Question

Two masses m and 2m are initially at rest and the separation between them is r. When they are released, find the velocities of the masses when separation between then becomes 0.5r

Answer 1

initial potential energy stored between m and 2m (P.Ei) = Gm.2m/r
Final potential energy stored between m and 2m ( P.Ef)= Gm.2m/(0.5r)
Let speed of mass m = V1
Speed of mass 2m = V2
Kinetic energy of m and 2m initially = 0 { because bodies are rest intially}
Kinetic energy of m and 2m finally = 1/2mV1² + 1/2 × 2m V2²
We know,
P.Ei + KEi = P. Ef + K.Ef
-Gm.2m/r + 0 =- Gm.2m/(r/2) + 1/2mV1² + 1/2 2mV2²
2Gm²/r = 1/2m(V1² + 2V2²)
4Gm/r = V1² + V2² ----------(1)

Use conservation of linear momentum ,
mV1 + 2mV2 = 0 => V1 = -2V2 --------(2)

Solve equations (1) and (2)
4Gm/r = 5V2² => V2 = √{4Gm/5r}
V1 = -2√{4Gm/5r} , here negative sign shows that first body moves in opposite direction of second body .