Why is momentum conserved in each case?

Discussion in 'Physics & Math' started by kingwinner, Nov 26, 2006.

  1. kingwinner Registered Senior Member

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    796
    I don't understand why momentum is consesrved in each of the 2 cases below:

    Case 1) A 10g ball and a 30g ball are placed in a tube with a massless compressed spring between them. When the spring is released, the 10g ball flies out of the tube at a speed of 6.0m/s. With what speed does the 30g ball emerge from the other end?

    Case 2) Bob sees a staionary cart 8.0m in front of him. He then accelerates steady at 1.0m/s^2, jumps on the cart, and roll down the street. Bob has a mass of 75kg and the cart's mass is 25kg. What is the cart's speed just after Bob jumps on?

    I know perfectly how to solve these 2 problems, just apply the momentum conservation law, the mechanical calculations are easy, but for momentum to be conserved, the system must have no net external forces. Now, it's like I'm blindly applying momentum conservation law without understanding it because I'm not sure whether momentum is conserved in each case, and if so, why it's conserved.

    Can somebody kindly explain WHY and HOW momentum is conserved in each case and in which "system" is the momentum conserved in each case? (please don't solve the problems because I know how to do so...) Any help is greatly appreciated!

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  3. przyk squishy Valued Senior Member

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    Do you know why momentum is conserved in general?

    (Reminder: think about F = dp/dt and Newton's third law)
     
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  5. §outh§tar is feeling caustic Registered Senior Member

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    4,832
    Momentum is conserved when no external forces act on the system. Specifically, momentum in the X direction is conserved when no external forces act on the system in the X direction. Verify that there are no such external forces in your cases.

    Once you understand this part, you can fiddle with what przyk said and establish it mathematically.
     
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  7. alain du hast mich Registered Senior Member

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    "Case 1) A 10g ball and a 30g ball are placed in a tube with a massless compressed spring between them. When the spring is released, the 10g ball flies out of the tube at a speed of 6.0m/s. With what speed does the 30g ball emerge from the other end?"

    Assuming you ignore friction and all that, it would be at 2m/s. So momentum left would be 10 x 6 and momentum right would be 30 x 2, so when you add them together it would have a combined momentum of 0

    "Case 2) Bob sees a staionary cart 8.0m in front of him. He then accelerates steady at 1.0m/s^2, jumps on the cart, and roll down the street. Bob has a mass of 75kg and the cart's mass is 25kg. What is the cart's speed just after Bob jumps on?"

    I am much too lazy to work out Bobs speed when he jumps, but the speed of the cart + Bob would be 3/4 of Bobs original speed, as it has 4/3 the mass
     
  8. Mosheh Thezion Registered Senior Member

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    2,650
    conserved????

    where else would it go?

    -MT
     
  9. kingwinner Registered Senior Member

    Messages:
    796
    I know that when the net external force on a system is zero, then the total momentum of the system is conserved. But I find that the cases I mentioned above seem to have many complicated forces before, during and after the interaction, how can I know if the net external force is zero or not?
     
  10. James R Just this guy, you know? Staff Member

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    32,982
    First, you need to define what "system" you're talking about. Then, determine what is inside the system and what is outside. Is there any net force acting on the system from the outside? If so, momentum won't be conserved. If not, momentum is conserved for the system.

    Let's take a look.

    Potentially relevant objects here include: the 10g ball, the 30g ball, the tube, the spring, the Earth.

    What do you take to be "the system" in this case? Which interactions are between internal parts of the system and which are external? Is there an external net force on the system?

    In this case, relevant objects are: Bob, the cart, the street.

    Which of these objects make up a "system"? Which system is most convenient to use to solve the problem? [Hint: the system with no external forces on it during the relevant parts of the motion will be the one for which momentum is conserved.]
     
  11. kingwinner Registered Senior Member

    Messages:
    796
    Hi James R

    For (1), I would pick the system to be the 2 balls
    For (2), I would pick the system to be Bob and the cart
    But in both cases, gravity is an external force, spring force, normal force are also external forces, and I am not sure if the net external force would be zero or not...
     
  12. kingwinner Registered Senior Member

    Messages:
    796
    Another problem relating to spring forces and Newton's Laws...
    3) "A spring is attached to a 2.0kg block. The other end of the spring is pulled by a motorized toy train that moves forward at 5.0cm/s. The spring constant is 50N/m and the coefficient of static friction between the block and the surface is 0.60. The spring is at its equilibrium length at t=0s when the train starts to move. When does the block slip?"

    This is a sample problem done in my textbook, in the solution it says "As the right end of the spring moves, streching the spring, the spring pulls backward on the train and forward on the block with equal strength..."

    The bolded part is what puzzles me. How is it possible that the forces have equal magnitude? I mean, what justifies that they have the same magnitude? I can't follow this logic. The spring is moving (maybe some parts are accelerating, too)...this is complicated...
     
  13. §outh§tar is feeling caustic Registered Senior Member

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    4,832
    Remember what I said, net external forces in a direction. In this case, gravity is offset by a normal force. And so 'momentum' in that direction is trivially conserved. But what you're interested in is momentum in the direction of motion. Are there any net external forces in the direction of motion?
     
  14. §outh§tar is feeling caustic Registered Senior Member

    Messages:
    4,832
    You can think of it by contradiction. If the spring pulled on the train harder than it pulled on the block (or vice versa) then there would be a net force (and also an acceleration) on the spring. But how can the spring be accelerating if the thing it is connected to is not accelerating? It is like saying you are accelerating but your shirt is not accelerating (to put it a little roughly).
     
  15. James R Just this guy, you know? Staff Member

    Messages:
    32,982
    kingwinner:

    But that would make the spring "external" to the system, so any forces the spring exerts on the balls would be external forces. Energy and momentum are no longer conserved for "the system".

    That is a reasonable choice for the part of the problem where Bob jumps onto the cart. The external forces due to the Earth are gravity and the normal reaction forces on Bob and the cart, but they cancel each other out, giving no net external force. Therefore, you can use conservation of momentum.

    Newton's Third Law: For every force, there is an equal and opposite force acting on a different object.

    If the train pulls on the spring with force F, then the spring pulls back on the train, also with force F.
     
  16. kingwinner Registered Senior Member

    Messages:
    796
    (1) So how can I solve this problem if momentum is not conserved?

    (3) "If the train pulls on the spring with force F, then the spring pulls back on the train, also with force F" <--I agree with this

    What I am having problem with is the following:
    "As the right end of the spring moves, streching the spring, the spring pulls backward on the train and forward on the block with equal strength..."
    Why are the magnitudes of the spring force on the train and the spring force on the block equal?
     
  17. James R Just this guy, you know? Staff Member

    Messages:
    32,982
    Pick a better system!

    Use the blocks plus the spring as the system. There is no net external force on that system, so total momentum is conserved.

    Well, they aren't equal as long as the spring is extending its length. Once the spring has reached a constant extended length, then they will be equal. Obviously, to extend the spring you need to exert a bigger force on one end than on the other end.
     
  18. kingwinner Registered Senior Member

    Messages:
    796
    (1) But so far I assumed massless spring approimation for every problem I did. So can the spring be considered out of the system? If not, how can I find the momentum of the spring

    (3) Why can't they be equal as the spring is extending its length? Shouldn't there be no difference in the 2 forces according to §outh§tar's arguement that the spring is NOT accelerating at any time?

    Thanks!
     
  19. §outh§tar is feeling caustic Registered Senior Member

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    4,832
    I didn't say the spring wasn't accelerating. I asked a rhetorical question for you to think about.

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    Lemme explain

    In the train example, the time derivative of kx is obviously not zero so the acceleration of the end of the spring with respect to some inertial frame (equilibrium, if you like) is inconstant. But that's not relevant..

    What I meant to say was that the end of the spring attached to the train was accelerating with respect to the equilibrium point but that the end of the spring attached to the mass wasn't. Think of a toy train attached to a spring which is attached to a 200 pound bowling ball. Obviously we can have the end of the spring attached to the toy train extending (moving) while the other end attached to the bowling ball remains stationary (friction accounted for). Hence the end attached to the bowling ball is not accelerating though forces are indeed acting on it.

    Let's see how to quantify the example I just gave:

    Until the bowling ball begins to slip, it obviously does not accelerate. Hence the bowling ball can be characterized as an inertial reference frame for the motion of the toy train and the other end of the spring. The end of the spring moves at a CONSTANT speed v (v = the speed of the train). Now how can the spring and, specifically, the train be moving at a constant speed be moving at a constant speed?

    The fancy answer is that the spring is being driven but I don't think you need differential equations to solve this problem..

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    In other words, when we say that the train is moving at constant speed, we mean it is moving at constant speed with respect to the inertial bowling ball. If there wasn't a spring attached to the train, it would be accelerating with respect to the inertial bowling ball. That's a subtle point which I think you missed.

    The main point here is that the force of the train is opposite to the spring force provided by the end of the spring attached to it. Note that I did not say 'equal'.

    The problem should be clearer now. On a spring moving at constant speed, we have the force of the toy train in the -x direction (with a magnitude of kx) and the force of the bowling ball in the x direction (provided by friction). Remember that friction increases up to uN in order to offset a change in position. I hope you can see what the back of your book meant now.

    Hence uN = kx when the 'resistance' of friction has been 'used up'.

    Kinematics should do the rest.
     
  20. kingwinner Registered Senior Member

    Messages:
    796
    3) When you say the spring is moving at constant speed, what does it mean? (at which point of the spring?) When it's stretching, will all parts of the spring be moving at constant speed?


    "As the right end of the spring moves, streching the spring, the spring pulls backward on the train and forward on the block with equal strength..."

    Main question 1: So why can't these 2 foces be equal strength as the spring is extending its length?

    Main question 2: And why will they be of equal magnitude only at the time just before the block slips?


    My textbook's solution is comparing this massles "spring" approximation to massless "string" approximation, but I am not that satisfied with this analogy...for a massless string, the forces pulling on each end of a string on 2 objects are always of equal magnitude (no matter you're accelerating or not), but for massless springs, it can stretch and is not totally the same as massless strings...

    Thanks for explaning!
     
  21. Sarkus Hippomonstrosesquippedalo phobe Valued Senior Member

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    8,435
    The spring is MASSLESS and can not have any momentum.

    This is why they gave it as massless - the same way that maths uses the infamous "inelastic" beams.

    So in effect, the ONLY two things within your system you need to be concerned with are the two balls.
     
  22. Farsight

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    3,492
    Kingwinner: kinetic energy and momentum are two different ways of quantifying a given mass moving at a given velocity. One is to do with "stopping distance", the other to do with "stopping time". The thing is, collisions involve two objects in contact for the same time. That's why momentum is conserved and kinetic energy is not. See ENERGY EXPLAINED for more:

    http://www.sciforums.com/showthread.php?t=60611
     

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