Could a radar pulse be considered particular in nature? I ask because I have a real problem with wave/particle duality (one of the biggest cop outs in the whole of physics, in my opinion)
A radar pulse is definitely a wave.
It is spatially bounded. (at least in one dimension!)
It has energy.
It is a 'quanta'.
Is this not particle-like?  Could a pulse of sufficiently high energy have mass in accordance with E=MC^2 and E=hf?
Apologies for the ignorance of a simple engineer. 
 A: In quantum mechanics the electromagnetic field interacts as quantized particles, photons. We ask if you can neglect the discrete theory for the classical continuous theory.
I'll use typical numbers found in wikipedia.
A typical radar pulse lasts $t=1 \mu s $ at a frequency of $f=3$ GHz.
I would estimate the maximum power of the pulse arriving at a $1 m^2$ cross section at less than $1  MW$. The energy of the pulse is therefore $E=P\times t=1 J$.
The energy of a quantum of light is given by 
$E_{quantum} = h f$, in this case $E_{quantum} \approx 10^{-24}J$
This means the radar pulse carries about $10^{24}$ discrete quanta, which means we can safely approximate it as a continuous wave.
A: Quanta of a electromagnetic wave or photon is the minimum amount of energy that can be transferred from one place to another. Increasing the energy of a pulse means you are increasing number of photons in the pulse and we can write 
$$E=nh\nu$$ 
Where equivalent mass can be written as 
$$E=pc$$
$$p=\frac {h\nu}{c}$$ 
and
$$p=mv$$
where $v=c$ is the velocity of light, n are number of photons, hence you can see that by increasing the energy you can not increase the equivalent mass of the photon as the frequency of the photon remains same, you are increasing just the numbers of the photons. 
Radars emit electromagnetic radiation in microwave range hence their photons has very small energy, which makes really hard to observe particle like behavior from them.
I hope this will help
EDIT: Regorusly one may argue that mass of photon is not physical entity. However in several situations one may define the effective momentum of the photon as 
$p=\frac{E}{c}$ and from here one may deduce $m=\frac{p}{c}$, This analogy is not correct regorusly but it works in several situations. It is like scalar diffraction theory which is not regorusly correct but works fine in several occasions
A: $E=mc^2$ is true only for a particle at rest, the photon is never at rest.
The full equation is 
$$E^2=p^2c^2+m^2c^4.$$
A photon has no mass, but two photons can meet to form an electron and a positron, see pair production.
