Modern Physics: Finding pressure using energy.

[baltimorebest]

Homework Statement

A beam of pions ($$\pi$$-mesons, with rest energy m_$$\pi$$ c²=140 MeV) consisting of 10¹⁰ pions/cm^2s moves with speed such that the total energy of each pion is 10^5 MeV. What is the pressure the beam exerts on a surface that absorbs it and brings it to rest?

Please note: For some reason my pi's are showing up strangly. The first one is supposed to say pi-mesons and the second one is supposed to be a subscript. Thanks.

Homework Equations

The Attempt at a Solution

I am not sure where to start. Any help would be appreciated.

 

[diazona]

Well, what do you know about pressure?

 

[baltimorebest]

Well I know that it is force per unit area.

 

[diazona]

OK, now remember that pressure is always calculated over a surface. To find pressure, you pick a surface, find the force on the surface, and divide by the area of the surface. Can you do that?

 

[baltimorebest]

I see what you are saying, but I am still unsure how to proceed with tis problem. I'm having trouble finding the relationship between pressure and energy. My idea is that I have to find the Force, and for that I need mass and acceleration. I can find mass from the rest energy, but the acceleration is what is confusing me.

 

[diazona]

No, actually you don't need mass and acceleration. There is another, more general form of Newton's law that you can use - do you know what it is?

 

[baltimorebest]

Is it F + U*dm/dt=m*dv/dt?

 

[diazona]

No, I'm thinking of something else, more general than that.

 

[baltimorebest]

Hmmm I'm not sure then. That was the most general I could think of.

 

[diazona]

Have a look at the Wikipedia page for Newton's second law. You'll find it there.

 

[baltimorebest]

F=dp/dt?

 

[diazona]

Yep, that's the one.

So, you need to pick a surface, find the force on that surface using F = dp/dt, and then divide the force by the area. Can you make any progress on that?

By the way, in this case, since the force is constant over time, the force will be equal to its average. That means that you can use either F_avg = Δp/Δt or F = dp/dt, whichever one makes more sense to you.

 

[baltimorebest]

Hmm ok I'll work on it tonight and let you know tomorrow afternoon.

 

[diazona]

OK; I probably won't be able to respond at that time (traveling without internet access) but someone else can take over.

 

[baltimorebest]

Ok so I wasn't able to make much progress on this problem. Can anyone else who reads this help me? Thanks in advance.

 

[zzzoak]

$$\Delta P =P(final)-P(initial)$$

 

[baltimorebest]

Hmm yeah I understand that but I am having trouble getting to the point where I can use that equation.

 

[zzzoak]

Consider only one particle moving to the wall.
Can you get
$$\Delta P$$
for this particle?

 

[baltimorebest]

Well I know P=gamma*m*v. I think I can find 'm' from the rest energy, and 'v' from finding first the Kinetic Energy, then solving for 'v.' Finally I think I can find gamma after I have all these other values. Is this correct?

 

[diazona]

Seems reasonable to me. You can try it, and if you run into any problems you can't track down yourself, post your work here and we'll help you check it.

 

[baltimorebest]

Hey diazona thanks for following up with me...

Ok so I tried to do what I had planned, but I think I ran into a small problem. I first solved for m and got 1.6E-9 kg. Then I used E=Gamma(mc^2) to solve for 'v'. I got a value of 2.99E8 which is very close to c. When I plug this back into the formula for gamma, my denominator was essentially 0, since the ratio of v/c was 1.

 

[diazona]

For one thing, the mass you found is wrong by several orders of magnitude. Check that calculation again.

To find γ, you don't necessarily have to solve for v first. In fact, for objects that move close to the speed of light, it's much easier (and gives you better precision) to find γ using some other formula, rather than calculating it from the speed, for exactly the reason you discovered. What other way do you have to calculate γ?

 

[baltimorebest]

For the mass I used mc^2=140E6 and that's how I got my value for m. Was there something wrong in my conversion to get standard units? I thought I should change the Mev into ev?

And for Gamma, should I just use the value that I found from E=gamma(mc^2)? Could I multiply that gamme by the mass and the velocity that I found to be very close to c to get the answer?

 

[diazona]

And for Gamma, should I just use the value that I found from E=gamma(mc^2)? Could I multiply that gamme by the mass and the velocity that I found to be very close to c to get the answer?

Yes, that's exactly what I would recommend.

For finding the mass, there's nothing wrong with converting the MeV into eV, but according to what you wrote, you completely dropped the units when you wrote the equation down! You can't do that. It should be
$$mc^2 = 140 \mathrm{MeV}$$
or
$$mc^2 = 140\times 10^6 \mathrm{eV}$$
Plug in a numerical value for c, with units, and solve for m, doing whatever unit conversions you need to along the way. (You will find that you need to convert the energy to a different unit, neither MeV or eV, unless you use a calculator that does the conversions for you)

 

[baltimorebest]

Ohhhh ok so I converted 140 MeV to 2.24E-11 Joules. Then I divided this by c^2 to get m=2.48E-28. I hope this is right...

I then took this value for 'm' and multiplied it by 'v' (2.99E8 m/s) and finally multiplied by gamma (714.3) to get 5.316E-17. I think/hope this should be 'p' for one meson?

 

[diazona]

What units are on the 2.48×10^-28 and on the 5.316×10^-17?

 

[baltimorebest]

Oh sorry again... kg and kg*m/s

 

[diazona]

Yep, looks good. Always do remember to include those units! (well whatever are the proper units for whatever quantity you're talking about)

So where does that leave you with the problem?

 

[baltimorebest]

Well I know that dP/dt is equal to F, and F is what I am looking for. However P appears to be a constant, so I think I am still missing something. You told me earlier I could think of it in terms of delta p/ delta t. I guess in this case the final 'p' is 0 since it is absorbed and brought to rest. I don't have any 't' values though.

 

[diazona]

OK, think about this: the meaning of the equation F = dp/dt is that the force on an object is equal to the rate of change of that object's momentum. Or if you prefer the F_avg = Δp/Δt form, the average force on an object is equal to the change in the object's momentum over a certain period of time. Either way, the first thing you need to do is to figure out what object you are using the equation for. In other words, F is the force on what?