Spring cannon question involving conservation of energy.

[MrMagoo22]

Homework Statement

Two children are playing a game in which they try to hit a small box on the floor with a marble fired from a spring-loaded cannon mounted on a table. The target box is 2.2m horizontally away from the edge of the table. The spring is compressed 1.1cm, but the ball falls short of the box by 27cm. How far does the spring need to be compressed to hit the box directly? It is assumed that friction will not affect this problem.

Homework Equations

Kinetic Energy: Ke = (1/2)mv^2
Elastic Potential Energy: ePe = (1/2)kx^2
Conservation of Mechanical Energy: Pe = -Ke
Kinematic Equations:
X = ViT + (1/2)at^2

The Attempt at a Solution

Started working with elastic potential energy first, realizing that I'd need to know the spring constant to get the potential energy, which would lead to the answer:

ePe = (1/2)k(0.011)
ePe = -Ke = -1/2mv^2
(.0055)k = (-1/2)mv^2

Wasn't working out past that point, too many unknowns, so I tried working with the trajectory of the shot with kinematics to get the initial velocity of the shot and work back:

Vix =
Vx =
Ax = 0
X = 1.92
T =
Viy = 0
Vy =
Ay = -9.81
Y =

None of the kinematics worked off that small amount of information, so I was stuck again. It feels like there isn't enough information available to get an actual answer (no mass for ball, no table height, no spring constant, really any of those would help.) but I may be overlooking something.

 

[gneill]

Try putting away your calculator until the very end and work symbolically. You will find that all the unknown values can be lumped together into a single constant, and that you won't even need that if you set up a ratio at the end (a bit like using PV = nRT to find changes in pressure and volume).

You'll have to make certain assumptions about the trajectory (for example, you might assume that the launch angle is horizontal, and that the table height is fixed at some value, say h).