A 22-g bullet traveling 240 m/s penetrates a 2.0-kg block of wood and emerges going 150 m/s. If the block is stationary on a frictionless su

Question

A 22-g bullet traveling 240 m/s penetrates a 2.0-kg block of wood and emerges going 150 m/s. If the block is stationary on a frictionless surface when hit, how fast does it move after the bullet emerges?

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Khang Minh 4 years 2021-08-06T01:26:27+00:00 1 Answers 341 views 0

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  1. Nho
    1
    2021-08-06T01:27:31+00:00
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    Answer:

    After the bullet emerges the block moves at 0.99 m/s

    Explanation:

    Given;

    mass of bullet, m₁ = 22 g = 0.022 kg

    initial speed of the bullet, u₁ = 240 m/s

    final speed of the bullet, v₁ = 150 m/s

    mass of block, m₂ = 2.0 kg

    initial speed of the block, u₂ = 0

    Let the final speed of the block = v₂

    Apply principles of conservation of linear momentum;

    m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

    0.022 x 240 + 2 x 0 = 0.022 x 150 + 2v₂

    5.28 = 3.3 + 2v₂

    5.28 – 3.3 = 2v₂

    1.98 = 2v₂

    v₂ = 1.98 / 2

    v₂ = 0.99 m/s

    Therefore, after the bullet emerges the block moves at 0.99 m/s

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