Question

Two small bodies of masses 10 kg and 20 kg are kept at 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.

Answer 1

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ANSWER::

Linear momentum of both bodies is 0 at initial state.

Gravitational force in internal.

Final momentum is also 0.

m₁v₁ = m₂v₂

10 x v₁ = 20 x v₂

v₁ = 2 v₂  ........................................................................ Equation (1)

Potential Energy is conserved.

Initial Potential Energy = ( - 6.67 x 10⁻¹¹ x 10 x 20 ) / 1 = -13.34 x 10⁻⁹ J

When separation is 0.5 m,

-13.34 x 10⁻⁹ + 0 = (-13.34 x 10⁻⁹) / (1/2) + (1/2) x 10 v₁² + (1/2) x 20 v₂² ........ Equation (2)

-13.34 x 10⁻⁹ = -26.68 x 10⁻⁹ + 5 v₁² + 10 v₂²

-13.34 x 10⁻⁹ = -26.68 x 10⁻⁹ + 30 v₂²

v₂² = 13.34 x 10⁻⁹ / 30

v₂² = 4.44 x 10⁻¹⁰

v₂ = 2.1 x 10⁻⁵ m/s

So , v₁ = 4.2 x 10⁻⁵ m/s

Hope it helps!

Answer 2

Heya Mate.............................

Here is your answer.............................

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since there is no external force on the system. So total energy will be conserved.

(Total energy) state l =( Total energy) state ll

==> -Gmm'/1 = 0.5 mv^2 +  0.5 mv * 2-Gmm'/0.5

==> -Gmm' = 1/2(mv^2 + m'v'^2) - 2Gmm'

==> 1/2(mv^2 + m'v'^2) = Gmm'

==> (mv^2 + m'v'^2) = 2Gmm'..............................(1)

Here v = velocity of smaller mass(m= 20kg)

v' = velocity of bigger mass( m' = 20kg)

Also momentum of the system will be conserve initial momentum = final momentum

==> O = mv - m'v ( Because both are moving in opposite direction)

==> mv = m'v'

$==> \frac{v}{v'} = \frac{m'}{m} = \frac{20}{10}$

==> v = 2v'......................................(2)

put in v = 2v' in eq. (1)

m(2v')^2 + m'v'^2 = 2G mm'

==>10 * 4v'^2 + 20v'^2 = 2*G (10)*(20)

==> 60v^2 = G 2*200

$==> v'^{2} = (20/3)G$

$= (20/3) * 6.67 * 10^{-11}$

$v' = \sqrt{6.66*6.67*10^{-11}$

$= 2.1 * 10^{-5}m/s$

so v = 2v'

       $=2*2.1*10^{-5} m/s$

$= 4.2*10^{-5} m/s (Answer)$

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Thanks..............................