# Rigid Body Dynamics R&D help!

by nbjsargent
Tags: body, dynamics, randd, rigid
 P: 789 $$x=x_0+v_0 t+\tfrac{1}{2}a t^2$$ and $$v=v_0+a t$$ x is the position at time t, $x_0$ is the position at time 0, v is the position at time t, $v_0$ is the velocity at time 0, a is the acceleration (force/mass). Lets say time zero is when the bullet first touches the slab, so $x_0$=0, a=3g = -96 ft/sec^2, $v_0$=4000 ft/sec. When the bullet stops, v=0, and you can use the second equation to get t, then substitute into the first equation to get x when the bullet stops. The energy of the bullet at impact is $\tfrac{1}{2}mv_0^2$ but I'm not sure how to translate that into deformation of the slab. I agree with Ryoko, solving the equations for a real solution will require some serious computational power. You probably need to semi-fake it here, use some simple model that gives reasonable results without a lot of calculation. You probably want to use the fact that the momentum of the bullet $\tfrac{1}{2}m[v_0,0,0]$ is conserved too, where now you have to use the vector momentum $[v_0,0,0]$. This will also be the momentum of the slab and all the pieces after the bullet stops.