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# A block of mass m is placed on a triangular block of mass M which in turn is placed on a horizontal surface. Assuming frictionless 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'.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

509 Points
10 years ago

you have written             mgh=1/2mu(square)  +1/2Mv(square)

this equation is correct but u(velocity of small block ) should be with respect to ground  and here you substituted with respect to bigger block....

so take u=relative velocity of small block with respect to big block