Prove that coefficient of restitution/resilience of perfectly elastic collision in one dimensions unity.

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Suppose two balls A and B of masses m

_{1}and m_{2}are moving initially along the same straight line with velocities u_{1}and u_{2}respectively.When u_{1}> u_{2}relative velocity of approach before collision,= u

_{1}– u_{2}Hence the two balls collide, let the collision be perfectly elastic. after collision, suppose v

_{1}is velocity of A and v_{2}is velocity of B along the same straight lien,(a) when v_{2}> v_{1}, the bodies separate after collision.Relative velocity of separation after collision

= v

_{2}– v_{1}Linear momentum of the two balls before collision.

= m

_{1}u_{1}+ m_{2}u_{2}Linear momentum of the two balls after collision.

= m

_{1}v_{1}+ m_{2}v_{2}As linear momentum is conserved in an elastic collision, thereforem

_{1}u_{1}+ m_{2}u_{2 }= m_{1}v_{1}+ m_{2}v_{2}………….(i)or m

_{2}(v_{2}− u_{2}) = m_{1}(u_{1}− v_{1}) …………(ii)Total K.E. of the two balls before collision= \(\frac{1}{2}m_1u^2_1+\frac{1}{2}m_1u^2_2……….(iii)\)

Total K.E. of the two balls after collision= \(\frac{1}{2}m_1v^2_1+\frac{1}{2}m_1v^2_2……….(iv)\)

As K.E. is also conserved in an elastic collision, there from (iii) and (iv),\(\frac{1}{2}m_1v^2_1+\frac{1}{2}m_1v^2_2=\frac{1}{2}m_1u^2_1+\frac{1}{2}m_1u^2_2\)

\(\frac{1}{2}m_2(v^2_2-u^2_2)=\frac{1}{2}m_1(u^2_1-v^2_1)\)

\(m_2(v^2_2-u^2_2)=m_1(u^2_1-v^2_1)……(v)\)

Dividing (v) by (ii),\(\frac {m_2(v^2_2-u^2_2)}{m_2(v_2-u_2)}=\frac{m_1(u^2_1-v^2_1)}{m_1(u_1-v_1)}\)

\(\frac{(v_2+u_2)(v_2-u_2)}{(v_2-u_2)}=\frac{(u_1+v_1)(u_1-v_1)}{(u_1-v_1)}\)

or v

_{2}+ u_{2}= u_{1}+ v_{1}or v

_{2}− v_{1}= u_{1}− u_{2}…(iv)Hence is one dimensional elastic collision relative velocity of separation after collision is equal to relative velocity of approach before collision.

From \(\frac{v_2-v_1}{u_2-u_1}=1\)

by definition,\(\frac{v_2-v_1}{u_2-u_1}=1\) = e =1