Question

Block X of mass M is attached to block Y of mass 2M by a light string that passes over a pulley of negligible friction and mass, as shown above. In which direction will the center of mass (COM) of the two-block system move after it is released from rest, and what is the magnitude of the acceleration a of block X ?

Answer 1

The correct answer is (A) Down and to the right; 2g/3.

1. Net Force: The only external forces acting on the system are the gravitational forces on each block, pulling them downward. Since block Y has a larger mass (2M), it experiences a greater downward force than block X. This creates a net downward force on the system.

2. Center of Mass (COM): The COM of the system is the weighted average of the positions of the two blocks, taking into account their masses. Due to the net downward force, the COM will also move downward.

3. Movement Direction: As block Y descends, it pulls on the string, causing block X to move to the right. Therefore, the overall movement of the COM will be down and to the right.

4. Acceleration of Block X:

- The tension in the string is the same throughout, so the force pulling block X to the right is equal to the force pulling block Y downward (minus its weight).

- Using Newton's Second Law (F = ma) for block X:

- Force = Tension = (2M - M)g = Mg

- Mass = M

- Acceleration (a) = F/m = Mg/M = g

- However, we need to consider the fact that block X is also moving downward due to gravity. This downward component of its acceleration is also g.

- Therefore, the total acceleration of block X is the vector sum of its horizontal and vertical accelerations, which gives us 2g/3 at an angle of 45 degrees (down and to the right).

So, the COM of the system will move down and to the right, and the magnitude of the acceleration of block X will be 2g/3.

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