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Momentum questions

  1. Dec 9, 2004 #1
    [SOLVED] momentum questions

    I've started reviewing topics for my semester final in my physics class and I realized that there were some fundamental questions about momentum in my textbook, which I couldn't really do...

    Here they are:

    -You are standing at the center of a stationary boat (1.8 m from shore) and you start walking towards the shore. Neglect frictional force on the boat from the water. Once you reach the end of the boat, you jump towards the shore. Does the motion of the boat change when you reach it?

    Common sense tells me no; but I'm not really sure. If there's no external force on the system, the common center of mass of the system should remain stationary and when you suddenly stop moving, the boat should stop moving as well. But doesn't the ground provide an external force in this system? Does this explain why the boat's motion is unaffected when you hit the ground (and is it)?

    -If all the cars in the world started driving eastward, what would happen to the length of the day?

    I assumed that the earth-cars system had no external forces on it. The earth normally rotates eastward; so if suddenly the cars of the earth-cars system started moving eastward, should the earth's rotation slow down, and days get longer? Is this correct?

    -The momentum of the center of mass of a system is the sum of the momenta of the individual particles, but the same is not true for kinetic energy. Why not?

    Is this because (if there is no external forces on the system) mechanical energy is constant, and since internal potential energy in a system can change, so can kinetic? Or am I going about this in a totally wrong way?

    -An asteroid 100 km in diameter is discovered to be on a collision course with Earth. In a show of international cooperation and mutual disarmament, nations lanch all their nuclear missisles at the asteroid and succeed in blowing it to smithereens. Will the center of mass of the asteroid still hit Earth?

    I think that the answer to this question is no, because if you think of the asteroid by itself, there is an external force on it (from the nuclear missles) and this can change the momentum of the center of mass. Is this incorrect?

    -An hourglass is inverted and placed on a scale. Compare scale readings a. before sand begins to hit the bottom; b. while sand is hitting the bottom; c. when all the sand is on the bottom.

    My answers were A. Simply the mass of the hourglass * 9.8 m/s^2; B. A larger value; C. the same answer as A. When the particles of sand hit the scale, don't they undergo a change in momentum (impulse) that has the same result as an actual force being applied to the scale over a small duration of time? When the sand particles are not hitting the scale, and there are no collisions taking place, I assumed weight was just mg and there was no impulse.

    Any help is appreciated in advance. :smile:
     
    Last edited by a moderator: Dec 9, 2004
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  3. Dec 9, 2004 #2

    Andrew Mason

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    This is a centre of mass / momentum conservation problem. The boat has to move away from the shore as you walk toward the shore in order to preserve center of mass (0 momentum). If you stop when you reach the end of the boat, the boat will stop. When you jump toward shore with speed s, the boat receives an impulse equal to your mass x s. By ground do you mean water? The boat is only affected by the water if there is friction, which there isn't.

    Your answer is correct. The earth would decrease its rotation slightly but only until the cars stopped. Eventually the cars would hit an ocean and stop at which time the earth's original rotation would resume.

    The reason is that kinetic energy is not a vector quantity but is a function of a vector quantity. It is a scalar function of the sum of all the momenta of the individual particles. To find KE of the system you have to add all the momentum vectors first to determine the speed of the centre of mass and then determine KE from that speed.

    Mathematically:
    [tex]K(|\vec a+\vec b+\vec c|) \ne K(|\vec a|) + K(|\vec b|) + K(\vec |c|)[/tex]
    where K(x) is the function [itex]K(v) = \frac{1}{2}mv^2[/itex]

    AM
     
  4. Dec 9, 2004 #3

    cepheid

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    Man, It's times like these that I realise I don't understand things/remember my first year stuff as well I would like :rofl:

    I think I can help you with that last part: Why is the momentum of the centre of mass of a system of particles equal to the sum of the momenta of the individual particles, but not so for the kinetic energy? First , let's define a coordinate system with three generalised position coordinates representing the components of the position vector

    [tex]\vec{r}_{j}[/tex]

    of the jth particle in the system. The components depend on the particular coordinate system you choose...if it's Cartesian, they are
    [tex](r_{jx}, r_{jy}, r_{jz})[/tex]

    But we're just using generalised coordinates...it doesn't matter what kind. Of course, the position of this particle is a function of time!

    [tex]\vec{r}_{j}(t)[/tex]

    Its time derivative (i.e. the velocity of the particle) is written as

    [tex] \dot{\vec{r}}_j[/tex]

    I went over the notation in detail, just in case it was different from what you are using. So, the definition of the centre of mass of the system of particles is:

    [tex] \vec{R}_{com} = \frac{\sum{m_j \vec{r}_j}}{\sum{m_j}} [/tex]

    The denominator is just the sum of the masses of the particles in the system--it is the total mass M. So how do we get the momentum of the centre of mass? We need its velocity. Differentiate both sides of the equation. Note that 1/M on the right hand side is a constant and comes outside the derivative. Furthermore, the derivative of a sum of functions is the sum of the derivatives of each, so the d/dt can go inside the summation symbol...and the derivative of a constant times a function is the constant times the derivative, so the d(mv)/dt of each particle becomes m*dv/dt (i.e. m*(r "dot")):


    [tex] \dot{\vec{R}}_{com} = \frac{1}{M} \sum{m_j \dot{\vec{r}}_j} [/tex]

    So that's the velocity of the centre of mass expressed in terms of the velocities of the individual particles. To obtain the momentum of the c.o.m., multiply its velocity by the total mass M:

    [tex] M \dot{\vec{R}}_{com} = \sum{m_j \dot{\vec{r}}_j} [/tex]

    What does this equation say? That the momentum of the centre of mass of the system of particles is equal to the sum of the individual momenta of the particles. QED!

    Mathematically, this turned out to be true because momentum depends linearly on velocity. Kinetic energy depends on the square of the velocity...it does not increase linearly with v. To see these effects explicity, perform the same procedure again. Don't worry about the (1/2)M...here is the velocity of the c.o.m *squared*:

    [tex] (\dot{\vec{R}}_{com})^2 = \left[\frac{1}{M} \sum{m_j \dot{\vec{r}}_j}\right]^2 [/tex]

    Right away you can see that when you square that sum on the right hand side, you're going to get a big mess involving cross terms: You're not just going to get the sum of the kinetic energies of each particle.

    Looking at your explanation, it sort of makes sense, but is overly complicated...you're saying that in a given object, the thermal motions of the constituent particles are random..they don't necessarily correspond to the "bulk" (macroscopic) motion of the object. Each particle is zipping around, and is attracted to its neighbours, so as it moves farther apart/closer to other particles, it's potential/kinetic energy increases/decreases, and none of these changes in internal energy are relevant to the energy of the macroscopic system...fair enough. I just think that this mathematical proof shows that even without resorting to examining the microscopic situation, we can still prove the statement. Even if the particles were all moving sort of in unison, the kinetic energy of the c.o.m. would still not be equal to the sum of the kinetic energies of the particles. We could both be horribly mistaken, so I'd appreciate others' input.
     
  5. Dec 9, 2004 #4

    Andrew Mason

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    I am afraid it will. But the theory is that all the small pieces will burn up and not reach the earth. The centre of mass cannot change (if you ignore the impact of the missiles themselves, which is reasonable given the size of the asteroid) because there is no external force applied: it is just the asteroid exploding.

    Your hourglass answers are all correct.

    AM
     
  6. Dec 9, 2004 #5
    Thanks so much for the help! ::smile::

    Cepheid, your mathematical explanation makes more sense to me than Andrew Mason's (no offence!). I think I understand the problem now.

    So the velocity of the center of mass does change, but it's so slight that it doesn't really make a difference?

    It would be unfortunate for earth, if that ever were to happen. Hopefully never in my lifetime. :smile:
     
  7. Dec 9, 2004 #6

    Andrew Mason

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    The impact of the missiles will slow it down a tiny amount but won't change its direction at all if the asteroid is moving directly toward the earth. In that case there is not even a slight change in the direction. If they intercept it while it is moving at an oblique angle, the impact of the missiles will nudge its direction a tiny amount but not enough to make it miss the earth.

    One strategy that could work would be to have the missiles detonate on the side of the asteroid but not explode it.

    AM
     
    Last edited: Dec 9, 2004
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