A 1470-kilogram truck moving with a speed of 27.0 m/s runs into the rear end of a 1010-kilogram stationary car. If the collision is complete

Question

A 1470-kilogram truck moving with a speed of 27.0 m/s runs into the rear end of a 1010-kilogram stationary car. If the collision is completely inelastic, how much kinetic energy is lost in the collision

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Đan Thu 4 years 2021-08-25T11:04:30+00:00 1 Answers 24 views 0

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  1. minhkhang
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    2021-08-25T11:06:18+00:00
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    Answer:

    \Delta KE = 218.375\ kJ

    Explanation:

    Given,

    Mass of the truck, M =  1470 Kg

    initial speed of the truck, u = 27 m/s

    mass of car, m = 1010 Kg

    initial speed of car, u’ = 0 m/s

    Final speed, V = ?

    Using conservation of momentum

    Mu = (M+m) V

    1470\times 27 = (1470 + 1010 ) V

    V = 16\ m/s

    Change in kinetic energy

    \Delta KE = \dfrac{1}{2} Mu^2 - \dfrac{1}{2}(M+m)V^2

    \Delta KE = \dfrac{1}{2}\times 1470\times 27^2 - \dfrac{1}{2}\times (1470 + 1010)\times 16^2

    \Delta KE = 218.375\ kJ

    Change in KE = \Delta KE = 218.375\ kJ

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