A bullet with a mass m b = 11.5 g is fired into a block of wood at velocity v b = 249 m/s. The block is attached to a spring that has a spri

Question

A bullet with a mass m b = 11.5 g is fired into a block of wood at velocity v b = 249 m/s. The block is attached to a spring that has a spring constant k of 205 N/m. The block and bullet continue to move, compressing the spring by 35.0 cm before the whole system momentarily comes to a stop. Assuming that the surface on which the block is resting is frictionless, determine the mass of the wooden block.

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Neala 4 years 2021-08-09T11:29:08+00:00 1 Answers 116 views 0

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  1. Vodka
    0
    2021-08-09T11:30:20+00:00
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    Answer:

    0.358Kg

    Explanation:

    The potential energy in the spring at full compression = the initial kinetic energy of the bullet/block system

    0.5Ke^2 = 0.5Mv^2

    0.5(205)(0.35)^2 = 12.56 J = 0.5(M + 0.0115)v^2

    Using conservation of momentum between the bullet and the block

    0.0115(265) = (M + 0.0115)v

    3.0475 = (M + 0.0115)v

    v = 3.0475/(M + 0.0115)

    plugging into Energy equation

    12.56 = 0.5(M + 0.0115)(3.0475)^2/(M + 0.0115)^2

    12.56 = 0.5 × 3.0475^2 / ( M + 0.0115 )

    12.56 = 0.5 × 9.2872/ M + 0.0115

    12.56 = 4.6436/ M + 0.0115

    12.56 ( M + 0.0115 ) = 4.6436

    12.56M + 0.1444 = 4.6436

    12.56M = 4.6436 – 0.1444

    12.56 M = 4.4992

    M = 4.4992÷12.56

    M = 0.358 Kg

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