Question

Two blocks of masses \( m \) and \( 2m \) are connected by a massless string which passes over a fixed frictionless pulley. If the system of blocks is released from rest, the speed of the centre of mass of the system of two blocks after a time of 5.4 s is (Acceleration due to gravity = 10 ms\(^{-2}\))

• A. 6 ms\(^{-1}\)
• B. 8 ms\(^{-1}\)
• C. 4 ms\(^{-1}\)
• D. 12 ms\(^{-1}\)

Hint:

For problems involving pulley systems, use Newton's second law for both blocks to find the acceleration. Then, use the kinematic equation to find the velocity of the center of mass.

Answer 1

The Correct Option is A

Step 1: The two blocks are connected by a string, so they will move with the same acceleration. Let the acceleration of the blocks be \( a \). The forces on the blocks are: For block \( m \): \[ T - mg = ma \] For block \( 2m \): \[ 2mg - T = 2ma \] Step 2: Adding these two equations: \[ 2mg - mg = 3ma \] \[ mg = 3ma \] \[ a = \frac{g}{3} = \frac{10}{3} = 3.33 \, \text{ms}^{-2} \] Step 3: The speed of the centre of mass after time \( t = 5.4 \, \text{s} \) is: \[ v = at = 3.33 \times 5.4 = 6 \, \text{ms}^{-1} \] Thus, the speed of the centre of mass is 6 ms\(^{-1}\).