En ingles
So in the sector of the side we have a sphere of mass A that is hanging in the wire of length R
and the sphere is about to be hit by mass mB and velocity VB so after getting hit by the particle
mB the particle sticks together with the sphere for a single body of total mass mA+mB and will
now gain a velocity of v2 and due to this velocity the body mA+mB in blue will be allow it to
rotate in the form of a circle of radius R and we want to know the minimun velocity of the
particle VB so the body mA+mB will complete a full circle like this, ok?, so the pointed line
should be the trayectory of the body mA+Mb, so we want to find this minimun velocity, ok soo
for starting solving this exercise first we can look at the conservations that can be used so lets
look at the scenario 1 to scenario 2, what kind of conservation we have, so besides of the
gravitation of force there is no extra internal force and the gravitation of force its only pointing
in the Y direction so there is no force acting on the X direction and we know that the particle
mB is pointing in the X direction we can have right before the collision and right after the
collision conservation of the linear momentum in the X direction, so here is linear Px notice we
cannot use conservation of energy from sceneario 1 to 2 because the particle mB sticks
together with the particle mA and this requires, and for the parts of mA and mB to be together
some of the mechanical energy of the scenario 1 has lost in the form of heat to get to scenario
2, so from scenario 1 to 2 we have conservation of linear momentum, so lets see what the
conservation of linear momentum gets us. So first lets observe the linear momentum from
scenario 1 as it will be equal to the linear momentum of the scenario 2, in scenario 1 we only
have the particle of mass mB moving with vB which is the quantity we want to find in the end
of the exercise and in scenario 2 we have the full particle mA+mB travelling with initial velocity
that can be called v2, ok?, we had the relation between, so from the conservation of the linear
momentum we have this relation between the initial velocity of particle B, we have the initial
velocity of body A+B which is equal to mA+mB divided by mB times v2, so this is what
conservation of linear momentum gives us, now lets go back to our sketch, so we had that
from scenario 2 to scenario 3 no mechanical energy is lost and so we have conservation of
mechanical energy, so I put here mechanical energy, ok?, but we don’t, but we cannot imply
conservation of linear momentum because from the moment the body is in the bottom until
the body reaches the top we have the action of gravitation of force and the action of the
tension of the bodies so we will loose some linear momentum due to the external forces doing
the whole time that goes from the bottom to the top of the trayectory, soo, let me put here a
2, we have conservation of mechanical energy, between scenario 2 and 3 so we have that the
mechanical energy of scenario 2 so the kinetic energy of scenario 2 plus the gravitational
energy 2 has to be equal to kinetic energy on scenario 3 plus the gravitational energy of
scenario 3, ok?, so we can always choose where our cero of the gravitational potency is so
because of this I choose it to be here in the initial position of mA+mB just to make the
calculations easier,ok?, so to that here It would be cero and we only had the kinetic energy so
the scenario 2 is mA+mB times v2 squared over 2 and this has to be equal to mA+mB v3
squared over 2 so the velocity of body mA+mB on the top of the trayectory plus the
gravitational potencial so let me read it down here, plus mA+mB times g, so times the height of
the trayectorys and since we choose it to be here the relative height, is cero to the current
point it will be 2 times the radious of this trayectory so the ball to be here the height we will
use that we will use to calculate the gravitational potencial would be 2 R, so times 2 R,ok?, we
can divide both sides of the equation mA+mB, ok?, so we have so that will have the relation
between the velocity at the bottom of the trayectory to the velocity at the top of the
trayectory, so v2 squared over 2 is equal to v3 squared ove 2 plus g times 2 R,ok?, now lets skip
