Solve Wedge-Block Problem: Velocity of Block

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AnwaarKhalid
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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?