Solving Block on a Wedge: Find Distance Slid by Triangle

[bmoore509]

Homework Statement

A triangular wedge 5 m high, 12 m base length, and with a 14 kg mass is placed on a frictionless table. A small block with a 5 kg mass (and negligible size) is placed on top of the wedge as shown in the figure below.
All surfaces are frictionless, so the block slides down the wedge while the wedge slides sidewise on the table. By the time the block slides all the way down to the bottom of the wedge, how far does the wedge slide to the right?

Homework Equations

m1v1+m2v2=m1v1f+m2v2f ?
tantheta=opp/adj
w=mg

The Attempt at a Solution

I really don't know where to go with this. So far I have
Theta = arctan (5/12) = 22.61986495

w=mgsintheta=5*9.8*sin(22.61986495)=49
So I know that the 49 there is the force that will be making the block go down the wedge. But what I don't know is what to do with that information. I'm stuck and confused.

 

[dacruick]

what you're going to want to do is calculate the x component of the normal force. This is the same force that is accelerating your block.

The "reaction force" to this force will push the wedge the other way. So basically you have a constant force acting upon both the block and the wedge. if you have force and mass then you have acceleration.

Now that you have the acceleration, you can find out the length of the hypotenuse of the wedge and there you have a distance travelled. Once you have accleration, distance, and initial velocity, you can find the time. Once you have found the time, you can use that and the accleration of the second block to find the distance.

I was very concise if you want me to elaborate on anything please ask.