Conservation of energy, how far must a spring be compressed

[sjcorona]

Two children are playing a game in which the try to hit a small box on the floor with a marble fired from a spring loaded gun mounted on a table. The target box is horizontal distance D=2.20 m from the edge of the table. Bobby compresses the spring 0.011m but the center of the marble falls 0.27 m short of the center of the box How far should Rhoda compress the spring to score a direct hit? Assume that neither the spring nor ball encounters friction in the gun.

So I know that the ball will leave the gun with initial kinetic energy of 1/2mv^2 = the 1/2 k Δx^2 of the springs potential energy. I tried isolating velocity and using that in my kinematic equation d=v(t)+1/2at^2 and solved for time thinking that maybe I could find the height of the table and find the right velocity from there then work backwards to spring potential energy, but that didn't work because I don't know k or m and am not sure if I as n the right track

 

[Doc Al]

By what factor must the initial velocity increase to reach the box?

Hint: Make use of the fact that initial velocity is horizontal in all cases.

 

[sjcorona]

New velocity= 1.14(d)/t I can plug that back into my kinetic=spring potential but I still have too may unknowns

 

[Doc Al]

New velocity= 1.14(d)/t I can plug that back into my kinetic=spring potential but I still have too may unknowns

Think in terms of ratios. To increase the velocity by a factor of 1.14, by what factor must Δx increase?

 

[sjcorona]

1.14

So are you saying that all I had to do to solve this was make a ratio of it all? That sounds far too easy

 

[Doc Al]

1.14

So are you saying that all I had to do to solve this was make a ratio of it all?

Yep.

That sounds far too easy

Sometimes a problem looks harder than it is.

 

[sjcorona]

Wow...thank you