a block of mass m is pushed towards movable wedge of mass nm and height h, with velocity u. All the surfaces are smooth. The minimum value of u for which the block reach the top of the wedge is
Dear Student ,
Now that every surface is smooth, no loss of energy will be observed.
Therefore the Minimum condition would be for the mass to climb the greatest height and come to rest, meaning at the top of the wedge and come to rest w.r.t the wedge. But the wedge would already have started moving because of the Impact. Therefore at the end, they move as a single entity with some velocity 'v', say.
Now conserve the Momentum,
mu = (M+m)*v....(1)
Conserve the energy,
0.5*m*u2 = 0.5*(M+m)*v2 +mgh....(2)
Solve both the equations (by eliminating 'v') and we can have the Minimum value of 'u'.
Regards
Now that every surface is smooth, no loss of energy will be observed.
Therefore the Minimum condition would be for the mass to climb the greatest height and come to rest, meaning at the top of the wedge and come to rest w.r.t the wedge. But the wedge would already have started moving because of the Impact. Therefore at the end, they move as a single entity with some velocity 'v', say.
Now conserve the Momentum,
mu = (M+m)*v....(1)
Conserve the energy,
0.5*m*u2 = 0.5*(M+m)*v2 +mgh....(2)
Solve both the equations (by eliminating 'v') and we can have the Minimum value of 'u'.
Regards