- #1

MathVoider

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- Homework Statement
- Two block, A and B, of identical mass are connected together by a massless string which is always under tension. They each are of a distance L from the pulley. The pulley has negligable size compared to the blocks. All kinds of friction can be ignored. Initially the block B is suspended by a string. The question is, when the aforementioned string is cut, which block reaches the wall/pulley before the other? In other terms which block travels the distance "L" the fastest.

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- Relevant Equations
- x

## My Solution

after the string is cut we have a system well defined by the following free body diagram:I argue that the acceleration of the block A in the x direction (##a_A^x##) is:

$$ F_A = T_1 $$

$$ ma_A^x = T_1 $$

$$ a_A^x = \frac{T_1}{m} $$

I argue that the acceleration of the block B in the y direction (##a_B^y##) is:

$$ F_B^y = T_2\cos(\theta)-F_g $$

$$ ma_B^y = T_2\cos(\theta)-mg $$

$$ a_B^y = \frac{T_2\cos(\theta)}{m} -g $$

I argue that the acceleration of the block B in the x direction (##a_B^x##) is:

$$ F_B^x = -T_2\sin(\theta) $$

$$ ma_B^x = -T_2\sin(\theta) $$

$$ a_B^x = -\frac{T_2\sin(\theta)}{m} $$

Now that we've got the equation making out of the way i can explain my resoning:

I state that ##T_1## and ##T_2## are egual because when the block B moves

**along**the direction of the string then the block A has to move by the same distance. As such we get:

$$ a_A^x = \frac{T_1}{m}$$ $$a_B^x = -\frac{T_1\sin(\theta)}{m} $$

if we take the absolute value we get:

$$ a_A^x = \frac{T_1}{m}$$ $$a_B^x = \frac{T_1\sin(\theta)}{m} $$

now I argue that ##a_A^x>a_B^x## because ##\sin(\theta)## goes from 0 to 1, since block A and B have the same distance from the pulley\wall, and they both start at rest, and block A has a greater accelleration then block B then block A will reach the pulley\wall faster.