A bullet is fired through a board 18.0 cm thick in such a way that the bullet’s line of motion is perpendicular to the face of the board. Th

Question

A bullet is fired through a board 18.0 cm thick in such a way that the bullet’s line of motion is perpendicular to the face of the board. The initial speed of the bullet is 420 m/s and it emerges from the other side of the board with a speed of 320 m/s. find
(a) the acceleration of the bullet as it passes through the board and (b) the total time the bullet is in contact with the board.

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Linh Đan 3 years 2021-08-30T11:31:40+00:00 2 Answers 3 views 0

Answers ( )

  1. thanhha
    0
    2021-08-30T11:33:25+00:00
    Reply

    d=18cm=0.18m thickness of board

    vi=420m/s speed before entering the board

    vf=320m/s speed when leaving the board

    vf²=vi²+2×a×d

    Acceleration a=(vf²-vi²)/2/d

    a=-205.5m/s².

    As expected, a is negative because the bullet is decelerated.

    vf=vi+a×t

    t=(vf-vi)/a=(320-420)/(-205.5)= 100/205.5=0.486s

    t=0.486s=486ms

  2. Latifah
    0
    2021-08-30T11:33:37+00:00
    Reply

    Answer:

    a) a=-205555.56\ m.s^{-2}

    b) t=4.865\times 10^{-5}\ s

    Explanation:

    Given:

    • thickness of the sheet, s=18\ cm=0.18\ m
    • initial speed of bullet before entering the board, u=420\ m.s^{-1}
    • final speed of the bullet after emerging form the board, v=320\ m.s^{-1}

    a)

    Using the equation of motion:

    v^2=u^2+2a.s

    320^2=420^2+2\times a\times0.18

    a=-205555.56\ m.s^{-2}

    b)

    Using the other form of equation of motion:

    v=u+a.t

    320=420-205555.56t

    t=4.865\times 10^{-5}\ s

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