Answer to Question #159922 in Classical Mechanics for Hussain Ousman Darboe

Question

A bullet after penetrating 3cm of a wall loses half of its speed. How far will the bullet penetrate the wall afterwards?

Answer 1

Loss of kinetic energy:

$\frac{mv^2}{2}-\frac{m(\frac v2)^2}{2}=Fx,$ where $x=3\cdot 10^{-2},$

$F=\frac{\frac{mv^2}{2}}{4\cdot10^{-2}}.$

Total:

$\frac{mv^2}{2}=F\Delta x,$

$\Delta x=\frac{\frac{mv^2}{2}}{F}=4\cdot 10^{-2},$

$\Delta x=x+x_1,$

$x_1=\Delta x-x=1\cdot 10^{-2}=1~cm.$