# What's the physical reason behind the increment of total kinetic energy of a system after an inelastic collision?

Suppose, an object of mass $m_2$ is at rest and an object of mass $m_1$ collides with the object of mass $m_2$ with a velocity $u$.

If the object of mass $m_1$ comes back in the opposite direction with the same speed it hit the other object, then the total kinetic energy of the system after the collision is:

$$\frac12 m_1 u^2 + \frac12m_2v^2$$

(let $v$ be the speed of the object of mass $m_2$ after the collision), which is greater than the system's total kinetic energy before the collision,

$$\frac12m_1u^2 + \frac12m_20^2$$

(the object of mass $m_2$ was at rest before collision, hence, its initial speed was $0$).

How does this increment of total kinetic energy of the system occur? What's the physical reason behind this?

• Have you applied the Law of Conservation of Momentum to this collision? If you do so, you'll find that the bodies can't have the velocities you say they have, after the collision. – Philip Wood Jul 4 '17 at 8:38
• Yes, they can. I have applied. – Hisab Jul 4 '17 at 9:02
• Hi @Hisab. I have added math formatting to your question and hope that all is still as intended. Otherwise, please correct me with an edit. You can do math formatting with \$...\$ and \$\$...\$\$. – Steeven Jul 4 '17 at 9:20
• okay.. @Steeven – Hisab Jul 4 '17 at 9:23
• In an elastic collision both momentum and kinetic energy are conserved: if you are finding that kinetic energy is not conserved then the collision is not, in fact, elastic. In other words: you've made a mistake. – tfb Jul 4 '17 at 10:18

If the object of mass $m_1$ comes back in the opposite direction with the same speed it hit the other object, then the total kinetic energy of the system after the collision is: $$\frac12m_1u^2+\frac12m_2v^2$$

Actually, this is not correct. Unless $v=0$ of course (which it would be for example when a ball hits a stationary wall). It is not possible for the object to bounce back with the same speed if the other object starts to move as well. They must share the energy present.

Otherwise, as you are pointing out, there is more energy in the system after than before. Some energy must have come into existence suddenly, which is impossible (from the energy conservation law) - energy must come from somewhere, so if it is not added to the system from the outside or from the inside (from fuel or deformations or alike) then the total energy in this system can't increase.

Edit - for the inelastic collision

For an inelastic collision, the difference from the above explanation is that there is energy added (or removed) to the system.

• When two cars crash, energy is absorbed to deform the metal. Total energy that goes as kinetic energy reduces.
• When two cars crash and one's fuel tank explodes, energy is absorbed to deform the metal but also much energy is released from the chemically stored energy in fuel. More energy is now going as kinetic energy than before the impact.

So, if you of some reason has a case where the speed of one object just before impact happens to equal it's speed right after impact, then the released energy just happens to be just enough for this to be equal. Had a tiny bit more energy been released/given to the system, then the speed might have been higher than before impact.

• The collision isn't elastic, it's inelastic. In an inelastic collision, total kinetic energy isn't conserved. The ball can bounce back with the same speed if $m_2 v = 2m_1 u$ . There's no problem with that. – Hisab Jul 4 '17 at 11:06
• @Hisab I see. The only addition to the answer is, that in that case, energy is added to the system. And this added energy just happens to be just right for maintaining the speed. See my edit. – Steeven Jul 4 '17 at 11:23

Sorry about my earlier comment: I'd misread your question. The collision is, of course, compliant with momentum conservation as long as $m_2 v = 2m_1 u$. In this case, there is a gain in total KE, equal to $\frac{1}{2}m_2 v^2$. This won't usually be the case, as you say. But it could be the case, for example if one of the bodies had a compressed spring inside it that was triggered to expand on collision, or a small explosive charge…

• I thought your law of conservation of momentum comment was appropriate. As per Steeven's answer, if the velocity of $m_1$ is $u$ before and after the collision; conservation of momentum requires $v = 0$. – JMac Jul 4 '17 at 10:50
• Steven's answer is wrong. The ball can bounce back with the same speed if $m_2 v = 2m_1 u$ – Hisab Jul 4 '17 at 11:05
• @Hisab That doesn't conserve momentum... The momentum before the collision was $m_1 u$, how does it become $2m_1 u + m_2 v$ after? – JMac Jul 4 '17 at 12:05
• The momentum after collision is : $-m_1u + m_2v$ . If $m_2v = 2m_1u$ , we get the momentum after collision is: $-m_1u + 2m_1u = m_1u$ – Hisab Jul 4 '17 at 12:13
• @Hisab I see what you're talking about now; but then that does violate energy conservation. You need to find values that satisfy the momentum equations and don't violate energy conservation. I think you should have included the $m_2 v = 2 m_1 u$ in the question itself. It would have been easier to explain why the reasoning falls apart. – JMac Jul 4 '17 at 13:05