Is cofficient of restitution equation valid for collisions involving more than 2 bodies? I couldn't solve a question and the only equation I was missing was of cofficient of restitution i.e. $e=\frac{v_{final2} - v_{final1}}{u_{intial1} - u_{initial2}}$ .
But the collision was among 3 objects. This was the figure given in the question: 
(elasticity given was 1, fwiw)
Won't the third body or the tension impact the velocities? Is coefficient of restitution equation generally valid when collision among 3 bodies takes place?
Like would the equation be applicable when masses are aligned this way:
Edit: the aforementioned question

Edit 2: I have mentioned more on what exactly I am having doubt in the comment section of 147875 sir's answer.
 A: The coefficient of restitution (e) is always between bodies (i.e., two bodies) rather than among bodies (i.e., more than 2 bodies). If you want to deal more than two bodies, always take two at a time along with appropriate components.
The coefficient of restitution (e) is defined as,
$$
e = \frac{\text{Relative velocity after collision}}{\text{Relative velocity before collision}}
$$
If the initial velocity of the center sphere is $u$ and let $u'$ be the velocity after collision and $v$ be the velocity of suspended spheres, which are constrained to move in horizontal direction. Then, coefficient of restitution (e) is given by
$$
e = \frac{v\cos(60^\circ) - u'\cos(30^\circ)}{u\cos(30^\circ)} \tag{1}
$$
for elastic collision, $e = 1$, you can substitute in (1) and to get
$$
\sqrt{3}(u + u')  = v \tag{2}
$$
In deriving the result above, we took the center sphere along with one of the bottom spheres. We can equation (2) along with equation due to impulse to solve the problem.
Edit 1:

Really thanks but I am still confused. What I really want to know is
that- the external force and the extra body would impact the velocity
just after impact of both the bodies but still the equation gives
correct results. Why didn't the equation fail in this situation? What
principle is it really based on? Like momentum conservation is not
applicable here but the equation is still valid. I have phrased the
question really badly

When solving the problem, we only took the primary collision and the consecutive collisions is not considered. Does the third body have any effect on the velocity of the other two? Yes it does, that's the reason, we took component only along one body and derived the result (1) and (2). Is momentum conservation applicable here? All the laws of elastic collision? Yes it does. We will derive this result below.
Conservation of Kinetic Energy gives,
$$
\begin{align}
\frac{1}{2}mu^2 &= \frac{1}{2}m{u'}^2 + 2 \times \frac{1}{2}m'v^2 \\
\implies m(u^2 - {u'}^2) &= m'v^2 \tag{3}
\end{align}
$$
conservation of momentum gives,
Alng $x$-axis is,
$$
0 = m'v\sin(30^\circ) - m'v\sin(30^\circ) \tag{4}
$$
Along $y$-axis is,
$$
\begin{align}
mu & = mu' + 2 \times m'v\cos(30^\circ) \\
\implies m(u - u') &= \sqrt{3}m'v \tag{5}
\end{align}
$$
Divide Eq(3) by Eq(5) gives,
$$
(u + u') = \frac{v}{\sqrt{3}} \tag{6}
$$
Now, if we look Eq(2) and Eq(6) are both same. Hence rules of elastic collision is applicable.
