A piece of clay stuck to a SHM oscillator

[Leo Liu]

Homework Statement

A quiz question

Relevant Equations

$E_i=E_f$

I don't get why the total mechanical energy is not conserved in this situation.
When the length of the spring reaches the maximum, the speed of the block is 0 and we have the following equation:
$$E=K+U=1/2mv^2+1/2kA^2,\text{where A is the amplitude} \implies E=1/2kx_{max}^2$$
I can't see why the amount of energy changes. Could someone please explain the reason to me? Thanks a lot.

 

[haruspex]

I can't see why the energy changes.

Read it more carefully. It says the ME of the falling clay is lost.

 

[Leo Liu]

Read it more carefully. It says the ME of the falling clay is lost.

Aha, I see--this is kinda tricky. Thank you.

 

[Delta2]

If we want to extent this a bit, i think the vertical momentum of the system is not conserved either.

 

[Leo Liu]

If we want to extent this a bit, i think the vertical momentum of the system is not conserved either.

Well, it is conserved if you consider the movement of the Earth :P

 

[Delta2]

Well, it is conserved if you consider the movement of the Earth :P

Well, yes if we include Earth into our system it is conserved but by system i meant the block+clay.

 

[hutchphd]

Homework Statement:: A quiz question
Relevant Equations:: $E_i=E_f$

I don't get why the total mechanical energy is not conserved in this situation.
When the length of the spring reaches the maximum, the speed of the block is 0 and we have the following equation:

I'm with you. No mention is made of the height of the drop...it could be made negligibly small in which case the problem is better. Badly stated problem by my reckoning.

 

[haruspex]

If we want to extent this a bit, i think the vertical momentum of the system is not conserved either.

Right, but it explicitly asks about horizontal momentum.

 

[Delta2]

Right, but it explicitly asks about horizontal momentum.

I have my doubt about conservation of horizontal momentum because we have the force of the spring acting in the x- direction which is at a maximum value btw during the collision.
I think "none of these" is the ultimate correct answer after all but of course one may argue that the horizontal momentum is approximately conserved.

 

[haruspex]

I have my doubt about conservation of horizontal momentum because we have the force of the spring acting in the x- direction which is at a maximum value btw during the collision.
I think "none of these" is the ultimate correct answer after all but of course one may argue that the horizontal momentum is approximately conserved.

I don't see how it affects horizontal momentum. The block is stationary when the clay sticks to it. No horizontal momentum immediately before impact and none immediately after.

 

[Delta2]

and none immediately after

Isn't the spring force , via its impulse, going to transfer horizontal momentum $\int_t^{t+\Delta t} kx(t) dt$ during the impact time interval $(t,t+\Delta t)$.

 

[haruspex]

Isn't the spring force , via its impulse, going to transfer horizontal momentum $\int_t^{t+\Delta t} kx(t) dt$ during the impact time interval $(t,t+\Delta t)$.

Yes, but in all such questions we take Δt to be arbitrarily small.

 

[haruspex]

Btw, the question could be clearer. It doesn’t specify whether it's the horizontal momentum of the block or of the block+clay system. If the latter, it is unnecessary to assume the block is stationary for the duration of the impact.

 

[Delta2]

Btw, the question could be clearer. It doesn’t specify whether it's the horizontal momentum of the block or of the block+clay system. If the latter, it is unnecessary to assume the block is stationary for the duration of the impact.

Judging by the answer i think it means the horizontal momentum of the clay.

 

[haruspex]

Judging by the answer i think it means the horizontal momentum of the clay.

The answer given is that neither the clay nor the block change in momentum.
Given the question but not the answer, I would have guessed it was asking about the block.