Particle Dynamics: Proving t(a) < t(b) for Two Masses Connected by a String

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Two blocks A and B, each of mass m, are connected by a string over a pulley, with block B initially held in place. Upon release, block A moves to the right, while block B swings down, leading to the goal of proving that the time taken for block A to reach the pulley, t(a), is less than the time taken for block B to hit the surface, t(b). The discussion involves deriving equations for the motion of both blocks, considering the tension in the string, which is not constant. Participants suggest using energy conservation and integrating to find the relationship between the accelerations of both blocks. The challenge remains to accurately connect the equations for blocks A and B to establish the inequality t(a) < t(b).
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Homework Statement


Two blocks A and B , each of same mass m are attached by a thin inextensible string through an ideal pulley .
Initially block B is held in position as shown in figure . Now the block B is released . Block A will slide to right and hit the pulley in time t(a) . Block B will swing and hit the surface in time t(b).
Assume the surface as frictionless.
Then prove that ,
t(a)<t(b)



The Attempt at a Solution



Let the tension in the string be T .
L=1/2(T/m)[t(a)^2]
How to get the time t(b) .
 

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hi nik jain! :smile:

the tension isn't constant, is it? :wink:

call the tension T(t),

do F = ma for block A, and F = ma and τ = Iα for block B …

what do you get? :smile:
 
Yes Tension is not constant.

Let the block moves a distance (x) .
1/2m[v(a)^2] = Tx (By energy conservation)
1/2m[v(a)^2] = mAx (A is the acceleration of block A)
1/2m[v(a)^2] = mv(a)(dv/dx)x
integrating it
v(a) = (2x)^1/2
again integrating it
t(a) = (L/2)^1/2

How to get the t(b) ?
Please make the equation as above .
And how the acceleration of block A relates to the acceleration of block B ?
 
nik jain said:
1/2m[v(a)^2] = Tx (By energy conservation)

(try using the X2 button just above the Reply box :wink:)

But T is not constant, so it's 1/2m[v(a)^2] = ∫ T dx
1/2m[v(a)^2] = mAx (A is the acceleration of block A)

But now you've lost T. :redface:

The whole point is to connect the A and B equations, using T.

For B, you will need two parameters, length and angle. :wink:
 
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