# Assumption in a momentum problem

1. Nov 12, 2015

### Mr Davis 97

I have the following problem: "A cannon of mass 5.80 × 103 kg is rigidly bolted to the earth so it can recoil only by a negligible amount. The cannon fires an 85.0 kg shell horizontally with an initial velocity of 551 m/s. Suppose the cannon is then unbolted from the earth, and no external force hinders its recoil. What would be the velocity of tan identical shell fired by this loose cannon? (Hint: In both cases assume that the burning gunpowder imparts the same kinetic energy to the system.)"

My question is, is that assumption from the hint at the end unrealistic? I understand that the amount force the gunpowder exerts might be the same, but the kinetic energy would not, since in the first part the cannon absorbs whatever force from the gunpowder comes its way, thus there should be less kinetic energy in the first part than in the second part where the cannon is unbolted. Thus, is the assumption just used for the sake of solving the problem, and not actually a realistic one?

2. Nov 12, 2015

### phinds

Just offhand, I'd say you have it backwards. The kinetic energy of the ball in the first instance will be caused by the full force of the gunpowder but in the second instance some of that force will be expended moving the cannon backwards and slightly less moving the ball forwards. The total magnitude of kinetic energy will be the same in both cases but the ball has more kinetic energy in the first case than in the second. Am I missing something?

3. Nov 12, 2015

### nasu

"absorbing force" is not a meaningful concept. Force is a measure of interaction, it is there in both cases. The effect of the force will be different in the two cases.
And the force itself may be different, as it depends on the how fast the volume of the gases changes .
Whereas the amount of energy released by burning the gunpowder, which depends on the properties of the powder used, it is the same.

Same as the caloric power of a fuel is the same, no matter how you use the heat released.

Edit. Same thing as in Phinds post. I was just thinking too long before posting. :)

4. Nov 12, 2015

### jbriggs444

None of the energy of the gunpowder goes into the rigidly mounted cannon. All of its delivered energy must go to the cannonball. To a first approximation, the delivered energy of the gunpowder in the non-rigidly mounted cannon is identical and is split between cannon and ball. However, it is not necessarily the case that the delivered energy will be exactly identical.

In the non-rigidly mounted cannon, a given pressure results in a greater rate of volume increase than in the rigidly mounted cannon. That is because both ball and cannon accelerate away. It is at least plausible that the more rapid decrease in pressure reduces the efficiency of the deflagration. The caveat in the original question prevents the wary student from wasting time diving into that possibility.

Last edited: Nov 12, 2015
5. Nov 12, 2015

### Mr Davis 97

Okay, so you're saying that the way the force interacts with ball and cannon changes in such a way as to make the kinetic energy the same in both cases; in the first case the cannon doesn't move so more of the force is delegated to the ball, and in the second case the cannon moves so less of the force is delegated to the ball. Even if this is correct, would it be correct in the real world to assume that the kinetic energy of the system is the same in both cases?

6. Nov 12, 2015

### PeroK

You need to do a calculation of KE loss when the Earth is involved. The Earth can absorb a significant (*) amount of momentum (e.g. when a ball bounces on the floor, reversing its momentum and delivering a significant (*) impulse to the Earth). But, if you calculate the KE transferred to the Earth, it's negligible.

(*) Significant in terms of the momentum of the ball.

You should think about why and then do a rough calculation to convince yourself.

7. Nov 12, 2015

### nasu

You use the term "force" in a way that is more appropriate for energy.
There is no such thing as an amount of force from which some is "delegated" to the ball and some to the cannon.
The force is an instantaneous measure of the interaction. Once the ball is out of cannon, it does not "have" any of the force from the gunpowder gases.
But it still has the energy.
The force is not useful for understanding this problem. It varies quickly and in a complicated way during the explosion of the gunpowder.
And there is force on the cannon in both cases. Maybe even larger in the first case, even though the cannon does not move.

8. Nov 12, 2015

### Mr Davis 97

I can see why momentum is not conserved in the first case: there is an external force on the ball-cannon system that changes the overall momentum of the system. It is conserved in the second case because the system is isolated. However, I am still not understanding how we can be sure that kinetic energy is the same in both cases. Intuition would tell me that the force of the gunpowder interacts with the cannon and the ball in the same way in both cases. The only difference is in the first case the force of the gunpowder does no work on the cannon while in the second case it does. This would make it seem that the ball would have the same kinetic energy regardless of either case, and the cannon would have kinetic energy depending on whether it is rigid or not. This would make it seem as though ball-cannon system has more kinetic energy in the second case than in the first case, even though the energy stored in the gunpowder is the same in both cases. Where am I going wrong? Does the gunpowder apply more force to the ball if the cannon is rigid?

9. Nov 12, 2015

### PeroK

Do the calculation in both cases and include the Earth!