0
$\begingroup$

Object A has mass $m_A$, and is travelling west with velocity $v_A$. Object B has mass $m_B$ and is travelling east with velocity $-v_B$. The objects collide in a perfectly inelastic collision so that the velocities of the two objects after the collision are the same. Prior to the collision, the objects are made of uniform, identical materials and shapes.

How much of the kinetic energy is absorbed by deformation of object A, and how much of the kinetic energy is absorbed by deformation of object B?

If there is not a unique solution to this question because of a lack of necessary constraints, please answer the question with some examples for the unspecified constraints.

EDIT: when I said same shapes, I meant they are both cylinders or prisms or spheres or something. The dimensions and masses of the shapes may be different.

$\endgroup$
11
  • 1
    $\begingroup$ Applying the conservation of energy gives us the energy of deformation of both object, however a portion of theoretical deformation energy has already been dissipated as heat, by knowing the properties of both materials and the geometry of object after collision we can estimate the stored strain energy. $\endgroup$ Nov 3 '18 at 7:38
  • $\begingroup$ @SolarMike last time I played snooker, I didn't get many inelastic collisions between the balls. $\endgroup$
    – alephzero
    Nov 3 '18 at 13:59
  • $\begingroup$ The stored energy could be anything between "100% in A", or "100% in B", (suppose A was a metal block and B was a lump of putty, or vice versa) and "no energy stored at all in A or B." For the last case two cubes with the colliding faces coated with a very thin layer of adhesive, so they stick together but there is no residual deformation after the impact. All the "excess kinetic energy" is dissipated as heat and disturbances in the surrounding air, not stored anywhere. $\endgroup$
    – alephzero
    Nov 3 '18 at 14:06
  • $\begingroup$ @Sam Farjamirad, how about the objects, prior to collision, are identical flat plates of steel bolted to 100mm-diameter, 4mm-thick tubes of 6061 aluminum with identical lengths adequate to absorb all the kinetic energy, and backed by blocks of steel of possibly varying mass. The aluminum tubes are uniaxially loaded and collapse in concertina mode. $\endgroup$ Nov 3 '18 at 14:28
  • 1
    $\begingroup$ In the case of the objects made out of the same material and same shape (which implies same mass) but different velocities, both will absorb the same amount of energy as they will experience the same force at the impact. $\endgroup$
    – user190081
    Nov 3 '18 at 16:07
0
$\begingroup$

This is one-dimensional travel (along a line). Applying conservation of energy and momentum for a perfectly elastic collision, we would find

$$ \left(m_A + m_B\right) v_{Af}^2 - \left(m_A v_{Ao}^2 + m_B v_{Bo}^2\right) = 0 $$

$$ \left(m_A v_{Af} + m_B v_{Bf}\right) - \left(m_A v_{Ao} - m_B v_{Bo}\right) = 0 $$

This is two equations and two unknowns (final velocities).

In a PERFECTLY inelastic collision, we have

$$ \left(m_A + m_B\right) v_{f}^2 - \left(m_A v_{Ao}^2 + m_B v_{Bo}^2\right) = 2\left(q - w - (m_A + m_B)\Delta\tilde{U}\right) $$

$$ \left(m_A + m_B\right)v_f - \left(m_A v_{Ao} - m_B v_{Bo}\right) = 0 $$

where $q$ is the heat loss, $w$ is work done on the object(s), and $\tilde{U}$ is the specific internal energy (J/kg) of the combined system. Solve the momentum equation for $v_f$. For an adiabatic collision of perfectly rigid objects, find the temperature change from

$$ \left(m_A + m_B\right) v_{f}^2 - \left(m_A v_{Ao}^2 + m_B v_{Bo}^2\right) = -2\left(m_A + m_B\right)\tilde{C}_V\Delta T $$

When energy is absorbed, $KE_f < KE_i$, and the temperature of the final object is greater than the initial temperature. When the solids are perfectly incompressible, $\tilde{C}_V = \tilde{C}_p$. For an isothermal collision of perfectly rigid objects,

$$ \left(m_A + m_B\right) v_{f}^2 - \left(m_A v_{Ao}^2 + m_B v_{Bo}^2\right) = 2q $$

Again, when energy is absorbed,$KE_f < KE_i$, and the process is exothermic (heat leaves). Finally, for perfectly plastic objects where all the energy is absorbed in work,

$$ \left(m_A + m_B\right) v_{f}^2 - \left(m_A v_{Ao}^2 + m_B v_{Bo}^2\right) = -2w $$

When the final kinetic energy is less than the sum of the two starting objects, the final object will experience work done ON it. You can pretend this is equivalent to $p\Delta V$ and say, the final object has a total volume that is less than the sum of the two starting objects.

The net kinetic energy change for object A is $(1/2)m_A (v_{f}^2 - v_{Ao}^2)$. Apply the same for object B.

As for energy absorbed by the objects individually, the answer is ambiguous. The objects cease to exist as individual objects at the moment of the collision. Any energy lost is absorbed by the final combined object, not by any one individually.

$\endgroup$
8
  • $\begingroup$ Oops! Fixed it. $\endgroup$ Nov 3 '18 at 17:45
  • $\begingroup$ When a bug hits your windshield, I guess you could say the bug has become part of the car, but it's pretty clear that the bug has taken on more strain energy than the car. $\endgroup$ Nov 5 '18 at 2:23
  • $\begingroup$ @JeffreyHood Your original problem has both A and B of the same material, presumably solids. The analysis that is needed to handle a collision of two different materials is an entire course on the mechanics of elastic, plastic, and brittle solids (let alone that we should then consider liquids and compressible gases). $\endgroup$ Nov 5 '18 at 3:28
  • $\begingroup$ That's true. For all I know your answer may very well be correct. If I knew the answer I wouldn't have asked the question. In the end, I'm trying to size beams for a vehicle crumple zone and am a bit out of my domain. Maybe I should be more specific? Otherwise, if the answer is ambiguous, would you say I would need to use another factor to select them other than the potential crushing energy? Do you have any way to make the question I wrote unambiguous? $\endgroup$ Nov 5 '18 at 4:17
  • $\begingroup$ @JeffreyHood for a car, as you now specify, if it hits a tree then the final velocity is not the same as the original velocity - except for that of the tree... $\endgroup$
    – Solar Mike
    Nov 5 '18 at 7:45
0
$\begingroup$

For Simplicity we assume the objects are moving on the same path but opposite direction, however it would be only a bit complex if they were not moving colinearly if one introduces vectors.

the momentum of particles before the collision and after are preserved.

$m_Av_A + m_Bv_B = (m_A +m_b)v_{(A+B)} $

So let's consider object A. It has changed it's velocity by the amount $v_A -v_{(A+B)}$

So the kinetic energy it has lost is $ E_k= \frac{ m \space[v_A -v_{(A+B)}]^2}{2}$

And same way we can calculate the loss for B.

$\endgroup$
2
  • $\begingroup$ I think you mean m * (v_A ^ 2 - v_(A+B) ^ 2) / 2. In any case, my question is not what the change in kinetic energy is, but rather where the lost kinetic energy goes. $\endgroup$ Nov 5 '18 at 2:29
  • $\begingroup$ @Jeffrey Hood, no i meant what i wrote. As to where the lost energy goes, to plastic deformation. Vibration of the two masses which will die out to heat radiation. rearranginging of friction loaded structural scaffolding of the material. $\endgroup$
    – kamran
    Nov 5 '18 at 5:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.