According to this paper authored by several employees from the Insurance Institute Highway Safety (IIHS), one test isn't better or worse than the other. They are different:
- full frontal tests gauge the ability to manage deceleration
- offset barrier tests gauge the ability to handle structural stress
Achieving low scores in full frontal barrier tests indicates that a vehicle's front end and restraint system can manage the forces associated with the high vehicle decelerations of full frontal crashes.
Low scores in offset barrier tests would indicate that the vehicle's front end structures are able to prevent intrusion even when the crash energy is concentrated on only part of the front structure.
On the other hand, crash tests performed by New Car Assessment Programs (NCAP) agencies around the world seem to indicate survivability is slightly worse in full-frontal tests than in offset tests.
There are three NCAP agencies that perform both tests, China's C-NCAP, Japan's J-NCAP and Korea's K-NCAP. I gathered statistics from J-NCAP (206 cars tested from 2001 to 2010) and C-NCAP (147 cars tested from 2007 to 2011). A summary of the scores (higher is better):
C-NCAP Full Frontal: 76%
C-NCAP 40% Offset: 88%
J-NCAP Full Frontal (Driver): 78%
J-NCAP 40% Offset (Driver): 82%
J-NCAP Full Frontal (Passenger): 81%
J-NCAP 40% Offset (Passenger): 91%
An interesting factor however is how vehicle length can impact the scores. There seems to be a stronger correlation between car length and test scores when full frontal tests are performed.
r = 0.406 for full frontal and r = 0.222 for 40% offset. (Pearson correlation coefficient)
r = 0.516, r = 0.353: still a stronger correlation when a full frontal test is performed.
r = 0.455, r = 0.088: unsurprisingly, there's nearly no correlation on the passenger's side since offset tests are done on the driver's side.
If the kinetic energy of both the ball decreases, how can their
velocities be equal?? The front one ( B ),from the beginning, had low KE; if
it decreases during the deformation, how can its velocity be equal to
the velocity of the rear ball ( A )?
In order to get a clear picture, let's consider the extreme case when the velocity of B = 0
Let's make a concrete example with numbers $m_A = 1, m_B = 2, M = 3$:
Suppose that:
$v_a = 6m/s$ and $v_b, p, E_k = 0 \rightarrow E_k = 0.5 * 6^2 = 18, p = 1 * 6 = 6, v_{cm} = p/M = 2$
Kinetic energy and momentum are conserved only in elastic collisions, but if the bodies stick together the collision is inelastic an only momentum is conserved:
After the collision velocity of A would be anyway lower as KE should be distributed among more mass, but some KE is lost in the crash. How much?
Momentum is conserved: $ p_{ab} = 6$ , from this datum you can calculate its velocity which now coincides with the velocity of center of mass:
$$v_{ab} = v_{cm}= \frac{6}{3} = 2$$ and $E_{AB} = 0.5 * 2^2 *3 = 6 \rightarrow E_A = 2 + E_B = 4$.
Some energy has been transferred to B (4 J), but two thirds of the kinetic energy(12 J) have been changed to other forms of energy. The general law of 'conservation of energy' has not been, anyway, violated
Velocity of center of mass is the same, although KE has changed. Note that momentum is conserved because we are assuming that on the surface of contact there is no friction.
I hope your main question has got an answer by now, velocities can and must be equal because AB is now one single body: the rear ball has decreased and the front ball has increased its v and the two values level out.
(This is not due to the loss of KE, even if it had been conserved the two bodies would have levelled their v to 3.464, but this would violate the principle of conservation of momentum that would have increased to 10.4)
As to the queries in your comments: when the bodies have reached the maximum deformation they will move at final and same v. It is impossible to determine how much of the amount of KE lost and transformed will be absorbed by each body, as this depends on the material they are made of: the more a body is deformable the more energy it will absorb
.. But what about the case when the front ball is moving?
It makes no difference! Just think of communicating vessels, once two bodies are joined and become a single body... energy, velocity and momentum level out and are unified.
Best Answer
when metal is crumpled, work is performed on it. Part of that work is dissipated as heat, which warms the metal slightly, as the metal yields under the applied stresses and deforms plastically. You can demonstrate this yourself by rapidly bending a piece of coat hanger wire back and forth, and then feeling it at the bend. The rest of the work input gets stored as strain energy in the deformed metal itself. this strain energy increases the hardness and subsequent yield strength of the deformed metal, which you can also demonstrate to yourself by trying to unbend the bent portion of the coat hanger: it takes less work to bend an unbent segment of the wire adjacent to the bent part than it does to unbend that bent part.