Consider a horizontal surface moving vertically upward with...

The relative velocity of the particle with respect to the surface, along the normal, is the approach speed of the particle.

${v}_{app} = 4m{s}^{- 1}$

The separation velocity will be

${v}_{sep} = e{v}_{app} = \dfrac{1}{2} \times 4 = 2m{s}^{- 1}$

∴ {\overset{̄}{v}}_{m, gr} = {\overset{̄}{v}}_{m, s} + {\overset{̄}{v}}_{s, gr} = 4\text{m/s}

$∫ Ndt = 4\left(2\right) + 2\left(2\right) = 12$

$\text{Now,}∫ \muN ⋅ dt = 2\left({v}_{xf} - 2\right)$

$= - \dfrac{1}{3} \times 12 = 2\left({v}_{xf} - 2\right)$

${\left({v}_{x}\right)}_{final} = 0$