Two masses m1 and m2 (m1 > m2) are suspended by an inextensible string which passes over a pulley of negligible mass (see figure). The mass m1 is initially at a height h1 and the two masses are initially at rest. What maximum height will reach m2 when the system is released? Give the result as a function of the givens of the problem.
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We will solve the problem in two steps. First, we will calculate the speed of the two masses when m1 reaches the ground, then we will calculate the additional height that m2 rises when the string is no longer tense.
The following figure shows the initial and final states of the system. In state A the two masses are at rest and m1 is at a height h1 from the ground. In the final state B the mass m1 has reached the ground and both masses have the same speed because they are linked by the string.
The mechanical energy of the system with the two masses is preserved during the motion, because the only force that acts on them is the weight (which is a conservative force) and the total work of the tensions for both sides of the string is canceled. Therefore we can write the principle of conservation of energy for the system of the two masses:
And we can isolate the speed from the previous equation:
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When the mass m1 reaches the ground, the string is no longer tense. But since m2 has a velocity, it will continue to rise until it stops. Next we are going to calculate the additional height h it rises until it stops.
State B (when mass m1 reaches the ground) and state C (when mass m2 has stopped) are represented in the following figure. Note that we have now chosen the position of mass m2 in state B as the zero height position.
The mechanical energy of mass m2 is conserved between states B and C, because the tension of the string is zero and the weight is a conservative force:
After isolating h and substituting the value of v calculated previously we obtain:
With respect to the ground, the mass m2 rises to a height h1 + h:
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