In an inelastic collision kinetic energy is lost, so how can momentum still be conserved?

7 answers & replies

• [+][–]Vivek Mishra12thSample· 5d ago

  Momentum is conserved when there is no net external force. In a collision the two bodies push each other with equal and opposite forces (third law) for a short time, so the total momentum of the pair cannot change. That has nothing to do with whether the bodies bounce or deform. KE can go into heat, sound, deformation. Momentum cannot "go" anywhere because it is a vector that the internal forces cancel in.

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  • [+][–]Yash Tiwari11thSample· 5d agoAuthor

    Samajh gaya bhai, momentum energy se alag hai. Internal forces cancel in momentum, but energy can transform. Thank you sab ko 🙏

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• [+][–]Tushar PawarDropperSample· 5d ago

  In inelastic collisions momentum is also not conserved, because some of it is lost as heat and sound, just like KE. Only in elastic collisions are both conserved.

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  • [+][–]Siddharth RaoDropperSample· 5d ago

    That's the common mistake. Heat and sound take away energy, not momentum. In every collision (elastic or inelastic) momentum is conserved as long as external forces are negligible during the collision time. Only the kinetic energy is lost in inelastic ones. Yash's numbers (a) show it exactly: momentum 6 before and 6 after, energy 18 to 6.

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    • [+][–]Tushar PawarDropperSample· 5d ago

      Correct, I mixed the two. Energy has many forms, momentum doesn't.

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• [+][–]Neel Chatterjee12thSample· 5d ago

  Coefficient of restitution $e$ is the ratio of the relative speed of separation to the relative speed of approach: $e=\frac{v_2-v_1}{u_1-u_2}$

  • $e=1$: elastic (relative speed unchanged)
  • $e=0$: perfectly inelastic (bodies move together after collision, no separation)
  • $0<e<1$: ordinary inelastic collision

  Check on your elastic one: $(4-(-2))/(6-0)=1$ ✓. Inelastic: $(2-2)/6=0$ ✓.

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• [+][–]Nisha Verma12thSample· 5d ago

  Two shortcut facts for MCQs: if the masses are equal and the collision is elastic head-on, the bodies exchange velocities. A light body hitting a very heavy body at rest bounces back with nearly the same speed, and the heavy one hardly moves. Both are limits of Yash's elastic formulas, put $m_1=m_2$ (then $v_1=0$, $v_2=u_1$) or $m_2\gg m_1$ (then $v_1\approx-u_1$).

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