Answer:
The centripetal force on body 2 is 8 times of the centripetal force in body 1.
Explanation:
Body 1 has a mass m, and its moving in a circle with a radius r at a speed v. The centripetal force acting on it is given by :
Body 2 has a mass 2m and its moving in a circle of radius 4r at a speed 4v. The centripetal force on body 2 is :
So, the centripetal force on body 2 is 8 times of the centripetal force in body 1.
Answer:
Explanation:
using the law of the conservation of the linear momentum:
where is the inicial momemtum and
is the final momentum
the linear momentum is calculated by the next equation
P = MV
where M is the mass and V is the velocity.
so:
where m is the mass of the proton and is the velocity of the proton after the collision,
is the mass of the nucleus and
is the velocity of the nucleus after the collision.
therefore, we can formulate the following equation:
m(270 m/s) = m + 14m
then, m is cancelated and we have:
270 = +
This is a elastic collision, so the kinetic energy K is conservated. Then:
and
Kf =
then,
=
here we can cancel the m and get:
=
now, we have two equations and two incognites:
270 = +
(eq. 1)
=
in the second equation, we have:
36450 = (eq. 2)
from this last equation we solve for as:
and replace in the other equation as:
270 = 14
so,
Vp is negative because the proton go in the -i hat direction.
Finally, replacing this value on eq. 1 we get: