That you say you have done bouncing, but cannot figure out this, plus that you talk about the 'angle' of the slope tells me you are using trigonometry to do the bouncing... *sigh*

The solution is incredibly simple, using simple vector operations.

You have the slope normal vector (As opposed to its angle)
You have the incident velocity.

The outgoing velocity is then: newVelocity = velocity - normal*(1+e)*dot(velocity, normal)
where e is the coefficient of restitution. For a perfectly elastic bounce e = 1, for a perfectly inelastic bounce (your 'slide') e = 0.

If you're struggling to compute the normal vector. Assuming you have two positions for the slope 'a' and 'b' that you're using to compute your angle like atan2(b.y-a.y, b.x-a.x), then the normal for this slope is instead: unit(a.y-b.y, b.x-a.x). The normal is a unit-length vector that points 'out' (or 'in' for these purposes, makes no difference to the result) of the surface, so the left side of a box will have normal (-1, 0) etc.

That you say you have done bouncing, but cannot figure out this, plus that you talk about the 'angle' of the slope tells me you are using trigonometry to do the bouncing... *sigh*

 

The solution is incredibly simple, using simple vector operations.

 

You have the slope normal vector (As opposed to its angle)

You have the incident velocity.

 

The outgoing velocity is then: newVelocity = velocity - normal*(1+e)*dot(velocity, normal)

where e is the coefficient of restitution. For a perfectly elastic bounce e = 1, for a perfectly inelastic bounce (your 'slide') e = 0.

 

If you're struggling to compute the normal vector. Assuming you have two positions for the slope 'a' and 'b' that you're using to compute your angle like atan2(b.y-a.y, b.x-a.x), then the normal for this slope is instead: unit(a.y-b.y, b.x-a.x)