# Homework Help: Bullet Dynamics problem

1. Oct 17, 2005

### vabamyyr

bullet enters a wall with speed v1 and goes out with speed v2, the wall is h units thick. During that time it experiences an impending force which is proprtional to v^2. I have to find the time the bullet goes out the wall.

I can take t0=0 v0=v1 x0=0

Using N2 law i get d(mv)/dt = -F..............mdv/dt = -m*v^2 (im not sure if im missing here something)
dv/dt = -v^2

dt = - dv/v^2 ===> t = (v1-v2)/v1*v2...this result is wierd since it doesnt contain h so i must have constructed the equation wrongly i guess.
I think that F depends on h since F can act only between [0; h], and maybe the answer should be multiplied with h and t = h*(v1-v2)/v1*v2...but it still doesnt look right to me....any opinions?

2. Oct 18, 2005

### ehild

Well, the force was said to be proportional to v^2.
So F= -kv^2, where k is a constant
and dv/dt=-k/m* V^2

[Quote}

dv/dt = -v^2
dt = - dv/v^2 ===> t = (v1-v2)/v1*v2...this result is wierd ....[/QUOTE]

Including that constant k and m the mass of the bullet, you get

t=(v1-v2)/(v1*v2)*m/k (A)

But you do not know m/k. You should get it from h, the distance travelled.

You are on the right track. F depends on the position of the bullet inside the wall through the velocity. So the acceleration canbe considered as function of the position, x.

a = dv/dt = dv/dx * dx/dt = dv/dx *v.
F= ma = - kv^2

m*dv/dx *v = - k*v^2,

or dv/dx = (-k/m)*v

Solve this differential equation with the condition that v=v1 at x=0 and v=v2 at x=h. You get an expression containing the unknown k/m which you can use to determine t from equation (A).

ehild

3. Oct 18, 2005