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Q.
If a ball of mass $4\, kg$ moving with velocity of $3\, m / s$ collides elastically head on with another ball of mass $6 \,kg$ which is at rest, then the ratio of speed of second ball to first after collision will be
Solution:
By conservation of linear momentum
$4 \times 3=4 v_{1}+6 v_{2}$
$6=2 v_{1}+3 v_{2}\,\,\,\,...(i)$
$e=\frac{v_{2}-v_{1}}{3}=1$
$v_{2}-v_{1}=3$
$2 v_{2}-2 v_{1}=6\,\,\,\,...(ii)$
By solving equation (i) $\& $ (ii)
$v_{2}=\frac{12}{5} \,m s ^{-1}$
$v_{1}=\frac{12}{5}-3$
$v_{1}=\frac{-3}{5} \,m s ^{-1}$
$\frac{v_{2}}{v_{1}}=\frac{\frac{12}{5}}{\frac{3}{5}}$
$\frac{v_{2}}{v_{1}}=4: 1$