# Mechanics - Elastic springs and strings

1. Mar 9, 2006

### mr bob

A particle P of mass m is attached to one end of a light elastic string of natural length L whose other end is attached to a point A on a ceiling. When P hangs in equilibrium AP has length $\frac{5l}{3}$. Show that if P is projected vertically downwards from A with speed $\sqrt(\frac{3gl}{2})$, P will come to instantaneous rest after moving a distance $\frac{10l}{3}$.

I thought about working all this out by finding the energies before and after the projection.

Before:-
$KE =\frac{3gl}{4}$
$GPE = 0$
$EPE = 0$

After:-
$KE = 0$
$GPE = -(y - 5/3L)g$ where y is the full length of stretched string.
$EPE = \frac{gl(Y- 5/3L)^2}{2L(Y- 5/3L)}$

i cant seem to get the anwer using these, are they correct?

Last edited: Mar 9, 2006
2. Mar 10, 2006