Using Atwood's Machine to Measure Time Intervals

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To use an Atwood's machine as a timer for a vertical fall of 1.30m in 10 seconds, the acceleration can be calculated using the formula a = (2h)/t², resulting in an acceleration of 0.026m/s². This acceleration can be related to the masses M and m using the equation a = (M - m)/(M + m) * g, where g is assumed to be 9.80m/s². By substituting the calculated acceleration into this equation, a relationship between the two masses can be established. The discussion emphasizes that while this relationship is crucial, no additional information is provided to determine specific values for M and m. The focus remains on deriving the mass relationship necessary for the timing mechanism.
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Using this site as a reference: http://hyperphysics.phy-astr.gsu.edu/hbase/atwd.html

Can somebody show me how to solve this:

Suppose you don't have a watch, or a timer, and you want to use an Atwood's machine as a timer to give an accurate time interval (for the vertical fall) of t = 10.0s. Suppose the vertical fall, y, is 1.30m. Determine values of M and m which would make t - 10s.

Some formulas

a = (2h)/t2
g= a x ((2M+m)/m)/m
 
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Find the acceleration using the formula you have.

Then set this acceleration = \frac{M-m}{M+m}*g (taking M as the heavier mass and m as the lighter mass of the atwood machine).

this will only give a relationship between M and m. is there any more information for the problem?
 
nope just that. I am guessing it might be sufficient just to have the relationship

i'm assuming g= 9.80
?
 

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