Does recoil affect speed of trajectory?

[BradP]

Does the mass of a cannon affect the speed at which the ball will leave the barrel? If you produce two equal explosions -- one in which the cannon sits on a frictionless surface, and one in which it is bolted to the ground -- will the cannonball fly further when it is bolted?

How would you evaluate that?

 

[cantgetaname]

If you produce two equal explosions -- one in which the cannon sits on a frictionless surface, and one in which it is bolted to the ground -- will the cannonball fly further when it is bolted?

How would you evaluate that?

Yes.
It can be proved by the conservation of linear momentum.
Taking the cannon and cannonball as the system, equate the sum of their initial linear momenta to final linear momenta, in both cases and compare.

 

[BradP]

I thought about that, but why do you use conservation of momentum and not kinetic energy? I figure if all of the energy goes into the cannon/cannonball system, the total kinetic energy will be the same in both cases.

Using conservation of energy: .5*m1*v1^2 + .5*m2*v2^2 = .5*m2*v3^2

Using conservation of momentum: m1*v1 + m2*v2 = m2*v3

But these equations are not equal, so they cannot both be correct. Furthermore, wouldn't the momentum of the system be zero since the cannon and ball are moving in opposite directions?

 

[cantgetaname]

Always remember the basic concept. Everything you use comes with certain assumptions, that are stated when you read them somewhere.
For conservation of momentum:

If the external force on a system is zero, there is no acceleration produced. The velocity therefore remains constant and consequently the momentum. (mass is constant)

So to apply this, ask yourself - Is the net force on the system zero?
For the conservation of energy:

If the work done on the system is zero, mechanical energy is conserved.

Again,to use this you need to be sure that the work done on the system is zero.

In the case of your question, if you neglect the friction between the cannon and the ground, there is no force on the system in the horizontal direction. So you can use conservation of momentum.
(In the case of it being bolted to the ground, you could assume cannon & Earth to be one object)

Now, think about it. Is the work done on the system really zero? Is energy conserved here?

NOTE: Work/Energy is a scalar and unlike momentum, won't cancel out.

 

[BradP]

Perhaps we should clarify the points in time when we are talking about conservation. When I said "conservation of energy", I meant that the energy that the explosion transfers to the cannon/ball is the same whether the cannon is bolted to the ground or not. I don't mean that the energy of the system is the same before the explosion as it is after the explosion (assuming you do not put chemical energy into the equation).

Anyway, for your questions, the net force on the system in either case is not zero. The explosion creates an impulse force. And therefore, work is also done during that impulse force. The only information of interest are the velocities immediately at the end of that impulse force, because friction and air resistance are negligible.

Therefore, assuming you use equal amounts of gunpowder in each case, the total system should have the same kinetic energy for either case. The scalar nature of energy is of course reflected in the equation, with velocity squared.

 

[cantgetaname]

1. If you take earth, cannon, ball as one system, there is never an external force on this system in any case.
2. The force due to the gunpowder is an internal non conservative force.
If instead of gunpowder, you had a spring to launch the ball, you could have done what you are trying to do right now. (as long as you account for the energy stored in the spring)

 

[cantgetaname]

And yes, even if you do assume that the exploding gunpowder 'imparts equal energies' in each case, you will have to take into account the backward velocity gained by the earth.

 

[BradP]

I don't see how it helps to include the spring as an internal force. You can draw the system boundary wherever you want, and if you exclude the spring, that tells you exactly how much energy is transferred into the moving objects, and thus their total kinetic energy.

Imagine you have two objects launched by the same spring. The object is the system. All forces are external. In each case, the spring is fully compressed a distance x, then released. The spring will transfer a total amount of energy into the object equal to F*x, or (1/2)k*x^2. You would set the kinetic energy of the object to that same potential energy for an object of either mass.

If the spring is simply compressed by at both ends instead of one (like a cannon and ball), it still is going to transform all of its potential energy into their kinetic energy. So why can you not analyze it like that?

 

[cantgetaname]

I've just realized that we've both committed some fundamental errors in this discussion.
First I will try to answer the original question.

When the cannon is free to move,assuming the gunpowder gives a certain fixed amount of energy (say E)...
Cannon (mass M) goes back with V, ball (mass m) forward with v
Using energy conservation: 0.5 M V^2 + 0.5m v^2 = E

Things change in the second case. Cannon can only move with the Earth and together they form an extremely large mass that acquires a negligible velocity.
Using energy conservation: 0.5 (M+mass of earth) V'^2 + 0.5m v'^2 = E

Using the two equations,
0.5 M V^2 + 0.5m v^2 = 0.5 (M+mass of earth) V'^2 + 0.5m v'^2

V' here can be expressed here in terms of v', using conservation of momentum.
From there it should be easy to see that v'>v.

Now. The equation you included in your second post (#3)...

Using conservation of momentum: m1*v1 + m2*v2 = m2*v3

...is wrong. Those are two completely different events you got there in one equation.

Momentum alone cannot be used to prove the result we need here. (so i was wrong with that too)
The assumption that the gunpowder releases a fixed amount of energy both times is also critical.

 

[cantgetaname]

If the spring is simply compressed by at both ends instead of one (like a cannon and ball), it still is going to transform all of its potential energy into their kinetic energy. So why can you not analyze it like that?

Sure we can, but in that case we will need to know the proportion in which the energy is imparted to each object.
That can only be found by the conservation of linear momentum.

 

[BradP]

You do not need to include the Earth in the system. Kinetic energy is calculated with respect to a stable reference point. In relation to my desk, my laptop has no kinetic energy. In relation to the moon, it has very large kinetic energy. Therefore, just consider the velocity of both the cannon and cannonball to be relative to the Earth's surface. I think that is simplest.

The cannon then has zero velocity, and equating the first event to the second event gives:

.5*m1*v1^2 + .5*m2*v2^2 = .5*m2*v3^2

.

(The cannon is free in the left side of the equation, and fixed in the right side)

I know that

m1*v1 + m2*v2 = m2*v3

relates the first event and second event as well. I am not saying it is correct, I have been trying to hold it up to criticism. Is that the type of equation you meant when you said to use conservation of linear momentum?

Anyway, I think we are making some progress. So suppose we have the mass of the cannon, the mass of the ball, and the speed of the ball immediately after it is fired while resting on a frictionless surface. We can use conservation of linear momentum in order to find the velocity of the cannon in this event.

But when the cannon is fixed, we have to equate the cannonball's kinetic energy to the total kinetic energy in the first event in order to solve for velocity. Do you agree?

 

[cantgetaname]

You do not need to include the Earth in the system. kinetic energy is calculated with respect to a stable reference point. In relation to my desk, my laptop has no kinetic energy. In relation to the moon, it has very large kinetic energy. Therefore, just consider the velocity of both the cannon and cannonball to be relative to the Earth's surface. I think that is simplest.

Agree.

I have been trying to hold it up to criticism. Is that the type of equation you meant when you said to use conservation of linear momentum?

I was wrong (and lost) when I said that. Conservation of momentum along can't lead us anywhere.

But when the cannon is fixed, we have to equate the cannonball's kinetic energy to the total kinetic energy in the first event in order to solve for velocity. Do you agree?

Sounds correct.

 

[BradP]

Okay, thanks for helping me think it through :)