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This is the question:

When cars are equipped with flexible bumpers, they will bounce off each other during low-speed collisions, thus causing less damage. In one such accident, a 1850 kg car traveling to the right at 1.60 m/s collides with a 1450 kg car going to the left at 1.10 m/s . Measurements show that the heavier car's speed just after the collision was 0.270 m/s in its original direction. You can ignore any road friction during the collision.

I solved this problem correctly with conservation of momentum: (m1v1 + m2v2)i = (m1v1 + m2v2)f. The result was 0.597 m/s.

However, when I attempted to use the equations for elastic collisions, shown here: enter image description here

I ended up with v2,f = 1.927 m/s. Why is the second equation in the picture wrong? I thought that if a collision is elastic, these equations always apply.

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  • $\begingroup$ They may cause less damage to the cars, but not necessarily to the occupants. In an inelastic collision the vehicles crumble absorbing energy that would otherwise be absorbed by the occupants. $\endgroup$
    – Bob D
    Nov 14, 2019 at 23:57
  • $\begingroup$ We have MathJax running on the site which means that you can write mathematics that will be neatly marked up using more or less LaTeX math-mode syntax. This is prefered to weither plain ascii math or (especially) images of math. $\endgroup$
    – dmckee --- ex-moderator kitten
    Nov 15, 2019 at 1:14
  • $\begingroup$ Follow the procedure I outlined in this answer to check your work. $\endgroup$
    – John Alexiou
    Nov 16, 2019 at 15:36
  • $\begingroup$ If this collision conserved energy, a final velocity would not be given. It could be calculated. (I would call this a non-elastic collision.) $\endgroup$
    – R.W. Bird
    Nov 16, 2019 at 18:42

2 Answers 2

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I suspect the discrepancy arises because the collision was not elastic.

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  • $\begingroup$ Is it because it is not completely, perfectly elastic? Because I thought that if two objects bounce off of one another then it is elastic. $\endgroup$
    – Zachery Thach
    Nov 14, 2019 at 23:14
  • $\begingroup$ Yes indeed. Maybe Zachery Thach does not know that an elastic collision means something very precise in Physics: a collision in which kinetic energy is conserved. Collisions between macroscopic bodies (like cars) are never properly elastic. $\endgroup$
    – Philip Wood
    Nov 14, 2019 at 23:15
  • $\begingroup$ Hi Zachery. There are degrees of elasticity. Perfect elasticity means that the objects will bounce off each other in a way that conserves their original combined KE. As the elasticity reduces, so do their recoil velocities. $\endgroup$
    – Marco Ocram
    Nov 14, 2019 at 23:18
  • $\begingroup$ @MarcoOcram At least in the books I learned from "elastic" is often reserved only for the perfect case and there are degrees of inelasticity. A collision where the particpants stick and move off as one is "completely inelastic", and all case that are neither elastics nor completely inelastic are described as "inelastic". (Special case: the ther is more kinetic energy in the final state you can use "super-elastic".) $\endgroup$
    – dmckee --- ex-moderator kitten
    Nov 15, 2019 at 1:12
  • $\begingroup$ @dmckee You're right. It would be better to speak of degrees of inelasticity. $\endgroup$
    – Marco Ocram
    Nov 15, 2019 at 6:56
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The equations for plastic collision with a coefficient of restitution $\epsilon$ are

$$\begin{aligned} v_1^f & = \frac{m_1 - \epsilon m_2}{m_1+m_2}v_1 + \frac{\epsilon m_2+m_2}{m_1+m_2} v_2 \\ v_2^f & = \frac{\epsilon m_1+m_1}{m_1+m_2} v_1 + \frac{m_2 - \epsilon m_1}{m_1+m_2} v_2 \end{aligned} $$

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