A particle X of mass m and initial velocity u collides with another...

For perfectly elastic collision,

total momentum before collision = total momentum after collision

i.e. $mu + \dfrac{3m}{4} \times 0 = m{v}_{x} + \dfrac{3m}{4}{v}_{y}$

$⇒ 4u = 4{v}_{x} + 3{v}_{y}... \left(i\right)$

Coefficient of restitution,

$e = \dfrac{\text{Velocity of separation after collision}}{\text{Velocity of approach before collision}} =\left({v}_{y} - {v}_{x}\right) \left({u}_{x} - {u}_{y}\right)$

$⇒ 1 = \dfrac{{v}_{y} - {v}_{x}}{u - 0}$ & [$∵ {u}_{x} = u$ and ${u}_{y} = 0$]

$u = {v}_{y} - {v}_{x}... \left(ii\right)$

From equation (i) and (ii), we get

${v}_{x} = \dfrac{u}{7}$ and ${v}_{y} = \dfrac{8u}{7}$

$∴ \dfrac{\lambday}{\lambdax} = \dfrac{\dfrac{h}{\dfrac{3}{4}m{v}_{y}}}{\dfrac{h}{m \times {v}_{x}}} = \dfrac{4}{3} \times \dfrac{vx}{vy} = \dfrac{4}{3}\dfrac{\dfrac{u}{7}}{\dfrac{8u}{7}} = \dfrac{1}{6}$