Question

A particle of mass m moves on the axis of a ring of radius r and a mass capital m if the particle at rest is released ftom p then its kinetic energy at centre will be t

Answer 1

$KE_f=\frac{G.m.M}{2R}$ will be its kinetic energy at center

Explanation:

Given :

mass of particle = m

radius of the ring, = r

mass of the ring = M

$\int{du}=-\int{\frac{G.m.dM}{2R} }$

where dM = elemental mass of the ring.

$U=-\frac{G.m.M}{2R}$

According to the conservation of mechanical energy:

$U+KE=constant$

$U_i+KE_i=U_f+KE_f$

Initially the particle was at rest.

$-\frac{G.m.M}{2R} +0=-\frac{G.m.M}{R}+KE_f$

$KE_f=\frac{G.m.M}{2R}$

TOPIC: kinetic energy, gravitational potential enenrgy

https://brainly.in/question/7671019