# Conservation of linear vs. angular momentum in two similar cases

I have a question that eludes my understanding:

Imagine we have a bullet of mass $$m$$ and a rigid pendulum with a bob of mass $$M$$ hanging from a rigid rod of negligible mass hanging from the ceiling from a pivot. The bullet has horizontal velocity and gets embedded.

*Case 1: The bullet hits the bob. It is my understanding that conservation of linear momentum here applies (I tried, it works).

*Case 2: The bullet hits at some point along the rod. Here conservation of linear momentum does not apply (it doesn't work) but conservation of angular momentum with respect to the pivot does.

Now I understand case 2 since the external forces have zero torque on the pivot, but I don't understand case 1 since the non-rotational dynamics for me looks exactly the same and the external force due to the pivot has an horizontal component, which should not allow linear momentum to be conserved. Could you please explain this rigorously?

• You can also treat case one with the conservation of angular momentum, and I don't see why you can not use linear momentum in case 2. Commented Jul 18 at 10:16
• Of course case 1 can be treated with conservation of angular momentum, but I tried case 2 with linear momentum and it won't work. Commented Jul 18 at 10:22
• @GiovanniPiacentini Show us what you tried that didn't work. Commented Jul 18 at 12:05
• Bar is $L$ long and bullet hits at midpoint: Linear m: $mv_i = mv_{1f} + Mv_{2f}$ Angular m: $m \left(\frac{L}{2}\right) v_i = (I_1 + I_2) \omega = \left( \frac{m L^2}{4} + M L^2 \right) \omega$ So: $\omega = \frac{m v_i}{2L \left(\frac{m}{4} + M\right)}$ Since $v = \omega r$, then: $m v_i = m \omega \left(\frac{L}{2}\right) + M \omega L = \left(\frac{m}{2} + M\right) L \omega=\left(\frac{m}{2} + M\right) L \left( \frac{m v_i}{2L \left(\frac{m}{4} + M\right)} \right)$ $\left(\frac{m}{2} + M\right) \left( \frac{m v_i}{2 \left(\frac{m}{4} + M\right)} \right) \neq m v_i$ Commented Jul 18 at 12:49
• Even if it did work, my question still stands since the linear momentum should not conserve when we have external forces, such as the force of the pivot on the bar. Commented Jul 18 at 12:54

It is a valid question, that in case 1 it seems that momentum shouldn't be conserved due to the horizontal force of the pivot point. The simple answer to this question is that, the pivot point does not exert a horizontal force on the rod+bob+bullet system.

Refer the image given below:

For clarity, imagine that the bullet hits the rod just below the pivot point. You will feel that the hinge force(blue force) will be towards right by intuition, which is correct. The bullet in this case will have lower angular momentum with respect to the pivot, thus resulting a lower angular speed, meaning the linear momentum is reduced. This is in accordance with the hinge force opposing the bullet's momentum direction, thus momentum is reduces

Take the case where the bullet hits below the bob(suppose there is a massless rod there), then the angular momentum of bullet is much higher than the previous case. This results in a higher angular speed, meaning that the momentum has actually increased. This means that the hinge force is towards left, in the same direction of momentum of bullet, thus the momentum of system increases.

In the case where bullet directly hits the bob, it is the bridging case between the above two cases. The force will be neither left nor right, it will be 0! The bullet and the bob are the same perpendicular distance from the hinge, meaning the bullet perfectly transfers its momentum to the bullet+bob system.

Note: I have directly talked about angular speed without using momentum of inertia, because the bob has some mass, it will take some of the angular momentum. You can also calculate it yourself and check the direction of force.

Tip: Try to think fundamentally first, then bring in the mathematics. It will make it much easier to understand different cases of physics.

• Thank you for the excellent explanation. Now I understand my mistake. I assumed that the force at the pivot point is always present and always in the same direction. Commented Jul 18 at 14:32

*Case 2: The bullet hits at some point along the rod. Here conservation of linear momentum does not apply (it doesn't work) ....

It does apply if you expand the system to include the mass of the block the pivot is attached to. If the block is bolted to the Earth, you can use the mass of the Earth for the pivot block mass but you cannot ignore it. While it tempting to treat the the pivot block mass as infinite and the velocity of the pivot block as zero after the collision this does not work (as you discovered). The Earth does acquire a finite amount of the kinetic energy. Imagine firing the bullet at block bolted directly to the Earth (no pivot). You could say the momentum >0 and after the impact when the bullet comes to rest inside the block and assume the final velocity is zero and the final momentum is also zero and falsely conclude that linear momentum is not conserved.

If the earth did acquire any velocity, Its kinetic energy is still negligible. It will have finite momentum, but as kinetic energy is 1/2 mv^2 = 1/2 pv, it tends to 0. – EagerToLearn >Commented

It does tend towards zero, but so does the final kinetic energy of the bullet and in that context the energy acquired by the Earth is non negligible.

Let's do a worked example for firing a bullet into a block attached to the Earth:

Mass of the Earth: 6E24 kgs
Mass of bullet: 0.02 kgs
Initial velocity of the bullet: 500 m/s
Initial linear momentum of the bullet: 10 kg m/s
Final velocity of the system: $$\approx$$ 1.666E-24 m/s
Initial Kinetic energy of the bullet: 2500 Joules
Final kinetic energy of the system: 8.333E-24 Joules
Final kinetic energy of the Earth $$\approx$$ 8.333E-24 Joules
Final kinetic energy of the bullet $$\approx$$ 2.77E-50 Joules
Final linear momentum of the bullet: 3.33E-26 kg m/s
Final linear momentum of the Earth: 10 kg m/s

As can be seen from the above calculations the final kinetic energy is tiny and being an inelastic collision most of the initial kinetic energy is lost as heat. However, it can also be noted that the kinetic energy acquired by the Earth is many orders of magnitude greater than the final kinetic energy of the bullet and treating the kinetic energy of the Earth as negligible introduces as large error. It can also be noted that over 99.99% of the initial linear momentum of the bullet is acquired by the Earth. The linear momentum acquired by the Earth most certainly cannot be treated as negligible.

• If the earth did aquire any velocity, Its kinetic enegry is still negligible. It will have finite momentum, but as kinetic energy is 1/2 mv^2 = 1/2 pv, it tends to 0. Commented Jul 18 at 16:04

Force at the pivot don't affected the linear momentum.

Lets look at the equations:

for the bullet mass, $$~m$$ $$m\,(v_m-v_0)=\lambda\tag 1$$ for the bob mass, $$~M~$$ $$M\,v_M=-\lambda\tag 2$$

and for the massless rod $$M\,L^2\,\omega=\rho\,\lambda\tag 3$$

additional equation is the conservation of the energy

$$m\,(v_m^2-v_0^2)+M\,v_M^2+M\,L^2 \,\omega^2=0\tag 4$$

with those 4 equations you obtain $$~v_m~,v_M~,\omega~,\lambda~$$

• $$~v_m~$$ bullet velocity after the collision
• $$~v_M~$$ bob velocity after collision
• $$~\omega~$$ rod angular velocity after collision
• $$~\int \lambda\,dt~$$ constraint force at the collision

If you add equation (1) and (2) you obtain the linear momentum

$$P=m\,(v_m-v_0)+M\,v_M=0$$

thus , it is always zero !