Dropping ball into spring (inelastic collision)

[Samei]

Homework Statement

: [/B]I asked a question on this topic before, but I want to make sure I know the material well. So, I looked up another question similar to it (and a little more complex) to check my understanding.

Here is the practice problem: A board with mass M, when placed into a coil spring (with spring constant k), compresses the spring in a distance of x. A ball of mass m is dropped from rest onto the board from a height h. The ball sticks into the board.

What is the restoring force made by the spring?

Bonus: What is the amplitude made by the spring?

Homework Equations

:[/B] Conservation of energy, conservation of momentum, F_spring = -kx. v_{b, final}² = 2ad

The Attempt at a Solution

: [/B]I am given spring constant k, board mass M, ball mass m, and height h.
I noticed that the ball sticks into the board, so it is an inelastic collision. Conservation of energy cannot be applied, but conservation of momentum can.

At the instant the ball touches the board, m(v_{b, final}) = (m + M)(v_2b)
I can find v_{b, final} by using the kinematic equation, v_{b, final}² = 2ad, with d = h and a = g since it is moving downwards. v_{b, final} is √(2gh).

Then, (v_2b) = (m*√(2gh))/(m + M))

After collision, I can now apply the conservation of energy.

But before that, I take note of how the board already tips the spring from equilibrium.
Mg = kx, x₀ is the first compression due to the board alone. (Mg)/k = x₀.

Now, for the energy. (0.5)(m+M)(v_2b)² = 0.5kx₁² + (m+M)gx₁
x₁ is the compression of the spring due to the ball and the board.
Solving the quadratic gives me x₁.

To find restoring force, F_spring = -k (x₀ + x₁)

For the bonus, amplitude is (0.5)((x₀ + x₁))

Is this right? Did I make errors along the way?
Thanks for all the help in advance!

 

[haruspex]

Now, for the energy. (0.5)(m+M)(v_2b)² = 0.5kx₁² + (m+M)gx₁
x₁ is the compression of the spring due to the ball and the board.

Need to be careful here. What is the change in PE of the spring as the board descends by x₁?

 

[Samei]

Need to be careful here. What is the change in PE of the spring as the board descends by x₁?

The change in PE of the spring is 0.5kx₁². But do you mean that it is supposed to be negative? That would make sense.

I have been reading some more examples, and I think I am going to do a minor revision.
I'm going to be more explicit, so any flaws in the logic can be corrected.

After impact of m and M, conservation of energy can be applied.
E₁ = E₂
KE₁ + PE₁ + SpringPE₁ = KE₂ + PE₂ + SpringPE₂

Substituting known values, I can now write this statement as
0.5(m+M)(v_2b)² + (m+M)gx₁ + 0.5(k)(x₀)² = 0.5(m+M)(0)² + (m+M)g(0) + 0.5(k)(x₁)²

When the ball and the board compress the spring, the final velocity will be zero, and essentially as it goes downward x₁, the "h" in PE₂ will also be zero.

This will be simplified as 0.5(m+M)(v_2b)² + (m+M)gx₁ + 0.5(k)(x₀)² = 0.5(k)(x₁)²
0.5(m+M)(v_2b)² + 0.5(k)(x₀)² = 0.5(k)(x₁)² - (m+M)gx₁
0 = 0.5(k)(x₁)² - (m+M)gx₁ - [0.5(m+M)(v_2b)² + 0.5(k)(x₀)²]

In here, x₁ includes x₀ right? Or is 0.5(k)(x₁)² really 0.5(k)(x₁ + x₀)²?
That quadratic looks even more complicated. Solving with only variables is going to be complex, unless it cancels somewhere. Are there other principles I may have missed?Also, for the bonus question: Amplitude is actually only equal to (x₀ + x₁).

 

[haruspex]

The change in PE of the spring is 0.5kx₁²

No it isn't. What is the PE of the spring before the board hits it? What is the PE when at extension x₀+x₁?

 

[Samei]

No it isn't. What is the PE of the spring before the board hits it? What is the PE when at extension x₀+x₁?

Before the ball hits it, the board already has compressed it x₀. So the SpringPE at that instant would be 0.5(k)(x₀)².
When it is extended, then it will be 0.5(k)(x₀ + x₁)².

So, change in SpringPE is 0.5(k)(x₀ + x₁)² - 0.5(k)(x₀)². This simplifies to:
0.5k((x₀ + x₁)² - (x₀)²)
0.5k((x₁² + 2x₀x₁))

In here, all variables are known except for x₁.

 

[haruspex]

Before the ball hits it, the board already has compressed it x₀. So the SpringPE at that instant would be 0.5(k)(x₀)².
When it is extended, then it will be 0.5(k)(x₀ + x₁)².

So, change in SpringPE is 0.5(k)(x₀ + x₁)² - 0.5(k)(x₀)². This simplifies to:
0.5k((x₀ + x₁)² - (x₀)²)
0.5k((x₁² + 2x₀x₁))

Yes. Continuing with your preceding post...

This will be simplified as 0.5(m+M)(v2b)2 + (m+M)gx1 + 0.5(k)(x0)2 = 0.5(k)(x1)2

Here you correctly included the initial spring PE on the left, but you' ve only got the change in extension on the right.
What equation do you have now?

Your answer for the amplitude is wrong. Please post your corrected energy equation and subsequent working.

 

[Samei]

Here you correctly included the initial spring PE on the left, but you' ve only got the change in extension on the right.
What equation do you have now?

Your answer for the amplitude is wrong. Please post your corrected energy equation and subsequent working.

I thought I was so close. Wouldn't the amplitude be equal to the distance the spring is compressed? From -kd = F_Spring?

Anyway, the new E₁ = E₂ will be as follows.

0.5(m+M)(v_2b)² + (m+M)gx₁ + 0.5k(x₀)² = 0.5k((x₁² + 2x₀x₁))
I am unsure how to solve for x₁ here. My approach is to expand the equation first.
0.5(m+M)(v_2b)² + (m+M)gx₁ + 0.5k(x₀)² = 0.5kx₁² + kx₀x₁
0 = 0.5kx₁² + kx₀x₁ - 0.5(m+M)(v_2b)² - (m+M)gx₁ - 0.5k(x₀)²
0 = 0.5kx₁² + [kx₀ - (m+M)g]x₁ - 0.5[(m+M)(v_2b)² - k(x₀)²]
So, this will be another quadratic. It looks very bad. :(

Substituting values from the initial post (#1),
0 = 0.5kx₁² + [Mg - mg - Mg]x₁ - 0.5m²[(2gh)/(m + M) - g²/k]

 

[haruspex]

0.5(m+M)(v2b)2 + (m+M)gx1 + 0.5k(x0)2 = 0.5k((x12 + 2x0x1))

You're not being consistent over the spring PE. On the left you have the initial spring PE, on the right the increase in its PE.
Either delete the initial PE on the left or change the right hand side to be the final spring PE.

Wouldn't the amplitude be equal to the distance the spring is compressed?

The amplitude is the maximum displacement above and below the equilibrium position. Where is the equilibrium position?

 

[Samei]

You're not being consistent over the spring PE. On the left you have the initial spring PE, on the right the increase in its PE.
Either delete the initial PE on the left or change the right hand side to be the final spring PE.

Oh, ok. I will change it right now.
0.5(m+M)(v_2b)² + (m+M)gx₁ = 0.5k((x₁² + 2x₀x₁))
0.5(m+M)(v_2b)² + (m+M)gx₁ = 0.5kx₁² + kx₀x₁
0 = 0.5kx₁² + [kx₀ - (m+M)g]x₁ - 0.5(m+M)(v_2b)²

Again, I will be using substitutions which I solved for from initial post (#1),
0 = 0.5kx₁² + [Mg - mg - Mg]x₁ - 0.5(2m²gh)/(m + M)
0 = 0.5kx₁² + [- mg]x₁ - (m²gh)/(m + M)

Slightly simpler, but still a complex quadratic for x₁.

With my calculations, this quadratic simplifies to [mg ± m √(g(1-2h)/(m+M)) ]/k
So, x₁ = [mg ± m √(g(1-2h)/(m+M)) ]/k

If this is correct, then restoring force would be k*[mg ± m √(g(1-2h)/(m+M)) ]/k, which cancels out the k.
This means that restoring force is [mg ± m √(g(1-2h)/(m+M)) ].

The amplitude is the maximum displacement above and below the equilibrium position. Where is the equilibrium position?

The equilibrium position is without m or M. It is the "zero" position and (x₁ + x₀) is the displacement going down or up. My thoughts were that at half an oscillation, the compression would represent the peak on a sinusoidal graph.

Here is my new attempt for the bonus: 2(x₁ + x₀)
Amplitude = 2[mg ± m √(g(1-2h)/(m+M)) ]/k + Mg/k
So amplitude is 2(Mg + [mg ± m √(g(1-2h)/(m+M)) ]) /k

 

[haruspex]

With my calculations, this quadratic simplifies to [mg ± m √(g(1-2h)/(m+M)) ]/k

1-2h makes no sense. It's dimensionally wrong. What would happen if h > .5 (in whatever units!)?
Your quadratic was dimensionally ok, so the error must be in solving it.

The equilibrium position is without m or M.

No, that's the spring's relaxed position. The equilibrium position for the oscillation is where M+m could sit at rest and not oscillate.

 

[Samei]

1-2h makes no sense. It's dimensionally wrong. What would happen if h > .5 (in whatever units!)?
Your quadratic was dimensionally ok, so the error must be in solving it.

I'll look at it again. I may have miscalculated or made a mistake on understanding my own handwriting (happens often).

No, that's the spring's relaxed position. The equilibrium position for the oscillation is where M+m could sit at rest and not oscillate.

The equilibrium position is then when kx_e = (m+M)g
and x_e is the distance from the zero position.
So, amplitude is then x₁ + x₀ - x_e

 

[haruspex]

The equilibrium position is then when kx_e = (m+M)g
and x_e is the distance from the zero position.
So, amplitude is then x₁ + x₀ - x_e

Yes.

 

[Samei]

Yes.

I'll look over everything again just to make sure.
Alright! Now, I think I have improved on this topic. Thanks again, haruspex!

 

[Samei]

Yes.

Actually, I have one more question about the amplitude of this problem if I can. I looked up other ways to find amplitude and I saw how most strategies were to use angular frequency and maximum velocity. Is this solution different?

It seems simple enough. But I am not sure if it is a shortcut, at least that I can apply here, maybe?

 

[haruspex]

Actually, I have one more question about the amplitude of this problem if I can. I looked up other ways to find amplitude and I saw how most strategies were to use angular frequency and maximum velocity. Is this solution different?

It seems simple enough. But I am not sure if it is a shortcut, at least that I can apply here, maybe?

Both methods work. Which is simpler depends on what facts you start with.

 

[Samei]

Both methods work. Which is simpler depends on what facts you start with.

Oh, ok. In this case, relating restoring force with amplitude is easier done the original way.
My class has not discussed that chapter yet though, so I may have to stick with the original.

Again, thanks a lot! I appreciate the help!