Differential Equation for deceleration of a bullet

[Unsilenced]

Homework Statement

A bullet of mass m strikes an amor plate with initial velocity v0. As the bullet burrows into the plate, its motion is impeded by a frictional force which is directly proportional to the bullet's velocity. There are no other forces acting on the bullet.

-Use Newton's Second Law to set up an IVP for the position of the bullet
-How thick should the plate be to stop the bullet?

Homework Equations

[None given with problem]

The Attempt at a Solution

I attempted to use a modified version of a harmonic oscillator formula, where there is no spring. Only the momentum of the object and the friction force resisting it.

m*d^2x/dt^2-kdx/dt=0.

However, in solving it I've run into a problem. While I got an equation for x, x(t)=v0*e^Rt/(R-1)-v0/(R-1) (where R=m/k), trying to use that to find the time at which the bullet stops yields

0=v0*R*e^Rt/(R-1)

e^x=/=0, regardless of x, so there is no possible time at which dx/dt=0.

Is my model flawed, or have I made some other error?

 

[haruspex]

m*d^2x/dt^2-kdx/dt=0.

Think about what that equation is saying. For a greater dx/dt, what happens to the acceleration?

 

[jbrussell93]

m*d^2x/dt^2-kdx/dt=0.

Check your sign on the friction term ;)

 

[Unsilenced]

Think about what that equation is saying. For a greater dx/dt, what happens to the acceleration?

Acceleration becomes more negative, that is, deceleration increases with velocity. So, I could make both terms negative?

Check your sign on the friction term ;)

Switching the signs in the initial equation only changes whether R is positive or negative. That doesn't solve the fact that my final equation is mathematically impossible.

 

[jbrussell93]

Acceleration becomes more negative, that is, deceleration increases with velocity. So, I could make both terms negative?

Yes, but making them both positive is probably less confusing. For example, the equation of motion is:
F=ma=-kv

Now bring them to one side, set a=dv/dt and solve for v(t). Then integrate again to solve for x(t)

Switching the signs in the initial equation only changes whether R is positive or negative. That doesn't solve the fact that my final equation is mathematically impossible.

Yes, you will have a negative exponential which approaches 0 as time goes to infinity. You should be able to solve it from there.

 

[Unsilenced]

Yes, you will have a negative exponential which approaches 0 as time goes to infinity. You should be able to solve it from there.

Ah, I see. So it is meant to be asymptotic? I thought it might be, but it didn't make sense from a physical interpretation, I.E that the bullet never stops moving.

 

[jbrussell93]

Ah, I see. So it is meant to be asymptotic? I thought it might be, but it didn't make sense from a physical interpretation, I.E that the bullet never stops moving.

Yes unless I'm missing something. A better example may be a boat which turns off its motor. How far does it drift before stopping? In this case the asymptotic relation is a bit more realistic. With the bullet-wood example, the k value is much larger so it approaches 0 very quickly. It may help to think of the "wood" instead as being jello :)

 

[Unsilenced]

Yes unless I'm missing something. A better example may be a boat which turns off its motor. How far does it drift before stopping? In this case the asymptotic relation is a bit more realistic. With the bullet-wood example, the k value is much larger so it approaches 0 very quickly. It may help to think of the "wood" instead as being jello :)

Yeah, that does make it a bit easier to visualize. In the case of a bullet hitting a metal plate, the medium tends to be denser than the projectile, and all sorts of deformation occurs. In real life this equation would look a bit more like this: http://i.imgur.com/2KA5jiv.gif :p


A non-deforming projectile in a fluid medium makes the result a bit more logical. The result I ended up getting was

xf=v0*m/k

Seems legit.

 

[jbrussell93]

The result I ended up getting was

xf=v0*m/k

Seems legit.

Yup that's what I calculated as well.