Stopping vertically oscillating mass by applying momentum?

[amiras]

Homework Statement

I came with this problem myself, so there may be not enough information given to solve it.

Imagine that mass M oscillating on the vertical spring with force constant k and amplitude A. Now let's say that the mass M is moving upward. Now imagine that some other object with mass m having speed v_m comes from the sky and collides with the oscillating mass and after collision sticks to it.

Homework Equations

Is it possible stop the mass M from oscillation? What position? What velocity of v_m?

The Attempt at a Solution

Can I apply conservation of momentum at every displacement of the spring assuming it happen in short time?

Applying it at equilibrium position would give: v_s*M - v_m*m = 0, where v_s is the speed of the mass M. But does the final momentum equal to zero means that both bodies stuck and do not move or that they exchanged momentum and still moving to opposite directions?

 

[nasu]

It depends on what you mean by "stop the mass M from oscillation".
It is possible to arrange the things so that after collision the two bodies have zero velocity.
However, unless this happens right in the equilibrium position (new one, with the extra body attached), the oscillations will start again. So you only stop it for an instant and then it restarts, with a different amplitude and period.

 

[amiras]

Yes that is exactly what I wanted.

Ok so imagine that d2 = (m+M)g/k is the new equilibrium position for the system.

After the collision the body must move just as much to stretch spring from current equilibrium position to the new. In other words all the left kinetic energy after collision must be used to stretch extra distance.

Initially spring stretched d1 = mg/k. So using 1/2(m+M)v^2 = 1/2k((d2)^2 - (d1)^2) Would give the right speed after collision.

Now we want after collision masses will move with this velocity v, so using conservation of momentum:

m*v_e + M*v_s = v(m+M), where v_e the speed of mass m, and v_s speed of mass M.

From here it is possible to calculate v_e what I wanted initially.

Is that OK?

 

[nasu]

Yes that is exactly what I wanted.

Ok so imagine that d2 = (m+M)g/k is the new equilibrium position for the system.

After the collision the body must move just as much to stretch spring from current equilibrium position to the new. In other words all the left kinetic energy after collision must be used to stretch extra distance.

You don't need any kinetic energy to stretch spring from current position to the new equilibrium. If after collision the system is not in the equilibrium position, there will be some force acting towards the equilibrium position (the net force is zero at equilibrium, by definition).