Elastic Collision in One Dimension

Consider two bodies moving initially along the line joining their centers as shown in the figure below:

Assume initial direction of motion to be positive. Also assume that

$u_1 > u_2$ (so that collision may take place). Conservation of momentum gives:

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

$...eq^n (i)$

$=> m_1(u_1-v_1)=m_2(v_2-u_2)$

$...eq^n (ii)$

In case of perfectly elastic collision, total kinetic energy also remains conserved i.e. KE will be same before and after the collision:

$\frac1 2 m_1u_1^2+\frac1 2m_2u_2^2=\frac1 2 m_1v_1^2+\frac1 2 m_2v_2^2$

$...eq^n (iii)$

$=> m_1(u_1^2-v_1^2)=m_2(v_2^2-u_2^2)$

$...eq^n (iv)$

Dividing

$eq^n(iv)$ by

$eq^n(ii)$ , we get:

$u_1+v_1=u_2+v_2$

$...eq^n (v)$

$=> (u_1-u_2)=(v_1-v_2)$

$...eq^n (vi)$

Thus, in 1-D perfectly elastic collision, "velocity of approach" before collision is equal to the "velocity of recession" after collision.

Now, let's multiply

$eq^n (vi)$ by

$m_2$ and subtract it from

$eq^n (ii)$ :

$(m_1-m_2)u_1+2m_2u_2=(m_1+m_2)v_1$

$...eq^n (vii)$

$v_1 = (\frac(m_1-m_2) (m_1+m_2)) u_1 + (\frac(2m_2) (m_1+m_2)) u_2$

$...eq^n (viii)$

Similarly, multiplying

$eq^n(vi)$ by

$m_1$ and adding it to

$eq^n(i)$ :

$2m_1u_1+(m_2-m_1)u_2=(m_1+m_2)v_2$

$...eq^n (ix)$

$v_2 = (\frac(2m_1) (m_1+m_2)) u_1 + (\frac(m_2-m_1) (m_1+m_2)) u_2$

$...eq^n (x)$

This is the derivation of final speeds of both objects after collision.

$eq^n(viii)$ and

$eq^n(x)$ give the final speed of

$m_1$ and

$m_2$ respectively. Care should be taken before applying these equations that these are valid only in case of perfectly elastic collisions.