Question

Two spherical asteroids have the same radius R. Asteroid 1 has a mass M and asteroid 2 has mass 2M. The two asteroids are released from rest with distance 10R between their centers.What is the speed of each asteroid just before they collide in G, M, and R.

Answer 1

Answer:

$v_{f2}= \sqrt{\frac{4GM}{15R}}$

$v_{f1}= -2v_{f2}= -\sqrt{\frac{16GM}{15R}}$

Explanation:

For this case we know the following info:

$M_1 = M$ represent the mass for the asteroid 1

$M_2 = 2M$ represent the mass for the asteroid 2

$R$ represent the radius for the two asteroids

$d_1 = 10R$ represent the initial distance between the centers

$v_i = 0m/s$ represent the initial velocity for the asteroids

The figure attached shows the situation for this case.

For this case we can use conservation of momentum, and we assume that we have to assume that the final velocities for both asteroid after the collison are different and lets say $v_{f1}$ and $v_{f2}$, if we apply conservation of momentum we got this:

$(M+3M)v_i = Mv_{f1} + 2Mv_{f2}$

And since $v_i =0$ we have:

$0 = M (v_{f1}+2v_{f2})$

$v_{f1}= -2v_{f2}$    (1)

Now we can apply conservation of energy and we have gravitational potential energy at the begin at the end andjust and kinetic energy at the end so we have this:

$U_i =U_f +K_f$

$-\frac{G M *2M}{10R}= -\frac{GM*2M}{2R} + \frac{1}{2} 2M (v^2_{f1}+ v^2_{f2})$

Now we can apply some algebra and we got this:

$2GM^2 (\frac{1}{2R} -\frac{1}{10R}) = \frac{1}{2}M (v^2_{f1}+ v^2_{f2})$

Now we can apply condition (1) and we got this:

$2GM^2 (\frac{8R}{20}) = \frac{1}{2}M ((-2v_{f2})^2 +2(v_{f2})^2)$

$\frac{4 GM^2}{5}= \frac{1}{2} M (6v^2_{f2})$

We can multiply both sides by 2 and cancel one M on both sides and we got:

$\frac{8GM}{5R} = 6 v^2_{f2}$

And we can solve for $v_{f2}$ and we got:

$v_{f2}= \sqrt{\frac{4GM}{15R}}$

Ansd usin condition (1) we got:

$v_{f1}= -2v_{f2}= -\sqrt{\frac{16GM}{15R}}$