A bullet of mass 50 g moving at 400 m`cdot s^(-1)` penetrates a wall with an average force of 4`xx 10 ^(4)` N. It comes out of the other side of the wall at 50 m`cdot s^(-1)`. Find the thickness of the wall.

Explanation

In the case of the 1st bullet,
m = 50 g = 5 `xx 10^(-3)` kg
As the average force = `4 xx 10^(4)`N,
a = average retardation = `(4 xx 10^(4))/(50 xx 10^(-3)) "m"cdot s^(-2)`
u = initial velocity = 400 m`cdot s^(-1)`
v = final velocity = 50 m`cdot s^(-1)`
Let the thickness of the wall = s.
From `v^(2) = u^(2) - 2`as.
s = `(u^(2) - v^(2))/(2a) = ((400^(2) - 50^(2)) xx 50 xx 10^(-3))/(2 xx 4 xx 10^(4)) = 0.0984`m
The second bullet cannot penetrate the wall, hence its final velocity should be 0, i.e., v = 0.
As it cannot cover the distance s , the maximum possible mass `m_(0)` corresponds to s = 0.0984 m.
Average retardation, a = `(F)/(m_(0)) = (u^(2) - v^(2))/(2s) = (u^(2))/(2s)`
or, `m_(0) = (2Fs)/(u^(2)) = (2 xx 4 xx 10^(4) xx 0.0984)/(400^(2)) = 0.0492` kg