Solving Elastic Collision of Blocks with Masses 12 kg and 6.8 kg

[anyone1979]

[SOLVED] Slide and Collide.

I need help getting started on this.

A block of mass 12 kg lies at a vertical height 7.3 m on a frictionless incline plane. It slides down the plane and collides elastically with a second block whose mass is 6.8 kg. Find the final velocities of the two blocks.

so far I got:
(1/2)M1V1^2 = (1/2)M1V2 ^2 + (1/2)M

V2 = ((M1V1^2) - (M1V1^2))/M2.

I am not sure where to start.

 

[azatkgz]

$$\frac{1}{2}M_1v_0^2=\frac{1}{2}M_1v_1^2+\frac{1}{2}M_2v_2^2$$

And also use conservation of momentum

$$M_1v_0=M_1v_1+M_2v_2$$

 

[anyone1979]

Thanks.
So the initial velocity of M1 at the top is zero right?
Do I need to get the the velocity while the mass is sliding down the incline?
If so, I will need to calculate the x direction, and then use the same velocity sliding down the incline as the initial velocity until it collides right?

 

[azatkgz]

In this question nothing extra is given,so we can use just

$$v_0=\sqrt{2gh}$$

 

[anyone1979]

Thanks, for the final equation, I got:

Vbf - Vaf = -(Vbi - Vai)

Is that right?