Conservation of energy

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?

Bystander
Homework Helper
Gold Member
When the two masses make contact they move as one.
I see one inelastic collision, and one elastic collision. Might be mistaken.

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

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

Stephen Tashi
$\frac{dx(0)}{dt} (M+m) = \sqrt{2gh} M$

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?

jim mcnamara
Does that reconcile the two methods?

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?

Stephen Tashi
Is it also possible to use conservation of energy to get the same result as the first method?

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

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

Stephen Tashi