A bullet of mass 5 g travels with a speed of 500 `ms^(-1)` .if it penetrates a fixed target which offers a constant resistive force of 1000 N to the motion of the bullet ,find ,(a) the intial Kinetic energy of the bullet penetrated before coming to rest,and (c)the speed with which the bullet emerges out of the target if target is of thickness 0.5 m.

Explanation

Given ,m=5 g=`5xx10^(-3)` kg,v=500 `ms^(-1)` F=1000 N
(a)Kinetic energy of the bullet `=(1)/(2)mv^(2)`
`=(1)/(2)xx(5xx10^(-3))xx(500)^(2)=625 J`
(b)Let the bullet penetrate through a distance S m in the target .
Work is obtained the initial kinetic energy of the bullet . `therefore` 1000 S=625
or S=625/1000 =0.625 m
Thus,the distance penetrated by the bullet =0.625 m
(C)Energy spent against the resistive force offered by the target in penetrating through it =1000 N `xx` 0.5 m =500 J
`therefore ` Kinetic energy left with the bullet on emerging out of the target =625 J-500 J=125 J
If the speed of bullet is now v,then kinetic energy =`(1)/(2)mv^(2)`
`therefore (1)/(2)mv^(2)=125 J` or `(1)/(2)xx(5xx10^(-3))v^(2)=125`
or `v.=sqrt((2xx125)/(5xx10)^(-3))=223.6 ms^(-1)`