= Solution
{c}
Take $+y$ upwards in the source diagram. Write $w$ for $B$'s incoming <speed>, so the initial <velocities> are
$$
\mathbf u_A=(-u,0),\qquad
\mathbf u_B=\left(\frac{\sqrt3}{2}w,\frac12w\right).
$$
\b[A contact-direction assumption is needed for a unique answer.] The diagram specifies the incoming <velocities> but does not specify the contact normal or an outgoing <angle>. First adopt the usual intended model: the <impulse> is horizontal, so neither particle changes its $y$ component of <velocity>. This is the smooth <oblique collision> model with a horizontal line of centres at impact. Bringing $A$ to rest in the $x$ direction then brings it entirely to rest.
Let $b$ be $B$'s outgoing $x$ component of <velocity>. <Conservation of momentum> and conservation of <kinetic energy> give
$$
-3u+\frac{\sqrt3}{2}w=b,
\qquad
3u^2+w^2=b^2+\frac14w^2.
$$
Put $q=(\sqrt3/2)w$. The second equation becomes $3u^2+q^2=(q-3u)^2$, so $q=u$. Therefore
$$
\boxed{w=\frac{2u}{\sqrt3},\qquad
\mathbf v_A=(0,0),\qquad
\mathbf v_B=\left(-2u,\frac{u}{\sqrt3}\right).}
$$
The outgoing <kinetic energies> are
$$
\boxed{K_A=0,\qquad K_B=\frac12m\left(4u^2+\frac{u^2}{3}\right)=\frac{13}{6}mu^2.}
$$
The outgoing <speed> of $B$ is $u\sqrt{13/3}$, directed above the negative $x$ axis at <angle> $\tan^{-1}[1/(2\sqrt3)]\simeq16.1^\circ$. The initial <kinetic energy> is also $(3/2+2/3)mu^2=13mu^2/6$, as required for an <elastic collision>.
For the <perfectly inelastic collision> with these same incoming <velocities>, both particles share a common outgoing <velocity>. <Conservation of momentum> gives
$$
\boxed{\mathbf V=\frac{3m\mathbf u_A+m\mathbf u_B}{4m}
=\left(-\frac u2,\frac{u}{4\sqrt3}\right),\qquad
|\mathbf V|=\frac{\sqrt{13}}{4\sqrt3}u.}
$$
Its direction is again $16.1^\circ$ above the negative $x$ axis. The separate translational <kinetic energies> of the constituents and their total are
$$
\boxed{K_A=\frac{13}{32}mu^2,\qquad K_B=\frac{13}{96}mu^2,\qquad
K_{\rm final}=\frac{13}{24}mu^2.}
$$
The loss of translational <kinetic energy> is
$$
\boxed{K_{\rm initial}-K_{\rm final}=\frac{13}{8}mu^2.}
$$
This becomes internal <energy>, for example deformation and heating; total <energy> is conserved. In the <elastic collision> no <kinetic energy> is lost and the particles separate with different <velocities>. In the <perfectly inelastic collision> relative motion disappears and the common <velocity> is the original <centre of mass> <velocity>.
The limitation of the printed data can be made explicit. Without the horizontal-<impulse> assumption, write the final <velocity> of $A$ as $(0,h)$. Then <conservation of momentum> determines
$$
\mathbf v_B=\left(\frac{\sqrt3}{2}w-3u,\frac12w-3h\right),
$$
and conservation of <kinetic energy> gives only
$$
\boxed{w(\sqrt3u+h)=2u^2+4h^2.}
$$
There is a family of possibilities. For a concrete second solution, $h=u/\sqrt3$ gives
$$
w=\frac{5\sqrt3}{6}u,\qquad
\mathbf v_A=\left(0,\frac{u}{\sqrt3}\right),\qquad
\mathbf v_B=\left(-\frac74u,-\frac{7\sqrt3}{12}u\right).
$$
These <velocities> conserve both components of <momentum> and conserve <kinetic energy>, and they satisfy the specified vanishing $x$ component for $A$, while requiring a different incoming <speed>. They can arise from a smooth contact normal proportional to $(1,1/\sqrt3)$. Thus the horizontal-<impulse> result is conditional, rather than uniquely implied by the printed statement. For any member of this family the corresponding sticking result remains well determined once $w$ is chosen:
$$
\mathbf V=\frac14\left(-3u+\frac{\sqrt3}{2}w,\frac12w\right),\qquad
K_{\rm final}=2m|\mathbf V|^2,\qquad
\Delta K=\frac{3m}{8}\left(\left[u+\frac{\sqrt3}{2}w\right]^2+\frac{w^2}{4}\right).
$$
The last expression is the lost relative <kinetic energy> in a <perfectly inelastic collision> of these two <point masses>.
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