Two identical beads each of mass m can slide freely on a ring of mass M lying on smooth horizontal surface. The masses are brought to diametrically opposite positions and both are imparted with...
Two identical beads each of mass m can slide freely on a ring of mass M lying on smooth horizontal surface. The masses are brought to diametrically opposite positions and both are imparted with same tangential speed v in same direction. Find the kinetic energy of the ring when beads are about to collide.
The momentum of a closed system is a constant (as the law of conservation of momentum states). The full momentum of a system is the sum of momentums of all systems' bodies. Note that momentum is a vector quantity, mass multiplied by its velocity, and it is conserved as a vector, including direction.
Our system consists of a ring and two beads. Consider an axis along the initial speeds of beads (they are parallel), and projection of velocities on this axis. Momentum is a constant vector, so the projection of momentum on any axis is also a constant. Also it is evident from the symmetry of the situation that a ring will move only along the same axis. Denote the speed of a ring just before the collision as `V.`
The starting momentum's projection is `0+mv+mv=2mv` (a ring is in rest). Just before the collision the beads' velocities are perpendicular to the chosen axis, so their projections are zero. Thus the momentum just before the collision is `MV.`
This way we obtained a simple equation, `2mv=MV,` or `V=(2m)/M v.` This is the answer.