Modifying a CE transistor Amp to a Single Ended Class B Amp
I think I should move the Q point on the DC load line to either Ic=Ic(sat) or Vce=Vcc to only allow half the AC signal to be outputted,

This answer involves setting the Q point to Ic=Ic(sat) rather than I(cutoff). Although it amplifies half the AC signal, in common emitter configuration, it does not have the nice quality of low quiescent power found in &quot;real&quot; class B amplifiers.
To amplify only a half-cycle of an input wave, you may set the Q-point of your amplifier on the saturation line of your transistor. &quot;A&quot; is an example of point on the saturation line. This may or may not qualify as a class-B amplifier, but it will give you a half-cycle output (but with a twist -- it is inverted).

[Image taken from Quara.com]
Although the diagram does not show it, in general, there will be multiple values of \$I_{B}\$ that will have characteristics that pass through that point. Choose the lowest value of \$I_B\$ that has a characteristic that passes through that point.
If your Q-point is set as described, if the input should raise the voltage at the base of your transistor, there will be no increase in \$I_C\$ and therefore no increase in \$V_{R2}\$, and hence no change in \$V_{out}\$. However, if the input should decrease the base voltage, there will be a decrease in \$I_C\$, and a corresponding drop in \$V_{R2}\$, and a rise in \$V_{out}\$.
Thus, because this is an inverting amplifier, &quot;negative&quot; half cycles on the base will cause positive half cycles on the output.
Here is an example circuit (not using your transistor, however):



simulate this circuit &ndash; Schematic created using CircuitLab
The waveform generated by this circuit looks like this:

A significant disadvantage of this design is that when the input is quiescent, the transistor is &quot;on&quot;. The opposite design is to set the Q-point of the amplifier at cutoff. The advantage of that design is that when the input is quiescent, the transistor is &quot;off&quot;. A disadvantage is, that for an inverting amplifier, the quiescent voltage will be high (if taken from the collector to ground), and will drop when an input is applied.