# Conservation of kinetic energy in perfectly Elastic collision

Suppose two bodies of mass $$m_1$$ and $$m_2$$ moving with velocity $$u_1$$ and $$u_2$$ and they collide (perfectly elastic collision) and after collision their velocities are $$v_1$$ and $$v_2$$.
Then by law of conservation of linear momentum. $$m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$$ And if I am given only this information is it possible to prove that the initial kinetic energy of the system is equal to final kinetic energy of the system, or the information is insufficient? Any help would be appreciated.

• Are you given all the specific values of initial velocities, final velocities, and masses? Are there any unknowns? Sep 27, 2021 at 22:26

By the very definition of an elastic collision, both kinetic energy and momentum are conserved. If it were any collision, you are certain that if no net external forces are acting on the system, the initial momentum equals the final momentum. However, only in perfectly elastic collisions is the kinetic energy conserved.

• does that mean there is no mathematical proof of the conservation of kinetic energy in perfectly elastic collision?
– user297948
Sep 27, 2021 at 16:59
• We define the perfectly elastic collision is one in which the kinetic energy is conserved. It is the basis of further analysis of this type of collision. Sep 27, 2021 at 17:02

The equation for the conservation of momentum does not imply that kinetic energy is equal before and after a collision - oftentimes, it is not. For any input parameters $$m_1, m_2, u_1, u_2$$, there are an infinite number of possible $$v_1$$ and $$v_2$$ that will satisfy the equation and the conservation of momentum. There are few solutions that will also satisfy the conservation of kinetic energy. The fact that a collision is elastic by definition implies conservation of kinetic energy, but the mere conservation of momentum does not imply conservation of kinetic energy (inelastic collisions conserve momentum but not KE).

• Thank you for your answer I understood the concept very clearly.
– user297948
Sep 27, 2021 at 17:08

Conservation of linear momentum doesn't equate to collision being elastic ,so no you can't prove energy conservation using that equation

• Sorry I forgot to mention that the collision is perfectly elastic.Also I need the mathematical proof of conservation of kinetic energy.
– user297948
Sep 27, 2021 at 16:55
• Suppose the particles were moving towards each other with u1 and u2 and after collision they move in opposite direction of their initial velocity with speed v1 and v2 . Then u1 + u2 = v1 + v2 (sorry for my bad editing ) use this equation along with conservation of momentum you can easily prove energy conservation. Sep 27, 2021 at 16:59
• you said that for elastic collision the coefficient of restitution becomes 1 but this is proved by the conservation of kinetic energy and that is my question.
– user297948
Sep 27, 2021 at 17:10
• I think I made it look confusing collision is elastic iff energy is conserved,this is by definition. About coefficient of restitution e = relative velocity of separation/ relative velocity of approach Sep 27, 2021 at 17:16

Let's suppose two bodies collide the total kinetic energy from the encounter is equal because of the elastic net conversion K = 1/2 m v2 which is adequate and satisfies the law of conservation in kinematics as well as in non-linear dynamics hence I would say the total energy is conserved throughout unless it were to be acted upon otherwise, so in technicality this would be a perfect frame that we discuss of to have it be conserved. I hope I gave you a bit more of a grasp on the coefficient and those parameters!

• Welcome to Physics! I don't think this answers the question, though; it seems to boil down to "kinetic energy may be conserved", but the question was asking whether it must be conserved. Sep 27, 2021 at 17:28

If for a system:$$\sum F_{external}=0$$ Then $$\sum_i(m_iv_i)=constant$$ Or simply $$m_1u_1+m_2u_2=m_1v_1+m_2v_2$$ ; for a two body system. But we cant conclude about $$KE=constant$$, simply from this equation.

So here we use concept of elasticity. In elastic collisions, kinetic energy remains constant while in inelastic collision it doesn't remains constant.

For checking nature of collision we use Newton's coefficient of restitution ($$e$$), which is defined as:

$$e=\left|\frac{\Delta v}{\Delta u}\right|$$ Here, $$\Delta v$$ is velocity of seperation (after collsion) and $$\Delta u$$ is velocity of approach (before collision).

So we can clearly see that $$e\in [0,1]$$.

So for, $$e=1$$, collision is perfectly elastic and kinetic energy remains conserved.

For other values of $$e$$, we have inelastic collsion, and energy is wasted in form of friction, heat, sound or other forms of energy. So, for our real non-idealistic world, $$e\in (0,1)$$.