# Conservation of Momentum problem — Firing a cannonball

• Ineedhelpwithphysics
In summary, the final velocity and direction of the cannon will be 4 m/s to the left because the momentum before the shot was 0 due to the opposite and equal reaction. This is calculated using the equation m1v1 + m2v2 = m1v1' + m2v2' and considering the cannon's mass of 50,000kg and the cannonball's mass of 20kg. In the second part of the conversation, it is discussed that the cannon's recoil velocity will not change significantly if it is fixed to the ground, as the extra mass of the Earth will have a negligible effect on the velocity.f

#### Ineedhelpwithphysics

Homework Statement
A 50,000kg Cannon fires a cannon ball of mass 20kg. If the final velocity of the cannon ball is 10,000m/s to the right, what will the final velocity and direction of the Cannon be?
Relevant Equations
m1v1 + m2v2 = m1v1' + m2v2'
So I am guessing the cannons final velocity will be 4 m/s to the left because there momentum before shot was 0 because of opposite and equal reaction so

50,000kg x -4 m/s + 20kg x 10,000 m/s = 0 ?

Homework Statement:: A 50,000kg Cannon fires a cannon ball of mass 20kg. If the final velocity of the cannon ball is 10,000m/s to the right, what will the final velocity and direction of the Cannon be?
Relevant Equations:: m1v1 + m2v2 = m1v1' + m2v2'

So I am guessing the cannons final velocity will be 4 m/s to the left because there momentum before shot was 0 because of opposite and equal reaction so

50,000kg x -4 m/s + 20kg x 10,000 m/s = 0 ?
Yes

• Delta2
Homework Statement:: A 50,000kg Cannon fires a cannon ball of mass 20kg. If the final velocity of the cannon ball is 10,000m/s to the right, what will the final velocity and direction of the Cannon be?
Relevant Equations:: m1v1 + m2v2 = m1v1' + m2v2'

So I am guessing the cannons final velocity will be 4 m/s to the left because there momentum before shot was 0 because of opposite and equal reaction so

50,000kg x -4 m/s + 20kg x 10,000 m/s = 0 ?
The answer to this particular problem is easy to guess. Can you guess the cannon's recoil velocity if the cannon had mass 48,340 kg and the final velocity of the 18.75 kg cannon ball 9,853 m/s to the right? If you are like me and cannot guess this one, what would you have to do instead to answer the question?

The answer to this particular problem is easy to guess. Can you guess the cannon's recoil velocity if the cannon had mass 48,340 kg and the final velocity of the 18.75 kg cannon ball 9,853 m/s to the right? If you are like me and cannot guess this one, what would you have to do instead to answer the question?
I would add the question what if the cannon is fixed to the ground?

I would add the question what if the cannon is fixed to the ground?
Then conservation of momentum doesn't hold for the system cannon+ball because there is the external force from the ground. However it holds for the system cannon+ball+ground.

Then conservation of momentum doesn't hold for the system cannon+ball because there is the external force from the ground. However it holds for the system cannon+ball+ground.
But would the cannonball go faster?

But would the cannonball go faster?
Well hm I don't know, if we consider that the ground has infinite mass, it will move with 0 speed but its momentum ##0\cdot\infty## seems undefined... Or what do you think

Well hm I don't know, if we consider that the ground has infinite mass, it will move with 0 speed but its momentum ##0\cdot\infty## seems undefined... Or what do you think
Since the cannon is firmly attached to the Earth, the cannon and the Earth recoil as one. Thus, the recoiling mass is that of the cannon plus the Earth. It is very large relative to the mass of the cannon but not infinite. Furthermore, the recoil momentum of the Earth+cannon system is not undefined; it is equal in magnitude and opposite in direction to the momentum of the cannonball.

• Delta2
But would the cannonball go faster?
Yes, but by a very very small amount. Given some fraction of the chemical energy of the charge that goes into the kinetic energies of the system is fixed whether the cannon is free to recoil or fixed to the ground, the velocities of the cannon labelled ##1## and the cannonball labelled ##2##, we have; $$v_2 = \sqrt{ \frac {2E } {m_2(1 + \frac{m_2}{m_1})}}$$

As ##m_1## gets larger, ##v_2## gets larger but not by much as it approaches the limit of ##\sqrt{\frac{2E} {m_2}}## as ##m_1 →∞##.

Given the original numbers of 20kg and 50,000kg we can take the ratio of the velocity ##v_2## with the mass of the cannon plus the Earth to the velocity of the cannon alone. It will be exactly ##\sqrt{1 + \frac{20}{50000}}## which is 1.0002.

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• Delta2
Yes, but by a very very small amount. Given some fraction of the chemical energy of the charge that goes into the kinetic energies of the system is fixed whether the cannon is free to recoil or fixed to the ground, the velocities of the cannon labelled ##1## and the cannonball labelled ##2##, we have; $$v_2 = \sqrt{ \frac {2E } {m_2(1 + \frac{m_2}{m_1})}}$$

As ##m_1## gets larger, ##v_2## gets larger but not by much as it approaches the limit of ##\sqrt{\frac{2E} {m_2}}## as ##m_1 →∞##.

Given the original numbers of 20kg and 50,000kg we can take the ratio of the velocity ##v_2## with the mass of the cannon plus the Earth to the velocity of the cannon alone. It will be exactly ##\sqrt{1 + \frac{20}{50000}}## which is 1.0002.
Yes, I had assumed that was the point of your question in post #4. Of course, you are assuming that the work done by the expanding gases at the point where the ball exits the cannon is constant, but I think that's provable.
But we should end this diversion. Going off topic on a thread is discouraged, though it's less of an issue once the OP's question has been resolved.

• Delta2 and bob012345