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 respectively
applying 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
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 respectively
applying 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
Abdul Qadeer , 14 Years ago
Grade 12
