Combined translational and angular motions Consider the following image below:



The blocks are identical and initially block 2 is held horizontally. The pulley is at the mid point of the rope initially. Block 2 is released from rest. The string is mass-less and taut. We need to find which blocks collides first. 


Now intuitively B-2 will collide with the wall And B-1 will collide the pulley. And so the String length from B-1 to pulley is not fixed. As well as the distance from pulley to B-2.  
Now consider a middle position as shown. B-2 will experience torque from gravity about the pulley.  We see that using newton's laws $$ Mg- T\cos \theta = Ma$$ and $$ T=Ma $$ Where a is the acceleration of B-1 to the right.
Now we can write the torque equations in terms of string length , and angle in makes with the vertical at that instant.  $$ Mg\sin \theta x = Mx^2\alpha $$ where x is length of string from B-2 to Pulley at the instant. Alpha is the angular acceleration with which B-2 goes.
Now everything is variable  $\theta,  x,  \alpha$. And I don't see how we can go proceed further from here. Since I can't see how proceeding from here would get the time.  
I was wondering if someone could conceptually make me 'see' which block will collide first. I sense the acceleration of B-1 is always more?
 A: Proposition. Block 1 reaches the pulley before Block 2 reaches the vertical edge of the support. 
Proof: Introduce the (standard stationary) inertial coordinate system $Oxy$ with origin $O$ at the point of the pulley, coordinate axis $Ox$ oriented  horizontally from left to right on the picture and coordinate axis $Oy$ oriented  vertically going upwards on the picture (this is the standard natural coordinate system). Block 1 moves only along $Ox$ and I denote its coordinate by $z$, which is the oriented distance (so negative for this problem) from Block 1 to the pulley. Block 2 moves in the $Oxy$ plane and I denote its coordinates by $(x, y)$ in $Oxy$ (here $y$ is negative).
Let the $z=z(t)$ be the motion of Block 1 and $x= x(t), \, \,y=y(t)$ be the motion of Block 2. Let $\vec{a}_1= (a, 0)$ be the acceleration of Block 1, where $a = a(t) = \ddot{z}(t)$. Let $\vec{a}_2 = (b, c)$ be the acceleration of Block 2, where $b = b(t) = \ddot{x}(t)$ and $c = c(t) = \ddot{y}(t)$. 
According to the problem's assumptions, the total force acting on Block 1 is $\vec{T}_1$ and the total force acting on Block 2 is $\vec{T}_2 + m\, \vec{g}$, where $|\vec{T}_1| = |\vec{T}_2| = T = T(t)$. If by $\theta = \theta(t)$ we denote the clockwise oriented angle between the horizontal direction $Ox$ and the direction of the rope from the pulley to Block 2, then 
$$ \vec{T}_1 = (T, 0) \,\, \text{ and } \,\, \vec{T}_2 = \big(- T \cos(\theta), \,\,  T \sin(\theta) \, \big)  $$ while the gravitational force is $$m\,\vec{g} = ( 0, -mg)$$ By the second law of Newton for Block 1 we have $$m \vec{a}_1 = \vec{T}_1$$
which component-wise can be written as
$$m \, (\ddot{z}, 0) = (T, 0) \, \, \text{ yielding the expression } \,\, m \, \ddot{z} = T$$ By the second law of Newton for Block 2 we have $$m \vec{a}_2 = \vec{T}_2 + m\, \vec{g}$$
which component-wise can be written as
$$m \, (\ddot{x}, \ddot{y}) = \big(-T \cos(\theta), \, T \sin(\theta)\, \big) + (0, -mg) $$ yielding the pair of expressions
$$m \, \ddot{x} = -T \cos(\theta) \,\,\text{ and } \,\, m \, \ddot{y} = T \sin(\theta) - m\, g$$ From all these three different equations, we are going to use two of them, namely
$$m \, \ddot{z} = T \,\, \text{ and } \,\,m \, \ddot{x} = -T \cos(\theta) $$ which are the same as 
$$\ddot{z} = \frac{T}{m} \,\, \text{ and } \,\, \ddot{x} = -\frac{T}{m} \cos(\theta) $$ Assume the motion takes place for $t$ in the time interval $[0, t_{end}]$, where $t=0$ is the initial moment when the two blocks are held horizontally and the pulley is the midpoint of the rope connecting them, while $t=t_{end}$ is when either Block 2 has reached the vertical edge of the support or Block 1 has reached the pulley, whichever comes first. When $t=0$ we have $\theta(0) = 0$ and when $t=\theta(t_{end}) \leq \frac{\pi}{2}$. Therefore $0 \leq \theta(t) \leq \frac{\pi}{2}$.
Lemma. the function $x(t) + z(t) > 0$ for all $t$ in the interval $(0,t_{end}]$.
Proof of the lemma: Differentiate the function $x(t) + z(t)$ twise with respect to $t$. This gives us
$$\frac{d^2}{dt^2} \big(\, x(t) + z(t) \,\big) = \ddot{x}(t) + \ddot{z}(t) = -\, \frac{T(t)}{m} \cos{\theta(t)}+ \frac{T(t)}{m} = \frac{T(t)}{m}\Big(1 - \cos{\theta(t)}\Big) > 0$$ for all $t \in (0,t_{end}]$ because for these values of $t$ the angle $\theta \in (0, \frac{\pi}{2}]$ and thus $0 \leq \cos{\theta} < 1$. Therefore the function $\dot{x}(t) + \dot{z}(t)$ is strictly increasing in the interval $[0, t_{end}]$ so for all $t \in (0, t_{end}]$ we have that
$$\dot{x}(t) + \dot{z}(t) > \dot{x}(0) + \dot{z}(0) = 0$$ because $\dot{x}(0) = \dot{z}(0)$ since by assumption the two blocks start their motion with zero initial velocity. Consequently, the function $x(t) + z(t)$, whose first derivative is $\dot{x}(t) + \dot{z}(t) > 0$ for $t \in (0, t_{end}]$, is also strictly increasing for all $t \in [0, t_{end}]$. Thus for $t \in (0, t_{end}]$
$$x(t) + z(t) > x(0) + z(0) = \frac{l_0}{2} - \frac{l_0}{2} = 0$$ In the latter expression, $l_0$ is the length of the rope and the facts that $x(0) = \frac{l_0}{2}$ and $z(0) = - \frac{l_0}{2}$ follow from the problem's assumption that at the beginning of the motion, the pulley is the midpoint of the straight segment connecting Block 1 and Block 2  
Continuing the proof of the main proposition: By the lemma we have that for all $t \in (0, t_{end}]$ 
$$x(t) > - z(t) = |z(t)| \geq 0$$ because of the way the coordinate system was introduced. In it the coordinate $z$ of Block 1 is always negative while the coordinate $x$ of Block 2 is always positive. Conseqeuently, at the end of the motion, when $t = t_{end}$ we reach the inequalities 
$$x(t_{end}) > |z(t_{end})| \geq 0$$ which imply that $x(t_{end}) \neq 0$ so at the end of the motion Block 2 would not have reached the vertical edge of the support. By the choice of $t_{end}$, i.e. it has to be the moment of time when either Block 1 reaches the pulley or Block 2 reaches the vertical edge of the support, we are left with the only option that at $t_{end}$ Block 1 has reached the pulley, while Block 2 is still swinging towards the vertical edge of the support. In other words $$z(t_{end}) = 0 \,\, \text{ while } \,\, x(t_{end}) > 0$$ 
Heuristic argument: Basically, if we project the motion of the two blocks on the $Ox$ axis, Block 1 moves with acceleration $\frac{T}{m}$ towards the pulley, while the projection of Bock 2 onto the $Ox$ axis moves with the smaller acceleration $\frac{T}{m}\cos{\theta}$ towards the pulley. Since both of the blocks start at the same distance from the pulley, namely the length of the rope $l_0$ divided by two, Block 1 will arrive at the pulley before the projection of Block 2 arrives at the pulley due to the fact that Block 1 has greater acceleration than Block 2.  
A: This question teaches two important and general lessons:


*

*If you don't know how to solve an answer, just get started and parametrize your ignorance.  

*Answer the question you're asked — no more and no less.
The OP points out ignorance of the time at which the first collision takes place, so let's just make up a symbol for this parameter: $\newcommand{\tc}{t_\mathrm{c}} \tc$.  We don't know what the tension in the rope is, but we've parametrized it by $T$ which is assumed to be equal everywhere along the rope and is implicitly a function of time.  Similarly, we don't know the angle that the rope makes on the right-hand side, but we've parametrized it by $\theta$ which is also implicitly a function of time.  Now, the distance an object moves is the integral of its velocity, which is the integral of its acceleration.  We only care about the horizontal distance moved by each block during this time, so we only need to integrate the horizontal velocity, and therefore the horizontal acceleration.  But we already know these accelerations in terms of our parameters:
\begin{align}
  a_1 &= \frac{T(t)}{M}, \\
  a_2 &= \frac{T(t)\, \sin\theta(t)}{M}.
\end{align}
[Here, I'm using the OP's convention that $\theta(0) = 90^\circ$ and decreases in time.]  So we can get the horizontal distance each block moves between the initial time $0$ and $\tc$:
\begin{align}
  d_1 &= \int_{0}^{\tc} \int_{0}^{t'} a_{1}\, dt\, dt' = \int_{0}^{\tc} \int_{0}^{t'} \frac{T(t)}{M}\, dt\, dt' \\
  d_2 &= \int_{0}^{\tc} \int_{0}^{t'} a_{2}\, dt\, dt' = \int_{0}^{\tc} \int_{0}^{t'} \frac{T(t)\, \sin\theta(t)}{M}\, dt\, dt'.
\end{align}
Now for $t>0$, we evidently have $\theta(t)>0$, which means that $\sin \theta(t) < 1$.  Therefore, we get to use the inequality properties of integrals to say that $d_2 < d_1$.  This tells us that B-2 didn't move as far as B-1.  But since the blocks have the same horizontal distance to travel to get to the collision point, that means B-1 collided first.
