Density

[steevey]

Is there a relationship between the density of an object and how much momentum or energy it transfers to another object per square inch/2.54centimeters?

[lemur]

momentum is the product of mass and speed, I think. It would help me to know the situation of energy-transfer you're thinking of specifically, though.

[steevey]

momentum is the product of mass and speed, I think. It would help me to know the situation of energy-transfer you're thinking of specifically, though.

 

Well, would an iron cube transfer more of either per square inch versus a sponge cube the same size?

[lemur]

Well, would an iron cube transfer more of either per square inch versus a sponge cube the same size?

transfer more of either what? What are the cubes transferring what to? Are they sliding on a carpet or floating in high orbit?

[swansont]

The amount of momentum transfer is purely a function of mass and speed. "Momentum per square inch" really has no meaning, because it still does not tell you the mass — it depends on the geometry. For a fixed geometry, surface area correlates with mass, and more mass will transfer more momentum to the target. For a fixed mass, however, changing the geometry will not change the momentum transfer. (I'm assuming a 1-D, head-on example, and ignoring the trivial case where you miss the target)

[pioneer]

The higher density object will impart greater momentum density. Momentum is MV. Density is M/D3, so momentum density would be M/D3 V. This would be a measure of the destructive force of the impact since it would create greater compression/shear stresses.

 

For example, a 1KG sponge versus a 1kg uranium ball, with both moving at 10 m/sec, so they have the same momentum. The sponge might leave a mark and push you back. But the higher momentum density of the uranium ball will hurt much more.

Edited November 28, 2010 by pioneer

[steevey]

The amount of momentum transfer is purely a function of mass and speed. "Momentum per square inch" really has no meaning, because it still does not tell you the mass — it depends on the geometry. For a fixed geometry, surface area correlates with mass, and more mass will transfer more momentum to the target. For a fixed mass, however, changing the geometry will not change the momentum transfer. (I'm assuming a 1-D, head-on example, and ignoring the trivial case where you miss the target)

 

Not what I'm saying exactly. Just forget the word momentum then and instead think about kinetic energy. Have you ever noticed that a ping pong ball thrown at you hurts a lot less than a rock thrown at you? If they both went at roughly the same speed, the rock would hurt more. Why? Because a rock has more atoms in it and therefore requires more energy to make it resist all the other forces otherwise holding it still to the ground. What I'm wondering though, is there a relationship between how much energy you put into that rock or pingpong ball, and how much energy it transfers to another object per square inch. Because they certainly don't have the same amount of atoms per square inch, so wouldn't it effect how much kinetic energy it transfers from a specific section vs something else that's more or less dense?

[swansont]

Energy transfer is a different issue. It depends on what kind of collision, because in any sort of inelastic collision there is kinetic energy lost to other forms. And again, the quantity of concern is mass, rather than density. Density will manifest itself if you constrain the geometry, or if it somehow affects the type of collision.

[pioneer]

Say you had a block of foam and a block of iron, both with the same mass. For any given velocity, both will have the same momentum. If we held a metal sheet in front of us, both blocks would feel about the same, when they hit with both imparting the same momentum. Say we remove the shield. If a flat side of each cube block hits you, with the same momentum, the iron would pack more punch, since the force would be concentrated on a smaller surface area.

 

In the next scenario, both blocks will hit you but using one of the very pointy apexes. The initial force will be the same for both, since the surface area is the same. However, since the foam is a weaker material than the iron, the sharp point of the foam will collapse absorbing energy, causing the surface area to increase. Less energy will go into you. The iron is stronger and would retain its sharp point sending almost all the energy into you.

 

Say we had a high enough velocity for both the iron and foam, such that the energy is not able to dissipate into the foam when it hits you. Now the foam and the iron would have a similar effect upon impact at the apex. The analogy is pressure washing with water. At high velocity even a liquid can't dissipate the collision energy and deform efficiently, making it abrasive. The foam would feel much tougher with all its momentum focused at a point longer.

Edited November 28, 2010 by pioneer

[ewmon]

FYI, you guys are talking about the the "cross-sectional density" of projectiles.

[swansont]

As I said, these are issues about geometry and type of collision, and not strictly density. I can make a low-density object that will give you something approximating an elastic collision and transfer a larger fraction of energy than something with higher density.

[John Cuthber]

" 'Momentum per square inch' really has no meaning"

True, but monentum per second per square inch is pressure. If you are talking about transfers then they take finite time so the concept isn't entirely meaningless- just muddled.

 

The original question is too poorly specified to be answerable.

A lump of iron and a lump of sponge differ in compressibility as well as density, which adds another complicating factor.

[Newbies_Kid]

Just forget the word momentum then and instead think about kinetic energy. Have you ever noticed that a ping pong ball thrown at you hurts a lot less than a rock thrown at you? If they both went at roughly the same speed, the rock would hurt more. Why? Because a rock has more atoms in it and therefore requires more energy to make it resist all the other forces otherwise holding it still to the ground. What I'm wondering though, is there a relationship between how much energy you put into that rock or pingpong ball, and how much energy it transfers to another object per square inch.

 

force and time are inversely proportional. An object with 100 units of momentum must experience 100 units of impulse in order to be brought to a stop. the greater the time over which the collision occurs, the smaller the force acting upon the object. Thus, to minimize the effect of the force on an object involved in a collision, the time must be increased. And to maximize the effect of the force on an object involved in a collision, the time must be decreased. In this case assuming that you're about to hit your belly with a rock and a pingpong ball at the same speed. You will feel getting hit by ping pong ball give more pain because the time duration of the impact is shorter than being hit by a rock. It similar when you throw an egg into a mattress instead letting it hit a wall.

Edited November 29, 2010 by Newbies_Kid

[Mr Skeptic]

Is there a relationship between the density of an object and how much momentum or energy it transfers to another object per square inch/2.54centimeters?

 

Sort of. If you throw something really fast, its density will be the dominant factor for the energy/momentum transfer speed (or if you prefer, how deep a hole it can poke into something). Hence one reason for the use of depleted uranium for ammunition. In addition, a denser object will generally lose less energy to drag on its way to the target.

 

If the object in question is soft enough or the speed slow enough, then the "softness" of the projectile is also an issue. A soft projectile will deform on impact, which will increase the surface area and lose kinetic energy dependent on the sort of forces involved in the deformation. Note that this is not necessarily better for the target; an unarmored human will take more damage from a bullet that deforms because it can leave a much larger hole, whereas the armor piercing bullet will pass right through them with less damage. As such, for soft targets extra surface area results in more momentum transfer because otherwise the bullet would go right through. This also relates somewhat to density, since less dense things are frequently (but not always) softer, eg a sponge or tennis ball vs a baseball, and so can deform more.

 

Another aspect of deforming is that such increases the time of contact, so lessens the instantaneous forces and energy transfer involved.

 

Finally, if you have an elastic collision, the result is (potentially) less energy transfer to the target and double the momentum transfer, however it is likely the target would be unharmed since damaging the target would make the collision inelastic.