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Bullet Dynamics problem

  1. Oct 17, 2005 #1
    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. jcsd
  3. Oct 18, 2005 #2


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    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


    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).

  4. Oct 18, 2005 #3
    thanks, i liked ur comments and u have been most helpful
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