Question

Derive the expression for coefficient of restitution during a head on elastic collision.​

Answer 1

Answer:

v 2−v 1=−e(u 2−u 1). This formula is Newton's law of restitution. The coefficient of restitution always satisfies 0≤e≤1. When e=0, the balls remain in contact after the collision.

Answer 2

Answer:

Explanation:

In an elastic collision kinetic energy is conserved, so

$\begin{aligned}\frac{1}{2} m_{1} u_{1}^{2}+\frac{1}{2} m_{2} u_{2}^{2} &=\frac{1}{2} m_{1} v_{1}^{2}+\frac{1}{2} m_{2} v_{2}^{2} \\m_{1} u_{1}^{2}+m_{2} u_{2}^{2} &=m_{1} v_{1}^{2}+m_{2} v_{2}^{2} \\m_{1} u_{1}^{2}-m_{1} v_{1}^{2} &=m_{2} v_{2}^{2}-m_{2} u_{2}^{2} \\m_{1}\left(u_{1}+v_{1}\right)\left(u_{1}-v_{1}\right) &=m_{2}\left(v_{2}+u_{2}\right)\left(v_{2}-u_{2}\right)......(i)\end{aligned}$

Now, according to conservation of linear momentum,

$\begin{gathered}m_{1} u_{1}+m_{2} u_{2}=m_{1} v_{1}+m_{2} v_{2} \\m_{1}\left(u_{1}-v_{1}\right)=m_{2}\left(v_{2}-u_{2}\right)\end{gathered}......(ii)$

Dividing equation 1 by 2 we get,

$\begin{gathered}u_{1}+v_{1}=u_{2}+v_{2} \\u_{1}-u_{2}=v_{2}-v_{1} \\\frac{v_{2}-v_{1}}{u_{1}-u_{2}}=1\end{gathered}$

As a result, we have an equation with a restitution coefficient of one. This demonstrates that if the collision is elastic, the result will be 1. Furthermore, when it approaches 1, both equations will be approximately met, contributing to the conservation of kinetic energy, which will improve the likelihood of a collision occurring in an elastic plane. As a result, it's a metric for elastic collision.