We're assuming (forcing) a perfectly inelastic collision.Heiwa wrote:Maybe A has velocity 10 m/s and C is stationary?
lol reading this after my last post from the "missing jolt found, film at 11" thread:
there should be a jolt, according to the NIST/bazant scenario of one storey of freefall.
sure, the little connections around the perimeter could act like mini hinges, MAYBE, but, not the mighty core. when the upper core chunk hits the lower bit, there should absolutely be a big jolt. i've never seen any two like objects collide and not slow by half (once again, excluding a house of cards).
in a pure elastic collision, nearly one hundred percent of kinetic energy is retained, but the direction of the kinetic force goes in the opposite direction (up, in this case).
in a purely inelastic collision, one hundred percent of kinetic energy is spent and the two objects merge.
so, if the objects merge in the crush zone, they have spent all their potential energy. if they react elastically, they are then exerting force upwards on the mass above them.
either way, it seems like double dipping to me to say that the kinetic energy is retained after the collisions because all of the potential/kinetic energy wasn't spent breaking connections and crushing concrete, file cabinets, desks, cubicles, etc. the retained kinetic energy after a collision is directed AGAINST the downward drive of the mass higher up.
either type of collision should show either a jolt (elastic) or a deceleration (inelastic).
my simple opinion.
there should be a jolt, according to the NIST/bazant scenario of one storey of freefall.
sure, the little connections around the perimeter could act like mini hinges, MAYBE, but, not the mighty core. when the upper core chunk hits the lower bit, there should absolutely be a big jolt. i've never seen any two like objects collide and not slow by half (once again, excluding a house of cards).
in a pure elastic collision, nearly one hundred percent of kinetic energy is retained, but the direction of the kinetic force goes in the opposite direction (up, in this case).
in a purely inelastic collision, one hundred percent of kinetic energy is spent and the two objects merge.
so, if the objects merge in the crush zone, they have spent all their potential energy. if they react elastically, they are then exerting force upwards on the mass above them.
either way, it seems like double dipping to me to say that the kinetic energy is retained after the collisions because all of the potential/kinetic energy wasn't spent breaking connections and crushing concrete, file cabinets, desks, cubicles, etc. the retained kinetic energy after a collision is directed AGAINST the downward drive of the mass higher up.
either type of collision should show either a jolt (elastic) or a deceleration (inelastic).
my simple opinion.
Are you sure ? In the perfect inelastic collision in my opening example, 50% of the KE was *spent*, but that does depend upon the ratio of masses, so it's not a fixed number in all cases.newton wrote:in a purely inelastic collision, one hundred percent of kinetic energy is spent
Do you mean momentum ?it seems like double dipping to me to say that the kinetic energy is retained after the collisions
i'm just going by the rules. in a perfectly inelastic collision, all the kinetic energy is absorbed in deformation. by that definition, rigid bodies cannot collide inelastically.femr2 wrote:Are you sure ? In the perfect inelastic collision in my opening example, 50% of the KE was *spent*, but that does depend upon the ratio of masses, so it's not a fixed number in all cases.newton wrote:in a purely inelastic collision, one hundred percent of kinetic energy is spent
Do you mean momentum ?[/quote]it seems like double dipping to me to say that the kinetic energy is retained after the collisions
well, what i'm trying to get at, is.... momentum is mass in motion in one direction. after a collision, assuming equal masses for simplicity, half the momentum is transfered to the object hit, and half is retained in the hitter, but now in the opposite direction.
conservation of momentum doesn't factor in direction of force.
so, when debunkers say that the kinetic energy has leftover power, they are ignoring the vector.
kinetic energy does have a direction.
Do you mean ALL KE of the whole system, or the *spent* portion ?newton wrote:in a perfectly inelastic collision, all the kinetic energy is absorbed in deformation.
Can rigid bodies collide elastically ?by that definition, rigid bodies cannot collide inelastically.
the whole system.
rigid bodies can only collide elastically.
that's my understanding, anyway.
rigid bodies can only collide elastically.
that's my understanding, anyway.
- OneWhiteEye6,304336,30433
Twice as strange: in an elastic collision between equal masses where one body has KE = 0.5mv^2 and the other KE = 0, after collision the first has KE = 0 and the other KE = 0.5mv^2 (= T). It took T units of energy to decelerate the first from v to a stop and T units of energy to accelerate the second from rest to v. But there only was ever T units of energy in the system, not 2T!femr2 wrote:I cried foul, and pointed out that the energy difference exactly matched the energy required to accelerate one body and decelerate the other, and that that was where the energy went.
Ah, but as it's an inelastic collision, both masses end up at the same velocity, so for the example ...OneWhiteEye wrote:Twice as strange: in an elastic collision between equal masses where one body has KE = 0.5mv^2 and the other KE = 0, after collision the first has KE = 0 and the other KE = 0.5mv^2 (= T). It took T units of energy to decelerate the first from v to a stop and T units of energy to accelerate the second from rest to v. But there only was ever T units of energy in the system, not 2T!femr2 wrote:I cried foul, and pointed out that the energy difference exactly matched the energy required to accelerate one body and decelerate the other, and that that was where the energy went.
We have 50J in the system pre-impact, and only 25J afterwards.
m1 decelerates from 10m/s to 5m/s, and m2 accelerates from 0m/s to 5m/s, instantly.
m1... 0.5 * 1 * 5^2 = 12.5 J (decelerate from 10 to 5)
m2... 0.5 * 1 * 5^2 = 12.5 J (accelerate from 0 to 5)
Add 'em up and we get 25 J, leading me to proclaim...aha...there is everything accounted for ! The energy is *spent* changing the velocities of the masses ! None *used* deforming my rigid masses, because they can't deform !
Looks like it makes sense, yes ?
And yet I was STILL missing the point.
But you saw in my example of perfect inelastic collision that only 50% of the system energy was converted into another form. The two masses joined together and both carried on at a lower velocity. You are saying that no matter how fast the first mass is going, that both masses should end up stationary after the collision. There's no parameters in the math which denote rigidity, so the math wouldn't change if the mass wasn't rigid. I *spent* only half the system energy in the collision. Did I do anything wrong ? I don't think the calcs are wrong. I think the CoM calc is right (which is where the change in velocity is calculated, with the assumption that an inelastic collision has taken place, and both masses therefore travel at the same velocity after the impact. Momentum must be conserved.) If both masses are stationary after the collision, momentum is not conserved.newton wrote:the whole system.
What must happen to the physical mass during an elastic collision ?rigid bodies can only collide elastically.
that's my understanding, anyway.
Can a rigid mass do that ?
Hm, funny mass C of 1 kg and velocity 10 m/s collides with stationary funny mass A also of 1 kg and applies its energy 50 J to A. After collision (C is glued to A and accelerates A) both A and C has same velocity 7.07 m/s and move as one mass of 2 kg glued together with energy 50 J.
Now - before collision/glueing together) C had momentum 10 kgm/s. After collision A+C (now one mass of 2 kg) have momentum 14.14 kgm/s (and energy 50J).
Imagine that A+C (2 kg at 7.07 m/s) collides with stationary D (1kg) and forms A+C+D (3kg) proceeding at 5.77 m/s (and 50 J energy). But the momentum is 17.31 kgm/s
And so on. When 100 masses A, C, D ... are glued together they proceed at 1 m/s velocity (50 J energy is still there) and the momentum is 100 kgm/s.
What do we learn from above? Evidently that a mass A of 1 kg moving at 10 m/s has same kinetic energy as a mass X of 100 kg moving at 1 m/s, i.e. 50 J, but that A and X have different momentums; A:s is 10 kgm/s and X:s is 100 kgm/s. It is simple physics. And the glue! It slows things down but adds momentum.
Now - before collision/glueing together) C had momentum 10 kgm/s. After collision A+C (now one mass of 2 kg) have momentum 14.14 kgm/s (and energy 50J).
Imagine that A+C (2 kg at 7.07 m/s) collides with stationary D (1kg) and forms A+C+D (3kg) proceeding at 5.77 m/s (and 50 J energy). But the momentum is 17.31 kgm/s
And so on. When 100 masses A, C, D ... are glued together they proceed at 1 m/s velocity (50 J energy is still there) and the momentum is 100 kgm/s.
What do we learn from above? Evidently that a mass A of 1 kg moving at 10 m/s has same kinetic energy as a mass X of 100 kg moving at 1 m/s, i.e. 50 J, but that A and X have different momentums; A:s is 10 kgm/s and X:s is 100 kgm/s. It is simple physics. And the glue! It slows things down but adds momentum.
this i fun!
heiwa, conservation of momentum is a law. momentum can't increase without adding energy.
one collision slows by half, another collision slows by one third. three times the mass, after three collisions, half speed after first, and then that speed is reduced by a ratio of 2/1 after the third impact, momentum is conserved because the increase in mass is the reciprocal of the decrease in velocity.
okay, 1kg going 10m/s hits 1 kilogram going 0 m/s in outer space. how perfectly rigid bodies act, is like newton's cradle. one stops, the other takes off at the same speed. if it can't deform, it is perfectly elastic.
if you allow a 50/50 elastic, inelastic response, then the objects cannot be rigid anymore, and must be allowed to deform. deformation is part of inelastic response. so, that is the scenario where two objects can be glued together and move away at half speed after the collision.
and, yes, if the two objects have a perfect inelastic collision, all the energy is spent in deformation, joining one object to the other and coming to a complete stop. (i'd love to see this collision in outer space! i can't imagine it.)
once again, at least that's my understanding.
heiwa, conservation of momentum is a law. momentum can't increase without adding energy.
one collision slows by half, another collision slows by one third. three times the mass, after three collisions, half speed after first, and then that speed is reduced by a ratio of 2/1 after the third impact, momentum is conserved because the increase in mass is the reciprocal of the decrease in velocity.
okay, 1kg going 10m/s hits 1 kilogram going 0 m/s in outer space. how perfectly rigid bodies act, is like newton's cradle. one stops, the other takes off at the same speed. if it can't deform, it is perfectly elastic.
if you allow a 50/50 elastic, inelastic response, then the objects cannot be rigid anymore, and must be allowed to deform. deformation is part of inelastic response. so, that is the scenario where two objects can be glued together and move away at half speed after the collision.
and, yes, if the two objects have a perfect inelastic collision, all the energy is spent in deformation, joining one object to the other and coming to a complete stop. (i'd love to see this collision in outer space! i can't imagine it.)
once again, at least that's my understanding.
My simple calc showed the post-impact velocity to be 5m/s. Where do you get 7.07m/s from ?Heiwa wrote:Hm, funny mass C of 1 kg and velocity 10 m/s collides with stationary funny mass A also of 1 kg and applies its energy 50 J to A. After collision (C is glued to A and accelerates A) both A and C has same velocity 7.07 m/s and move as one mass of 2 kg glued together with energy 50 J.
But as you know, momentum must be conserved for both elastic and inelastic collision.Now - before collision/glueing together) C had momentum 10 kgm/s. After collision A+C (now one mass of 2 kg) have momentum 14.14 kgm/s (and energy 50J).
But your momentum has increased. CoM has been violated.
That an error has been made, and the principle of Conservation of Momentum has been violated.Imagine that A+C (2 kg at 7.07 m/s) collides with stationary D (1kg) and forms A+C+D (3kg) proceeding at 5.77 m/s (and 50 J energy). But the momentum is 17.31 kgm/s
And so on. When 100 masses A, C, D ... are glued together they proceed at 1 m/s velocity (50 J energy is still there) and the momentum is 100 kgm/s.
What do we learn from above?
The physics are simple, but the problem I didn't *get* when I didn't apply the physics properly is the real purpose of this thread. I'm not trying to deceive anyone here, but I think it's useful for us each to work through it ourselves. You can't add momentum from nowhere. There's no problem with my numbers, it's the definition of the model and the constraints which create a paradox. Have a look at the last few posts by OWE.It is simple physics. And the glue! It slows things down but adds momentum.
temporary deformation? compression/expansion return to equilibrium aka vibration?femr2 wrote:
What must happen to the physical mass during an elastic collision ?rigid bodies can only collide elastically.
that's my understanding, anyway.
Can a rigid mass do that ?
it's interesting, because the title of the thread has "in the real world". yet, there are no perfectly rigid bodies in the real world, and no perfectly elastic collisions.
Are you absolutely sure that a rigid body can behave correctly during an elastic collision ?newton wrote:okay, 1kg going 10m/s hits 1 kilogram going 0 m/s in outer space. how perfectly rigid bodies act, is like newton's cradle. one stops, the other takes off at the same speed. if it can't deform, it is perfectly elastic.
What work is done ?
I agree, but I removed the *anymore*, as even an elastic response is a problem for rigid bodies.if you allow a 50/50 elastic, inelastic response, then the objects cannot be rigid
Absolutely.deformation is part of inelastic response
That would violate CoM.if the two objects have a perfect inelastic collision, all the energy is spent in deformation, joining one object to the other and coming to a complete stop.
