Elastic impact, final positions with friction

Assumptions: the arc is smooth; the origin of the x-axis is the point B (the lowest point of the arc, where the 0.4 kg particle rests); $\mu_k=0.3$ acts on the horizontal surface; $g=9.81\ \text{m/s}^2$.

Speed of the 0.5 kg particle at B

Drop in height $h=0.25\sin30^\circ=0.125$ m.

$$\tfrac12m_1u^{2}=m_1gh\ \Rightarrow\ u=\sqrt{2(9.81)(0.125)}=1.566\ \text{m/s}$$

Elastic direct impact ($e=1$)

$m_1=0.5$ kg, $m_2=0.4$ kg (initially at rest). Momentum and restitution:

$$0.5(1.566)=0.5v_1+0.4v_2,\qquad v_2-v_1=1.566$$ $$v_1=\frac{m_1-m_2}{m_1+m_2}u=\frac{0.1}{0.9}(1.566)=0.1740\ \text{m/s}$$ $$v_2=\frac{2m_1}{m_1+m_2}u=\frac{1.0}{0.9}(1.566)=1.740\ \text{m/s}$$

Both move in the +x direction, and $v_2>v_1$, so they do not collide again.

Sliding distances (work-energy with friction)

$$\tfrac12mv^{2}=\mu_kmg\,d\ \Rightarrow\ d=\frac{v^{2}}{2\mu_kg}$$ $$d_1=\frac{0.1740^{2}}{2(0.3)(9.81)}=0.00514\ \text{m},\qquad d_2=\frac{1.740^{2}}{2(0.3)(9.81)}=0.514\ \text{m}$$

Answer: the 0.5 kg particle stops $0.0051$ m (5.1 mm) from the origin and the 0.4 kg particle stops $0.514$ m from the origin, both on the positive x-side.