RingNebula57
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Hello everyone, I have this problem at which I don't understand the solution. So we have an inclined plane and a block of mass m standig on it , at it's highest point. Knowing the angle of the inclined plane(alpha) , the height of the inclined plane, the friction coefficient between the 2 objects (miu) and that is the same with the friction coefficient of the horizontal plane( the inclined plane is fixed) , and that the horizontal plane and the inclined plane ARE NOT CONNECTED to each other we have to dermine the distance traveled by the block after leaving the inclined plane.
The thing that I didn't understand is the conservation of momentul on the horizontal axis. If we say that the velocity of the block , before entering the horizontal surface is v ( that can be found via cnoservation of total energy of the system) , than I would say that, because we consider the time for the object to "hop" from the inclined plane to the horizontal one almost instantaneous, the momentum on the vertical direction cancels and so the velocity at the beginning of the horizontal plane would be v'=v * cos(alpha) . What the solution is saying is that this speed is v'= v* (cos(alpha) - (miu)* sin(alpha) )
Why?
Thank you
The thing that I didn't understand is the conservation of momentul on the horizontal axis. If we say that the velocity of the block , before entering the horizontal surface is v ( that can be found via cnoservation of total energy of the system) , than I would say that, because we consider the time for the object to "hop" from the inclined plane to the horizontal one almost instantaneous, the momentum on the vertical direction cancels and so the velocity at the beginning of the horizontal plane would be v'=v * cos(alpha) . What the solution is saying is that this speed is v'= v* (cos(alpha) - (miu)* sin(alpha) )
Why?
Thank you