Bullet velocity using impulse-momentum

In summary, the mathematical model suggested for the variation in pressure inside the 10-mm-diameter barrel of a rifle as a 25-g bullet was fired is:p(t)=(950MPa)e^{-t/(0.16ms)}
  • #1
JJBladester
Gold Member
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2

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:

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

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



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

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.

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

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

This yields a number around -29 billion... Not even close to 520 m/s.
 
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  • #2
Yes, because you have to exercise in solving integrals :)
 
  • #3
Quinzio said:
Yes, because you have to exercise in solving integrals :)

[tex]\int e^u=u'e^u[/tex]

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

Then, [itex]u'=\frac{-1}{.00016}[/itex]

I know how to integrate, man. I'm not sure if you're trying to help me or trying to prove something...
 
  • #4
JJBladester said:
[tex]\int e^u=u'e^u[/tex]

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

Then, [itex]u'=\frac{-1}{.00016}[/itex]

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.
 
  • #6
Checked.
But you still need to review your work.

I just solved the problem and got the correct answer ~8.18%
 
  • #7
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

[tex] m_1*v_1 +\int F(t)dt = m_2*v_2[/tex]

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

[tex]\int e^{f(t)} dt[/tex]

with

[tex]\int k*e^u du[/tex]

However, you aren't solving for k correctly. Try solving for dt as a function of du.
 
  • #8
[tex]u=\frac{-1}{.00016}t[/tex]

[tex]du=\frac{-1}{.00016}dt[/tex]

[tex]dt=-.00016du[/tex]

[tex]\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}[/tex]

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.

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

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

[tex]mv_1+\int_{0}^{t}F(t)dt=mv_2[/tex]

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

Any ideas, because it feels like I'm running in circles.
 
  • #9
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
cjl said:
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:

[tex]Pressure=\frac{Force}{Area}[/tex]

[tex]\therefore Force=\left (Pressure \right )\left (Area \right )[/tex]

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

[tex]mv_1=\int_{0}^{t}F(t)dt=mv_2[/tex]

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

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
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
Got it! Thanks cjl! :)

[tex]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[/tex]

[tex]error=\left |\frac{theoretical-experimental}{theoretical} \right |*100=[/tex]




[tex]\left |\frac{520-477.5}{520} \right |*100=8.18[/tex]
 
Last edited:

FAQ: Bullet velocity using impulse-momentum

1. How is bullet velocity affected by impulse and momentum?

The bullet velocity is affected by both impulse and momentum. Impulse is the change in momentum, which is a product of mass and velocity. Therefore, as the impulse increases, the momentum and ultimately the bullet velocity also increases.

2. Is bullet velocity the same as bullet speed?

No, bullet velocity and bullet speed are not the same. Bullet velocity refers to the rate of change in the bullet's position, while bullet speed refers to the distance traveled by the bullet in a specific amount of time.

3. How does the mass of the bullet affect its velocity?

The mass of the bullet plays a crucial role in determining its velocity. According to the principle of conservation of momentum, the momentum of the bullet before and after firing should be the same. Therefore, a heavier bullet will have a lower velocity compared to a lighter bullet with the same amount of impulse.

4. What are the units of measurement for bullet velocity?

The units of measurement for bullet velocity are typically meters per second (m/s) or feet per second (ft/s). However, some sources may also use kilometers per hour (km/h) or miles per hour (mph). It is essential to check the units of measurement being used to avoid confusion.

5. Can bullet velocity be calculated using other factors besides impulse and momentum?

Yes, bullet velocity can also be calculated using other factors such as the bullet's initial and final positions, acceleration, and time. However, impulse and momentum are the most commonly used factors since they directly affect the bullet's velocity and are relatively easy to measure.

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