First slide

Gravitaional potential energy

Question

A point P lies on the axis of a ring of mass M and radius a, at a distance a from its centre C. A small particle starts from P and reaches C under gravitational attraction only. Its speed at C will be

Moderate

A

$\large \sqrt \frac {2GM}{a}$
B

$\large \sqrt {\frac {2GM}{a}\left ( 1-\frac {1}{\sqrt2} \right )}$
C

$\large \sqrt {\frac {2GM}{a}\left ( \sqrt 2-1\right )}$
D

zero

Solution

Let mass of particle be m
$\large u_i=-\frac {GMm}{\sqrt 2a}$ , $\large u_f=-\frac {GMm}{a}$
K.E. = u_i - u_f(Increase in K.E. = decrease in P.E.)
$\large \frac 12mV^2=\frac {-GMm}{\sqrt2 a}+\frac {GMm}{a}$
$\large \frac{{{V^2}}}{2} = \frac{{GM}}{a}\left( {1 - \frac{1}{{\sqrt 2 }}} \right)$
$\large V = \sqrt {2\frac{{GM}}{a}\left( {1 - \frac{1}{{\sqrt 2 }}} \right)}$

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