Register to reply 
A question for which I *should* know the answerby ptabor
Tags: None 
Share this thread: 
#1
Nov2905, 02:09 PM

P: 108

suppose you're firing a cannon in space:
of course, assuming a stationary cannon in the begining, the momentum of the ball must be equal and opposite that of the cannon itself. Now let's say I double the mass of the cannon, of course that is going to affect the velocity of the cannon, but an undergrad just asked me if the velocity of the cannon ball would also change. he claims that if you increase the mass of the cannon, the velocity of the projectile must also increase. I'm not so convinced that his reasoning is correct. I can see that it could possibly be true, but I don't see why it would necessarily be so. my hunch is that the velocity of the projectile would be (ignoring gravitational interactions) constant and that the velocity of the cannon must change to conserve momentum. If I am in error here, please light the way. 


#2
Nov2905, 02:59 PM

P: n/a

If the mass of the cannon ball is called
[tex]m_1[/tex], and the mass of the cannon is [tex]m_2[/tex], then the impulse that acts on the cannon ball must be equal, but negative than that acting on the cannon, so... [tex]m_1v_1 = m_2v_2[/tex] (v1 is the velocity of the cannon ball, v2 is the velocity of the cannon). so if you change the value of any of the mass or velcities, then the value of velocity of the cannon must change since they are dependant on each other. Another way to think of it is to say that the position of the centre of mass of the two objects stays in the same place overall. (I think that all that's right). I hope that is explains your question. 


#3
Nov3005, 11:29 AM

P: 108

What the kid stipulated, and what I didn't believe, is that if you change the mass of the cannon you will change the velocity of the outcoming bullet.
however, after doing the math I am convinced this is correct although it still seems counterintuitive to me. After all, we aren't changing the amount of the explosive used to provide the impulse, and not changing the mass of the bullet, so why should it go faster after receiving the same force? 


#4
Nov3005, 11:42 AM

P: 1,017

A question for which I *should* know the answer



#5
Nov3005, 11:51 AM

Sci Advisor
P: 5,095

Ptabor,
You and El are correct. The main thing to mention is that the cannon ball and force to propel the ball would remain constant. In that case the only thing that can change would be the velocity that the cannon recoils. If you don't hold something constant, how can you do a direct comapre between the two? They're not the same systems any longer. 


#6
Nov3005, 11:59 AM

P: 108

that's what my intuition told me, but the math shows it to be incorrect. unless, of course, my math is wrong. If you spot and error, please tell me as this would reconcile my conscience.
[tex] m_b1v_b1 = m_c1v_c1 [/tex] this is conservation of momentum. [tex] \frac {1}{2}m_b1v_b1^2 + \frac {1}{2}m_c1v_c1^2 = E [/tex] conservation of E Now let's say we change the mass of the cannon: [tex] m_b1v_b2 = m_c2v_c2 [/tex] [tex] \frac {1}{2}m_b1v_b2^2 + \frac {1}{2}m_c2v_c2^2[/tex] Now we can algebrate for v_b2: [tex] v_b2^2 = v_b1^2 + \frac {m_c1}{m_b1}v_c1^2  \frac {m_c2}{m_b1}v_c2^2 [/tex] going back to our expressions for the momentum, we can substitute v_c2 and v_c1, after a little bit of algebra I find: [tex] v_b2^2(1+ \frac{m_b1}{m_c2}) = v_b1^2(1 + \frac {m_b1}{m_c1}) [/tex] now let's say that originally the mass of the bullet is [tex]\frac{1}{10} [/tex] of the mass of the cannon, and then we double the mass of the cannon (so the bullet is now [tex]\frac {1}{20} [/tex] of the mass of the cannon). solving for v_b2 I get [tex] 1.023v_b1 [/tex] as far as I can see I haven't made any mistakes  although the answer disturbs me. 


#7
Nov3005, 01:32 PM

P: 8

perhaps considering the limit as the mass of the cannon becomes very large will clarify things?



#8
Nov3005, 01:41 PM

P: 108

yeah, I realize that in the large limit that the results will be virtually the same. I had considered that and of course took solace in it, but what surprises me is that it should make any sort of difference at all.
I had initially believed that the velocity of the bullet was completely independent of the mass of the cannon, which is evidently not the case. Although in the large limit this dependence is washed out. Actually, the whole question came up because the kid asked what would happen if instead of a cannon we used a laser. Clearly the velocity of the outgoing photon is not going to change, and his question is what would give. to me, it seems fairly obvious that the frequency of the emitted light would have to change (the energy of the outgoing light is hf). Am I simply bent out of shape for nothing? 


#9
Nov3005, 02:31 PM

P: 7

You could go about it this way:
You have two masses, m_1 and m_2, compressing a spring between them. Take it in three different scenarios; 1) m1 >> m2 2) m1 = m2 3) m1 << m2 In scenario 1, mass 1 will remain almost stationary while mass two moves off with significant velocity. In scenario 2, the masses will move off with equal velocity In scenario 3, mass 2 will remain almost stationary, mass 1 will move off. Because it's the same spring in each scenario (not counting the mass or inertia of the spring), you know you have a constant energy potential and therefore a consistent force action against the masses each time. Instead of viewing one as 'firing' the other, you could simply view it as it really is: gunpowder explosion pushes each object away with equal momentum. Let's take it a step further, and assume the spring is elastic in the sense that no energy is lost. We should, at some point, be able to establish a ratio between the original energy of the system and the masses to find velocity. I did it, but I don't have time to post the whole thing right now. Tomorrow. 


Register to reply 
Related Discussions  
Is this the answer to this question?  Introductory Physics Homework  1  
Please help me answer this question.  Introductory Physics Homework  9  
Please answer my question  Introductory Physics Homework  2 