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I'd like to find the maximal extension of a spring. The spring has a spring constant [itex]k[/itex]. There's a mass [itex]m[/itex] connected on the spring. From a height [itex]h[/itex] above the initial location of mass [itex]m[/itex] another mass [itex]M[/itex] 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:

[itex](M+m)\frac{d^2x(t)}{dt^2}+kx(t)=(M+m)g[/itex]

Which results in [itex]x(t) = A \sin(\omega t) + B \cos(\omega t) + \frac{(M+m)g}{k}[/itex] with [itex]\omega=\sqrt{k/(M+m)}[/itex].

With boundary conditions [itex]x(0)=0[/itex] and [itex]\frac{dx(0)}{dt} (M+m) = \sqrt{2gh} M[/itex] the coefficients [itex]A[/itex] and [itex]B[/itex] become [itex]\frac{M\sqrt{2gh}}{(M+m)\omega}[/itex] and [itex]-\frac{(M+m)g}{k}[/itex].

The maximum of [itex]x(t)[/itex] is [itex]\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}[/itex]

**Method 2:**

Conservation of energy:

[itex]Mgh + (M+m)g\Delta=k\Delta^2/2[/itex]

which results in

[itex]\Delta=\sqrt{\frac{2Mgh}{k} + \frac{(M+m)^2g^2}{k^2}} + \frac{(M+m)g}{k}[/itex]

The first term under the square root differs. Why?

Thanks in advance