[Experts only 2nd repost] What is terminal velocity of the puck?

Question

The puck slides along inclined plane.
Initial velocity of the puck is in horizontal direction
(i.e. at right angle to plane inclination)
and has absolute value v0 = 1m/s.
The angle of inclination is adjusted so, that coefficient
of kinetic friction is exactly equal to tan(a).

What is terminal velocity v1 of the puck?

This is probably all my fault. See picture at
http://alexandersemenov.tripod.com/puck/index.htm

Answer 1

1m/s

Although your question is very confusing (velocity at right angle to plane inclination???). The biggest hint that 1m/s is the right answer is the fact that there is no other infor: length of slide or inclined plane...

Answer 2

I am assuming that the puck is sliding down an plane that is inclined at an angle w/r/t the horizontal plane. I will label the angle of inclination th instead of a, since I use a to model acceleration.

To make sure the angle is correct, a right triangle with the hypotenuse as the inclined plane, would have th as the angle subtended by the hypotenuse and the horizontal side.


Looking at a fbd of the puck, it is acted on by two forces

gravity
m*g*sin(th)

and kinetic friction

m*g*cos(th)*u where
u=tan(th)

Also, since the puck is at terminal velocity, there is a balance between the frictional force and the gravitational force

that is m*a=m*0

so

m*g*sin(th)-m*cos(th)*g*u=0
note that the condition for this to be true is when
u=tan(th)

which is easy to prove:

m*g*sin(th)=m*cos(th)*g*u
dividing out mass and gravity
sin(th)/cos(th)=u
note that

tan(th)=sin(th)/cos(th)
u=tan(th)

so, to answer the question, the terminal velocity is equal to the initial velocity since
v(t)=vi+a*t
since a=o
v(t)=vi

v1=vi=1 m/s

j