Homework Help: Conservation of momentum - Vertical spring

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1. Mar 6, 2017

danielbaker453

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
Initial Momentum of system = Final Momentum of system
Ui+Ki=Uf+Kf

3. The attempt at a solution
Since the spring is initially in equilibrium,
kxi = Mg
using this we can find the spring constant.
Also, at the point of maximum elongation,
kxf=(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

2. Mar 6, 2017

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?

3. Mar 6, 2017

danielbaker453

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

4. Mar 6, 2017

BvU

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

Tom's faster, but we agree.

5. Mar 6, 2017

TomHart

Is there any mention of damping?

6. Mar 6, 2017

danielbaker453

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

7. Mar 6, 2017

BvU

Kinetic energy zero doesn't mean equilibrium position !

8. Mar 6, 2017

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.

9. Mar 6, 2017

danielbaker453

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

10. Mar 6, 2017

BvU

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

11. Mar 6, 2017

danielbaker453

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

12. Mar 6, 2017

BvU

Not in the collision. Afterwards, yes.

13. Mar 6, 2017

TomHart

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.

14. Mar 6, 2017

danielbaker453

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

15. Mar 6, 2017

BvU

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

16. Mar 6, 2017

BvU

That's the equilibrium position.

17. Mar 6, 2017

TomHart

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?

18. Mar 6, 2017

danielbaker453

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

19. Mar 6, 2017

TomHart

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

20. Mar 6, 2017

danielbaker453

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

21. Mar 6, 2017

danielbaker453

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

22. Mar 6, 2017

BvU

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

23. Mar 6, 2017

danielbaker453

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

24. Mar 6, 2017

haruspex

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

25. Mar 7, 2017

danielbaker453

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.