Bullet velocity using impulse-momentum

[JJBladester]

Homework Statement

The following mathematical model was suggested for the variation in pressure inside the 10-mm-diameter barrel of a rifle as a 25-g bullet was fired:

p(t)=(950MPa)e^{-t/(0.16ms)}

where t is expressed in ms. Knowing that it took 1.44 ms for the bullet to travel the length of the barrel and that the velocity of the bullet upon exit was measured to be 520 m/s, determine the percent error introduced if the above equation is used to calculate the muzzle velocity of the rifle.

Answer:
8.18%

Homework Equations

mv_1+\int_{t_1}^{t_2}F(t)dt=mv_2

Percent error = [|actual-experimental|/|actual|]*100

The Attempt at a Solution

We are given the experimental v₂ and are asked to find the "actual" v₂ using the mathematical model, then the percent error between them.

\left ( .025 \right )\left ( 0 \right )+\int_{0}^{.00144}950 \cdot 10^6e^{-t/.00016}dt=\left (.025 \right )\left (v_2 \right )

v_2=\frac{\frac{-950 \cdot 10^6}{.00016}e^{-.00144/.00016}}{.025}

This yields a number around -29 billion... Not even close to 520 m/s.

 

[Quinzio]

Yes, because you have to exercise in solving integrals :)

 

[JJBladester]

Yes, because you have to exercise in solving integrals :)

\int e^u=u'e^u

And that's what I did. I let u=\frac{-1}{.00016}t

Then, u'=\frac{-1}{.00016}

I know how to integrate, man. I'm not sure if you're trying to help me or trying to prove something...

 

[Quinzio]

\int e^u=u'e^u

And that's what I did. I let u=\frac{-1}{.00016}t

Then, u'=\frac{-1}{.00016}

I know how to integrate, man. I'm not sure if you're trying to help me or trying to prove something...

Now I'm sure you have to exercise. :)
Don't get offended, we are learning and passing time here. And I'm helping you until proved the opposite.

 

[JJBladester]

http://integrals.wolfram.com/index.jsp?expr=e^%282x%29&random=false

Check it. If we have e^(2x) and we integrate, we end up with [e^(2x)]/2

That is the same way I went about the problem I'm solving here.

I hate trying to do homework when I'm sick as a dog.... :(

 

[Quinzio]

Checked.
But you still need to review your work.

I just solved the problem and got the correct answer ~8.18%

 

[cjl]

OK, so I did the problem, and it definitely works out. A couple of comments:

1) Check your solution method. In your first post, you stated that

m_1*v_1 +\int F(t)dt = m_2*v_2

However, when you plug in the numbers, you're not actually integrating the force. Before you integrate, you need to solve for the force acting on the projectile.

2) You're messing up the u-substitution slightly in post 3. You're close, but you are making a crucial error. You're trying to replace

\int e^{f(t)} dt

with

\int k*e^u du

However, you aren't solving for k correctly. Try solving for dt as a function of du.

 

[JJBladester]

u=\frac{-1}{.00016}t

du=\frac{-1}{.00016}dt

dt=-.00016du

\int_{0}^{t}950\cdot 10^6e^{-t/.00016}=(950\cdot10^6)(-.00016)\int_{0}^{t}e^udu=(950\cdot10^6)(-.00016)e^{-t/.00016}

Ok, so above I've found the integral of the pressure function of the rifle. Below I solve for the pressure at 1.44ms from the original function.

F(t)=950\cdot 10^6e^{-t/.00016} \to F(.00144)=950\cdot 10^6e^{-.00144/.00016}=117239

Now I'm really lost. I try going back to the initial impulse-momentum equation:

mv_1+\int_{0}^{t}F(t)dt=mv_2

v_2=\frac{\int_{0}^{t}F(t)dt}{m}=\frac{(950\cdot10^6)(-.00016)e^{-.00144/.00016}}{.025}-.750

Any ideas, because it feels like I'm running in circles.

 

[cjl]

OK, now you're using the correct u-substitution (unlike before). There's still a couple of problems though. You still aren't actually integrating the force. You need to write an expression relating the force to the pressure, and then integrate that (rather than directly integrating the pressure as you are now).

Also, why are you solving for F(0.00144)? It's never needed for the solution.

 

[JJBladester]

You need to write an expression relating the force to the pressure...

I see... I was mistakingly thinking pressure was force, but it's really force/area. The relationship:

Pressure=\frac{Force}{Area}

\therefore Force=\left (Pressure \right )\left (Area \right )

F(t)=p(t)\left ( A \right )=\left [950\cdot 10^6e^{-t/.00016} \right ]\left [\pi (.005)^2 \right ]

mv_1=\int_{0}^{t}F(t)dt=mv_2

v_2=\frac{\left (950\cdot 10^6 \right )\left (\pi (.005)^2 \right )(-.00016)e^{-t/.00016}}{.025}

Plugging in 1.44ms (.00144s) into the equation above for v₂ yields -0.0589, not even close to 520 m/s. Did I mess up the math again?

 

[cjl]

Now you're extremely close. You performed the integral correctly, and that is the correct solution to the integral. However, you have to plug in both of the integration limits and take the difference (since at t = 0, e^-t/0.00016 is equal to 1, not 0)

 

[JJBladester]

Got it! Thanks cjl! :)

v_2=\frac{\pi(.005)^2\left (950\cdot 10^6 \right )(-.00016)\left [e^{-.00144/.00016}-e^{0/.00016} \right ]}{.025}=477.5m/s

error=\left |\frac{theoretical-experimental}{theoretical} \right |*100=




\left |\frac{520-477.5}{520} \right |*100=8.18