Showing Instaneous Rest of System When m2 Falls

[r162]

Homework Statement

The system is released (m2) from rest and mid AB
How can I show that he system comes to instaneous rest when m2 has fallen a distance
of (4am1m2)/(4m1^2-m2^2)?

Homework Equations

The Attempt at a Solution

 

[r162]

http://img204.imageshack.us/img204/9970/001lp.jpg

 

[kerimek]

To show that the system comes to instantaneous rest when m2 has fallen a distance of (4am1m2)/(4m1^2-m2^2), we can use the principle of conservation of energy. Initially, the system has only potential energy, given by m2gh, where m2 is the mass of the falling object, g is the acceleration due to gravity, and h is the height from which m2 is released. As m2 falls, this potential energy is converted into kinetic energy, given by (1/2)m2v^2, where v is the velocity of m2.

At the point where m2 has fallen a distance of (4am1m2)/(4m1^2-m2^2), we can say that all of the potential energy has been converted into kinetic energy, and the system has reached its maximum velocity. This is because this distance is equal to the maximum height that m2 can reach, given by (m2v^2)/(2g). Therefore, at this point, the system has no more potential energy to convert into kinetic energy, and m2 will come to instantaneous rest.

We can also use the equations of motion to show this. From the initial position of m2, we can calculate the time it takes for m2 to reach the given distance using the equation h = (1/2)gt^2. Once we have the time, we can use the equation v = gt to calculate the velocity of m2 at that point. This velocity will be equal to the maximum velocity of the system, and since it is the only velocity in the system, the system will come to rest at this point.

In conclusion, the system will come to instantaneous rest when m2 has fallen a distance of (4am1m2)/(4m1^2-m2^2), either by the principle of conservation of energy or by using the equations of motion.