Acceleration when motion is constrained by rope The following is a question form HC Verma (Newton's Laws of Motion). I am not able to understand that if we consider the body of mass $M$ to move downwards with acceleration $a$, then why would the body with mass $2M$ experience acceleration of $a/2$.
If the body $M$ is accelerating downwards with acceleration $a$ then shouldn't body $2M$ have the same acceleration as they both are connected?

 A: When pulley $B$ moves towards pulley $A$ assume mass $2M$ has moved a distance $x$.
In that same time mass $M$ has moved a distance $y= 2x$.
Differentiating the expression $y= 2x$ twice with respect to time gives the required relationship.
$\dfrac {d^2 y}{dt^2} = 2 \dfrac{d^2x}{dt^2}$, ie acceleration of mass $M$ = $2 \times $ acceleration of mass $2M$.
A: When mass M is moving downward by x length the 2AB(as there are two AB lengths) rope is moving x thus each rope is moving x/2 thus the mass 2M is moving x/2.Thus acceleration of mass M is twice of 2M.

A: What you need to understand is how they are connected. If they are connected as in the following picture:

Then they would accelerate and move with the same velocity since they are forced to by the rope. One could call this a movement by 1:1. Now notice that in your arrangement they are not connected in 1:1. They are connected as 1:2, when the right block moves a distance x then the left block will only have moved a distance $\frac{x}{2}$ and hence the acceleration is $\frac{a}{2}$ for the left block and $a$ for the right block.
Fun side note. Notice that the mass of the blocks are irrelevant. As long as we know how fast (or how far) the either one moves then we know what the speed (or distance) of the other one is since they are strictly related and constrained in motion by the rope.
A: See that:
$2AB+AM=L$ and the length of the string remains constant.
Differentiating twice we get
$$-2a_B+a_m=0$$ Note that length AB is decreasing so we get a negative sign on $a_B$ term.
$a_{2m}=a_b=\frac{a_m}{2}$
