# How to compare the relative change in momentum of two blocks attached by a pulley?

## 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 (1) the acceleration of A will be more than that of B (2) the acceleration of A will be less than that of B (3) the sum of rate of changes of momentum of A and B is greater than the magnitude of F (4) the sum of rate of changes of momentum of A and B is equal to the magnitude of F

Mar 22, 2018

See below.

#### Explanation:

Calling $C$ the string anchor point we have

$L = \left({x}_{B} - {x}_{A}\right) + \left({x}_{B} - {x}_{A}\right) + \left({x}_{B} - {x}_{C}\right)$ or

$L = 3 {x}_{B} - 2 {x}_{A} - {x}_{C}$ so

$0 = 3 {\ddot{x}}_{B} - 2 {\ddot{x}}_{A}$ because ${x}_{C} = {C}^{t e}$ then

${\ddot{x}}_{A} = \frac{3}{2} {\ddot{x}}_{B}$ ---> (1)

Now solving the system

{(t_A = 2 t_B), (t_C = 2 t_B), (t_B + t_C = m ddotx_B), (F - t_A = m ddotx_A), (ddotx_A = 3/2 ddotx_B):}

where ${t}_{i} , i = \left\{A , B , C\right\}$ are string tensions, solving for ${t}_{i} , {\ddot{x}}_{A} , {\ddot{x}}_{B}$ we obtain

{(),(t_A -> (4 F)/13), (t_B -> (2 F)/13), (t_C -> (4 F)/13), (ddotx_A -> (9 F)/(13 m)), (ddotx_B -> (6 F)/(13 m)):}

and comparing $F$ with $m {\ddot{x}}_{A} + m {\ddot{x}}_{B}$ we have

$F < \frac{15}{13} F$ ----> (3)