# Relation between Coefficient of Restitution and K.E

I know that Coefficient of Restitution(e) is defined as $$e = \frac{\mathbb{velocity\,of\,separation}}{\mathbb{velocity\,of\,approach}}$$ $$=\frac{v_2-v_1}{u_1-u_2}$$ Also for perfectly elastic collision $$e = 1$$ i.e. all kinetic energy is restored and in perfectly inelastic collision $$e=0$$ i.e. no kinetic energy is restored.
So is there any direct relation (equation,formula) between Coefficient of restitution and Kinetic energy.
Also I wrote a program in 10 min to create a graph of ($$e$$,% decrease in K.E.). I think it should go to 100% at 0 but even at $$e=10^{-10}$$ the % decrease in K.E. doesn't even go 10%.(It is only precise up to 11 decimal places) So my question is what is the relation between Coefficient of Restitution and Kinetic energy.
Edit: As stated by many answers that,$$\mathbb{\Delta K.E. = (1-e^{2})K.E._i}$$Thus I have created a newly modified graph according to above result. So is there something wrong as I don't think that my calculation is wrong because I have followed the procedure.
After taking the parameters $$m_1$$,$$m_2$$,$$u_1$$,$$u_2$$;I found $$v_1$$ and $$v_2$$ using the equations $$m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$$ and $$e = \frac{v_2-v_1}{u_1-u_2}$$.
After this I found initial kinetic energy $$K.E._i = \frac{1}{2}m_1u_1^{2} + \frac{1}{2}m_2u_2^{2}$$ and $$K.E._f = \frac{1}{2}m_1v_1^{2} + \frac{1}{2}m_2v_2^{2}$$Thus, percent decrease $$\mathbb{percent\,decrease} = \Biggl(\frac{K.E._i - K.E._f}{K.E._i}\Biggr)100$$So am i doing something wrong.

• I think the decrease in KE depends on the ratio of masses, or the reduced mass of the system. Sep 30, 2021 at 6:43
– user297948
Sep 30, 2021 at 6:44
• I elaborated in my answer below. Sep 30, 2021 at 7:15
• Thank you to all for their answers. I found every answer interesting. I couldn't accept all the answers (the green tick) but I do would accept if I could.
– user297948
Sep 30, 2021 at 8:36

I think that you have the wrong idea about what the term perfectly inelastic means when it comes to the loss of kinetic energy.

Whereas the term perfectly elastic does mean that kinetic energy is conserved, perfectly inelastic does not necessarily mean that all the kinetic energy is lost.

For example if two objects, originally moving in the same direction, collide and stick together, the kinetic energy cannot become zero as linear momentum has to be conserved.
However, if the objects are moving in opposite direction and the magnitude of their linear momentums is the same then after sticking together the kinetic energy is zero.

If you write down the equation which defines the coefficient of restitution and that for the conservation of linear momentum you can derive an equation for the loss of kinetic energy which depends on the initial and final velocities of the colliding objects and the masses of those two objects.

To obtain your graph you must have assumed some initial conditions which meant that after the collision the linear momentum of the two objects which were stuck together was not zero, hence the loss kinetic energy was not zero.
A lump of chewing gum hitting a wall and sticking to it is an example of the coefficient of restitution being equal to zero but the chewing gum, wall (and Earth) do have some kinetic energy as a result of such a collision.

• As ever, I will be interested as to why I was given a down vote. Sep 30, 2021 at 9:17

I use those 3 equations \begin{align*} &m_1\,(v_{1f}-v_{1i})=dp\\ &m_2\,(v_{2f}-v_{2i})=-dp\\ &v_{2f}-v_{1f}=-\epsilon(v_{2i}-v_{1i}) \end{align*} you have 3 equations for 3 unknowns $$~ v_{2f}~,v_{1f} ~,dp$$ the kinetic energy $$~K~$$is :

\begin{align*} &K_{i}=\frac 12 m_1 v_{1i}^2+\frac 12 m_2 v_{2i}^2\\ &K_{f}=\frac 12 m_1 v_{1f}^2+\frac 12 m_2 v_{2f}^2\\\\ &\text{those from the solution of the 3 equations you get the result }\\\\ &\frac{K_i-K_f}{K_i}= {\frac {m_{{2}}m_{{1}} \left({v_{1i}} -{v_{2i}} \right) ^{2} \left(1- {\epsilon }^{2} \right) }{ \left( m_{{1}}{{v_{1i}}}^{2}+m_{ {2}}{{v_{2i}}}^{2} \right) \left( m_{{1}}+m_{{2}} \right) }} \end{align*}

• index f final
• index i initial
• v velocity
• m mass
• dp impulse
• $$\epsilon~$$ coefficient of restitution ($$~0\le\epsilon\le 1$$)
• $$\epsilon=0~$$ elastic collision
• $$\epsilon=1~$$ plastic collision
• I think that quantity dp is change in linear momentum of the particular body.
– user297948
Oct 1, 2021 at 7:20
• \begin{align*} &\text{here is the theory behind: at the time that the collision accrue , apply Newton second law}\\ &m_1\,\frac{dv_1}{dt}=F_c\\ &m_2\,\frac{dv_2}{dt}=-F_c\\ &\text{where $~F_c~$ is the constraint force, intergrading the equations you obtain}\\ &m_1\int_{v_{1i}}^{v_{1f}}\,dv_1=\int F_c\,dt=dp\\ &m_2\int_{v_{2i}}^{v_{2f}}\,dv_2=-\int F_c\,dt=-dp\\ \end{align*}
– Eli
Oct 1, 2021 at 11:38
• dp is the impulse, I correct it
– Eli
Oct 1, 2021 at 11:43
• could you please explain how did you solve the equations because on solving them myself I am getting $$(1-e)^{2}$$ instead of $$(1-e^{2})$$, where $e$ is the coefficient of restituiton.
– user297948
Oct 1, 2021 at 14:35
• @AdityaSingh I use Maple program and this is the solutions of the equations \begin{align*} &v_{f1}={\frac { \left( e \left( v_{{{\it i2}}}-v_{{{\it i1}}} \right) +v_{{{ \it i2}}} \right) m_{{2}}+m_{{1}}v_{{{\it i1}}}}{m_{{1}}+m_{{2}}}} \\ &v_{f2}={\frac { \left( \left( -v_{{{\it i2}}}+v_{{{\it i1}}} \right) e+v_{{{ \it i1}}} \right) m_{{1}}+m_{{2}}v_{{{\it i2}}}}{m_{{1}}+m_{{2}}}} \\ &dp={\frac {-m_{{1}}m_{{2}} \left( -v_{{{\it i2}}}+v_{{{\it i1}}} \right) e-m_{{1}}m_{{2}} \left( -v_{{{\it i2}}}+v_{{{\it i1}}} \right) }{m_{{1 }}+m_{{2}}}} \end{align*}
– Eli
Oct 1, 2021 at 15:49

For clarity I used $$v_1$$, $$v_2$$ and $$m_1$$, $$m_2$$ for the initial velocity and mass of each particle.

After the impact the velocities are

\begin{aligned} v_1^\star & = v_1 - \frac{J}{m_1} & v_2^\star & = v_2 + \frac{J}{m_2} \end{aligned}

where the impulse $$J$$ is defined in terms of the reduced mass and the restitution $$\epsilon$$

$$J = -(1+\epsilon) \frac{1}{\frac{1}{m_1} + \frac{1}{m_2} } (v_2-v_1)$$

The way I calculate the ratio of KE after vs before, I come up with the following

$$\lambda = \frac{\rm KE^\star}{\rm KE} = \frac{ \epsilon^2\,m_1 m_2 (v_2-v_1)^2 + (m_1 v_1 + m_2 v_2)^2 }{(m_1+m_2) (m_1 v_1^2 + m_2 v_2^2) }$$

You can see that when $$\epsilon=1$$, then $$\lambda=1$$, and kinetic energy is preserved.

The minimum ratio $$\lambda$$ occurs when $$\epsilon=0$$ and it is equal to

$$\lambda_{\rm min} = \frac{ (m_1 v_1 + m_2 v_2)^2 }{(m_1 + m_2) (m_1 v_1^2 + m_2 v_2^2)}$$

it is interesting to see total linear momentum squared on the numerator, and the denominator equals total mass times two times KE.

Specifically if one of the masses is at rest then

$$\lambda = \begin{cases} \frac{m_1}{m_1+m_2} & v_2 = 0 \\ \frac{m_2}{m_1+m_2} & v_1 = 0 \\ \end{cases}$$

So the ratio of KE depends on the ratio of masses.

There is special case, when total momentum is zero, $$m_1 v_1 + m_2 v_2 = 0$$. This leads to $$\lambda = \epsilon^2$$

So when measured from the barycenter of the system the loss is $$1-\epsilon^2$$, and on any other coordinate system the loss is closer to $$1$$ as $$\lambda\gg 0$$ based on the analysis above.

Using, few equations, we can find the decrease in kinetic energy. First we conserve momentum: $$m_1u_1+m_2u_2=m_1v_1+m_2v_2$$

Lets assume, $$v_2>v_1\,\,,u_1>u_2$$ for sake of simplicity (to avoid taking modulus in restitution equation).

$$e=\frac{v_2-v_1}{u_1-u_2}$$

Clearly, we need to find actually, $$\Delta KE=\frac{1}{2}m_1(u_1^2-v_1^2)+\frac{1}{2}m_2(u_2^2-v_2^2)$$

By first and second equations, we can find value of $$v_1,v_2$$.

$$v_1=\frac{m_1u_1+m_2u_2-em_2(u_1-u_2)}{m_1+m_2}$$ $$v_2=\frac{em_2(u_1-u_2)-(m_1u_1+m_2u_2)}{m_1+m_2}$$

So, you can put these values in third equation and calculate for yourself, the $$\Delta KE$$. But its very cumbersome to find general equation, so i suggest to use some initial values of $$m_2,m_2,u_1,u_2$$ or do some assumptions like $$m_1=m_2$$ or $$u_1=u_2$$.

Feel free to point out any calculation error that might have crept in.

Remember kinetic energy in two body problems; $$T=\frac{1}{2}\mu v_{rel}^2+\frac{1}{2}(m_1+m_2)v_{cm}^2$$ The second term can be rewritten as $$\frac{p_{total}^2}{2m}$$ and conservation of momentum gives us $$\Delta T=\frac{1}{2}\mu v_{rel}^2-\frac{1}{2}\mu u_{rel}^2$$ and using definition of $$e=-\frac{v_{rel}}{u_{rel}}$$ we get that $$\Delta T=-(1-e^2)\cdot\frac{1}{2}\mu u_{rel}^2$$ Now if you work in the centre of mass frame $$u_{cm}=v_{cm}=0$$. So the initial kinetic energy in this frame becomes $$T_{0{\small{/CM}}}=\frac{1}{2}\mu u_{rel}^2$$ . Note that /CM means in the frame of centre of mass. We get $$\Delta T=-(1-e^2)T_{0\small{/CM}}$$