1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Another simple collision problem

  1. Dec 28, 2006 #1
    A sphere of mass M ("Particle 1") collides with one side of a dumbbell ("Particle 2"), which consists of two spheres each of mass M/2 separated by a rod, whose mass we will ignore. We will treat the spheres as being point particles. Particle 1 is moving at speed V in the +Z direction, and the direction of the collision impact is perpendicular to the dumbbell, and on a line going through the center of mass of that side of the dumbbell (Sphere A).

    Assume no loss of energy due to heat of compression (internal friction), and a purely elastic, instantaneous collision.

    The question is, what are the resulting movements of the two particles (sphere and dumbbell, each of mass M)?

    Relevant equations are:

    Momentum p = mv

    Kinetic energy K.E. = 1/2 mv2

    These equations should apply both to the linear components and the angular components, since we're treating the spheres as point particles which will revolve around each other once the dumbbell starts spinning, and we can substitute the instantaneous linear velocities for the angular velocity (and dispense with the dumbbell length and 2 Pi*Omega).

    Attempted solution
    First consider what happens during the instant of collision. I believe we can treat the collision as involving only Particle 1 and Sphere A (half of the dumbbell), because initially Sphere B remains stationary (and merely rotates a bit), while Sphere A moves solely in the +Z direction, along with Particle 1.

    If this is the case, using conservation of momentum and energy, we can calculate that Sphere A moves at 4/3 V, and Particle 1 moves at V/3 (both in the +Z direction).

    The hard part is what comes next. I figure that Sphere A starts pulling Sphere B along with it, since they're connected by a rod, and that the dumbbell's subsequent motion involves linear velocity in the +Z direction as well as rotational motion in the plane defined by the initial velocity and the dumbbell rod. Using both conservation equations, and the quadratic formula, I came up with seemingly nonsensical values for the linear and rotational velocities: one was 0.911 V (OK so far), but the other was -0.244 V. While I can at least assign relevance to a linear velocity of -0.244 (in the -Z direction), there's no way that either particle would move that way, and there's no meaning to be attached to a negative rotational speed.

    Analytically, I would think that we could imagine that the initial movement of the dumbbell, when Sphere A moves at 4V/3 and Sphere B is stationary, consists of a 2V/3 linear (Center of Mass) velocity combined with a 2V/3 instantaneous "tip" angular velocity. The dumbbell spins, and with each rotation, as A is moving forward and B backward, their respective speeds in our frame of reference are 4V/3 and 0. But, while momentum is conserved that way, half the energy is lost somehow, because we've halved the speed, and when we square those halves, we end up with a fourth which, though doubled (being equal in both linear and angular energy), still is missing 50% of the original energy.

    Where am I going wrong?
    Last edited: Dec 28, 2006
  2. jcsd
  3. Dec 28, 2006 #2


    User Avatar
    Science Advisor
    Homework Helper

    You are seem to have forgotten about conservation of angular momentum.

    Your argument about what happens to the other sphere at the instant of impact sounds right, but using conservation of linear and angular momentum and conservation of energy gives you 3 equations for 3 unknowns so you don't need to think out what happens from first principles.
  4. Dec 28, 2006 #3
    Thanks, but I'm not sure I understand you. Are you saying that angular momentum has to be *separately* conserved? If so, that would imply that the dumbbell can't start spinning, because there was no initial spinning motion, which I'm sure cannot be right.

    Also, what are the three unknowns, anyway? Once the collision occurs, I can forget about Particle 1, because it goes on its merry way at speed V/3 in the +Z direction. At that point, I think I've just got the linear and angular velocities of Particle 2 (the dumbbell) to consider, and I've got two equations to describe their relations. Using the one, I can solve for one unknown in terms of the other (using the momentum equation), and substitute that into the 2nd (energy) equation (which is where the quadratic formula came into play), to solve it.
    Last edited: Dec 28, 2006
  5. Dec 28, 2006 #4


    User Avatar
    Science Advisor
    Homework Helper

    Yes angular momentum is seperately conserved.

    If you measure angular momentum about the point of collision, then you are right, the A.M. measured about that point before the collision was zero. After the collision there are two components which add up to zero: the rotation of the dumbell about its own CG, and the the moment of linear momentum of the dumbell's CG.

    Alternatively, if you measure about the centre of the dumbell, the incoming mass has an angular momentum about that point before the collision (and a different angular momentum aftre the collision, of course).

    The speed of Particle 1 after the impact was one of the three unknowns. (The other two are the linear and angular velocities of the dumbell, of course). You found it without considering angular momentum, because you argued that the other end of the dumbbell doesn't move. If the collision had been at an arbitrary point on the rod, then both ends of the dumbbell would have moved and your special argument would not work.
  6. Dec 29, 2006 #5
    Dear AlephZero,

    I can see that, formally, one might consider there to be angular momentum in that sense, but since Particle 1 isn't connected (except at the instant of collision) to that point on Particle 2, I'm not sure what, in reality, this would mean. Are you saying that the formal (imaginary?) reduction of angular momentum of Particle 1 about that point is then transformed into the real angular momentum of Particle 2? That would, at least, give me a separate way to calculate Particle 2's angular velocity.

    I'd always thought of angular momentum as involving real spinning objects, not some mathematical construct that assigns angular momentum to every point in space, even if that space is filled with rectilinearly moving, non-rotating objects.

    In any case, since you seem to think my analysis of this special case is valid, and I'm pretty sure my math is correct, why am I coming up with what seem to be nonsensical answers regarding Particle 2's linear and angular velocities? Perhaps I should detail all the steps of that calculation... (I've already gone through it twice, though, with the same results.) First, I'll try your approach, and see what I get.

  7. Dec 29, 2006 #6


    User Avatar
    Science Advisor
    Homework Helper

    Well, the angular momentum is assigned the objects, not to the space itself!
    Think of angular momentum as analgous to the moment of a force.

    Moment of force = force times perpendicular distance from a point.

    Angular momentum = "moment of momentum" = linear momentum times perpendicular distance from a point. It doesn't matter whether the momentum is part of a spinning object, or a single particle.

    Example: for a dumbell, length L rotation speed w mass M at each end:

    Relative to the centre, the velocity of the ends is wL/2. The linear momentum of each mass is MLw/2. The moment of momentum for both particles is 2.MLw/2.L/2 = (1/2)ML^2w = Iw.

    Whether the particles are joined by a massless rod or not doesn't change either the linear or the angular momentum of the system.
    Last edited: Dec 29, 2006
  8. Dec 31, 2006 #7
    But wouldn't you say that Particle 1 has to actually be connected to, and moving around, that point, rather than just moving past it, for there to be angular momentum? After all, for there to be a moment of force, there has to be contact between the object pushing and the object being pushed.

    If there doesn't have to be a connection to or motion around that point, why pick that point at all? Why not pick the center of Sphere B, which is the point where, in fact, we'd decided that Sphere A pivots around initially anyway, for the brief instant of the collision? Or why not any other point in space?

    But, doing it your way, it seems that I come up with the same conclusion as from my initial motion analysis. Particle 1 begins with linear momentum MV, ends with MV/3, leaving 2MV/3 for Particle 2. Looked at from the standpoint of angular momentum, when Particle 1 is just about to contact Particle 2, it "has" angular momentum MV (which is just the same as the linear momentum), and when it slows to 1/3 V, its angular momentum also decreases by 2/3 MV.

    My motion analysis concluded that the center of mass of Particle 2 (the dumbbell) ended up with 2/3 V linear speed, and its two ends, Spheres A and B, 2/3 V instantaneous angular speed. This gives the right amount of energy, but too much momentum.

    Or, looking at it your way, Particle 1 had initial linear momentum of MV, as well as angular momentum of MV (which still seems bogus to me, since it really is moving only in a straight line). In that case, perhaps it all balances, since the angular momentum of Particle 1 still has 1/3 MV, and Particle 2 has the other 2/3 MV. But, then the energy would be off (if I consider Particle 1 to have angular motion after the collision, when it's still moving rectilinearly).

    I'm hopelessly confused!
  9. Dec 31, 2006 #8


    User Avatar
    Science Advisor
    Homework Helper

    A force has a moment about any point in space. It has a different moment about different points. This is a useful concept, e.g. you can often solve statics problems (which don't involve any actual rotation of anything) by taking moments of the same force about different points. Those points don't have to be on any the objects. For example it's sometimes useful to take moments about the point in space where some of the force vectors intersect which makes their moments of those particular forces zero, of course).

    A couple is a special case of two equal and opposite forces, and in that case the SUM of the moments of the two forces about every point in space is the same, and is called "THE moment of the couple" because it is a unique number.

    The same applies to momentum and angular momentum. The angular momentum of an object is usually different when measured about different points, unless it is rotating and its centre of gravity is not moving.

    As an example of "angular momentum of a particle not rotating about anything", how would you solve problems like a child running (tangential to its circumference) towards a stationary playground roundabout and jumping on it? There is obviously some angular momentum after the "impact" because the roundabout is rotating. Angular momentum is conserved, so where was it before the impact? The only place it could possibly be was in the (linear) motion of the child (not connected to anything rotating yet) because nothing else is moving.

    In that problem, you can't use conservation of energy because the "impact" is inelastic so energy is not conserved. You can't use conservation of linear momentum, because there is an unknown force at the pivot of the roundabout which changes the linear momentum of the system.

    But if you use conservation of angular momentum about the centre of the roundabout, the moment of that unknown force is zero and

    mv.r for the child before impact = m.rw.r + Iw for the child + roundabout after impact

    (m = mass of child, v = initial velocity of child, r = radius of roundabout, I = moment of inertia of roundabout, w = angular velocity after impact).

    In this example there is only one point about which it's useful to take the angular momentum - i.e. the point where the unknown force is applied. In your dumbell example, there are no external forces so you can take angular momentum about any point.
    Last edited: Jan 1, 2007
  10. Jan 1, 2007 #9


    User Avatar
    Science Advisor
    Homework Helper

    First off, apologies for making a mistake in my "roundabout" example (now corrected). I was trying to see it from your point of view and thought you could do it using conservation of linear momentum, but you can't.

    Back to your problem: I agree with your answers, "particle 1" has velocity v/3, the particle it collides with 4v/3, the other end of the dumbbell 0, the centre of the dumbbell has velocity 2v/3 and the rotation speed of the dumbbell is 4v/3L (if it's of length L).

    The energy is conserved if you calculate it correctly.

    If you take the your "two masses" approach, the energy after impact is

    1/2.m.(v/3)^2 for particle 1
    1/2.(m/2).4v/3)^2 for particle 2
    0 for the other end of the dumbbell (zero velocity).

    That is 1/2.m.(v^2/9 + 8v^2/9) = 1/2mv^2.

    Doing it in terms of the motion of the CG of the dumbell, the energy is

    1/2.m.(v/3)^2 for particle 1
    1/2.m.(2v/3)^2 for the translation of the dumbell (total mass, velocity of CG)
    1/2.(1/4)mL^2.(4v/3L)^2 for the rotation of the dumbbell (I = (1/4)mL^2)
    which sum up to (1/2)mv^2, as it should.

    Sorry, I can't follow your argument there, though you already know you came to the wrong conclusion. Did you forget about the (1/2)Iw^2 energy? you didn't seem to mention it anywhere.
  11. Jan 1, 2007 #10
    Thanks, AlephZero! I like the example of the child jumping onto a roundabout. I'm still not clear on the utility of saying linear and angular momentum are separately conserved, when it seems they obviously are interconvertible, and one can easily go between the two by taking the instantaneous (linear) velocity of the parts on the spinning object (idealized points in my example). I guess I can see, though, that as long as one doesn't count one motion as *both* a linear and angular momentum, it's just a matter of bookkeeping, and one should get the right answer. (For some reason I thought you wanted me to count Particle 1's pre-collision imaginary angular motion around the dumbbell's center of mass as both linear *and* angular momentum.)

    I think you may have erred, though, in your above energy calculation of the case occurring long after the collision, i.e., after the dumbbell has begun rotating, because it'd rotate about its center of mass, and not about the center of Sphere B, correct? Well, you could think of it that way *immediately* after the impact, but then B wouldn't be moving at all, so one wouldn't even need to consider its mass. [Besides, I think I have to conclude, from the momentum consideration below, that the dumbbell's center of mass cannot move at 2V/3 after the collision, but must move at something less. IF we were to glue Sphere B directly onto Sphere A just after the collision occurred, *then*, having twice the mass, it'd have to move at half the speed (and there'd be no rotation or angular momentum to consider, of course). But part of that momentum for our dumbbell goes into the spinning motion, and thus cannot also produce a 2V/3 linear velocity for the latter.]

    Let's now look at those momentum calculations. Before the collision, Particle 1 has p=MV. Afterwards, it has MV/3. Immediately after the impact, we can say that Particle 2's momentum lies solely in Sphere A, with a velocity of 4V/3, giving us a p = M/2 * 4V/3 = 2/3 MV. Thus, it adds up at that point. But after it moves a bit and drags Sphere B along with it, rotating in the process, its momentum now consists of the linear momentum of its center of mass, plus its angular momentum. I simply must be wrong to think that the dumbbell's center of mass moves at 2V/3, because if that were the case, it'd have a linear momentum of 2/3 MV, which would account for all the momentum, and any angular momentum would then be an excess. (See above.) But, what linear velocity will it have?

    Using the values obtained from the quadratic equation, where I solved for the linear and angular velocities of the dumbbell, it seems that 0.244 V would have to correspond to the linear velocity (since 0.911V is obviously too much, since it must be less than 2V/3), and the -0.911V value would have to correspond to the instantaneous angular velocity, i.e., the velocity measured in units of linear velocity, not radians per second (and, I'll interpret the negative sign as indicating a CW or CCW direction of rotation).

    Let's see if it checks out. The linear momentum would be 0.244MV, and the angular would be 2 * M/2 * -0.911V = -0.911MV, so the total would be -2/3 MV, not the required +2/3 MV. (I think the quadratic equation actually yielded values of 0.911 and -.244, and I thought I could reverse the negative sign to make it fit the analysis, but apparently not.)

    The energy of Particle 2 would be:

    Linear: 1/2 M (0.244V)2 = 0.029768 MV2

    Angular: 1/2 M (-0.911V)2 = 0.41496 MV2

    Total = 0.44473 MV2

    Along with the 1/18 MV2 (=0.055555 MV2) left in Particle 1, this yields the required 1/2 MV2. But, something about my interpretation of the negative sign is screwy (or I've messed up elsewhere). Like I said previously, there's no way the dumbbell would move in the negative Z direction after being impacted by a particle going in the +Z direction, so I'm still baffled.

    As far as the 1/2 I omega^2 equation, I was using the above-mentioned, instantaneous "linear-angular" velocity, rather than having to introduce the dumbbell length L and speed in radians per unit time.
    Last edited: Jan 1, 2007
  12. Jan 2, 2007 #11


    User Avatar
    Science Advisor
    Homework Helper

    Linear momentum is conserved because (1) force = rate of change or momentum and (2) internal forces are equal and opposite.

    Angular momentum is conserved because (1) moment of force = rate of change of angular momentum and (2) the moments of equal and opposite forces about any point is also equal and opposite.

    Linear and angular momentum are both useful for the same reason that forces and moments are both useful. A body free to move in a 2-d plane has THREE independent motions, not two - translation in X and Y, and rotation. The things that affect these are the component of force in X and Y, and the moment of the forces. In general, you need to consider all three.

    No, I didn't err, because I calculated the energy two different ways (considering each particle separately, and then considering two particles as a rigid body) and both ways gave the same answers and both showed that energy is conserved. I did use the moment of inertia about the CG, and the linear motion of the CG, when calculating the energy of the rigid body - because that's the only way to do it right.

    ... but then the collision would be inelastic, so energy would not be conserved. And the motion of the rest of the system would also be different from the elastic collision case.
  13. Jan 2, 2007 #12


    User Avatar
    Science Advisor
    Homework Helper

    Okay .... I think I finally figured out what's confusing you.

    Think about the motion of a rigid body translating and rotating with constant velocity. We both agree the KE of the whole body is (1/2)Mv^2 + (1/2)Iw^2, the linear momentum is Mv and the angular momentum about the CG is Iw, right?

    BUT... you can't then try to apply those equations to PARTS of the body. They only apply to the whole.

    For example, think of a wheel rolling along a plane (without slipping). The KE and momentum of the whole wheel are constant.

    But if you think about one particle on the rim of the wheel, its energy and momentum isn't constant. In fact its velocity varies between 0 (when it's touching the road) and 2v (when it's at the top).

    The idea of a rigid body is useful because it means you don't have to worry about all that detail in every problem.

    In your problem with only three masses, there's a choice. You can treat it as a particle and a rigid body. Or, at the time of the impact, but not for other times, you an treat it as 3 particles.

    You seem to be trying to mix and match the two methods when interpreting the results. If you treat the dumbbell as a rigid body, it has "translational" and "rotational" KE which add up to the total energy of the body. If you treat it as two particles, then at the time of impact it happens that one particle is at rest and the other is moving, and all the KE and linear momentum is in the moving particle. At other times the energy and momentum split between the particles will be different.

    You seem to be arguing that it doesn't make sense for one particle to have the same energy as the whole body - or the same energy as the "translation" energy of the whole body - or something like that. You can't make that sort of argument.

    The rigid body formulas only tell you about the whole body, not about the separate parts of it.

    And the particle analysis only tells you what happens at the time of impact, and nothing about about other times.
    Last edited: Jan 2, 2007
  14. Jan 2, 2007 #13
    Oops, my mistake. I saw the "4V/3" in the "4V/3L" of your dumbbell angular momentum calculation, and somehow thought you were considering the initial velocity of Sphere A (4V/3). Now I see that you were considering the velocity as 2V/3, but divided it by radius L/2 to get angular velocity in radians per unit time, and (2V/3)/(L/2) = 4V/3L.

    Interesting. Can we zap Sphere A well after the collision, and after it gets far away from Particle 1 (it moves at 4V/3, while Particle 1 moves at only V/3), with a virtually massless amount of "spiderman-glue," the other end of which is attached to our stationary Sphere B, which we have moved to be directly behind Sphere A, and then say that the glue eventually pulls B along in a purely elastic way? Let's imagine that a split second after the Spidey-glue attaches to A, it solidifies into a rigid, virtually massless rod. I would think that their combined velocity would then have to be 2V/3, since their combined mass was M and A's original momentum was 2MV/3.

    If that's reasonable, then I would think we must conclude that the dumbbell can't both twirl (at any non-zero omega) *and* move linearly at 2V/3.

    RE your subsequent post, I'll have to re-think my analysis, and see if I'm doing any improper mixing and matching.
  15. Jan 3, 2007 #14
    Ah, AlephZero, I think I found my problem! It was a conceptual one. I was thinking of momentum as if it were energy -- a scalar, rather than a vector quantity. When I realized that all the linear momentum initially in Particle 1 was used up by the linear momentum of Particles 1 and 2, I figured there was no "oomph" left to create any other motions. But "oomph" is energy, not momentum. I'll now think about momentum as being more of a bookkeeping quantity, in which one can, for example, create lots of twirling motion in one direction out of nothing, so long as one also creates an equal amount of twirling motion in the opposite direction. So long as there's enough energy to do the latter, everything is kosher.

    Incidentally, besides being of use in this problem (giving us another independent equation that can help us solve for the variables), is momentum also of use in inelastic collisions? In other words, is momentum conserved in inelastic collisions, while energy is not?
  16. Jan 3, 2007 #15


    User Avatar
    Science Advisor
    Homework Helper

    Yes. Momentum is conserved any time there are equal and opposite forces, because force = rate of change of momentum. And the action and reaction forces between two objects are always equal and opposite, according to Newton.
  17. Jul 16, 2007 #16
    Last edited: Jul 17, 2007
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Another simple collision problem