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

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

(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

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Answer 1 · 2018-03-22T10:34:51

See below.

Calling $C$ $C$ the string anchor point we have

$L = (x_{B} - x_{A}) + (x_{B} - x_{A}) + (x_{B} - x_{C})$ $L = (x_B-x_A)+(x_B-x_A)+(x_B-x_C)$ or

$L = 3 x_{B} - 2 x_{A} - x_{C}$ $L = 3x_B-2x_A-x_C$ so

$0 = 3 ddot x_B-2 ddot x_A$ because $x_C = C^(te)$ then

$ddot x_A = 3/2 ddot x_B$ ---> (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 $t_i, i = {A,B,C}$ are string tensions, solving for $t_i, ddotx_A,ddot x_B$ 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

$F < 15/13 F$ ----> (3)