Collisions: Elastic vs inelastic

[lendav_rott]

So there are a bunch of assignments in physics built on the conservation of momentum law where a bullet of some mass, hits a target of some mass, neglecting friction find the velocity at which the target starts moving. That is all very simple in case of an inelastic collision, all the energy of the bullet is transformed to the target+bullet mass, but what happens when it's a real scenario?

Assuming we know the mass of the bullet m_b and the power behind the rifle - the bullet gains an initial velocity of v_initial. The target is, say, a metal sphere of mass m_s at a distance of 200m - How do we calculate the aerial friction that slows down the bullet, assuming there is no wind to considerably change its direction?

At last the bullet hits the target at a velocity of v_final. How much energy does the sphere exactly gain?
The bullet hits the sphere and bounces back,there is likely a dent in the sphere and the bullet is deformed, therefore some of the kinetic energy is transformed into mechanical energy and heat. Since the bullet bounces back I would assume it has something to do with Hooke's law, where the surface of the sphere is acting like a spring. How much energy is consumed by deformation and the spring? Is there a way to know how much of the energy the sphere "gets to use"? To what extent can we use the conservation of momentum in this scenario? What are all the elements we have to consider?

 

[voko]

The flight of the bullet is studied in external ballistics. What happens when the bullet hits the target is studied in terminal ballistics. Things get hairy fairly quickly as soon as you start looking at the details.

 

[Enigman]

So there are a bunch of assignments in physics built on the conservation of momentum law where a bullet of some mass, hits a target of some mass, neglecting friction find the velocity at which the target starts moving. That is all very simple in case of an inelastic collision, all the energy of the bullet is transformed to the target+bullet mass, but what happens when it's a real scenario?

Assuming we know the mass of the bullet m_b and the power behind the rifle - the bullet gains an initial velocity of v_initial. The target is, say, a metal sphere of mass m_s at a distance of 200m - How do we calculate the aerial friction that slows down the bullet, assuming there is no wind to considerably change its direction?

Assuming bullet's spherical: http://en.wikipedia.org/wiki/Stokes'_law

At last the bullet hits the target at a velocity of v_final. How much energy does the sphere exactly gain?

Depends on coefficient of restitution: http://en.wikipedia.org/wiki/Coefficient_of_restitution

The bullet hits the sphere and bounces back,there is likely a dent in the sphere and the bullet is deformed, therefore some of the kinetic energy is transformed into mechanical energy and heat. Since the bullet bounces back I would assume it has something to do with Hooke's law, where the surface of the sphere is acting like a spring. How much energy is consumed by deformation and the spring?

Hooke's law doesn't apply after you reach yield strength:
http://en.wikipedia.org/wiki/Stress–strain_curve
After which elastic deformation gives way to:
http://en.wikipedia.org/wiki/Plastic_deformation#Plastic_deformation

Is there a way to know how much of the energy the sphere "gets to use"?

KE should be given by coefficient of restitution, if that's what you mean.

To what extent can we use the conservation of momentum in this scenario? What are all the elements we have to consider?

Momentum should be conserved regardless of the type of collision.

 

[AlephZero]

Assuming bullet's spherical: http://en.wikipedia.org/wiki/Stokes'_law

Stokes' law does not apply to this. It is only relevant for very slow fluid flow.
These are better links:
http://en.wikipedia.org/wiki/Drag_(physics)
http://www.grc.nasa.gov/WWW/k-12/rocket/shaped.html

 

[Enigman]

Forgot about the laminar flow clause... thanks for the correction;
I seem to be ticking with the IQ of a tick today...