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Suppose that the mass B, at some instant, is at a height h above the relaxed un-stretched length of the spring as shown in the figure.. Suppose that the vertical velocity of B is v at t

Irodov Problem 1.153

As soon as the spring is released it

pushes the mass B upwards When B moves high enough, it will stretch the spring causing it to pull mass A upwards Finally when spring is stretched enough, it will have enough tension to overcome mass A weight and lift it up The problem can be solved easily with conservation of

energy

Suppose that the mass B, at some instant, is at a height h above the relaxed (un-stretched) length

of the spring as shown in the figure Suppose that the vertical velocity of B is v at this instant and mass A has not left the ground yet Then the total energy of the system is the sum of i) the

potential energy stored in the spring due to its extension given by , ii) the kinetic energy

of mass B given by and iii) the potential energy gained by mass B during its motion

upwards given by (since it has gained a height of since its initial position when the thread was intact) The initial energy content of the system is only that stored in the compressed spring given by Since energy is conserved we have,

There are three forces acting on mass A, i) the force of gravity mg pulling it down, ii) the tension

in the spring and iii) the normal reaction from the ground N Since mass A is resting on the

ground it has no acceleration

When, mass A is about to leave the ground, the normal reaction becomes 0 and so we have,

a) When the initial compression in the spring is just enough for mass A to leave the ground, the maximum height to which mass B will rise will be equal to the extension needed in the spring

to lift A off the ground given by (3) At its highest point, mass B will have its velocity v=0 in (1).

So using this condition and substituting (3) in (1) we have,

Ignoring the negative possibility we have,

b) If the initial compression is , mass B will still be moving upwards when mass A

lifts off the ground (at ) , i.e v>0 We can compute this vertical velocity v by using equation (1) as,

The velocity of the center of mass of the system when mass A is about to leave the ground is

given by,

The kinetic energy of the CG when mass A is about to leave the ground is given thus by,

Before mass A has left the ground, if mass B moves x units upwards, the CG will move x/2 units upwards By the time mass A is about to leave the ground, mass B has moved up, so the CG would have moved up a height of The total energy of the CG just before

mass A leaves the ground is sum of its kinetic energy and its potential energy and is thus given

by,

This energy ECG must be conserved at all time When the CG reaches its maximum height, its velocity will be zero Suppose that the maximum height the CG attains is H, then it would have gained a potential energy of 2mgH and kinetic energy will be 0 So applying conservation of

energy we have,