A 2.0-kg object moving with a velocity of 5.0 m/s in the positive x direction strikes and sticks to a 3.0-kg object moving with a speed of 2

Question

A 2.0-kg object moving with a velocity of 5.0 m/s in the positive x direction strikes and sticks to a 3.0-kg object moving with a speed of 2.0 m/s in the same direction. How much kinetic energy is lost in this collision

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Doris 3 days 2021-07-22T07:16:55+00:00 1 Answers 0 views 0

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    2021-07-22T07:18:28+00:00

    Answer:

    5.4 J.

    Explanation:

    Given,

    mass of the object, m = 2 Kg

    initial speed, u = 5 m/s

    mass of another object,m’ = 3 kg

    initial speed of another orbit,u’ = 2 m/s

    KE lost after collusion = ?

    Final velocity of the system

    Using conservation of momentum

    m u + m’u’ = (m + m’) V

    2 x 5 + 3 x 2 = ( 2 + 3 )V

    16 = 5 V

    V = 3.2 m/s

    Initial KE = \dfrac{1}{2}mu^2 + \dfrac{1}{2}m'u'^2

                  = \dfrac{1}{2}\times 2\times 5^2 + \dfrac{1}{2}\times 3 \times 2^2

                  = 31 J

    Final KE = \dfrac{1}{2} (m+m')V^2 = \dfrac{1}{2}\times 5 \times 3.2^2 = 25.6 J

    Loss in KE = 31 J – 25.6 J = 5.4 J.

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Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )