Which block reaches the wall faster?

  • #1
MathVoider
2
0
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.

Scroll down for the diagram:
Relevant Equations
x
normale.PNG

My Solution​

after the string is cut we have a system well defined by the following free body diagram:

problema.PNG


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.

Question​

I want to know if my reasoning is correct or if I have forgotten/misused something.
 
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  • #2
MathVoider said:
$$ a_A^x = \frac{T_1}{m}$$ $$a_B^x = -\frac{T_1\sin(\theta)}{m} $$
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.
I do not see anything missed, although it seems like more work than is required.

I would have liked to see the point more clearly and directly made that $$|a_A^x| = \frac{T_1}{m} > \frac{T_1 \sin \theta}{m} = |a_B^x|$$
Personally, I would have taken a different approach. Reason first that the horizontal position of the center of mass will accelerate in the direction of the horizontal net force from the pulley. That being a rightward (or briefly zero) net external horizontal force, the center of mass must be to the right of the wall when one of the two blocks first reaches the wall. Clearly, this can only happen if block A reaches the wall first.
 
  • #3
jbriggs444 said:
I do not see anything missed, although it seems like more work than is required.

I would have liked to see the point more clearly and directly made that $$|a_A^x| = \frac{T_1}{m} > \frac{T_1 \sin \theta}{m} = |a_B^x|$$
Personally, I would have taken a different approach. Reason first that the horizontal position of the center of mass will accelerate in the direction of the horizontal net force from the pulley. That being a rightward (or briefly zero) net external horizontal force, the center of mass must be to the right of the wall when one of the two blocks first reaches the wall. Clearly, this can only happen if block A reaches the wall first.
Thank you
 

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