Why the coefficient of restitution cannot be 1

This is going to be a mechanics post from your usual maths blogger (plot twist!). In this post I will be trying to prove algebraically and physics-ally that the coefficient of restitution cannot be 1. Don’t worry, these confusing words will be defined. If you find a flaw in this argument please contact us or make another post proving it in a better way!

Newton’s law of restitution:

Speed of separation of particles divided by the speed of approach of particles = e

Or algebraically, for an object A of initial velocity u_A and final velocity v_A and for an object B of initial velocity u_B and final velocity v_B, e = (v_B-v_A)/(u_A-u_B). This is because we’re assuming the objects wont pass through each other.

e is the coefficient of restitution, it doesn’t have a unit and is between 0 and 1 (like all coefficients).

For this argument let’s assume conservation of momentum holds true. This means that for two particles A and B with masses m_A and m_B respectively, and initial velocity u_A and u_B respectively, and final velocity v_A and v_B, m_Au_A + m_Bu_B = m_Av_A + m_Bv_B. In other words total momentum (which is equal to mass multiplied by velocity) before an event is equal to total momentum after an event.

We will also define kinetic energy.

Let’s assume work done equals force times distance, written as W=Fs. Using F=ma, this means that W=mas. Using suvat equations (which we have proved in a previous post) a=(v-u)/t and s=t(v+u)/2. When substituting W=mas becomes W=m(v-u)(v+u)t/2t which simplifies to W=½mv² – ½mu².

So as we previously stated, m_Au_A + m_Bu_B = m_Av_A + m_Bv_B. Rearrange to get the terms of object A on one side and B on the other, then factorise the masses out to get m_A(u_A-v_A) = m_B(v_B-u_B).

Let’s assume for the argument that there is no loss in kinetic energy. This means that ½ m_Au_A² + ½ m_Bu_B² – (½ m_Av_A² + ½ m_Bv_B²)=0, multiplying by 2 and rearranging to get terms of object A on one side and B on the other gives m_Au_A² – m_Av_A² = m_Bv_B² – m_Bu_B². Factorise the masses out and factorise the square terms by doing difference of two squares to give m_A(u_A-v_A)(u_A+v_A) = m_B(v_B-u_B)(v_B+u_B). As we stated before conservation of momentum is m_A(u_A-v_A) = m_B(v_B-u_B). Let m_A(u_A-v_A) = m_B(v_B-u_B) = p. Substituting into our equation for kinetic energy gives us m_A(u_A-v_A)(u_A+v_A) = p(u_A+v_A) = m_B(v_B-u_B)(v_B+u_B) = p(v_B+u_B). We can divide by p on both sides which gives us u_A+v_A = v_B+u_B. Rearranging gives u_A-u_B = v_B-v_A which when dividing by the left side on both sides gives us 1 = (v_B-v_A)/(u_A-u_B). This is the coefficient of restitution!

So whenever the loss in kinetic energy is 0, the coefficient of restitution is 1. This goes the other way too, if the coefficient of restitution is 1 the loss of kinetic energy is 0. In practicality, however, there will always be a loss in kinetic energy in any collision (provided that there is no other energy entering the system). So since there is always a loss in kinetic energy, the coefficient of restitution can never equal 1.

In summary, assuming momentum is conserved, that work done = force × distance, and that there is always a loss in kinetic energy in a collision, we can prove algebraically that the coefficient of restitution can never equal 1.

By Helen Youssef

Reviewed by Sam Stuart