There are three known forms of uranium. These forms are called _____ of each other.
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Answer:
The magnitude of the final velocity is approximately 0.526 m/s in approximately the direction of 8.746° East of South
Explanation:
The given collision parameters are;
The kind of collision experienced by the four velcro-lined air-hockey disk = Inelastic collision
The mass of the first disk, m₁ = 50.0 g
The velocity of the first disk, v₁ = 0.80 m/s West = -0.8·i
The mass of the second disk, m₂ = 60.0 g
The velocity of the second disk, v₂ = 2.50 m/s North = 2.5·j
The mass of the third disk, m₃ = 100.0 g
The velocity of the third disk, v₃ = 0.20 m/s East = 0.20·i
The mass of the fourth disk, m₄ = 40.0 g
The velocity of the fourth disk, v₄ = 0.50 m/s South = -0.50·j
Therefore, the total initial momentum of the four velcro-lined air-hockey disk, is given as follows;
= m₁·v₁ + m₂·v₂ + m₃·v₃ + m₄·v₄ = 50.0×(-0.80·i) + 60.0×(2.50·j) + 100 × (0.20·i) + 40.0 × (-0.50·j)
∴ = -40·i + 150·j + 20·i - 20·j = -20·i + 130·j
∴ = -20·i + 130·j
By the law of conservation of linear momentum, we have;
= -20·i + 130·j
Therefore, given that the collision is perfectly inelastic, the disks move as one after the collision and the four masses are added to form one mass, "m", m = m₁ + m₂ + m₃ + m₄ = 50.0 + 60.0 + 100.0 + 40.0 = 250.0
∴ m = 250.0 g
Let, "v" represent the final velocity of the four disks moving as one after the collision
We have;
= m × v = 250.0 × v = -20·i + 130·j
∴ v = -20·i/250 + 130·j/250 = -0.08·i + 0.52·j
The final velocity of the four disks after collision, v = -0.08·i + 0.52·j
The magnitude of the final velocity, = √((-0.08)² + (0.52)²) ≈ 0.526
≈ 0.526 m/s
The direction of the final velocity, θ = arctan(0.52/(-0.08)) ≈ -81.254°
The direction of the final velocity, θ ≈ -81.254° which is 8.746° East of South