A 30-g bullet is fired with a horizontal velocity of 450 m/s and becomes embedded in block B, which has a mass of 3 kg. After the impact, bl

Question

A 30-g bullet is fired with a horizontal velocity of 450 m/s and becomes embedded in block B, which has a mass of 3 kg. After the impact, block B slides on 30-kg carrier C until it impacts the end of the carrier. Knowing the impact between B and C is perfectly plastic and the coefficient of kinetic friction between B and C is 0.2, determine (a) the velocity of the bullet and B after the first impact, (b) the final velocity of the carrier.

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Dulcie 5 years 2021-08-13T22:04:18+00:00 1 Answers 61 views 0

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  1. Ben
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    2021-08-13T22:06:17+00:00
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    Answer:

    (a) The velocity of the bullet and B after the first impact is 4.4554 m/s.

    (b) The velocity of the carrier is 0.40872 m/s.

    Explanation:

    (a) To solve the question, we  apply the principle of conservation of linear momentum as follows.

    we note that the distance between B and C is 0.5 m

    Then we  have

    Sum of initial momentum = Sum of final momentum

    0.03 kg × 450 m/s = (0.03 kg + 3 kg) × v₂

    Therefore v₂ = (13.5 kg·m/s)÷(3.03 kg) = 4.4554 m/s

    The velocity of the bullet and B after the first impact = 4.4554 m/s

    (b) The velocity of the carrier is given as follows

    Therefore from the conservation of linear momentum we also have

    (m₁ + m₂)×v₂  = (m₁ + m₂ + m₃)×v₃

    Where:

    m₃ = Mass of the carrier = 30 kg

    Therefore

    (3.03 kg)×(4.4554 m/s) = (3.03 kg+30 kg) × v₃

    v₃ = (13.5 kg·m/s)÷ (33.03 kg) = 0.40872 m/s

    The velocity of the carrier = 0.40872 m/s.

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