  v₃x = 10

pfy = p₁y + p₂y + p₃y = (m*v₁y + m*v₂y + 3m*v₃y) = m*(0) + m*(-30) + 3m*v₃y

⇒   pfy = -30m + 3m*v₃y

if

piy = pfy     ⇒    0 = -30m + 3m*v₃y    ⇒  v₃y = 10

then we have

v₃ = (10) i + (10) j

and its module can be obtained as follows

v₃ = √(v₃x² + v₃y²) = √(10² + 10²) = 10√2

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