Answer:
The original speed of the mess kit is 3.373 m/s
Explanation:
Given;
initial mass of mess kit, m = 3.6 kg
mass of mess kit after splitting, m₁ & m₂ = 1.8 kg
direction of the first half of mess kit, v₁ = 1.9 m/s, due north
direction of the second half of mess kit, v₂ = 6.0 m/s, 15° north of east
Apply principle of conservation of linear momentum
mv = m₁v₁ + m₂v₂
where;
v is the the original speed of the mess kit
vector component of the first final velocity given = v₁cos90i + v₁sin90j
vector component of the second final velocity given = v₂cos15°i + v₂sin15°j
mv = m₁v₁ + m₂v₂
mv = m₁(v₁cos90i + v₁sin90j) + m₂(v₂cos15°i + v₂sin15°j)
mv = m₁v₁(cos90i + sin90j) + m₂v₂(cos15°i + sin15°j)
v = m₁/m [v₁(cos90i + sin90j)] + m₂/m [v₂(cos15°i + sin15°j)]
v = 1.8/3.6 [1.9(cos90i + sin90j)] + 1.8/3.6 [6(cos15°i + sin15°j)]
v = 0.5(1.9j) + 0.5(5.796i + 1.553j)
v = 0.95j + 2.898i + 0.7765j
v = 2.898i + 1.7265j
[tex]v = \sqrt{v_x^2 + v_y^2} \\\\v = \sqrt{2.898^2 + 1.7265^2} \\\\v = 3.373 \ m/s[/tex]
Therefore, the original speed of the mess kit is 3.373 m/s
