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
Answers
Answered by
2
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 .
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 .
Similar questions