\(\sigma=\frac{3B-2G}{2G+6B}\)
\(\sigma=\frac{6B+2G}{3B-2G}\)
\(\sigma=\frac{9BG}{3B+G}\)
\(B=\frac{3\sigma-3G}{6\sigma+2G}\)
E = 2G(1 + σ) ….(1)
E = 3B(1 – 2σ) ….(2)
\(1=\frac{2G}{3G}(\frac{1+\sigma}{1-2\sigma})\)
⇒ 3B – 6Bσ = 2G + 2Gσ
⇒ 3B – 2G = σ (2G + 6B)
\(\sigma=(\frac{3B-2G}{2G+6B})\)
So, the correct answer is (A) : \(\sigma=\frac{3B-2G}{2G+6B}\)
According to elastic collision, the kinetic energy of the system will remain constant which means there will be no change in the kinetic energy of the system before and after the collision. It also goes along with the conservation of momentum.
Examples of Elastic Collision
According to inelastic collision, the kinetic energy of the system is not conserved, unlike inelastic collision. The kinetic energy is lost as it gets debauched in other forms of energy like heat, sound, etc, or is absorbed by the body. But they go after the conservation of momentum, like an elastic collision.
Examples of Inelastic Collision
Read More: Elastic and Inelastic Collision