New amplitude of a spring-mass system when a 2nd mass falls on the first

[songoku]

Homework Statement

A mass m is connected to a spring of spring constant k and performs simple harmonic motion with amplitude A on a horizontal surface. When the spring is stretched by A/2, another mass m falls vertically on first mass. What is the new amplitude of motion?

Relevant Equations

Simple harmonic motion

Conservation of momentum?

I am confused about conservation of momentum for this question.

I take the system as two objects, each having mass m. For first object (the one performing simple harmonic motion), there is net external force acting on it in horizontal direction (restoring force of spring). For second object (the one falling vertically), there is net external force acting on it (its weight).

1. The momentum in vertical direction is not conserved because initially (before they stuck), there is initial vertical momentum but after they stuck together, the vertical momentum is zero. But how about momentum in horizontal direction. I think momentum in horizontal direction is also not conserved because there is net external force in that direction. Or we can think it is also conserved because before and after they stuck there are net external force (and assuming these forces have equal magnitude)?

2. Do we need to consider momentum to solve this question?

Thanks

 

[kuruman]

Yes, you need to conserve horizontal momentum otherwise you will not be able to solve the problem. You may assume that the collision is completed in time $\Delta t$ much smaller than the period of oscillations so the horizontal momentum of the system before and after the collision is the same and so is the external force of the spring. In other words, assume that the oscillating mass magically doubles suddenly when its displacement is $A/2$.

 

[songoku]

Yes, you need to conserve horizontal momentum otherwise you will not be able to solve the problem. You may assume that the collision is completed in time $\Delta t$ much smaller than the period of oscillations so the horizontal momentum of the system before and after the collision is the same and so is the external force of the spring. In other words, assume that the oscillating mass magically doubles suddenly when its displacement is $A/2$.

What I know about law of conservation of momentum is "the total momentum of a system is constant if no net external force acting on the system".

So the law should be: ""the total momentum of a system is constant if there is no change in net external force acting on the system"? Or maybe I miss some basic concept of momentum?

Thanks

 

[Delta2]

What I know about law of conservation of momentum is "the total momentum of a system is constant if no net external force acting on the system".

So the law should be: ""the total momentum of a system is constant if there is no change in net external force acting on the system"? Or maybe I miss some basic concept of momentum?

Thanks

No the law of conservation of momentum is fine as it is.

For this problem the momentum is not conserved neither in the vertical or the horizontal direction:

• in the vertical direction you have the reaction from the "floor" that increases during the collision time $\Delta t$ and its impulse "neutralizes" the (vertical momentum of the falling mass + the impulse of the weight of the falling mass)
• in the horizontal direction you have the external force from the spring that will impart horizontal impulse $\Omega=\int_{t_0}^{t_0+\Delta t} kx(t)dt$ during the collision time $\Delta t$ and this will make the horizontal momentum to not be conserved

So you will not be able to solve this problem using conservation of momentum. However if we assume that the impulse $\Omega$ is small enough (in comparison with the momentum that the oscillating mass has at the time instant $t_0$ that the collision starts and in comparison with the impulse of the force that arise during the collision) then conservation of momentum in the horizontal direction holds approximately and so you can use it to solve the problem .

 

[songoku]

No the law of conservation of momentum is fine as it is.

For this problem the momentum is not conserved neither in the vertical or the horizontal direction:

• in the vertical direction you have the reaction from the "floor" that increases during the collision time $\Delta t$ and its impulse "neutralizes" the (vertical momentum of the falling mass + the impulse of the weight of the falling mass)
• in the horizontal direction you have the external force from the spring that will impart horizontal impulse $\Omega=\int_{t_0}^{t_0+\Delta t} kx(t)dt$ during the collision time $\Delta t$ and this will make the horizontal momentum to not be conserved

So you will not be able to solve this problem using conservation of momentum. However if we assume that the impulse $\Omega$ is small enough (in comparison with the momentum that the oscillating mass has at the time instant $t_0$ that the collision starts) then conservation of momentum in the horizontal direction holds approximately and so you can use it to solve the problem .

I see.Thank you very much kuruman and delta2

 

[kuruman]

Just to summarize what @Delta2 said in post #4, the second mass has zero initial horizontal velocity when the first mass is moving horizontally while experiencing an external force of magnitude $kA/2$. The approximation is that when the first and second masses reach their new common horizontal velocity, the external force on the first mass is still $kA/2$. There is a subtle difference between what this implies and what you inferred from it,

So the law should be: ""the total momentum of a system is constant if there is no change in net external force acting on the system"? Or maybe I miss some basic concept of momentum?

When there is an external force acting on the system, momentum is not conserved, period. What we are asserting here is that the collision is completed so fast that the external force, although present, has no effect on the system while the collision is taking place. When a force has no effect on a system, it's as if the force does not exist. Therefore, fully aware that this is only an approximation, we suspend the external force during the very short collision-time interval which then allows us to conserve momentum and solve the problem.

 

[Lnewqban]

Deleted.

 

[kuruman]

The velocity of the previously oscillating mass was zero at point A/2, exactly when the addition of masses happened, according to the original text of the problem.

I am not sure why you are saying this. If the second mass didn't drop on it, it would keep on going until reaching full amplitude at $x=A$. The original text of the problem says that the amplitude is $A$.

 

[Lnewqban]

I am not sure why you are saying this. If the second mass didn't drop on it, it would keep on going until reaching full amplitude at $x=A$. The original text of the problem says that the amplitude is $A$.

That is correct.
Hoping that the problem was simplified that way, I misread the part stating "When the spring is stretched by A/2".
Editing previous post.
Apologies.