# Bullet velocity using impulse-momentum

1. Oct 16, 2010

1. The problem statement, all variables and given/known data

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.

8.18%

2. Relevant equations

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

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

3. 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.

2. Oct 16, 2010

### Quinzio

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

3. Oct 16, 2010

$$\int e^u=u'e^u$$

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...

4. Oct 16, 2010

### Quinzio

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.

5. Oct 16, 2010

6. Oct 16, 2010

### Quinzio

Checked.
But you still need to review your work.

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

7. Oct 16, 2010

### 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.

8. Oct 17, 2010

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

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

$$dt=-.00016du$$

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

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.

9. Oct 18, 2010

### 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.

10. Oct 18, 2010

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?

11. Oct 18, 2010

### 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)

12. Oct 18, 2010

Got it! Thanks cjl!!! :)

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

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

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

Last edited: Oct 18, 2010