Conservation of momentum - Vertical spring

[danielbaker453]

Homework Statement

A block of mass 200g is suspended through a vertical spring. The spring is stretched by 1.0 cm when the block is in equilibrium. A particle of mass 120g is dropped on the block from a height of 45 cm. The particle sticks to the block after the impact. Find the maximum extension of the spring. Take g=10 m/s^2.

Homework Equations

Initial Momentum of system = Final Momentum of system
U_i+K_i=U_f+K_f

The Attempt at a Solution

Since the spring is initially in equilibrium,
kx_i = Mg
using this we can find the spring constant.
Also, at the point of maximum elongation,
kx_f=(M+m)g
using this we can find the maximum elongation.
The answer I get is 1.6 cm which is wrong (the correct answer is 6.1 cm). When I looked at the solution, I saw that conservation of momentum was applied first and then conservation of energy. But, I don't see why the momentum can be conserved. There is a net external force on the system because of gravity.
Could someone explain to me what is wrong with my approach and also why momentum conservation can be applied in this case?
Here is a diagram of this problem I drew

 

[TomHart]

You have an inelastic collision when the particle strikes the block so conservation of energy does not apply. Momentum is conserved during the collision, however.

If you drop a weight from some height onto a spring, do you think it will directly go to the equilibrium position?

 

[danielbaker453]

No. Not directly but only after the kinetic energy of the system becomes zero.

 

[BvU]

There is a net external force on the system because of gravity.

At the moment of collision, the net external force is zero: gravity is canceled by the force the ceiling exerts. But you are correct: the conservation of momentum is indeed not applicable over a time interval > 0. Only at the moment of collision, which is probably what they do in the book solution (that I can't see) to calculate the kinetic energy of the block + particle after the inelastic collision

[edit]Tom's faster, but we agree.

 

[TomHart]

No. Not directly but only after the kinetic energy of the system becomes zero.

Is there any mention of damping?

 

[danielbaker453]

At the moment of collision, the net external force is zero: gravity is canceled by the force the ceiling exerts. But you are correct: the conservation of momentum is indeed not applicable over a time interval > 0. Only at the moment of collision, which is probably what they do in the book solution (that I can't see) to calculate the kinetic energy of the block + particle after the inelastic collision

[edit]Tom's faster, but we agree.

That is

Is there any mention of damping?

No. I've typed the entire question as it is.

 

[BvU]

No. Not directly but only after the kinetic energy of the system becomes zero

Kinetic energy zero doesn't mean equilibrium position !

 

[TomHart]

Have you ever stepped onto one of the old mechanical scales? Don't you notice that the reading does not go immediately to the final weight. It tends to oscillate above and below the final weight that it settles on. (Due to friction, it eventually stops oscillating.) However, in your problem, since there is no mention of damping, I think it's safe to assume that there is no damping.

 

[danielbaker453]

Kinetic energy zero doesn't mean equilibrium position !

But why not? Wouldn't K.E. be zero when the spring is at maximum elongation?

 

[BvU]

It would. But you answered that in response to a question on equilibrium position. Which is probably what you calculated in post #1.

 

[danielbaker453]

Have you ever stepped onto one of the old mechanical scales? Don't you notice that the reading does not go immediately to the final weight. It tends to oscillate above and below the final weight that it settles on. (Due to friction, it eventually stops oscillating.) However, in your problem, since there is no mention of damping, I think it's safe to assume that there is no damping.

So since there is no damping, can energy be conserved?

 

[BvU]

Not in the collision. Afterwards, yes.

 

[TomHart]

So since there is no damping, can energy be conserved?

Once you use conservation of momentum to figure out the velocity of the particle/block combination, energy will be conserved forever after that point.

@BvU: Great picture above.

 

[danielbaker453]

Also, could anyone please explain what was wrong with my approach?

 

[BvU]

Picture stolen from Wiki, I'm afraid - no credit to me.

 

[BvU]

Also, could anyone please explain what was wrong with my approach?

Also, at the point of maximum elongation,
kxf=(M+m)g

That's the equilibrium position.

Show your work in detail.

 

[TomHart]

Also, could anyone please explain what was wrong with my approach?

If you set the particle on the block very lightly - as opposed to dropping it from some height - don't you think the maximum extension would be different?

 

[danielbaker453]

If you set the particle on the block very lightly - as opposed to dropping it from some height - don't you think the maximum extension would be different?

Yes! If it is dropped from a height the extension will be more as compared to when it is placed lightly.

 

[TomHart]

Yes! If it is dropped from a height the extension will be more as compared to when it is placed lightly.

Yes, so you need another tool other than F=kx.

 

[danielbaker453]

Also, here is a link to the solution I referred to

 

[danielbaker453]

Yes, so you need another tool other than F=kx.

And that would be energy conservation for which the K.E. can be figured out using momentum conservation!

 

[BvU]

And that would be energy conservation for which the K.E. can be figured out using momentum conservation!

Correct. You have been somewhat wrongfooted because in the video they freeze the motion at the lowest point.

 

[danielbaker453]

Thank you both! (BvU and TomHart)
I finally understand this problem.
Here is my work for anyone who is curious...

 

[haruspex]

Thank you both! (BvU and TomHart)
I finally understand this problem.
Here is my work for anyone who is curious...
View attachment 114176

You have posted an incorrect solution. Is that what you intended?

 

[danielbaker453]

You have posted an incorrect solution. Is that what you intended?

Yes. I realize that my solution is wrong. This is not what I intended when I said I understand this. I meant to conserve momentum first and then energy. This is to show what I did wrong to anyone who is curious. Sorry for the confusion.