Question

Two blocks A and B of equal mass are connected by a light inextensible taut string passing over two light smooth pulleys fixed to the blocks. The parts of the string not in contact with the pulleys are horizontal. A horizontal force F is applied to the block A as shown. There is no friction, then

• A. the acceleration of A will be more that of B
• B. the acceleration of A will be less than that of B
• C. the sum of rate of changes of momentum of A and B is greater than the magnitude of F.
• D. the sum of rate of changes of momentum of A and B is equal to the magnitude of F.

Answer 1

Answer: (D)

the sum of rate of changes of momentum of A and B is equal to the magnitude of F.

Answer 2

Let $v_A$ and $v_B$ be the velocities of blocks A and B, respectively. The sum of the rate of changes of momentum of A and B is given by:

$\frac{d}{dt}(m v_A) + \frac{d}{dt}(m v_B) = m a_A + m a_B$

The net external force on the system consisting of blocks A and B in the horizontal direction is $F$. According to Newton's second law for the system, the net external force equals the rate of change of total momentum:

$F = \frac{d}{dt}(m v_A + m v_B) = m a_A + m a_B$

Therefore, the sum of the rate of changes of momentum of A and B is equal to the magnitude of F.