Conservation of Momentum vs Constant k for an Ideal Spring

[BlazenHammer]

Homework Statement

A horizontal spring block system of mass 2 kg executes SHM. When the block is passing through its equilibrium position ,an object of mass 1 kg is put on it and two move together.
Calculate the new Amplitude of oscillation:

Relevant Equations

The usual SHM and conservation eqns. , nothing special!

I encountered a weird conflict between my thought process and that of author's solution in book:

The common viewpoint of both of us were invoking conservation of energy of this SHM system

But the author proceeds to solve it using conservation of momentum, taking the new mass added to system as internal to the system
(I felt uneasy here, not sure if I understand this because if I recall correctly , due to a similar derivation of rocket equations in past where adding of new mass kind of acts like/leads to existence of certain force interactions)

$2V = 3V'$

The author proceeds to find the the initial energy of the system and equates it to the energy when velcoity and mass has changed(Conservation of energy)

and arrives at $A' =\sqrt{\frac{2}{3}}A$

What I did was the invoke the conservation of energy, just keeping in mind that the angular velocity of SHM will change due to addition of new mass but $k$ of the spring remains constant

$E= \frac{mω^2A^2}{2}$
$E= \frac{m\frac{k}{m}A^2}{2}$

for intiial energy m = 2 and final energy expression m = 3

Solving the eqns gives that the new Amplitude remains same as the intital one?

$A' = A}$
Now since conservation of momentum is very fundamental priniciple so either I might be wrong in my assumption or the author might be applying the principle incorrectly which I'm not able to spot as I don't remeber my basics throroughly

 

[collinsmark]

Homework Statement: A horizontal spring block system of mass 2 kg executes SHM. When the block is passing through its equilibrium position ,an object of mass 1 kg is put on it and two move together.
Calculate the new Amplitude of oscillation:
Homework Equations: The usual SHM and conservation eqns. , nothing special!

I encountered a weird conflict between my thought process and that of author's solution in book:

The common viewpoint of both of us were invoking conservation of energy of this SHM system

But the author proceeds to solve it using conservation of momentum, taking the new mass added to system as internal to the system
(I felt uneasy here, not sure if I understand this because if I recall correctly , due to a similar derivation of rocket equations in past where adding of new mass kind of acts like/leads to existence of certain force interactions)

$2V = 3V'$

The author proceeds to find the the initial energy of the system and equates it to the energy when velcoity and mass has changed(Conservation of energy)

and arrives at $A' =\sqrt{\frac{2}{3}}A$

What I did was the invoke the conservation of energy, just keeping in mind that the angular velocity of SHM will change due to addition of new mass but $k$ of the spring remains constant

$E= \frac{mω^2A^2}{2}$
$E= \frac{m\frac{k}{m}A^2}{2}$

for intiial energy m = 2 and final energy expression m = 3

Solving the eqns gives that the new Amplitude remains same as the intital one?

$A' = A}$
Now since conservation of momentum is very fundamental priniciple so either I might be wrong in my assumption or the author might be applying the principle incorrectly which I'm not able to spot as I don't remeber my basics throroughly

The action of adding the 1 kg mass is equivalent to an inelastic collision. So you can't use conservation of energy to analyze that particular part. Some of the energy in original spring-mass system is converted to heat when the 1 kg mass is tacked on.

Conservation of energy applies before the mass is added, and again later after the mass is added, but not at the moment the mass is added. So you'll need to rely on conservation of momentum to determine the new velocity, immediately after adding the 1 kg mass.

 

[Andrew Mason]

Homework Statement: A horizontal spring block system of mass 2 kg executes SHM. When the block is passing through its equilibrium position ,an object of mass 1 kg is put on it and two move together.
Calculate the new Amplitude of oscillation:

The author should have made it clear that the additional mass was stationary with respect to the lab frame at the moment it was added. If the 1 kg mass was moving in the lab frame at the same velocity as the 2 kg block when they were put together, conservation of energy, as well as conservation of momentum, would apply.

AM