Two identical rods have spheres of equal radii attached to one end. Rod A has a
ID: 1415046 • Letter: T
Question
Two identical rods have spheres of equal radii attached to one end. Rod A has a hollow sphere attached to it, while the rod B's sphere is solid, and both spheres have the same mass. The rods are both pivoted about their free ends at a common point, and are released from a horizontal orientation so that the spheres swing down. The release of the rods are timed such that the two spheres collide head-on (nearly) at the vertical. If the spheres stick together at the collision, which of the following are true? A was released before B, and A's motion will change direction on impact. A was released before B, and B's motion will change direction on impact. A was released after B, and A's motion will change direction on impact. A was released after B, and B's motion will change direction on impact. Both rods were released at the same time, and the rods will stop on impact. Both rods were released at the same time, and the rods will stop on impact.Explanation / Answer
as both the spheres have same mass and held at same height, they have same potential energy at the beginning.
just before collision, they again have the same potential energy.
as change in potential energy is converted to kinetic energy, both the spheres will have same kinetic energy just before collision.
kinetic energy=0.5*mass*speed^2
as both the spheres have same mass, then they will have same speed just before collision.
now, as there is no external force acting on the spheres, total momentum before and after the collision will be conserved.
as before collison, the spheres have momentums in opposite direction and as momentum=mass*speed, so spjheres will have equal and opposite momentums
==>total momentum of the system before collision=0
hence total momentum after collision=0
as the masses stick together, speed of the system=0
hence option E is correct.
they are released at the same time and rods will stop after impact.
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