Block and wedge. Analysis of the center of mass

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SUMMARY

The discussion focuses on calculating the displacement of a wedge (mass M) when a block (mass m) slides down it, utilizing the concept of center of mass. The derived formula for the displacement D is D = H m cot("alpha") / (M + m). Participants emphasize that since no external horizontal forces act on the system, the center of mass in the x-direction remains constant. A more efficient method involves determining the initial and final positions of the center of mass to find the displacement without calculating individual accelerations.

PREREQUISITES
  • Understanding of center of mass equations
  • Familiarity with kinematics equations
  • Basic knowledge of pseudoforces in physics
  • Concept of frictionless surfaces in mechanics
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  • Study the derivation of center of mass equations for two-body systems
  • Learn about pseudoforces and their applications in non-inertial reference frames
  • Explore advanced kinematics problems involving inclined planes
  • Investigate the effects of friction on the motion of blocks and wedges
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Students of physics, particularly those studying mechanics, as well as educators looking for effective methods to teach center of mass concepts in dynamic systems.

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Homework Statement


A block of mass m rests on a wedge of mass M and height H which, in turn, rests on a horizontal table. All the surfaces are frictionless.
If the system starts at rest with the block on the top of the wedge, find the displacement of the wedge during the time the block descends to the bottom.
Consider "alpha" as the inclination angle of the wedge related to the horizontal surface.


Homework Equations


Center of Mass equations and concepts


The Attempt at a Solution


Using pseudoforces i have first encountered the accelerations of the block and wedge, then the time taken by he block to slide down the wedge and then applying the kinematics equations to the wedge i encountered the displacement , which is D = H m cot("alpha")/(M+m)

BUT I'm sure that there is a easier method to solve this problem using the concept of center of mass(since there is no horizontal external force acting on the system the horizontal component of the center of mass position should be at rest, right?), but I'm stuck.
Could someone help me(or teach me) to find a easier way to get the answer without have to calculate all the accelerations as i have made in the long solution above.

Thanks in advance
 
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centre of mass(of the entire two objects system) in x direction before and after should remain at same place. so you need to find an x-adjustment to the wedge such that the new position combined with the block's new position will give you the same CM as the CM for the intial positions.
1. find initial CM by first work out location of CM for each mass: block and wedge
2. combine the two CMs to find the CM of system
3. work out final location of block
4. assume x amount of shift by wedge, combine with block's new position and demand it to equal to initial CM for system, and solve for x.
 

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