Question

Two small bodies of masses 10 kg and 20 kg are kept a distance 1.0 m apart and released, assuming that only mutual gravitational forces are acting find the speeds of the particles when the separation decreases to 0.5 m.

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Solution

The linear momentum of 2 bodies is 0 intially. since gravitational force is internal, final momentum is also zero. So, (10kg)v1=(20kg)v2 orv1=2v2 Since P.E.is conserved, Initial P.E. =−6.67×1011×10×201=−13.34×10−9J When separation is 0.5 m ⇒13.34×10−9+0=−13.34×10−612+(12)×10v21+(12)×20v22⇒−13.34×10−9=26.68×10−9+5v21+10v22⇒−13.34×10−9=26.68×10−9+30v22⇒v22=−13.34×10−930=4.44×10−10⇒v2=2.1×10−5m/sSo,v1=4.2×10−5m/s

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