A ring of radius R = 8m is made of a highly dense-material. Mass of the ring is m R = 2.7 × 10 9 kg distributed uniformly over its circumference. A particle of mass (dense) m p = 3 × 10 8 kg is placed on the axis of the ring at a distance x 0 = 6m from the center. Neglect all other forces except gravitational interaction. Determine speed (in cm/sec.) of the particle at the instant when it passes through center of ring.:
Text Solution
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(9cm/s)
Sol. According to conservation of momentum –
m R V R = m P v P
where v R and v P are speed of ring and particle in opposite direction, when particle reaches center of ring.
2.7 × 10 9 v R = 3 × 10 8 v P
V P = 9v R
By conservation of energy
– 
+ 0 = – 
+ 1/2 m R V R 2 + 1/2 m P v p 2
GM R × M P 
= 
v P 2 
6.67 × 10 -11 × 2.7 × 10 9 × 3 × 10 8 
= 
v P 2 
v P = 9 cm /sec.
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