A 6.4 kg block with a speed of 5.4 m/s collides with a 12.8 kg block that has a speed of 3.6 m/s in the same direction. After the collision,

Question

A 6.4 kg block with a speed of 5.4 m/s collides with a 12.8 kg block that has a speed of 3.6 m/s in the same direction. After the collision, the 12.8 kg block is observed to be traveling in the original direction with a speed of 4.5 m/s. (a) What is the velocity of the 6.4 kg block immediately after the collision

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Kiệt Gia 5 years 2021-07-14T20:17:46+00:00 1 Answers 13 views 0

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  1. quangkhai
    0
    2021-07-14T20:19:07+00:00
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    Answer:

    The velocity of the 6.4 kg block immediately after the collision is 3.6 m/s in the original direction.

    Explanation:

    Given;

    mass of first block, m₁ = 6.4 kg

    initial speed of first block, u₁ =  5.4 m/s

    mass of second block, m₂ = 12.8 kg

    initial speed of second block, u₂ = 3.6 m/s

    final speed of second block, v₂ = 4.5 m/s

    To determine the final speed of 6.4 kg block immediately after the collision, we apply principle of conservation linear momentum;

    Total momentum before collision = Total momentum after collision

    m₁u₁  +  m₂u₂ = m₁v₁  + m₂v₂

    where;

    v₁ is the final speed of 6.4 kg block

    v₂ is the final speed of 12.8 kg block

    Substitute the given values of m₁, u₁, m₂,  u₂ , v₂ and calculate  v₁

    6.4 x 5.4  +  12.8 x 3.6 = 6.4v₁  +  12.8 x 4.5

    34.56 + 46.08  = 6.4v₁ + 57.6

    80.64 = 6.4v₁  +  57.6

    80.64 – 57.6 = 6.4v₁

    23.04 = 6.4v₁

    v₁ = 23.04 / 6.4

    v₁  = 3.6 m/s

    Therefore, the velocity of the 6.4 kg block immediately after the collision is 3.6 m/s in the original direction.

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