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A vessel at rest explodes, breaking into three pieces. two pieces, having equal mass, fly off perpendicular to one another with the same speed of 60 m/s. One goes along the negative x-axis and the other along the negative y-axis. The third piece has three times the mass of each other piece. What are the direction and magnitude of its velocity immediately after the explosion? What is the angle with respect to the x-axis? What is the magnitude of its velocity?

Respuesta :

jolis1796

Answer:

v₃ = 28.2842 m/s

∅ = 45°

Explanation:

Given info

vi = 0 m/s

m₁ = m₂ = m

m₃ = 3m

mi = m₁ + m₂ + m₃ = m + m + 3m = 5m

v₁x= - 60 m/s

v₁y= 0 m/s

v₂x= 0 m/s

v₂y= - 60 m/s

We can apply the Principle of Conservation of Momentum as follows

pix = pfx   ⇒   mi*vix = m₁*v₁x + m₂v₂x + m₃*v₃x

⇒  5m*(0) = m*(-60) + m*(0) + 3m*v₃x

⇒  0 = -60*m + 3m*v₃x     ⇒    v₃x = 20 m/s (→)     (I)

piy = pfy   ⇒   mi*viy = m₁*v₁y + m₂v₂y + m₃*v₃y

⇒  5m*(0) = m*(0) + m*(-60) + 3m*v₃y

⇒  0 = -60*m + 3m*v₃y    ⇒    v₃y = 20 m/s (↑)     (II)

then

v₃ = √(v₃x² + v₃y²)

⇒  v₃ = √((20 m/s)² + (20 m/s)²) = 20√2 m/s = 28.2842 m/s

is the magnitude of its velocity immediately after the explosion

∅ = tan⁻¹(v₃y / v₃x)

⇒ ∅ = tan⁻¹(20 m/s / 20 m/s) = tan⁻¹(1) = 45°

is the direction of its velocity immediately after the explosion (the angle with respect to the x-axis)