$\therefore$ J_N = M(v_N + 8)                                               .....(i)

J_f = µJ_N = M(–v_t + 6)                                          .....(ii)

and, Jf_R = I_CM (ω + ω₀)                                     .....(iii)

For wedge just after collision

$\therefore$ J_N sin 37° – J_f cos 37° = 2Mv

⇒ J_N = 6 Mv                                                      .....(iv)

and from Newton's formula

$\frac{3\mathrm{v}}{5}+{\mathrm{v}}_{\mathrm{N}}=-\mathrm{e}(0-{\mathrm{u}}_{\mathrm{N}})$                            .....(v)

from (i), (iv) and (v),

$\mathrm{v}=\frac{20}{11}\mathrm{m}/\mathrm{s}\to {\mathrm{andv}}_{\mathrm{N}}=\frac{32}{11}\mathrm{m}/\mathrm{s}$

and ${\mathrm{J}}_{\mathrm{N}}=\frac{120\mathrm{M}}{11}$

from (ii), ${\mathrm{v}}_{\mathrm{t}}=\frac{26}{11}\mathrm{m}/\mathrm{s}$

Now, Sphere's time of flight till ground for first time

$\mathrm{T}=\frac{2{\mathrm{u}}_{\mathrm{y}}}{\mathrm{g}}=\frac{2}{\mathrm{g}}({\mathrm{v}}_{\mathrm{N}}\cos 37\mathrm{°}-{\mathrm{v}}_{\mathrm{t}}\cos 53\mathrm{°})=\frac{2}{11}\sec$

$\therefore$ Distance moved by wedge during this duration = ${\mathrm{v}}_{\mathrm{t}}=\frac{40}{121}$ m