**edit 1 $\to$ 2 ** This might or might not be the answer... (but I believe the answer is **no.** to your Q1) :

Is there always some direction in which to shoot the cue ball so that some ball goes into a pocket (is sunk), and although other balls also may be sunk, the cue ball itself is not (so there is no scratch)?

This is also not an answer you seek, however this is a furtherance of a comment I made above, where I agree with Daniel Litt that with enough pockets, the cue ball will ultimately scratch. In fact, as long as there is more than zero pockets, it is always very likely that the cue ball will ultimately scratch, except for precisely finicky situations. I shall construct such an example.

Also, with Joseph's conditions, **no momentum is lost** with each ball-to-ball contact, thus the white ball will always have to keep moving. In fact, it's almost an analog (real, in $\mathbb{R}^2$) analogue to a **discrete stochastic matrix**: if there are any elements of the Markov chain which are *absorbing states*, then ultimately everything will be absorbed. Even if there is only one pocket (or just the standard six), the cue ball will ultimately end up in one of the pockets for this Q's criteria, unless you set it up just right.

The pockets are analogous to the aborbing states in the stochastic matrix of the Markov chain.

The exception to the absorbing states' always absorbing all possibilities occurs if there is a cycle in the Markov chain. Here is an example where you end up with an **infinite loop**, equivalent to a *cycle* in a *Markov chain*.

Your pool table is deterministic with length $y_t$ (along the long axis of the table) and width $x_t$, with pockets at the four corners and along the sides at $x=0$ and $x=x_t$

Set the cue ball up at $(x_c, y_c)$, with $x_c$ greater than half-the diameter of the ball + the width of the corner pockets $\times \sqrt(2)$

set one more ball $B$ at $(x_c, y_b)$ such that the cue ball and this ball are aligned at their x-position and non-overlapping in their y-position ($y_c - y_b \gt$ diameter of balls $+ \varepsilon$ for very tiny $\varepsilon$)

place all other balls at $(x_d, y_{something})$ such that $x_c - x_d \gt$ diameter of balls $+ \varepsilon$, and these other balls are not overlapping, e.g. $y_i - y_j \gt$ diameter of balls $+\varepsilon$ for each different $i,j$ pair of two balls in the remaining balls.

give the cue ball an impulse such that it moves with $v_x = 0$ and $v_y \gt 0$

*(infinite loop starts here)*

now the cue ball will move up until it hits the ball $B$, which occurs for the first time at $(x_c, y_b - \textrm{diameter} \div 2)$ as the point of contact for the two balls

momentum is transferred to the ball $B$ which moves with no motion along the x-axis, and positive along the y-axis until it hits the far end of the table

$B$ reflects back until it hits the cue ball, imparting a negative y-velocity to it

cue ball will hit the near wall and reflects back to hit ball $B$

*ad infinitum*

So this doesn't answer your question, but does prove that a scenario exists in a deterministic perfect momentum-conserving billiards game such that an oscillatory cycle is entered and never ends.

So with your conditions that we can make the rolling friction arbitrarily small, we'd end up scratching even if a configuration were found that was capable of sinking all fifteen balls and then the cue ball would be alone on a table moving until it fell into one of the pockets. Unless we could adjust $\mu$ for each starting configuration so that after sinking all fifteen balls, the cue ball glides to a stop. But that's just setting up too many parameters and free variables. This is why physicists recommend that we not live in a world without friction.

**(edit 2 bold claim retracted** by strikethroughs) Actually, I've reread your *revised* question, ~~and I think that this actually is the answer you seek, and the answer is ~~**no**. There is no shot that you can make in the game on the table you have defined such you will sink the specified ball and also not scratch. Your conditions make it such that if you do not enter an infinite loop, the cue ball will keep moving. To hit a ball into a pocket, the cue ball will cycle around and find a changed world, and will most likely end up scratching ultimately. and if you allow rolling friction $\mu_R$ to be a specifiably small value, perhaps the answer is yes. But the condition of allowing 14 other balls in the mix turns it into an extended type of problem like the 3-body problem for gravitational attraction, with wacky chaotic dynamics that are not easily tractably gotten around. ...

chaosthat occurs in pool is a result of the fact that even with best possible effort, racking up the $15$ balls into their initial configuration within the triangle is an imperfect activity. The balls are not all perfectly in contact; their hexagonal-packed lattice is not perfectly formed; inter-ball spacing is inconsistent, etc. You should clarify whether you mean to ask this question limited to an initial configuration of the balls+cue ball in standard place (and specify their coordinates), or for any arbitrary possible configuration of the 15 balls + the cue ball. – sleepless in beantown Oct 31 '10 at 2:23mouth. Corner Pocket Mouth: between 4.5 [11.43 cm] and 4.625 inches [11.75 cm] Side Pocket Mouth: between 5 [12.7 cm] and 5.125 inches [13.0175 cm] (The mouth of the side pocket is traditionally ½ inch [1.27 cm] wider than the mouth of the corner pocket.)" – Joseph O'Rourke Oct 31 '10 at 2:25