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27 December, 15:29

A bullet of mass 0.1 kg traveling horizontally at a speed of 100 m/s embeds itself in a block of mass 3 kg that is sitting at rest on a nearly frictionless surface. (a) What is the speed of the block after the bullet embeds itself in the block? v = m/s (b) Calculate the kinetic energy of the bullet plus the block before the collision: Ki = J (c) Calculate the kinetic energy of the bullet plus the block after the collision: Kf = J

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  1. 27 December, 15:43
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    (a) the speed of the block after the bullet embeds itself in the block is 3.226 m/s

    (b) the kinetic energy of the bullet plus the block before the collision is 500J

    (c) the kinetic energy of the bullet plus the block after the collision is 16.13J

    Explanation:

    Given;

    mass of bullet, m₁ = 0.1 kg

    initial speed of bullet, u₁ = 100 m/s

    mass of block, m₂ = 3 kg

    initial speed of block, u₂ = 0

    Part (A)

    Applying the principle of conservation linear momentum, for inelastic collision;

    m₁u₁ + m₂u₂ = v (m₁ + m₂)

    where;

    v is the speed of the block after the bullet embeds itself in the block

    (0.1 x 100) + (3 x 0) = v (0.1 + 3)

    10 = 3.1v

    v = 10/3.1

    v = 3.226 m/s

    Part (B)

    Initial Kinetic energy

    Ki = ¹/₂m₁u₁² + ¹/₂m₂u₂²

    Ki = ¹/₂ (0.1 x 100²) + ¹/₂ (3 x 0²)

    Ki = 500 + 0

    Ki = 500 J

    Part (C)

    Final kinetic energy

    Kf = ¹/₂m₁v² + ¹/₂m₂v²

    Kf = ¹/₂v² (m₁ + m₂)

    Kf = ¹/₂ x 3.226² (0.1 + 3)

    Kf = ¹/₂ x 3.226² (3.1)

    Kf = 16.13 J
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