How is a ballistic pendulum affected by two objects with different masses? Imagine that bullet A (made of lead) has twice the mass of bullet B 
It is shot at the same muzzle velocity as bullet B onto a ballistic pendulum; all other initial variables are the same: size and mass of brass ring, size of bullet, length of the pendulum rod, etc. 
What would be different about the final state of the apparatus, i.e. the angle of the pendulum? 
Can someone explain how two objects with different masses can have the same initial velocity? 
 A: 
Can someone explain how two objects with different masses can have the same initial velocity?

Since the kinetic energy $E_{kin}=m\cdot v^2/2$ you need more energy = more gunpowder to get the same velocity for a heavier bullet. 

What would be different about the final state of the apparatus, i.e. the angle of the pendulum?

The pendulum would swing higher, since the higher kinetic energy provided by the heavier bullet would be transformed into a higher potenial energy when the pendulum gets kicked up.
A: In the ballistic pendulum, it is the momentum that matters (that is conserved).
For a pendulum with mass $M$, being hit by a bullet with mass $m$ and velocity $v$, the velocity $v_1$ immediately after the impact is (from conservation of momentum):
$$v_1 = \frac{m}{m+M}v$$
And when you have an object with twice the mass but the same velocity, you get the pendulum velocity $v_2$ after impact:
$$v_2 = \frac{2m}{2m+M}v$$
This velocity allows you to calculate the energy after the collision (always less than the initial energy - because it's an inelastic collision). 
$$E = \frac12 (M+m) v_1^2$$
The pendulum will stop deflecting when all the kinetic energy has been turned into potential energy:
$$(M+m) g \ell (1-\cos\theta) = \frac12 (M+m) v_1^2$$
A little rearranging will show you that the heavier bullet will cause a greater deflection of the pendulum. Exactly how much greater will depend on the mass of the pendulum itself.
$$1 - \cos\theta = \frac{(M+m)v_1^2}{2(M+m)g\ell} = \frac{\left(\frac{m}{m+M}v\right)^2}{2g\ell}$$
Now assuming that the deflection $\theta$ is small so we can approximate $\cos\theta = 1-\frac12 \theta^2$, then
$$\theta = \frac{1}{\sqrt{2g\ell}}\frac{m}{M+m}v$$
As you can see, for $m << M$ the deflection angle will be proportional to mass (and impact velocity). As the mass of the bullet gets larger, the deflection will be slightly less than proportional to $m$.
As for your final question:

Can someone explain how two objects with different masses can have the same initial velocity?

Have you ever driven down a motorway next to a large truck? Different objects, different masses, same velocity. Not sure I understand where you are going with that question...
A: Initial velocity is provided by an external force acting on the object to move it from rest. Their initial velocity can be the same, however since the mass is different their acceleration will be different.
A: In equillibrium state the initial velocity of all bodies is equal .bcoz of both the bullets are at rest so their initial velocities are also equal whether both have same or different mass.
