Q.Arrive at an expression for elastic collision in one dimension and discuss various cases.

Step 1. Set up. Masses $m_1,m_2$ move along a line with initial velocities $u_1>u_2$ and final velocities $v_1,v_2$. Momentum conservation: $m_1u_1+m_2u_2=m_1v_1+m_2v_2$, i.e. $m_1(u_1-v_1)=m_2(v_2-u_2)$ ... (i).

Step 2. Kinetic energy conservation (elastic): $\tfrac12m_1u_1^2+\tfrac12m_2u_2^2=\tfrac12m_1v_1^2+\tfrac12m_2v_2^2$, i.e. $m_1(u_1^2-v_1^2)=m_2(v_2^2-u_2^2)$.

Step 3. Factorise both sides using $a^2-b^2=(a-b)(a+b)$: $m_1(u_1-v_1)(u_1+v_1)=m_2(v_2-u_2)(v_2+u_2)$. Dividing this by equation (i) cancels the common factor $m_1(u_1-v_1)=m_2(v_2-u_2)$, leaving $u_1+v_1=u_2+v_2$, i.e. $u_1-u_2=v_2-v_1$ -- relative speed of approach equals relative speed of separation.

Step 4. Solving this together with equation (i) gives the final velocities explicitly: $v_1=\dfrac{m_1-m_2}{m_1+m_2}u_1+\dfrac{2m_2}{m_1+m_2}u_2$ and $v_2=\dfrac{2m_1}{m_1+m_2}u_1+\dfrac{m_2-m_1}{m_1+m_2}u_2$.

Step 5. Special cases. (i) Equal masses ($m_1=m_2$): $v_1=u_2,\,v_2=u_1$ -- velocities exchange.

(ii) Equal masses, target at rest ($u_2=0$): $v_1=0,\,v_2=u_1$ -- first body stops, second takes its speed.

(iii) Very light body hits heavy, stationary target ($m_1\ll m_2,\,u_2=0$): $v_1\approx-u_1,\,v_2\approx0$ -- light body rebounds. …

Combine momentum and kinetic-energy conservation (factorising with the difference-of-squares identity) to derive v1, v2, th …

• Deriving only the final formulas without discussing any of the requested special cases. …