A 3.6 kg mess kit sliding on a frictionless surface explodes into two 1.8 kg parts, one moving at 1.9 m/s, due north, and the other at 6.0 m

Question

A 3.6 kg mess kit sliding on a frictionless surface explodes into two 1.8 kg parts, one moving at 1.9 m/s, due north, and the other at 6.0 m/s, 15° north of east. What is the original speed of the mess kit?

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Gia Bảo 4 years 2021-08-05T03:30:51+00:00 1 Answers 20 views 0

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  1. thuthuy
    0
    2021-08-05T03:32:29+00:00
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    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

    v = \sqrt{v_x^2 + v_y^2} \\\\v = \sqrt{2.898^2 + 1.7265^2} \\\\v = 3.373 \ m/s

    Therefore, the original speed of the mess kit is 3.373 m/s

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