Conservation of momentum applies when net force is zero. Suppose that there is a system of a canon and a canonball. Total momentum of the system is zero before canonball is fired. Now canonball is fired from the canon, and in frictionless cases, horizontal-axis momentum of the whole system would be preserved.

Now friction between ground and canon is added. What happens to the momentum of the whole system in this case?

I am confused because for inelastic collsion of the case when a bullet hits a block, it is often said that momentum is preserved during the collision (though not after the collision) even when friction exists between the bullet and the block.

Can I say that friction is negligible at the moment of canonball being fired because $F\Delta t = \Delta p$ would be small as $\Delta t$ is very small?

  • $\begingroup$ Are you talking about the friction between cannonball and the cannon or the friction between cannonball and the target or the cannon and ground? $\endgroup$
    – Eiver
    Oct 7 '13 at 7:27
  • $\begingroup$ cannon and ground. $\endgroup$
    – user29157
    Oct 7 '13 at 7:30

The momentum of the whole system is still conserved -- it's just that when you add friction between the cannon and the ground, you have to include the ground (and in fact the whole planet that it's attached to) as part of "the system". When the cannon ball flies off in one direction, the cannon is pushed in the other. But due to friction the cannon pushes in turn on the Earth, which is not fixed to anything, so its velocity changes very slightly.

However, the mass of the Earth is very large, so the change in velocity is minute. A typical cannonball would have a mass of the order $10\:\mathrm{kg}$. Assuming a muzzle velocity of $100\:\mathrm{m/s}$, the ball gains about $10^3\:\mathrm{kgm/s}$ of momentum. The Earth's momentum therefore changes by the same amount in the opposite direction, but since the Earth's mass is around $6\times 10^{24}\:\mathrm{kg}$, this corresponds to a velocity change of $10^3/6\times 10^{24} \sim 10^{-22}\:\mathrm{m/s}$, which is so small it's effectively unmeasurable.

In fact, the above is a slight oversimplification. The Earth isn't completely solid, so what actually happens is that the ground near the cannon will start moving first, and then set other parts of the Earth moving very shortly afterwards. The result is seismic waves, which echo around the planet until they eventually get dissipated. If the cannon is big enough these can be measured from very far away - potentially from the other side of the world. But once all the waves have died down, the end result is the same: the Earth's velocity has changed by a very small amount. Similar considerations can be made regarding angular momentum about the Earth's centre, i.e. firing the cannon will also slightly change the Earth's rotation.

Of course, this tiny velocity change will be exactly cancelled out once the shot hits something and transfers its momentum back to the Earth through the same process.


Yes. If you consider a simplistic case where at one instant everything is at rest and at the second instant the cannonball shoots in one direction and the cannon recoils in the opposite direction, then as far the cannonball is concerned, it is no longer in contact with the cannon or Earth. So momentum is preserved. Later the cannon will convert some of its kinetic energy to heat due to friction and some to change the motion of Earth (very tiny change). Meanwhile the cannonball would continue on its way until it hits something.


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