Solving Sliding Sideways: Block Speed After Long Time

[Identity]

A block is placed on a plane at an angle of \theta. It is given an initial sideways speed of v which has no component up or down the plane. The coefficient of friction between the block and plane is \mu = \tan\theta. What is the speed of the block after a long time?


Using normal analysis i figured that
a_{down}=mg(\sin{\theta}-\mu\cos{\theta})=0
and
a_{across}=-\mu mg\cos{\theta}=-mg\sin{\theta}
So that the block eventually comes to a stop.

But the solutions say that the speed the block loses going sideways is converted into downwards motion... how does this work?

 

[Naty1]

what are you trying to do??

what is "sideways" ? if there is no velocity component up or down the incline?, then any velocity must be orthogonal to it...

 

[Identity]

I'll copy the question and solution down for the sake of clarity (Source: Introduction to Classical Mechanics by David Morin)

Q:
"A block is placed on a plane inclined at an angle \theta. The coefficient of friction between the block and the plane is \mu=\tan\theta. The block is given a kick so that it initially moves with speed v horizontally along the plane (that is, in the direction perpendicular to the direction pointing straight down the plane). What is the speed of the block after a very long time?"

A:
"The normal force from the plane is N=mg\cos\theta. Therefore, the friction force on the block is \mu N = (\tan\theta)(mg\cos\theta)=mg\sin{\theta}. This force acts in the direction opposite to the motion. The block also feels the gravitational force of mg\sin\theta pointing down the plane.
Because the magnitudes of the friction force and the gravitational force along the plane are equal, the acceleration along the direction of motion equals the negative of the acceleration in the direction down the plane. Therefore, in a small increment of time, the speed that the block loses along its direction of motion exactly equals the speed that it gains in the direction down the plane. Letting v be the total speed of the block, and letting v_y be the component of the velocity in the direction down the plane, we therefore have

v+v_y=C

where C is constant. C is given by its initial value, which is v+0=v. The final value of C is v_f+v_f=2v_f (where v_f is the final speed of the block), because the block is essentially moving straight down the plane after a very long time. Therefore,

2v_f=v \Rightarrow v_f = \frac{v}{2}"I still can't understand how the block changes direction :(
If friction is acting against one direction does that mean that it can't simultaneously prevent the block going down the plane?