How to Calculate Spring Compression for a Direct Hit in a Projectile Game?

[azure kitsune]

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 gun that is mounted on a table. The target box is 2.20 m horizontally from the edge of the table. Bobby compresses the spring 1.10 cm, but the center of the marble falls 27.0 cm short of the center of the box. How far should Rhoda compress the spring to score a direct hit? Assume that neither the spring or the ball encounters friction in the gun.

Homework Equations

Elastic potential energy = $$1/2 \cdot kx^2$$

(kinematics equations)

The Attempt at a Solution

Let x be the distance that the person compresses the spring. Let d be the horizontal distance the marble travels from the end of the table. (So we are trying to find an x such that d = 2.20 m).

Through a lot of crazy math, involving projectile motion, breaking into x,y components, conservation of energy, etc., I ended up with:

$$d = x * \sqrt{ \dfrac{ 2 k \Delta y } { m g } }$$ (where k = spring constant, delta y = height of the table, m = mass of object, g = gravitational acceleration)

The important thing about that result was that d and x are directly proportional. Since they give you one pair of (x,d), I could find the constant of proportionality to solve that d = 2.20 m when x = 1.25 cm.

I was wondering if there was any easier way to show that x and d were directly proportional?

 

[Redbelly98]

EDIT: Disregard this post. I didn't read the question clearly.

One problem I see with your equation is: d should not depend on either the height of the table or on gravity, since the marble is moving on a level horizontal surface.

Just what type of force is causing the marble to come to a stop?

 

[nasu]

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 gun that is mounted on a table. The target box is 2.20 m horizontally from the edge of the table. Bobby compresses the spring 1.10 cm, but the center of the marble falls 27.0 cm short of the center of the box. How far should Rhoda compress the spring to score a direct hit? Assume that neither the spring or the ball encounters friction in the gun.

Homework Equations

Elastic potential energy = $$1/2 \cdot kx^2$$

(kinematics equations)

The Attempt at a Solution

Let x be the distance that the person compresses the spring. Let d be the horizontal distance the marble travels from the end of the table. (So we are trying to find an x such that d = 2.20 m).

Through a lot of crazy math, involving projectile motion, breaking into x,y components, conservation of energy, etc., I ended up with:

$$d = x * \sqrt{ \dfrac{ 2 k \Delta y } { m g } }$$ (where k = spring constant, delta y = height of the table, m = mass of object, g = gravitational acceleration)

The important thing about that result was that d and x are directly proportional. Since they give you one pair of (x,d), I could find the constant of proportionality to solve that d = 2.20 m when x = 1.25 cm.

I was wondering if there was any easier way to show that x and d were directly proportional?

There is nothing to break into components. The initial velocity is horizontal, right?
The range is d=v*t (v=initial speed).
The time is constant, no matter what is the initial speed. (t=sqrt(2h/g); h-height of table, constant)

From conservation of energy, v= x*sqrt(k/m)
So d proportional to v prop to x or in full equations
d=v*t= x *sqrt(k/m) * sqrt(2h/g)

 

[azure kitsune]

Wow, that's what I did, except it took me 1 1/2 page somehow, and then I couldn't understand what I wrote. Thanks for the help.