Answer:
[tex]v_1 =\sqrt{\dfrac{16GM}{15R}}[/tex]
[tex]v_2 =\sqrt{\dfrac{4GM}{15R}}[/tex]
Explanation:
given,
mass of asteroid 1 = M
mass of asteroid 2 = 2M
radius of two asteroid = R
Distance between the asteroid = 10 R
Speed of the asteroid before collision = ?
using conservation of momentum
M u + 2M u' = M v₁ + 2 M v₂
initial speed of asteroid is equal to zero
0 = v₁ + 2 v₂
v₁ = -2 v₂
using conservation of momentum
initial potential energy is converted into potential energy and the kinetic energy of both the asteroids.
[tex]\dfrac{GM(2M)}{10R}=\dfrac{GM(2M)}{2R}+\dfrac{1}{2}Mv_1^2 + \dfrac{1}{2}(2M)v_2^2[/tex]
[tex]\dfrac{GM(2M)}{10R}-\dfrac{GM(2M)}{2R}=\dfrac{1}{2}M(-2v_2)^2 + \dfrac{1}{2}(2M)v_2^2[/tex]
[tex]6v_2^2 = \dfrac{8GM}{5R}[/tex]
[tex]v_2 =\sqrt{\dfrac{4GM}{15R}}[/tex]
now,
[tex]v_1 =-2\sqrt{\dfrac{4GM}{15R}}[/tex]
[tex]v_1 =\sqrt{\dfrac{16GM}{15R}}[/tex]
hence, the velocity of asteroid are
[tex]v_1 =\sqrt{\dfrac{16GM}{15R}}[/tex]
[tex]v_2 =\sqrt{\dfrac{4GM}{15R}}[/tex]