A bullet, of mass 50 g and of initial speed `400 m*s^(-1)`, penetrates a wall against an average force of `4xx 10^4`N. it comes out with a speed of `50 m*s^(-1)`. What is the thickness of the wall ?
Another bullet with a lesser mass, but with the same initial velocity , penetrates the wall but is unable to come out . what is the maximum possible mass of the second bullet ?

Explanation

In the first case `v_1 =400 m*s^(-1) , v_2 =50 m*s^(-1)` ,
`F=4xx10^4 N , m=50 g =0.05 kg`
Let the thickness of the wall be x .
Since,, the work done against the force is the change in kinetic energy of the bullet,
`1/2 m(v_1^2-v_2^2)=F*x or, x=(m(v_1^2-v_2^2))/(2F)`
`or, x=(0.05xx(400^2-50^2))/(2xx4xx10^4)=984.4 xx10^(-4) m=9.844cm`
In the second case, let the mass of the bullet be `m_1` As `v_2 =0`
`1/2 m_1v_1^2=F*x`
`therefore m_1-(2F*x)/(v_1^2) =(2xx4xx10^4xx984.4xx10^(-4))/((400)^2) kg`
`492.2 xx10^(-4) =49.22 g`
Hence , the maximum possible mass of the second bullet is 49.22 g.