Two particles of masses $20 k g$ and $10 k g$ are initially at a distance of $1.0 m$ . Find the speeds of the particles when the separation between them decreases to $0.5 m$ , if only gravitational forces are acting.

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Solution

Since only internal forces acts, center of mass must remain stationary,

$20 v_{1} + 10 v_{2} = 0$

$v_{2} = - 2 v_{1}$

Also, conserving total energy of the system,

$\frac{1}{2} 20 v_{1}^{2} + \frac{1}{2} 10 v_{2}^{2} - \frac{G (10) (20)}{0.5} = - \frac{G (10) (20)}{1}$

$30 v_{1}^{2} = 200 (6.67 X 10^{- 11})$

$v_{1} = 2.11 X 10^{- 5} m / s$

$| v_{2} | = 2 | v_{1} | = 4.22 X 10^{- 5} m / s$

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Two particles of masses 20kg and 10kg are initially at a distance of 1.0m. Find the speeds of the particles when the separation between them decreases to 0.5m, if only gravitational forces are acting.

Since only internal forces acts, center of mass must remain stationary,20v1+10v2=0v2=−2v1Also, conserving total energy of the system,1220v21+1210v22−G(10)(20)0.5=−G(10)(20)130v21=200(6.67X10−11)v1=2.11X10−5m/s|v2|=2|v1|=4.22X10−5m/s

Two particles of masses $20 k g$ and $10 k g$ are initially at a distance of $1.0 m$ . Find the speeds of the particles when the separation between them decreases to $0.5 m$ , if only gravitational forces are acting.