# Conservation of energy

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1. Dec 29, 2014

### Dan112

Hi.

I'd like to find the maximal extension of a spring. The spring has a spring constant $k$. There's a mass $m$ connected on the spring. From a height $h$ above the initial location of mass $m$ another mass $M$ falls. When the two masses make contact they move as one. I found two methods to calculate that extension but the solution differs slightly.

Method 1:
Solve a differential equation of Newton's law and use conservation of momentum in the boundary conditions:
$(M+m)\frac{d^2x(t)}{dt^2}+kx(t)=(M+m)g$
Which results in $x(t) = A \sin(\omega t) + B \cos(\omega t) + \frac{(M+m)g}{k}$ with $\omega=\sqrt{k/(M+m)}$.
With boundary conditions $x(0)=0$ and $\frac{dx(0)}{dt} (M+m) = \sqrt{2gh} M$ the coefficients $A$ and $B$ become $\frac{M\sqrt{2gh}}{(M+m)\omega}$ and $-\frac{(M+m)g}{k}$.
The maximum of $x(t)$ is $\Delta=\sqrt{A^2+B^2}+\frac{(M+m)g}{k} = \sqrt{\frac{2ghM^2}{(M+m)k}+\frac{(M+m)^2g^2}{k^2}}+\frac{(M+m)g}{k}$

Method 2:
Conservation of energy:
$Mgh + (M+m)g\Delta=k\Delta^2/2$
which results in
$\Delta=\sqrt{\frac{2Mgh}{k} + \frac{(M+m)^2g^2}{k^2}} + \frac{(M+m)g}{k}$

The first term under the square root differs. Why?

2. Dec 29, 2014

### Bystander

I see one inelastic collision, and one elastic collision. Might be mistaken.

3. Dec 29, 2014

### Dan112

Okay. So if they move together I should use the differential equation. Thank you

4. Dec 29, 2014

### Stephen Tashi

I prefer:

$\frac{1}{2}(M+m)x'(0)^2 = Mgh$
$\omega A = x'(0) = \frac{1}{\omega} \sqrt{ \frac{2Mgh}{M+m} }$
$A = \sqrt{\frac{M+m}{k} } \sqrt{ \frac{2Mgh}{M+m} } = \sqrt{\frac{2Mgh}{k}}$

Does that reconcile the two methods?

5. Dec 30, 2014

### Dan112

Yes it does. Thank you. It results in the same answer as the second method. But if I'm correct then both masses will move separately if that method is used. I didn't realise I was mixing elastic and inelastic collisions. Is it also possible to use conservation of energy to get the same result as the first method?

6. Dec 30, 2014

### Stephen Tashi

Did you mean the conservation of momentum? The second method used the conservation of energy.

7. Dec 31, 2014

### Dan112

Oh sorry. Yes I mean conservation of momentum. I tried to combine both conservation of momentum and energy to give me the extension like this:
$\sqrt{2gh}M=(M+m)v$
$(M+m)v^2/2=k\Delta^2/2$
Which results in:
$\Delta=\sqrt{\frac{2ghM^2}{k(M+m)}}$

I'm obviously mixing conservation of energy and momentum and inelastic and elastic collisions in all these calculations which of course is wrong.

8. Dec 31, 2014

### Stephen Tashi

The momentum of a system is conserved where there are no external forces. To embed the problem in a system where there are no external forces, you'd have to add the earth to the to the picture so that its gravity is not an external force to the system. Considering the earth as part of the system, the initial momentum of the system can be taken as zero. It remains zero. The momenta of the indiviual masses in the system cancel out.

Likewise, the kinetic energy of a system need not be conserved when external forces are acting. I suppose you can consider the collision in this problem to be inelastic if include the spring as part of the system that contains mass m, but the force of gravity is an external force to the system of the two masses and the spring, so that by itself says that kinetic energy need not be conserved.

Given the exernal forces (when we don't consider the earth as part of the system) it's impressive that the conservation of total energy still works. It would also work if we used an inverse square law for gravity instead of modeling it as a constant. However, what would happen if gravity was a time-varying force? Would this imply energy going in and out of the system? It's an interesting question to think about.