# Dropping a block on another block which is performing SHM

Hamiltonian
Homework Statement:
A block is performing SHM with an amplitude ##A## and angular frequency ##\omega##, if an identical block is dropped on the original block while it passes through its mean position find the common velocity of the two blocks, assume friction to be sufficient.
Relevant Equations:
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at the mean position of the original block, its velocity is ##V = \omega A##
once the new block is dropped we can conserve momentum in the horizontal direction $$m\omega A = (2m)v$$ $$v = \frac{\omega A}{2}$$
where ##v## is the common velocity of the blocks.

but if instead of conserving linear momentum I try to conserve energy to find the common velocity
$$(1/2)\omega^2 m A^2 = (1/2)(2m)v^2$$
$$v = \frac {\omega A}{ \sqrt {2}}$$
why is conserving energy to find the common velocity of the blocks wrong here? is it because the friction has done work on the upper block to increase its velocity from zero? but instantaneously the block has no displacement so ##W_{friction} = 0##

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Delta2

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why is conserving energy to find the common velocity of the blocks wrong here? is it because the friction has done work on the upper block to increase its velocity from zero?
Yes.

Call the oscillating block ‘A’ and the block that is dropped ‘B’.

B is accelerated to the common velocity over a finite time-period (because the frictional force between A and B is finite). During this time-period, B is unavoidably sliding against A and the friction causes heating.

This is very similar to a simple 1D inelastic collision, where 2 objects move towards each other and stick together, with some kinetic energy (unavoidably) causing heating.

Delta2 and Hamiltonian
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where 2 objects move towards each other and stick together
And in such cases energy goes into internal vibrations, dissipated as heat and sound.

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B is accelerated to the common velocity over a finite time-period (because the frictional force between A and B is finite). During this time-period, B is unavoidably sliding against A and the friction causes heating.
I think that the idea of "unavoidable sliding" is needlessly specific. Sliding can be avoided, for example, if the falling block has a metal pin that fits snugly in a hole drilled in the bottom block. Saying that mechanical energy is converted into internal energy would be sufficient.

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I think that the idea of "unavoidable sliding" is needlessly specific. Sliding can be avoided, for example, if the falling block has a metal pin that fits snugly in a hole drilled in the bottom block. Saying that mechanical energy is converted into internal energy would be sufficient.
I know what you mean. In fact I thought about that specific point when writing the reply.

The diagram clearly shows the geometry for the collision - 2 flat parallel faces which will meet. Sliding with friction is (as far as I can see) the only possible process for this particular setup.

The OP (Post #1) had written “but instantaneously the block has no displacement”. I interpreted this as implying that the OP believed the block reached its final velocity instantly. I wanted to correct this by explaining the underlying mechanism. Simply saying “mechanical energy is converted into internal energy“ would not have achieved this.

The fact that sliding would occur at all might not be obvious (to a learner). I wanted to emphasise that sliding with friction was both present and indeed unavoidable in the system described.
Edit: And as a result, acceleration would take some finite time.

Delta2
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It is also interesting to note that if the block were dropped at maximum spring extension, energy is indeed conserved before and after the collision.

rsk and Delta2
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Sliding can be avoided, for example, if the falling block has a metal pin that fits snugly in a hole drilled in the bottom block.
@Steve4Physics effectively covered that by mentioning 1D inelastic collisions.
But even in the purely frictional case there need be no sliding if we allow a little deformation.
During the brief impact, the normal force is ##N(t)##. If the vertical speed just before impact is u and the final horizontal speed is v and there is no sliding then ##mu=\int N(t).dt##, ##mv\leq\int N(t)\mu_s.dt##, whence ##\mu_s\geq \frac vu##.

Steve4Physics and Delta2
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@Steve4Physics effectively covered that by mentioning 1D inelastic collisions.
But even in the purely frictional case there need be no sliding if we allow a little deformation.
During the brief impact, the normal force is ##N(t)##. If the vertical speed just before impact is u and the final horizontal speed is v and there is no sliding then ##mu=\int N(t).dt##, ##mv\leq\int N(t)\mu_s.dt##, whence ##\mu_s\geq \frac vu##.
Yes, I agree. My objection was strictly to "unavoidably sliding" being too narrow and specific.

Steve4Physics
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@kuruman and @haruspex - you have convinced me - "unavoidably sliding" was excessive!

At the microscopic level, there could be interlocking bumpy surfaces such that the interlocking prevents sliding.

There would then be some sort of pattern of distortion propagating through each block while its velocity changes.

We can combine @kuruman ’s comment in in Post #6 that “if the block were dropped at maximum spring extension, energy is indeed conserved” with @haruspex ’s limit for ##\mu_s## in post #7. This then suggests something interesting - that there is only a range of positions during the SHM where sliding occurs (if indeed sliding occurs at all!).

Hamiltonian
It is also interesting to note that if the block were dropped at maximum spring extension, energy is indeed conserved before and after the collision.
I am not able to see why some amount of energy won't be stored as internal energy. At the maximum spring extension, the original block is instantaneously at rest but the new block is still making an inelastic collision with the original block so shouldn't energy still be stored as internal energy?

or is it in fact not an inelastic collision and the new block can also be placed gently on the original block and then friction will start to accelerate it?

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At the maximum spring extension, the original block is instantaneously at rest but the new block is still making an inelastic collision with the original block so shouldn't energy still be stored as internal energy?
Yes, but that's the KE that was in the vertical motion of the dropped block. Up until now, we have only been concerned with the KE in the horizontal motion of the lower block.

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The vertical motion has nothing to do with it. If instead the descending block were placed in the path of the moving block, an elastic collision would send it careening to the right with ##\omega A## and stop the the oscillating block instantly. No sliding is necessary, just glue to make an inelastic collision. Sliding not required, just an inelastic collision and eventual dissipation of the energy to other degrees of freedom.

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send it careening to the right
Or wherever the ships' hulls in need of cleaning were hauled up.

Hamiltonian
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Or wherever the ships' hulls in need of cleaning were hauled up.
@Hamiltonian299792458 :
To careen originally meant to clean the hull of a ship. Secondarily, to heel a ship over in order to clean it. Thirdly, of a ship, to heel over due to wind and waves.
The modern usage, of any thing or anyone to move at great speed or wildly seems to be a confusion with "career".

Hamiltonian
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@Hamiltonian299792458 :
The modern usage, of any thing or anyone to move at great speed or wildly seems to be a confusion with "career".
As in "your career is on a fast track"?

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As in "your career is on a fast track"?
"Career" comes from the Latin for chariot, so its original meaning in English, as a verb, was to move swiftly. Thence, as a noun, to refer to a person's life course, then more narrowly to the employment aspect.