NEET-XII-Physics
11: Gravitation
- #7Two 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.Ans : Consider a system of two bodies. The initial linear momentum of the system is zero as the bodies were initially at rest when they were released.
Since the gravitational force is an internal force and the net external force on the system is zero, so by the law of conservation of linear momentum, the final momentum of the system will also be zero.
So, 10`` \times ``v1 = 20`` \times ``v2
⇒ v1 = 2v2 ...(i)
Applying the law of conservation of energy, we have:
Initial total energy = final total energy ...(ii)
Initial total energy`` =\frac{-6.67\times {10}^{11}\times 10\times 20}{1}`` + 0
= -13.34 × 10-9 J ...(iii)
When the separation is 0.5 m, we have:
Final total energy `` =\frac{-13.34\times {10}^{-9}}{1/2}+\left(\frac{1}{2}\right)\times 10{v}_{1}^{2}+\left(\frac{1}{2}\right)\times 20{v}_{2}^{2}...\left(iv\right)``
From (iii) and (iv), we have:
-13.34 × 10-9 = 26.68 × 10-9 + `` 5{v}_{1}^{2}+10{v}_{2}^{2}``
⇒-13.34 × 10-9 = 26.68 + 10-9 + `` 30{v}_{2}^{2}``
`` \Rightarrow {v}_{2}^{2}=\frac{-13.34\times {10}^{-9}}{30}`` = 4.44 × 10-10
⇒ v2 = 2.1 × 10-5 m/s
∴ v1 = 4.2 × 10-5 m/s
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