# Elastic Potential Energy of a marble

1. Jul 10, 2007

### rum2563

1. The problem statement, all variables and given/known data
A horizontal spring, of force constant 12 N/m is mounted at the edge of a lab bench to shoot marbles at targets on the floor 93.0 cm below. A marble of mass 8.3 X 10 ^ -3 kg is shot from the spring, which is initially compressed a distance of 4.0 cm. How far does the marble travel horizontally before hitting the floor?

2. Relevant equations
E-elastic = 1/2kx^2
E-gravity = mgy
E-kinetic = 1/2mv^2

3. The attempt at a solution

The given information:
m = 8.3 X 10 ^ -3 kg
k = 12 N/m
x = 4.0 cm --> 0.04 m
delta Y = 93 cm --> 0.93 m

In the start, the elastic energy and the kinetic energy should be the same. So for some reason, I think I have to find the velocity.
Ee = Ek
1/2kx^2 = 1/2mv^2
1/2 (12) (0.04)^2 = 1/2 (8.3 X 10 ^ -3)v^2
0.0096 = 0.00415 X v^2
v^2 = 2.31
v= 1.52 m/s

So now that I have velocity, I think I can find the horizontal distance.

IMPORTANT NOTE:
Our teacher told us that we cannot use the equation delta X = vi^2 / g X sin2theta. So please anyone help me in doing in this question without the use of this projectile motion equation.

2. Jul 10, 2007

### rum2563

3. Jul 11, 2007

### Gear300

Today is tomorrow, but I will help...just wait for my next response

4. Jul 11, 2007

### Gear300

Alright, with the velocity in place you have found the initial velocity (considering the ball started going down after it was released). Now you need to find, lets say...the time (t) as to how long it takes to reach the limit of hitting the floor, so lets first consider the ball's vertical motion (y). Using the equation
s = (1/2)at^2 + v0t in which s = .93m, v0 = 0m/s [the ball was traveling horizontally at 1.52m/s, but it is relatively dropping vertically for sin(0) = 0] and a = 9.8m/s^2 (vertical component - acceleration of gravity)...you'll be able to find the time it takes right before hitting the floor. Now, since we're neglecting air resistance (I'm assuming), the velocity of the vector component x is the same at any time, so just use x = x0 + avg(v)t, in which the avg(v) is simply 1.52m/s at any time. x0 is 0 and t is the value you solved for earlier. The answer is the horizontal distance traveled.

Last edited: Jul 11, 2007