Answer:
[tex]v=339.3217\,m.s^{-1}[/tex]
Explanation:
mass of the bullet, [tex]m= 2.5\,g[/tex]
muzzle velocity, [tex]v=380\,m.s^{-1}[/tex]
total thickness of the board on which the target is taken, [tex]s=1.5 \times 25.4 \,mm= 38.1\times 10^{-3}\,m [/tex]
average stopping force by the board, [tex]F=960 \,N[/tex]
We know,
[tex]F=m.a[/tex]
[tex]960=2.5\times 10^{-3}\times a[/tex]
[tex]a= 3.84\times 10^5 \,m.s^{-2}[/tex]
This "a" is the deceleration of bullet in the board material.
Now using the equation of motion,
[tex]v^2=u^2-2a.s[/tex]
-ve sign, because a is deceleration
[tex]v^2=380^2-2\times 3.84\times 10^5\times 38.1\times 10^{-3}[/tex]
[tex]v=339.3217\,m.s^{-1}[/tex]
is the velocity when it exits the board material.
