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
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(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

1 Answer
Mar 22, 2018

Answer:

See below.

Explanation:

Calling #C# the string anchor point we have

#L = (x_B-x_A)+(x_B-x_A)+(x_B-x_C)# or

#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)