A $10$ kg ball and a $20$ kg ball approach each other with velocities $20$ m/s and $10$ m/s respectively. What are their velocities after collision if the collision is perfectly elastic.

Let the velocities of $10kg$ & $20kg$ ball before collision are $u_1=20m/s$(+X axis) & $u_2=-20m/s$( -X axis) and after collision $v_1\&v_2$ respectively. Than by the principal of momentum conservation: -

$m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2$

$(10\times 20)+(20\times (-20\left)\right)=(10\times v_1)+(20\times v_2)$

$v_1+2v_2=-20$ ---------------------(i)

Since the collision is perfectly elastic (e = 1)

$e=(v_2-v_1)/(u_1-u_2)=1$

$v_2-v_1=20-(-20)=40$------------------------(ii)

On adding (i) & (ii): -

$3v_2=20$

$44v_2=6.67m/s$

$v_1=6.67-40=-33.33m/s$

The $10kg$ ball will travel along $X-axis$ with a velocity of $33.33m/s$ & the $20kg$ ball will travel along $+Xaxis$ with a velocity of$6.67m/s.$