A bullet shot into a door vs. a bullet shot into a suspended block I’m learning about angular momentum right now and I’ve been given this problem from University Physics, $13$e by Young and Freedman:

A door $1.00 \;\text{m}$ wide, of mass $15 \;\text{kg}$, can rotate freely about a vertical axis through its hinges. A bullet with a mass of $10 \;\text{g}$ and a speed of $400 \;\text{m/s}$ strikes the center of the door, in a direction perpendicular to the plane of the door, and embeds itself there. Find the door’s angular speed.

I’ve been told that you cannot use conservation of linear momentum in this problem because external forces are acting on this system (bullet + door). As far as I can tell, the external force in question is that exerted by the pivot on the door.
However, I’m not seeing the difference between this question and one regarding a ballistic pendulum, which we do use conservation of linear momentum (and conservation of mechanical energy) to solve. In both cases, a bullet is being shot at something that rotates around a point. In the case of a ballistic pendulum, are there not external forces acting on the system? Why is conservation of linear momentum valid in that case?
 A: In the ballistic pendulum case, it is assumed that all the mass of the block is concentrated at the point of bullet impact.  In the case of the door, the mass is distributed over the entire door and, since the door is assumed rigid (and constrained at the hinges), different parts of the door will have to have different tangential velocities.
A: The difference is because the door can transmit a force from the hinges which is in the horizontal direction whereas a vertical string cannot exert a horizontal force.
The only force available to the string is its tension and that acts along the string. 
The string can only exert a horizontal component of force when it is not vertical.
Suppose the door hinges were such that during the collision they did not constrain the door moving horizontally but immediately after did, that is, there was a bit of slack in the hinges.
So during the collision you could use linear momentum conservation but then immediately after there would be an impulse on the door due to hinges which would mean that the linear momentum of the door would change again. 
Using angular momentum conservation about the hinges means that the torques applied by the hinges are zero.
A: The 2 systems are equivalent. In one case you have a simple pendulum, in the other you have a compound pendulum. The fact that the pendulum string is flexible does not enter into the calculation. It could be replaced by a light rigid rod without affecting the outcome. The only difference then is the distribution of mass about the pivot point, which I think is the sole reason why conservation of linear momentum works for the simple pendulum but not the door.
In both cases you should use the conservation of angular momentum about the pivot point, not the conservation of linear momentum. Assuming that the force on the string or door from the pivot point acts towards the pivot (a central force) then there is no change in the angular momentum of the system during the instantaneous collision.
In both cases the initial angular momentum is $mvL$ where $L$ is the distance of the bullet from the pivot point. The final angular momentum is $J\omega_0$ where $J$ is the moment of inertia about the pivot and $\omega_0$ is the angular velocity immediately after the bullet is embedded, which is assumed to happen instantaneously.
For the simple pendulum, in which the mass $M$ of the pendulum is concentrated at distance $L$ from the axis, you have
$mvL=J\omega_0=(ML^2+mL^2)\omega_0$
$mv=(M+m)V$...(*)
where $m,M$ are the masses of the bullet and block and $v,V$ are the linear velocities of the bullet immediately before and of the bullet-and-block immediately after the collision. I have also used the fact that $V=\omega_0 L$. In this case the conservation of angular momentum is equivalent to the conservation of linear momentum. 
For the compound pendulum, for which the mass is distributed like a rod up to a length $2L$ from the axis, you have
$mvL=J\omega_0=(\frac13M(2L)^2+mL^2)\omega_0=(\frac43ML^2+mL^2)\omega_0$
$mv=(\frac43M+m)V$
which is not the same as the result using the conservation of linear momentum in eqn (*) above. 
[However, the conservation of linear momentum can be applied if the bullet strikes the door at the Centre of Percussion. (Thanks to Andrew Morton for pointing this out in his comment below.) The door swings on its hinge with the same period as a simple pendulum of the same mass concentrated at the CoP. The moment of inertia of the door can be written as $J=Mk^2$ where $k$ is the distance between the hinge and the CoP. So if the bullet strikes the door at the CoP then the conservation of angular momentum gives the same result as the conservation of linear momentum :
$mvk=(Mk^2+mk^2)\omega_0$
$mv=(M+m)V$
where now $V=\omega_0 k$ is the velocity of the CoP immediately after the collision.]
Both the simple and compound pendulums rotate about a fixed pivot. This motion can be decomposed into an instantaneous linear motion of the CM and a rotation about the CM. For the simple pendulum, the pivot is far outside of the bob, so the rotation of the bob about its CM is negligible compared with the motion of the CM. To a good approximation the impact results only in linear motion of the CM, so it is modelled well as a 1D linear collision. For the compound pendulum, the pivot point is not far from the CM compared with the size of the door, so the rotational motion about the CM is significant compared with the motion of the CM. The impact results in rotational as well as linear motion, so it cannot be approximated as a 1D linear collision.
A: Suppose there is some play in the door hinge, such that in the first small increment of time after the bullet has hit the door, the entire door can travel in the same direction the bullet did, without rotating.
However, now the door starts to rotate instead of moving linearly. Why can that be? It can only because the hinge pulls back on its side of the door to make the door's edge stay where it is.
This force on the door eats up some of the linear momentum it got from the bullet, so analysing the situation in terms of linear momentum is going to be complex.
Fortunately, since the door+bullet system only interacts with its environment through the hinge (and we're assuming the hinge is well-oiled and doesn't transmit any torque), its angular momentum around the hinge is preserved. This allows us to analyse the situation relatively easily using angular momentum.
In the pendulum case (where we treat the pendulum bob as a point) either linear momentum (in the direction of the bullet's travel) or angular momentum ought to work, and both should give the same result. In this case it is somewhat easier to use linear momentum, so that's the calculation that will usually be presented.
Conservation of mechanical energy is not valid in either of the situations, because some energy is lost as heat and stress when the bullet embeds itself in the target, deforming it.
