Conservation of Energy child's toy

1. Nov 24, 2008

chrispat

1. The problem statement, all variables and given/known data

A child's toy consists of a block that attaches to a table with a suction cup, a spring connected to that block, a ball, and a launching ramp. The spring has a spring constant k, the ball has a mass m, and the ramp rises a height y above the table, the surface of which is a height H above the floor.

Initially, the spring rests at its equilibrium length. The spring then is compressed a distance s, where the ball is held at rest. The ball is then released, launching it up the ramp. When the ball leaves the launching ramp its velocity vector makes an angle theta with respect to the horizontal.

Throughout this problem, ignore friction and air resistance.

With what speed will the ball hit the floor?
Express the speed in terms of k, s, m, g, y, and/or H.

2. Relevant equations

Ki+Ui=Kf+Uf

3. The attempt at a solution

0.5ks^2+0.5m(vi^)2+mg(hi)=0.5m(vf)^2+mg(hf)

vf=sqrt[(0.5ks^2)+(0.5m(vi)^2)+0.5m]

where vi=sqrt[(ks^2-2mgy)/m]

2. Nov 24, 2008

dpowd

chrispat,
The equations in your attempt are all correct statements in that you have appropriate expressions for the quantities vi, Ui - spring, UI - grav, etc. However, when are you declaring your initial values? That is, when during the experiment are you setting time = 0? That decision is entirely up to you, with no wrong answer. However, there are some points in the experiment where it makes better sense to set t = 0, because it allows you to set one of the terms in Ui + Ki = Uf + Kf equal to zero.
There's no reason to have non-zero terms for both Ui - spring and Ki. If you declare your starting time so that you can zero one of these terms out, it will make the problem easier.

3. Nov 25, 2008

chrispat

Ok so I took t=0 to be when the ball is at its highest point.

0.5m(vi)^2+mg(hi)=1/2m(vf)^2

(vi)^2+2gh=vf^2

Substituting for vi:

[ks^2-2mgy/m]^2+2gh=vf^2