# Dissipated Energy from Falling Object using Lagrangian [closed]

A plate of mass $M$ moves horizontally with initial speed $v$ on a frictionless table. An object of mass $m$ is dropped vertically onto it from the height $h$ and smashes. How much energy is dissipated in this process?

I have been taught E-L formulas, but don't see how they could be applied to the question at hand.

In the y direction, the falling object gains energy due to gravitational potential and eventually smashes on the plate. Since the table is frictionless, I imagine that energy in the x-direction would be conserved. Thus, I first imagined that the dissipated energy came only from mgh. Of course, this answer doesn't make use of the E-L formula and I was told that I am overlooking something.

I have absolutely no idea how to approach this and hate to be so vague, but I'm at a complete loss. Any push in the right direction would be appreciated!

## closed as off-topic by ACuriousMind♦, Kyle Kanos, JamalS, Danu, Waffle's Crazy PeanutMar 20 '15 at 3:22

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• For starters, energy is not a vector, so there is no "energy in the x-direction". There's just energy. It seems to me conservation of momentum would be something to consider, but that still wouldn't require Euler-LaGrange, so I must be missing something too. – Sean Mar 18 '15 at 13:47

You are overthinking, it looks like a simple inelastic collision.

Before:
Vertical (potential energy):
$E_{plate} = 0$, $E_{object} = mgh$
Horizontal (kinetic energy):
$E_{plate} = \frac{1}{2}Mv^2$, $E_{ball} = 0$

After:
Vertical:
The object smashes, not bounces, so its entire potential energy dissipates
Horizontal:
The plate now has a combined mass of $M+m$, the momentum is conserved
$Mv = (M+m)v'$
Kinetic energy of "Plate with smashed object" $E' = \frac{1}{2}(M+m)v'^2$
The total dissipated energy is the difference between "before" and "after", $\frac{1}{2}v^2\left(\frac{Mm}{M+m}\right) + mgh$