Answer
a) $v_m=\sqrt{5gR}\left(\dfrac{m+M}{m } \right)$
b) $v_m=\sqrt{5gR}\left(\dfrac{m+M}{2m } \right)$
Work Step by Step
First of all, we need to find the minimum speed that allows any block to circle the loop-the-loop without falling off. And to find this velocity, we need to analyze the forces exerted on it while it is on the top.
As we see in the figure below, there are two forces exerted on this object at the top, the normal force and its own weight.
Thus, the minimum speed for the object to complete this vertical loop occurs when the object is barely touching the rails which mean that the normal force is approaching zero.
So that,
$$\sum F_y=mg=ma_r$$
Thus,
$$\color{red}{\bf\not} mg=\color{red}{\bf\not} m\dfrac{v^2}{R}$$
Therefore,
$$v=\sqrt{gR}\tag 1$$
Now we need to make the system has this speed at the top of this vertical loop and we assume that the system (the object plus the Earth) is isolated since the rails are frictionless.
This means that the energy at the highest point is equal to the energy at the beginning at the lowest point.
$$E_{top}=E_{bottom}$$
$$U_{fg}+K_f=U_{ig}+K_i$$
a) when the two blocks stick together after the collision,
$$(m+M)g(2R)+\frac{1}{2}(m+M)v^2=0+\frac{1}{2}(m+M)v_{1}^2$$
where $v_1$ is their speed after the collision.
Plugging $v$ from (1);
$$2(m+M)gR+\frac{1}{2}(m+M)gR=\frac{1}{2}(m+M)v_{1}^2$$
$(m+M)$ is a common factor.
$$2.5 gR =\frac{1}{2} v_{1}^2$$
Solving for $v_1 $;
$$v_1 =\sqrt{5gR}\tag 2$$
Now we can use the conservation of momentum to find $v_1$ in terms of $v_m$.
$$p_{ix}=p_{fx}$$
$$mv_m=(m+M)v_1$$
Thus,
$$v_1=\dfrac{mv_m}{m+M}$$
Plugging into (2);
$$\dfrac{mv_m}{m+M}=\sqrt{5gR} $$
$$\boxed{v_m=\sqrt{5gR}\left(\dfrac{m+M}{m } \right)}$$
b) when the collision is perfectly elastic (By the same approach as we did above);
$$U_{fg}+K_f=U_{ig}+K_i$$
$$Mg(2R)+\frac{1}{2}Mv^2=0+\frac{1}{2}Mv_1^2$$
where $v_1$ is their speed after the elastic collision.
Plugging $v$ from (1);
$$2\color{red}{\bf\not} MgR+\frac{1}{2}\color{red}{\bf\not} MgR= \frac{1}{2}\color{red}{\bf\not} Mv_1^2$$
$$ 2.5gR = \frac{1}{2} v_1^2$$
Solving for $v_1 $;
$$v_1 =\sqrt{5gR} $$
which is the same result as (2) above.
Now we need to use the conservation of momentum to find $v_1$ in terms of $v_m$, where we know that the speed of the stationary block after the elastic collision is given by
$$v_1=\dfrac{2mv_m}{m+M}$$
Plugging into (2);
$$\dfrac{2mv_m}{m+M}=\sqrt{5gR} $$
Therefore,
$$\boxed{v_m=\sqrt{5gR}\left(\dfrac{m+M}{2m } \right)}$$
where $v_ m$ here for elastic collision is half $v_ m$ for inelastic collision.
