How far will the spring be compressed before the masses come to rest?

[jsr219]

A block of mass 5kg slides without friction at a speed of 8m/s on a horizontal table surface until it strikes and sticks to a mass of 4kg attached to a horzontal spring. The spring has a constant of k= 2000 n/m. The spring is in turn attached to a wall. How far will the spring be compressed before the masses come to rest.

the answers are
a.) .4
b.) .54
c.) .3
d.) .02
e.) .67

the correct answer is a

My main question is am I tackling the problem right ?

I know the equation for a spring is 1/2kx^2 = -w
I know that work = delta k
so i set 1/2mv^2=1/2kx^2
and i solved for x getting sqrt(mv^2/k) = x

I assume that the 5kgs is a force acting upon the 4kg

so i plug in in my values which are sqrt(5(8)^2/2000) = sqrt(320/2000) = .4


now although I arrived at the technical right answer, I'm not all sure I did the problem right. I very new to Potiental energy, springs, and what not

Please help me if I'm wrong, and please correct me
We can't learn without mistakes and constructive criticism
Thank you !

 

[issacnewton]

yes, your approach is right. but don't say things like

I assume that the 5kgs is a force acting upon the 4kg

kg is the unit of mass ,not force.

the key idea here is conservation of energy. think of energy as the currency of nature.
like the money in your bank account is always accounted for, the energy in nature is
always accounted for, always conserved. so the kinetic energy of the first block gets
converted into the potential energy of the spring and so the blocks come to rest.