Prove that coefficient of restitution/resilience of perfectly elastic collision in one dimensions unity.
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Please briefly explain why you feel this question should be reported.
Please briefly explain why you feel this answer should be reported.
Please briefly explain why you feel this user should be reported.
Suppose two balls A and B of masses m1 and m2 are moving initially along the same straight line with velocities u1 and u2 respectively.
When u1 > u2 relative velocity of approach before collision,
= u1 – u2
Hence the two balls collide, let the collision be perfectly elastic. after collision, suppose v1 is velocity of A and v2 is velocity of B along the same straight lien,
(a) when v2 > v1, the bodies separate after collision.
Relative velocity of separation after collision
= v2 – v1
Linear momentum of the two balls before collision.
= m1u1 + m2u2
Linear momentum of the two balls after collision.
= m1v1 + m2v2
As linear momentum is conserved in an elastic collision, therefore
m1u1 + m2u2 = m1v1 + m2v2 ………….(i)
or m2(v2 − u2) = m1(u1 − v1) …………(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 v2 + u2 = u1 + v1
or v2 − v1 = u1 − u2 …(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