Two pulleys and same masses - velocity?

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The discussion revolves around calculating the velocity of two equal masses suspended by pulleys after they are released and separated by a distance h. The conservation of energy principle is applied, equating initial potential energy to final kinetic and potential energies. It is established that as one mass moves down, the other moves up, with their distances traveled being in a 2:1 ratio. The final velocity of the leftmost mass is derived to be v = sqrt(8/15gh), confirming the relationship between their speeds based on the distances moved. The analysis emphasizes the connection between the masses' velocities and the constraints of the inextensible string.
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Homework Statement



If two equal masses are suspended as in the following, and they are both initially at rest, after they are released, what is the velocity of the leftmost mass at the instant the separation between the two masses is h ? Assume frictionless pulleys and extensionless strings.

yPOxJ.png


Homework Equations



KE(i) + PE(i) = KE(f) + PE(f)

The Attempt at a Solution



Set x as the height initially. For every extension of the string immediately left, the pulley system pulls the second mass up half the distance, so the left cord will have traveled 2/3h and the right cords will have moved 1/3h, for a total separation of h.

mgx + mgx = mg(x+1/3h) + mg(x-2/3h) + 1/2*m*v1^2 + 1/2*m*v2^2

We don't know what the velocity of each mass is, correct? We also know that each of the tensions a/b/c or T, and there's a gravitational force downwards with magnitude mg.
 
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The answer, apparently is v = sqrt(8/15gh)
 
help?
 
You know what the speed of each is relative to the other. Therefore you only have one unknown in your energy balance. Answer is correct.
 
LawrenceC said:
You know what the speed of each is relative to the other. Therefore you only have one unknown in your energy balance. Answer is correct.

How are the speeds related? What do we use to determine their relation?
 
You know how far each has traveled from its starting point...if one mass went twice as far (which you have already determined) in the same amount of time, how must their speeds relate given that the cord does not streach?
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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