A 3.02 kg object initially moving in the positive x direction with a velocity of 4.63 m s collides with and sticks to a 2.14 kg object initi

Question

A 3.02 kg object initially moving in the positive x direction with a velocity of 4.63 m s collides with and sticks to a 2.14 kg object initially moving in the negative y direction with a velocity of 2.99 m s Find the final components of velocity of the composite object Indicate the direction with the sign of your answer

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Adela 3 years 2021-08-24T14:27:27+00:00 1 Answers 28 views 0

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  1. thienhuong
    0
    2021-08-24T14:28:51+00:00
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    Answer:

    v = (2.71î + 1.24j) m/s

    Magnitude of v = 2.98 m/s

    Direction of v = 24.6°

    Explanation:

    Let the final velocity of the composite of both objects after collision be v.

    According to Newton’s second law of motion, in collisions, momentum is normally conserved

    Momentum before collision = Momentum after collision

    Momentum of object 1 before collision = (3.02)(4.63î) = (13.9826 î) kgm/s

    Momentum of object 2 before collision = (2.14)(2.99j) = (6.3986 j) kgm/s

    Momentum of both objects after collision = (3.02 + 2.14)(v) = 5.16v

    Momentum balance

    (13.9826î + 6.3986j) = 5.16v

    v = (2.71î + 1.24j) m/s

    Magnitide = √(2.71² + 1.24²)

    Magnitude of v = 2.98 m/s

    Direction of v = tan⁻¹ (1.24/2.71)

    Direction of v = 24.6°

    Hope this Helps!!!

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