Solve Wedge-Block Problem: Velocity of Block

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SUMMARY

The discussion centers on calculating the velocity of a triangular block (mass M) when a smaller block (mass m) slides down its frictionless surface. The user applies conservation of momentum and energy principles, leading to the equation V = (m/M)√(2Mgh/(M+m)). However, the user fails to incorporate the angle of inclination (theta) into their calculations, which is crucial for deriving the correct relationship between the blocks' velocities. Clarification is sought on the role of the angle in the derivation.

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  • Understanding of Newton's laws of motion
  • Familiarity with conservation of momentum
  • Knowledge of conservation of energy principles
  • Basic trigonometry related to angles of inclination
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  • Review the derivation of velocity in inclined planes using trigonometric functions
  • Study the impact of angles on momentum conservation in two-body systems
  • Explore energy conservation in non-frictional systems
  • Investigate the dynamics of block systems on inclined surfaces
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Homework Statement



A block of mass m is placed on a triangular block of mass M which in turn is placed on a horizontal surface. Assuming friction less surfaces find the velocity of the triangular block when the smaller block reaches the bottom end.The angle of inclination of the plane is 'theta'.

Homework Equations

The Attempt at a Solution


my approach to the problem...when the small block reaches the bottom...momentum in x direction is conserved...mu=Mv...where u and v are absolute velocities of the smallerb block bigger block respectivelyapplying conservation of energy...mgh=1/2mu(squared)+1/2mV(squared)solving the above two equations...V=m/M(root of){2Mgh/M+m}but the answer is complex in terms of angle...please explain where I am wrong
 
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AnwaarKhalid said:

Homework Statement



A block of mass m is placed on a triangular block of mass M which in turn is placed on a horizontal surface. Assuming friction less surfaces find the velocity of the triangular block when the smaller block reaches the bottom end.The angle of inclination of the plane is 'theta'.

Homework Equations

The Attempt at a Solution


my approach to the problem...when the small block reaches the bottom...momentum in x direction is conserved...mu=Mv...where u and v are absolute velocities of the smallerb block bigger block respectivelyapplying conservation of energy...mgh=1/2mu(squared)+1/2mV(squared)solving the above two equations...V=m/M(root of){2Mgh/M+m}but the answer is complex in terms of angle...please explain where I am wrong

I do not see the angle in your derivation. What exactly is your u?
 

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