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In inelastic collisions, the kinetic energy of the system is not conserved however momentum is still conserved.

How is this possible?

If momentum is conserved than we can write the following equation:

$$m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$$

Where $u$ represents initial velocity and $v$ represents final velocity.

If square the velocity on both sides in each term. Why doesn't the equality hold?

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    $\begingroup$ Are you asking about the cartoon scenarios where two objects stick together or something else? Inelastic collisions, in the introductory sense, tend to be dealt with in multiple steps. Regardless, inelastic is kind of a catch-all phrase that implies the collision generated some other kind of energy, taken from the loss of kinetic energy (e.g., heat). $\endgroup$
    – honeste_vivere
    Commented Feb 2 at 15:08
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    $\begingroup$ you can't separately square velocities on each side. For example, $2\cdot2=1\cdot4$ but $2\cdot2^2=8\neq16=1\cdot4^2$. Unless I misunderstood what you meant by that? $\endgroup$
    – jng224
    Commented Feb 2 at 15:10
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    $\begingroup$ Total energy is conserved in inelastic collision, but kinetic energy is not conserved. The missing kinetic energy is converted to thermal energy (i.e., heat) by plastic deformation of the colliding objects. $\endgroup$
    – Solomon Slow
    Commented Feb 2 at 15:26
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    $\begingroup$ Mathematically, the equation is only valid if you square the entire left and right side, not just the velocities on each side $\endgroup$
    – Bob D
    Commented Feb 2 at 15:30

3 Answers 3

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If the energy is conserved, we have $$m_1u_1^2+m_2u_2^2-m_1v_1^2-m_2v_2^2=0.\tag{1}$$ Your equation: $$m_1u_1+m_2u_2=m_1v_1+m_2v_2.\tag{2}$$ Now taking the square: \begin{align} (m_1u_1+m_2u_2)^2&=(m_1v_1+m_2v_2)^2\tag{3}\\ m_1^2u_1^2+m_2^2u_2^2+2m_1m_2u_1u_2&=m_1^2v_1^2+m_2^2v_2^2+2m_1m_2v_1v_2\tag{4}\\ m_1^2u_1^2+m_2^2u_2^2-m_1^2v_1^2-m_2^2v_2^2&=2m_1m_2(v_1v_2-u_1u_2)\tag{5} \end{align} Equation (5) is different from (1) in two ways. Firstly, each mass term in the left hand side of (5) is squared, which makes it an entirely different equation (unless $m_1=m_2$). Secondly, the right hand side of (5) now contains another term which is generally not zero. This term appears from the cross term of squaring. In other words $(a+b)^2\neq a^2+b^2$.

Side note: we are always allowed to change to a reference frame where the total momentum is zero. By taking $u'_1=u_1-U$ where $U=(m_1u_1+m_2u_2)/(m_1+m_2)$. In this frame, the total energy is zero after the collision.

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    $\begingroup$ What I don't understand is that momentum is given by P = mv and K.E by K.E = 1/2mv^2 I know inelastic collisions talks about the momentum being conserved for the whole system and not just one object, but since the velocity of both the objects change then why doesn't the total momentum? Some amount of energy was lost in an inelastic collisions so velocities of both the objects is now less than their initial velocities so momentum of each individual object decreases, thereby decreasing the overall momentum. $\endgroup$
    – Sher
    Commented Feb 4 at 5:57
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    $\begingroup$ Changing the momenta to ensure momentum conservation is easy. Just take $\Delta p_1=-\Delta p_2$ $\endgroup$
    – AccidentalTaylorExpansion
    Commented Feb 5 at 7:15
  • $\begingroup$ An example where momentum is conserves but energy is not is $u_1=u, u_2=0$ and then $u_1=u_2=u/2$ (assuming masses are equal). Here $\Delta p_1=-mu/2$ and $\Delra p_2=mu/2$. $\endgroup$
    – AccidentalTaylorExpansion
    Commented Feb 5 at 7:18
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In inelastic collisions, the kinetic energy of the system is not conserved however momentum is still conserved. How is this possible?

Think of them compressing a mutual one-way sticky spring (that only compresses and doesn't expand).

Let's say total momentum $L = m_1 v_1 + m_2 v_2$ where $v_1$ and $v_2$ are opposite in sign and the two masses are approaching one another.

Then, after the inelastic collision, the "new" (combined) object will have mass $M = m_1 + m_2$ and velocity $v$ = $L/M$. The new kinetic energy of this mass will be therefore be $$E_2 = \frac{L^2}{2M} < E_1 = \frac{m_1v_1^2}{2} + \frac{m_2v_2^2}{2}$$ with the difference $E_1 - E_2$ stored as elastic potential energy inside the spring.

So, all the pre-collision energy is still conserved (partly as kinetic energy and partly as spring elastic potential energy).

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  • $\begingroup$ If the two objects stick together, then the total momentum should be zero right? If lets suppose the two objects were approaching each other with different initial momentum so the total initial momentum is not zero then in this case momentum is not conserved? Or maybe this type of collision is not possible? $\endgroup$
    – Sher
    Commented Feb 4 at 5:46
  • $\begingroup$ "If the two objects stick together, then the total momentum should be zero right?" Not unless the total momentum of the system is zero (for example, if the two objects that collided and stuck together had equal but opposite momenta). "If lets suppose the two objects were approaching each other with different initial momentum so the total initial momentum is not zero then in this case momentum is not conserved?" Momentum would always be conserved. So momentum before and after the collision is still the same. "Or maybe this type of collision is not possible?" Why would it not be possible? $\endgroup$
    – ManRow
    Commented Feb 4 at 6:46
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Kinetic energy, during a collision, is lost, for eg., as sound energy. If there is no dissipation of energy in any manner (sound/heat/light) we call it an elastic collision, and then you can conserve the kinetic energy for all colliding particles.
Momentum is conserved for a system when there is no net external force on the system. If three balls were to collide at a point, you can't conserve the momentum of any two balls.
Momentum and energy are two very different mutually exclusive quantities, even though they are modeled with the same components. I can see why you feel they should follow the same rules, but they don't.

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