Suppose I have an ideal spring with a block of mass M attached to it at one end and the other end of the spring on a wall. The spring is executing SHM. I now gently place a small mass m on the block M when the block is in the equilibrium position.

Using Conservation of momentum I can imply that the energy of the system has decreased. I cannot understand what force is responsible for this energy loss. The only forces here are gravity(cancelled out by Normal force) , spring force(internal) and friction(internal). Since internal forces cannot do work ,how is the energy of the system changing?

(Spring is assumed to be ideal with no damping)

  • $\begingroup$ Perhaps I don't understand the question. How/why do you apply conservation of momentum without conserving mass at first? $\endgroup$ – Alchimista Mar 25 '19 at 8:37
  • $\begingroup$ The small block is given to be slowly placed on the bigger one (when the bigger one is in its equilibrium position), implying that no force is applied on the system hence conservation of momentum was applied. $\endgroup$ – Vaishakh Sreekanth Menon Mar 25 '19 at 9:42
  • $\begingroup$ But you change the mass of system, ie you change the system. Why do you want to conserve something of it if it is not the same?.... $\endgroup$ – Alchimista Mar 25 '19 at 9:45
  • $\begingroup$ What I meant is, if we take the velocity of Mass M before placing m as u and after placing it as v. We will have the relation Mu=(M+m)v. $\endgroup$ – Vaishakh Sreekanth Menon Mar 25 '19 at 9:47
  • $\begingroup$ No. It is like magic mass creation. Or You would still have to accelerate m, and this is energy taken from the potential of the spring. $\endgroup$ – Alchimista Mar 25 '19 at 14:41

You have a collision between a moving mass $M$ and a stationary mass $m$ with $M>m$.

You have put a constraint on the collision viz both masses move off together.

Now suppose that constraint was not there and the collision was elastic what would have happened?
Momentum would have been transferred from mass $M$ to mass $m$ and mass $m$ would be moving faster than mass $M$ which would mean that mass $m$ was not attached to mass $M$.

Is it possible to have an elastic collision and at the same time mass $m$ moves with mass $M$?
One could drill a horizontal hole in mass $M$ and push mass $m$ into that horizontal hole at the appropriate time.
Mass $m$ would now suffer repeated elastic collisions with the top and bottom of the hole whilst moving with mass $M$ ie it would oscillate inside the hole.
In this way no kinetic energy (mechanical energy) has been lost.

In practice the masses might suffer permanent deformation which would require work to be done to break bonds; there will be an increase the temperature of the masses (heat has been generated) and sound may be produced all contributing to the loss of kinetic energy (mechanical energy) of the system.

Without the horizontal hole and if the collision is inelastic there needs to be a mechanism to keep the two masses together and whatever that mechanism is there will be a loss of mechanical energy from the system.


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