Solving Physics Lab Cube Sliding Problem

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SUMMARY

The discussion focuses on solving the physics problem of a small cube sliding down a frictionless incline and colliding with a larger cube at the bottom. The small cube, with a calculated velocity of 4.8 m/s using the formula v=[2gh]^(1/2), collides with a larger cube of half its mass, resulting in the larger cube moving at 2.4 m/s after the collision due to momentum conservation. The height of the incline is 30 cm, and the total height from the ground is 120 cm, leading to the conclusion that the larger cube lands 30 cm away from the table's edge.

PREREQUISITES
  • Understanding of kinematic equations, specifically v=[2gh]^(1/2)
  • Knowledge of momentum conservation principles in inelastic collisions
  • Familiarity with basic physics concepts such as kinetic and potential energy
  • Ability to perform calculations involving height and distance in projectile motion
NEXT STEPS
  • Study the effects of friction on inclined planes in physics problems
  • Learn about different types of collisions, focusing on elastic vs. inelastic collisions
  • Explore the concept of projectile motion and how to calculate landing distances
  • Investigate the role of energy conservation in mechanical systems
USEFUL FOR

Students studying physics, educators teaching mechanics, and anyone interested in understanding motion dynamics and collision analysis.

dboychuk
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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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