Mechanics> Gravitation...
2 masses of mass 10 kg and 20 kg are kept 1 meter apart and released assuming that only mutual gravitional forces are acting . Find the speed of particles at separation decreases 0.5 meter
1 AnswersDecrease in gravitational P.E of system=Increase in K.E
G.10.20/0.5=1/2.10.v^2+1/2.20.u^2 (v-speed of mass 10 kg.,u-speed of mass 20 kg.)
=>400G=5v^2+10u^2..................(1)
Now only internal conservative force is acting thus velocity of C.O.M=0.
v(com)=(10v+20u)/10+20
=>v=2u(taking magnitudes only)..........(2)
solving we get,v=6*10^(-5)m/s
u=3*10^(-5)m/s (approx. values)
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