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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)?

Momentum p = mv

Kinetic energy K.E. = 1/2 mv

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).

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?

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 mv

^{2}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?

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