Irodov Problem 1.140

When the thread holding the mass to the wall is burnt, the free hanging mass drags the mass on the horizontal plane to the right. This elongates the spring which starts to resist the movement of the mass with a restoring force F. During the process, the spring also inclines at an angle to the vertical and F is inclined with the vertical at an angle. The vertical component of F tends to lift the mass upwards and will do so when its strong enough to resist the weight of mass mg.


Suppose that the elongation of the spring is x at the break-off point so that the total length of the spring is now l+x. Then, we have,




There are three forces acting in the vertical direction on the mass on the horizontal plane i) the normal reaction N, ii) the component of the force exerted by the spring and iii) the gravitational force mg. The mass has no acceleration along this direction prior to the break-off from surface. Further, at break-off the normal reaction N=0. So we have,











Now we need to determine the mass's velocity just before break-off. We shall solve this using the conservation of energy principle. The hanging mass, as it descends, looses potential energy. This loss in potential energy is converted into the i) energy stored in the spring and ii) the kinetic energy of the two masses. Just before break-off, the horizontal mass moves a distance,




and this is the same distance that the hanging mass falls.

Let the velocity of the masses just before break-off be v. Then by conservation of energy we have,

Irodov Problem 1.139

The figure depicts what happens in the problem. The mass m starts its journey from the top and slides downwards until it reaches the end of the rubber cord. At this point the mass has gained a lot of momentum and so when it hits the end it elongates the rubber cord with its impact and weight by a length making the total length . The question asks the value of .


The problem can be easily solved by using conservation of energy. The rubber cord exerts a restoring force of when the rubber band is elongated by (similar to a spring) and the energy stored in the rubber cord will be . At this point the mass has fallen a height of and thus lost a potential energy of . Also the mass would have stopped moving at this point so it has no kinetic energy. Since energy is always conserved the potential energy lost by the mass must be now stored in the elongated rubber cord. In other words,









If the mass m were not dropped and simply placed at the bottom of the rubber cord, it would elongate the rubber cord by a length of mg/k - this would be the equilibrium point of the rubber cord with mass. Since it was dropped from a height, the mass will exhibit simple harmonic motion about the equilibrium point. The + and - signs in the solution represent the maximum and minimum elongations experienced by the rubber cord during this simple harmonic motion. The maximum elongation is thus simply the solution with the + sign.