Solving Physics Lab Cube Sliding Problem

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A small cube slides down a frictionless incline and collides with a second cube of half its mass at the bottom. The height of the incline is 30 cm, and the table height is 90 cm, leading to a total height of 120 cm for calculations. The velocity of the small cube at the bottom is calculated to be 4.8 m/s using the formula v=[2gh]^(1/2). After the collision, the small cube comes to rest, and the velocity of the larger cube is determined to be 2.4 m/s through momentum conservation principles. The discussion concludes that the larger cube lands 30 cm away from the table's edge.
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In a physics lab, a small cube slides down a frictionless incline as shown, and collides at the bottom (where it is now moving horizontally) with a cube that is only one-half its mass. If the incline is 30 CM high and the table is 90 CM off the floor. where does each cube land?

the teacher gave us that the big cube lands 30 CM away. Thats all, does anyone know how to solve this?
 
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calculate the velocity of cube at bottom of incline with the relation...

v=[2gh]^1/2

apply momentum conservation principle after that..
both k.e. and p.e. will come into play...

tell me if you are still not able to solve it..
 
A:v of small cube=4.8m/s using v=[2gh]^1/2 [here, the incline is not useful at all ;except for calculating speed of cube]

where h = 90+30= 120cm>>1.2 m;[height from ground is taken into consideration]

M AND M/2;

After collision the small cube comes to rest.

M/2*4.8=M*V>>V=2.4m/s [MOMENTUM CONSERVATION][inelastic collision]

>>V^2 =U^2 +2AS
>>s = 30 cm

The assumptions made are:

1.The big cube starts moving from the edge of the table.
2.The table is smooth.
 
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