# 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?

• I think you can use conservation of momentum to work out the speed of the centre of the door, and then use $v=\omega r$ to find the angular frequency, $r$ being the perpendicular distance from the centre to the edge of the door with the hinges. Oct 15, 2016 at 12:57
• Like there's a force from the hinge, and there's there's a force from gravity, but if they didn't cancel the door would be moving, it might even fall off its hinges. I mean unless the pivot is causing friction with the motion of the door? Or it's tightened to provide a sort of damping force? Oct 15, 2016 at 13:06

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.

• I see. But what of the external force part? Oct 15, 2016 at 12:07
• Why don't you model it and see what you get. Assume that angular momentum is conserved, and then determine the change in linear momentum integrated over the door. Oct 15, 2016 at 13:17
• No, this is wrong. The mass distribution is not relevant, it's much simpler. If the impact line does not hit the center of mass, the ballistic pendulum will start to rotate -- but this does not concern conservation of momentum, which will still be valid. The relevant assumption is, that the string exerts no horizontal (i.e. in the direction of the impact) force -- whereas the door hinges do.
– Ilja
Oct 15, 2016 at 21:50

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.

• Could you explain a little more about where the pivot force comes from before the collision? I know I mentioned the force in my question, but I don’t really understand why it’s there. Or does the pivot force come during the collision? In that case, isn’t there a horizontal component of tension in the string during the collision as well? Oct 16, 2016 at 2:15
• The force at the pivot/hinge comes about during the collision because the bullet has given the door some linear momentum and the hinge has constrained the motion of one end of the door. A string cannot exert a force at right angles to itself and so when it is vertical can exert no horizontal force. Oct 16, 2016 at 6:30
• I'm sorry, but I still don't understand. So before the bullet hits the door, there is no force from the pivot, but when it does, there is a force to constrain one end of the door. However, before the bullet hits the block, there is no force in the horizontal direction yes because the string is vertical, but when the bullet hits, does the string not constrain the block similarly? Isn't the string is no longer vertical? Oct 16, 2016 at 13:03
• If the inability of the string to exert a horizontal force is the explanation, then the answer should be different if, instead of a hinge, the door is connected to the pivot by a very short but flexible string. The door would then be a pendulum with a very short string and very long weight (instead of long string and short weight). Would conservation of linear momentum apply in this case, as it does for the simple pendulum? I don't think it would. Oct 16, 2016 at 15:10

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.

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.

• This does not answer the question as to why you can use linear momentum conservation in one case and not the other. Oct 15, 2016 at 13:35
• Conservation of linear momentum should not be used in either case. It 'works' for the simple pendulum because the mass of the simple pendulum can be assumed to be concentrated at the point at which the bullet is embedded. Oct 15, 2016 at 14:24
• @sammygerbil can't the door being rigid be treated instantaneously after the impact to behave as it its mass is concentrated at a point. I mean if you dropped a bowling ball on a trampoline it's fairly clear why this approximation wouldn't work, but the door is rigid and stays rigid after impact with the bullet embedded Oct 15, 2016 at 14:44
• Yes, the door could be treated as a point particle in 1D if it is struck at its CM and if it is not hinged at one end. Otherwise it must be treated as a 2D object (rigid body) which can rotate as well as possibly translate. Oct 15, 2016 at 14:56
• @sammygerbil Could the door also be treated as a point particle in 1D if it was hinged and struck at the centre of percussion? Oct 16, 2016 at 13:29