Momentum vs Work: Find Out in Super Hero Experiment

[JFS321]

All,

I started this strange little thought experiment this morning, but I'm not quite sure what I'm missing. Here's the problem I solved (this is not homework, BTW): Find the speed at which a super hero (mass=76.0 kg) must fly into a train (mass = 19,537 kg) traveling at 35 m/s to stop it ... ANSWER: ~9,000 m/s.

What I'm wondering is this: How does a problem such as this relate to work and time? Am I correct in saying that there is (technically) no work done because the kinetic energy of the superhero and train cancel each other completely (ignoring that lost to sound, heat, etc.), and the collision takes place in such a short time that the distance is "0"?

Secondly, is it possible to calculate force and time in this problem? I'm assuming time would be extremely small because the force is so large (I'm thinking about the impulse equation here).

Any help clearing up my mental state is much appreciated!

 

[kuruman]

You can relate this to work done, say on the train by the superhero, if you make additional assumptions about how far the train travels before it stops or how long it takes for it to stop. With either assumption you can use the kinematic equations to find the average acceleration (assuming it is constant) and from that anything else you want. You also need to consider what happens to the train and its passengers if it is hit by a superhero traveling at Mach 27.

 

[jfizzix]

The work done on the train is equal to its change in kinetic energy, which is a lot considering the train is now at rest.

The time it takes to bring the train to rest is critical, as a ballistic collision would probably be a lot more dangerous on a train than a car crash.
All the energy has to go somewhere, and with the superhero traveling at 9 km/s (fun fact, the escape velocity from Earth is 11.2 km/s), and both the hero and the center of mass of the train being at rest after, much of this energy is likely transferred into internal energy of the train such as deformed metal, broken glass, twisted seats and shattered bones (in order of increasing catastrophe).

If the superhero is only flying, then he is applying some kind of force to the air around him to propel him forward, in which case a lot of energy would be dissipated as heat in the turbulence generated as the hero pushes with as much force possible on the train. This may well create a rocket-like fireball coming out the back of the superhero, which would be really cool, but is not often talked about in similar superhero exploits. Depending on the physics of how the superhero flies (maybe he bends space to make an antigravity field?) that would determine what's likely to happen when he starts pushing on the train.

 

[anorlunda]

If we simplify the problem to neglect the time in contact, or the acceleration, then we can use a simple momentum balance (presuming that the two velocities are in exactly opposite directions).

m₁v₁=m₂v₂
So if object 1 is superman, then
v₁=v₂*(m₂/m₁)

Which is about 9000 m/s as you said in the OP.

 

[Phylosopher]

-- I always thought about this, but I did not discuss it before with others. I will take the opportunity and express my thoughts. I would like to hear your opinions on it too.

I think your calculation uses the following: ##\Delta p=0##. Also, you assume everything to be ideal.

Work definition is: ##W= \int_{s_{0}}^{s_{1}} F \cdot ds## , no change in distance means no work, no matter what ##F## is.
You can relate the problem to impulse, which have no relation to distance ##I= \int_{t_{0}}^{t_{1}} F dt = mv_{2}-mv_{1}##, but have a relation to time.

With all that being said:
Let's just agree that the two objects do not move after contact and their momentum cancels out giving them both ##0## velocity. Well, this means they are still in contact. What time are we referring to in the impulse equation? We refer to the time in which action happen; time when quantities change. Precisely velocity.

Additionally, I believe if you assumed that the two objects do not change their position after contact (thus no work), then you can not say that the time for their velocities to approach ##"0"## is not ##"0"## (i.e Instantaneous change), because this will only happen if both changed position after contact, which contradicts with the proposed assumption.

Then, what is the time used for impulse equation, if it is ##0##? Will, I would like to say it is the time for deformation. Both the train and the superhero deform with contact. This deformation takes time to happen and change position of the object "internally". This would be the measure of impulse, that relates the initial and the final velocity. Thus in the big picture, there is impulse, there is time and there is work.

You might like to add an additional assumption: No deformation. Both impulse and work will equal ##0##. No work, no impulse but there is a change in velocity. But I can not see it in the impulse equation: ##I= \int_{t_{0}}^{t_{1}} F dt = mv_{2}-mv_{1}##.

I can see no escape in saying that the assumptions them selves are wrong. Unless there is a way to find the impulse relation written eariler with ##\Delta t=0##

-Maybe a proposition can be done with a Dirac delta though?

------------------------
Will. What I wrote is probably wrong! But I wanted to share my thoughts.

 

[Dale]

Am I correct in saying that there is (technically) no work done because the kinetic energy of the superhero and train cancel each other completely

No. Both the KE of the hero and the train are positive. They add, they do not cancel out.

Secondly, is it possible to calculate force and time in this problem? I'm assuming time would be extremely small because the force is so large (I'm thinking about the impulse equation here).

It is possible, but you would have to make some material model of the situation. Of course, you would model the hero as being completely incompressible, but you could model the train as a very large spring with some specified spring constant, or as a dashpot if you prefer.

 

[jbriggs444]

All,

I started this strange little thought experiment this morning, but I'm not quite sure what I'm missing. Here's the problem I solved (this is not homework, BTW): Find the speed at which a super hero (mass=76.0 kg) must fly into a train (mass = 19,537 kg) traveling at 35 m/s to stop it ... ANSWER: ~9,000 m/s.

Now calculate the super hero's kinetic energy just prior to the collision. Compare to the train's kinetic energy just prior to the collision.

Which is a bigger disaster? The train crashing into a brick wall? Or a superhero stopping it?

 

[Chestermiller]

No. Both the KE of the hero and the train are positive. They add, they do not cancel out.

It is possible, but you would have to make some material model of the situation. Of course, you would model the hero as being completely incompressible, but you could model the train as a very large spring with some specified spring constant, or as a dashpot if you prefer.

Yes. I think you identified the real question and associated mechanism here: Where does all the kinetic energy go? It certainly isn't decreased by work, since no work is done on either body (assuming the collision plane does not move). But it can be taken up by elastic deformation energy of the train and super guy if the collision is ideally elastic (and then released again as kinetic energy). But, if the deformation is not elastic, part of the energy is used to increase the internal energy of the bodies and surroundings, in terms of higher temperatures, which eventually gets dissipated (over the broader surroundings). The latter is the effect modeled by the dashpot.

 

[JFS321]

Thanks to all. I think my original mistake was idealizing the problem -- i.e., the superhero and train magically stop instantaneously. It's more intuitive now to think of the real-world effects of deformation, etc. and how those clearly relate to work and energy.