How Much Pressure Is Needed to Launch a Marble at 5 m/s?

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Homework Help Overview

The original poster is attempting to determine the pressure required to launch a marble at a speed of 5 m/s using a compressed air cannon. The setup involves a PVC cylinder and a quick release valve, with specific dimensions provided. The discussion revolves around the physics of pressure, force, and energy in relation to the marble's launch.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss using energy equations and the gas law to relate pressure and volume. Some suggest using force equations to derive acceleration and velocity, while others express uncertainty about the applicability of calculus in this context. Questions arise regarding the changing pressure during the marble's launch and how that affects the calculations.

Discussion Status

The discussion is ongoing, with participants exploring different approaches and questioning the assumptions behind their reasoning. Some guidance has been offered regarding the use of force and energy equations, but there is no consensus on a definitive method to solve the problem without calculus.

Contextual Notes

Participants note the lack of formal calculus knowledge, which is seen as a barrier to accurately modeling the problem. There is also mention of the need for assumptions about how pressure changes during the launch process.

Fifty
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I need to make a cannon that fires a marble. It must be accurate and adjustable enough to hit targets of varying distance and through varying obstacles. For propellant, I want to use compressed air, for which I have an apparatus (valve from a bike tire attached to a PVC cylinder roughly fifteen centimetres long. The cylinder is about five centimetres wide. A quick release valve attached to a string will be the trigger mechanism.

The question is, how much pressure should I load the tank up with to shoot a 5.4 g marble at around 5 metres per second?

My first thought was to use energy: E = Volume(V) x Pressure(P), but I am not sure this is the correct way to do this, mainly because I haven't actually learned this equation in physics class, but I came up with it based on unit analysis during chemistry class. I am not sure what assumptions are made by the equation for example.

My best attempt:

Atmospheric pressure in my classroom is about 14.7 PSI (we measured this in our chemistry class just across the hall) and all of the compressed air will be discharged after every shot. Or rather, enough air will leave the cylinder such that the pressure inside the tank is equal to the pressure outside the tank. If the seal is totally air tight and the pressure perfectly equalizes, the Kinetic energy of the marble after leaving the barrel should be equal to the volume of the cylinder times the difference in pressure (initial pressure in tank minus atmospheric pressure).

Is this correct? Again note that I haven't actually learned this formally.
 
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The gas equation is PV=RT, where R is a constant.

I think the best approach is to use the force equation to get an acceleration and then integrate this over the distance of the barrel.

If the marble has a cross sectional area A cm and the pressure differential is P, then the acceleration is

a = F/m = AP/m. The velocity is then V = (1/2) a t^2

I hope this helps.
 
Mentz114 said:
The gas equation is PV=RT, where R is a constant.

I think the best approach is to use the force equation to get an acceleration and then integrate this over the distance of the barrel.

I haven't learned any calculus yet.

If the marble has a cross sectional area A cm and the pressure differential is P, then the acceleration is

a = F/m = AP/m. The velocity is then V = (1/2) a t^2

I hope this helps.

But won't the acceleration change as pressure is released? Initially, the air will push with maximum force, but as the volume increases, won't the pressure the air exerts decrease and thus the force the air exerts? This means I can't use linear equations to solve for the final velocity. I get the idea from my teachers that this can be done with calculus, but I have not learned calculus yet.
 
Fifty said:
I haven't learned any calculus yet.



But won't the acceleration change as pressure is released? Initially, the air will push with maximum force, but as the volume increases, won't the pressure the air exerts decrease and thus the force the air exerts? This means I can't use linear equations to solve for the final velocity. I get the idea from my teachers that this can be done with calculus, but I have not learned calculus yet.
I can see that not having calculus is a difficulty here. You could try getting an approximate result by assuming the pressure falls linearly or quadratically and finding an average. But I can't see how to get a more accurate estimate without integrating. I'll think about it.

Maybe someone else has a better idea ?
 

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