Calculating Projectile Work and Force in a Compressed Helium Launcher

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The RAPTOR, a compressed helium projectile launcher developed by Brookhaven National Laboratory, utilizes high-pressure helium to propel projectiles at speeds exceeding 1600 m/s, significantly quieter than traditional concrete-cutting tools. The discussion focuses on calculating various physical parameters related to the projectile's motion, including work done, average force, impulse, time in the barrel, recoil speed, and energy required to split concrete. Participants clarify misconceptions about the mass and acceleration of the projectile, emphasizing the importance of the work-energy theorem and impulse-momentum principles. The calculations reveal that the work done on the projectile is approximately 2,300 J, and the average force exerted is around 1,150 N. The conversation highlights the complexities of the physics involved and the need for precise definitions and approaches in problem-solving.
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Homework Statement


Scientists at Brookhaven National Laboratory in New York in conjunction with Brooklyn Union Gas (now a division of Keyspan Energy) are developing a compressed helium projectile launcher called the RAPTOR (short for "rapid cutter of concrete"). The original technology behind the gas gun began in the 1980s as part of an anti-missile research program. Now instead of shooting down missiles in midair, the RAPTOR will be used to shoot tiny metal projectiles at the ground to cut concrete like a jackhammer. The device works by rapidly compressing helium gas from its storage tank pressure of 2 atmospheres to an unbelievable 1000 atmospheres in a fraction of a second. The resulting shock wave blasts the 1.8 g projectiles (about the same mass as a .22 caliber bullet) out the barrel of the gun at roughly 1600 m/s (more than twice the muzzle velocity of a high-powered rifle). The main benefit of this technology is that it is much quieter than conventional concrete cutters -- 85 dB for the RAPTOR compared to 125 dB for a jackhammer. The last reported prototype (RAPTOR III) was 2.0 m long, weighed 120 kg, and was able to split a 10 cm thick slab in seven shots. Determine …

1. the work done by the compressed helium on a projectile,
2. the average force of the compressed helium on a projectile,
3. the impulse delivered to a projectile,
4. the time a projectile spends in the barrel,
5. the recoil speed of the gun,
6. the height to which the gun would jump, and
7. the minimum energy needed to split the concrete slab.

Homework Equations


W = Fd
FΔt = mΔv
(m1+m2)v = m1v1' + m2v2'
F=ma

The Attempt at a Solution


1. w=Fd w=(120kg)(9.8m/s2)(2.0m) = 2352 J (is 9.8 the acceleration?)
2. F=ma F=(120kg)(9.8m/s2) = 1180 N
3. 1180N(Δt) = 120kg(1600 m/s)
4. Δt = F/mΔv
5. (0.0018kg + 120kg)(0m/s) = (0.0018kg)(1600m/s) + (120kg)(v2')
-120kg(v2') = 2.88kgm/s
v2' = -0.024 m/s
6.
7.

I'm going to try and work each of these out one at a time, so let's start with number 1! Or whichever the first one I got wrong is.
 
Last edited:
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I think you are missing the first question.

The projectile is what they are asking about, not the Raptor itself.
 
1. (0.0018 kg)(9.8m/s2)(2.0m) = 0.0353 J
 
Woopy said:
Determine … 1. the work done by the compressed helium on a projectile,

1. is 9.8 the acceleration?
No, this is not the acceleration of the projectile. Think work-energy theorem.



Woopy said:
Determine … 2. the average force of the compressed helium on a projectile,

2. F=ma F=(120kg)(9.8m/s2) = 1180 N
Again, this is not the acceleration, nor is this the mass of the projectile. Think definition of applied mechanical work.



Woopy said:
Determine … 3. the impulse delivered to a projectile,

3. 1180N(Δt) = 120kg(1600 m/s)
I have no idea what this is supposed to mean. Hint: what is the initial momentum of the projectile? What is the final momentum of the projectile?



Woopy said:
Determine … 4. the time a projectile spends in the barrel,

4. Δt = F/mΔv
EDIT: Oh, yeah, this looks good.



Woopy said:
Determine … 5. the recoil speed of the gun,

5. (0.0018kg + 120kg)(0m/s) = (0.0018kg)(1600m/s) + (120kg)(v2')
Looks good.
 
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1. 1/2(.0018kg)(1600m/s)²= 2,304 J
 
2. W=Fd

W=(120kg)(a)(2.0m)

what is a??
 
Woopy said:
2. W=Fd

W=(120kg)(a)(2.0m)

what is a??
You don't need a here; just F. You calculated W in part 1, and d is given in the problem. BTW, that is not the mass of the projectile.
 
2,300 J = F(2.0m)
F=1,150 N
 
3. Impulse delivered, my best guess is the use of the impulse-momentum theorem, which is FΔt = mΔv

(1150N)(Δt)=(.0018kg)(1600 m/s)
1150N(Δt)=2.88 kgm/s
Δt= 0.0025s

But that would be more appropriate for question #4, so what exactly does "impulse delivered" mean, what is the appropriate unit for the answer?
 
  • #10
What is impulse? If you think about the precise mathematical meaning of impulse (and the hint that I put in my first post), then the answer should become clear. And you are correct: your suggestion is more appropriate for part 4. I think you're trying to make it harder than it is. You know how to do this, you just haven't realized it yet.
 
  • #11
3. (1150N)(.0025s)=2.88kgm/s
4. 2.88kgm/s / 1,150N = .0025s

I don't understand why question 4 wasn't asked before question 3, however.
 
  • #12
There is sometimes (always) more than one way to approach a problem.
 

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