Answer: C. Total momentum is conserved as a vector. The other piece must have momentum components $Mu$ and $-\alpha Mv$ so its speed is

$$\frac{\sqrt{(Mu)^2+(\alpha Mv)^2}}{(1-\alpha)M}.$$

Dividing by $u$ gives $\dfrac{\sqrt{1+\alpha^2(v/u)^2}}{1-\alpha}$

A: assigns mass $\alpha M$ rather than $(1-\alpha)M$ to the other piece.
B: adds the perpendicular momentum components as scalar magnitudes.
D: uses only the $x$-component and omits the momentum needed to cancel the first piece’s perpendicular component.
Remember: conserve momentum in each perpendicular direction before finding a resultant speed.