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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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Danboghiu66

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

Answer:

a) [tex]a=-205555.56\ m.s^{-2}[/tex]

b) [tex]t=4.865\times 10^{-5}\ s[/tex]

Explanation:

Given:

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

a)

Using the equation of motion:

[tex]v^2=u^2+2a.s[/tex]

[tex]320^2=420^2+2\times a\times0.18[/tex]

[tex]a=-205555.56\ m.s^{-2}[/tex]

b)

Using the other form of equation of motion:

[tex]v=u+a.t[/tex]

[tex]320=420-205555.56t[/tex]

[tex]t=4.865\times 10^{-5}\ s[/tex]

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