What Is the Maximum Stretch of a Spring When a Ball Is Dropped on It?

[port31]

Homework Statement

A massless pan hangs from a spring that is suspended from the ceiling.
when empty the pan is 50cm below the ceiling. If a 100g clay ball is placed gently
on the pan, the pan hangs 60cm below the ceiling. Suppose the clay ball
is dropped from the ceiling onto an empty pan. What is the pans distance
from the ceiling when the spring reaches its maximum point.

The Attempt at a Solution

So first I find the spring constant
$mg=kx$
k=9.8 N/m
Now we find how much the spring stretches when the clay ball drops on it.
$mgh=\frac{kx^2}{2}$
h=.5m
when I solve for x i get x=.31m
so we add this to .5m and we get 81cm
the answer in my book is 93cm which is basically off by 10cm from my answer
Do I need to add 10cm to mine to account for the weight of the clay ball.
but it seems that is taken into account with the potential energy.

 

[ehild]

The ball falls x more than 50 cm.

ehild

 

[port31]

ok thanks for your post so
it needs to be $mg(h+x)=\frac{kx^2}{2}$

 

[ehild]

ok thanks for your post so
it needs to be $mg(h+x)=\frac{kx^2}{2}$

Right.

ehild

 

[sirisha kotik]

its already has an extension of 50 cm so it wud be mg(h +x)= kx^2/2

 

[azizlwl]

Can i reason it this way as calculated by OP?

If i assume it to be on frictionless surface and given a kinetic energy of mgh.
The spring will extend to an equation of mgh=1/2(kx²)
The weight will produce additional 10cm(given)
Thus total height is 31+10+50.

Any flaws in my thinking.
Thank you.