Question

Two manned satellites approach one another at a relative velocity of v = 0.250 m/s, intending to dock. The first has a mass of mı = 4.00 x 10² kg and the second a mass of m2 = 7.50 x 10 kg. If the two satellites collide elastically rather than dock, what is their final relative velocity? 

final relative velocity: _______ m/s 

Answer 1

1 Collision is elastic Coefficient of Restitution Velocity of separation Velocite of approach 1 - Velocity of separation 0.250 o final relative 0.250 ml Velocity

Answer 2

Because the problem only gives the relative initial speed between the two satellites, you are free to choose one of those satellites to be at rest initially. In the figure, assume the satellite on the left has mass m1" role="presentation" style="box-sizing: border-box; display: inline; line-height: normal; word-spacing: normal; overflow-wrap: normal; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">?1 and is moving at velocity v" role="presentation" style="box-sizing: border-box; display: inline; line-height: normal; word-spacing: normal; overflow-wrap: normal; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">? to the right, whereas the satellite on the left has mass m2" role="presentation" style="box-sizing: border-box; display: inline; line-height: normal; word-spacing: normal; overflow-wrap: normal; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">?2 and is stationary.

An elastic collision conserves both the total momentum and the total kinetic energy, and therefore

ptot,i=ptot,fm1v1,i+m2v2,i=m1v1,f+m2v2,fm1v+0=m1v1,f+m2v2,fm1v=m1v1,f+m2v2,f" role="presentation" style="box-sizing: border-box; display: inline; line-height: normal; font-size: 18px; word-spacing: normal; overflow-wrap: normal; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">?tot,i?1?1,i+?2?2,i?1?+0?1?=?tot,f=?1?1,f+?2?2,f=?1?1,f+?2?2,f=?1?1,f+?2?2,f$$$$

and

Ktot,i=Ktot,f12m1v1,i2+12m2v2,i2=12m1v1,f2+12m2v2,f2m1v2+0=m1v1,f2+m2v2,f2m1v2=m1v1,f2+m2v2,f2" role="presentation" style="box-sizing: border-box; display: inline; line-height: normal; font-size: 18px; word-spacing: normal; overflow-wrap: normal; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">?tot,i12?1?21,i+12?2?22,i?1?2+0?1?2=?tot,f=12?1?21,f+12?2?22,f=?1?21,f+?2?22,f=?1?21,f+?2?22,f$$$$

All the variables in the last expressions for ptot" role="presentation" style="box-sizing: border-box; display: inline; line-height: normal; word-spacing: normal; overflow-wrap: normal; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">?tot and Ktot" role="presentation" style="box-sizing: border-box; display: inline; line-height: normal; word-spacing: normal; overflow-wrap: normal; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">?tot are known except for the final speeds of the first and second satellite, v1,f" role="presentation" style="box-sizing: border-box; display: inline; line-height: normal; word-spacing: normal; overflow-wrap: normal; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">?1,f${\text{,}}_{}$ and v2,f" role="presentation" style="box-sizing: border-box; display: inline; line-height: normal; word-spacing: normal; overflow-wrap: normal; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">?2,f respectively. Solve the ptot" role="presentation" style="box-sizing: border-box; display: inline; line-height: normal; word-spacing: normal; overflow-wrap: normal; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">?tot$p_{\text{tot}}$ expression for v1,f" role="presentation" style="box-sizing: border-box; display: inline; line-height: normal; word-spacing: normal; overflow-wrap: normal; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">?1,f, and then substitute it into the Ktot" role="presentation" style="box-sizing: border-box; display: inline; line-height: normal; word-spacing: normal; overflow-wrap: normal; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">?tot expression to find v2,f" role="presentation" style="box-sizing: border-box; display: inline; line-height: normal; word-spacing: normal; overflow-wrap: normal; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">?2,f

v1,f=m1v−m2v2,fm1" role="presentation" style="box-sizing: border-box; display: inline; line-height: normal; font-size: 18px; word-spacing: normal; overflow-wrap: normal; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">?1,f=?1?−?2?2,f?1$$$$

So,

m1(m1v−m2v2,fm1)2+m2v2,f2=m1v2(m12v2−2m1m2vv2,f+m22v2,f2)+m1m2v2,f2=m12v2(m1m2+m22)v2,f2+(−2m1m2v)v2,f+0=0" role="presentation" style="box-sizing: border-box; display: inline; line-height: normal; font-size: 18px; word-spacing: normal; overflow-wrap: normal; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">?1(?1?−?2?2,f?1)2+?2?22,f(?21?2−2?1?2??2,f+?22?22,f)+?1?2?22,f(?1?2+?22)?22,f+(−2?1?2?)?2,f+0=?1?2=?21?2=0$$$$

Using the quadratic formula,

v2,f=2m1m2v+4m12m22v2−4(m1m2+m22)(0)2(m1m2+m22)=2m1m2v+4m12m22v22m2(m1+m2)=4m1m2v2m2(m1+m2)=2m1vm1+m2" role="presentation" style="box-sizing: border-box; display: inline; line-height: normal; font-size: 18px; word-spacing: normal; overflow-wrap: normal; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">?2,f=2?1?2?+4?21?22?2−4(?1?2+?22)(0)⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√2(?1?2+?22)=2?1?2?+4?21?22?2⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√2?2(?1+?2)=4?1?2?2?2(?1+?2)=2?1??1+?2$$$$

or

v2,f=2m1m2v−4m12m22v2−4(m1m2+m22)(0)2(m1m2+m22)=2m1m2v−4m12m22v22m2(m1+m2)=02m2(m1+m2)=0" role="presentation" style="box-sizing: border-box; display: inline; line-height: normal; font-size: 18px; word-spacing: normal; overflow-wrap: normal; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">?2,f=2?1?2?−4?21?22?2−4(?1?2+?22)(0)⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√2(?1?2+?22)=2?1?2?−4?21?22?2⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√2?2(?1+?2)=02?2(?1+?2)=0$$$$

Now substitute these possibilities for v2,f" role="presentation" style="box-sizing: border-box; display: inline; line-height: normal; word-spacing: normal; overflow-wrap: normal; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">?2,f into the expression for v1,f" role="presentation" style="box-sizing: border-box; display: inline; line-height: normal; word-spacing: normal; overflow-wrap: normal; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">?1,f.

v1,f=m1v−m2(2m1vm1+m2)m1=(1−2m2(m1+m2))v=(m1+m2−2m2m1+m2)v=(m1−m2m1+m2)v" role="presentation" style="box-sizing: border-box; display: inline; line-height: normal; font-size: 18px; word-spacing: normal; overflow-wrap: normal; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">?1,f=?1?−?2(2?1??1+?2)?1=(1−2?2(?1+?2))?=(?1+?2−2?2?1+?2)?=(?1−?2?1+?2)?$$$$

or

v1,f=m1v−m2(0)m1=v" role="presentation" style="box-sizing: border-box; display: inline; line-height: normal; font-size: 18px; word-spacing: normal; overflow-wrap: normal; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">?1,f=?1?−?2(0)?1=?$$$$

The solution with v1,f=v" role="presentation" style="box-sizing: border-box; display: inline; line-height: normal; word-spacing: normal; overflow-wrap: normal; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">?1,f=? and v2,f=0" role="presentation" style="box-sizing: border-box; display: inline; line-height: normal; word-spacing: normal; overflow-wrap: normal; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">?2,f=0 are the initial conditions, so ignore this solution because it would imply the two satellites do not interact at all. Use the first solution to find the final relative speed.

vfinal=v1,f−v2,f=(m1−m2m1+m2)v−2m1m1+m2v=(m1−m2−2m1m1+m2)v=((−1)(m1+m2)m1+m2)v=−v" role="presentation" style="box-sizing: border-box; display: inline; line-height: normal; font-size: 18px; word-spacing: normal; overflow-wrap: normal; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">?final=?1,f−?2,f=(?1−?2?1+?2)?−2?1?1+?2?=(?1−?2−2?1?1+?2)?=((−1)(?1+?2)?1+?2)?=−?$$$$

So, the final relative speed after the collision is just −0.250" role="presentation" style="box-sizing: border-box; display: inline; line-height: normal; word-spacing: normal; overflow-wrap: normal; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">$0.250$ m/s, which means they are separating at the same relative speed at which they approached each other.

It may seem that you did a lot of algebra to arrive at a very simple answer. It is often the case in physics that the amount of math involved depends on how the problem was set up. You may want to think about whether there is a way of setting this problem up that would reduce the amount of work involved in finding the answer. For example, you arbitrarily picked one of the satellites to define the "rest frame" used for analyzing the problem. Would some other choice of rest frame simplify the analysis?

Answer 3

Because the problem only gives the relative initial speed between the two satellites, you are free to choose one of those satellites to be at rest initially. In the figure, assume the satellite on the left has mass m1" role="presentation" style="box-sizing: border-box; display: inline; line-height: normal; word-spacing: normal; overflow-wrap: normal; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">?1$m_1$ and is moving at velocity v" role="presentation" style="box-sizing: border-box; display: inline; line-height: normal; word-spacing: normal; overflow-wrap: normal; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">?$v$ to the right, whereas the satellite on the left has mass m2" role="presentation" style="box-sizing: border-box; display: inline; line-height: normal; word-spacing: normal; overflow-wrap: normal; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">?2$m_2$ and is stationary.

An elastic collision conserves both the total momentum and the total kinetic energy, and therefore

ptot,i=ptot,fm1v1,i+m2v2,i=m1v1,f+m2v2,fm1v+0=m1v1,f+m2v2,fm1v=m1v1,f+m2v2,f" role="presentation" style="box-sizing: border-box; display: inline; line-height: normal; font-size: 18px; word-spacing: normal; overflow-wrap: normal; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">?tot,i?1?1,i+?2?2,i?1?+0?1?=?tot,f=?1?1,f+?2?2,f=?1?1,f+?2?2,f=?1?1,f+?2?2,f$$\begin{array}{rl}{p}_{\text{tot,i}}& =p_{\text{tot,f}}\\ m_1{v}_{1,\text{i}}+m_2{v}_{2,\text{i}}& =m_1{v}_{1,\text{f}}+m_2{v}_{2,\text{f}}\\ m_{1}v+0& =m_1{v}_{1,\text{f}}+m_2{v}_{2,\text{f}}\\ m_{1}v& =m_1{v}_{1,\text{f}}+m_2{v}_{2,\text{f}}\end{array}$$

and

Ktot,i=Ktot,f12m1v1,i2+12m2v2,i2=12m1v1,f2+12m2v2,f2m1v2+0=m1v1,f2+m2v2,f2m1v2=m1v1,f2+m2v2,f2" role="presentation" style="box-sizing: border-box; display: inline; line-height: normal; font-size: 18px; word-spacing: normal; overflow-wrap: normal; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">?tot,i12?1?21,i+12?2?22,i?1?2+0?1?2=?tot,f=12?1?21,f+12?2?22,f=?1?21,f+?2?22,f=?1?21,f+?2?22,f$$\begin{array}{rl}{K}_{\text{tot,i}}& =K_{\text{tot,f}}\\ \frac{1}{2}{m}_1{v}_{1,\text{i}}^2+\frac{1}{2}{m}_2{v}_{2,\text{i}}^2& =\frac{1}{2}{m}_1{v}_{1,\text{f}}^2+\frac{1}{2}{m}_2{v}_{2,\text{f}}^2\\ m_1{v}^2+0& =m_1{v}_{1,\text{f}}^2+m_2{v}_{2,\text{f}}^2\\ m_1{v}^2& =m_1{v}_{1,\text{f}}^2+m_2{v}_{2,\text{f}}^2\end{array}$$

All the variables in the last expressions for ptot" role="presentation" style="box-sizing: border-box; display: inline; line-height: normal; word-spacing: normal; overflow-wrap: normal; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">?tot$p_{\text{tot}}$ and Ktot" role="presentation" style="box-sizing: border-box; display: inline; line-height: normal; word-spacing: normal; overflow-wrap: normal; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">?tot$K_{\text{tot}}$ are known except for the final speeds of the first and second satellite, v1,f" role="presentation" style="box-sizing: border-box; display: inline; line-height: normal; word-spacing: normal; overflow-wrap: normal; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">?1,f$v_{\text{1,f}}$ and v2,f" role="presentation" style="box-sizing: border-box; display: inline; line-height: normal; word-spacing: normal; overflow-wrap: normal; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">?2,f$v_{\text{2,f}}$, respectively. Solve the ptot" role="presentation" style="box-sizing: border-box; display: inline; line-height: normal; word-spacing: normal; overflow-wrap: normal; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">?tot$p_{\text{tot}}$ expression for v1,f" role="presentation" style="box-sizing: border-box; display: inline; line-height: normal; word-spacing: normal; overflow-wrap: normal; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">?1,f$v_{\text{1,f}}$, and then substitute it into the Ktot" role="presentation" style="box-sizing: border-box; display: inline; line-height: normal; word-spacing: normal; overflow-wrap: normal; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">?tot$K_{\text{tot}}$ expression to find v2,f" role="presentation" style="box-sizing: border-box; display: inline; line-height: normal; word-spacing: normal; overflow-wrap: normal; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">?2,f$v_{\text{2,f}}$.

v1,f=m1v−m2v2,fm1" role="presentation" style="box-sizing: border-box; display: inline; line-height: normal; font-size: 18px; word-spacing: normal; overflow-wrap: normal; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">?1,f=?1?−?2?2,f?1$$v_{1,\text{f}}=\frac{m_{1}v-m_2{v}_{2,\text{f}}}{m_1}$$

So,

m1(m1v−m2v2,fm1)2+m2v2,f2=m1v2(m12v2−2m1m2vv2,f+m22v2,f2)+m1m2v2,f2=m12v2(m1m2+m22)v2,f2+(−2m1m2v)v2,f+0=0" role="presentation" style="box-sizing: border-box; display: inline; line-height: normal; font-size: 18px; word-spacing: normal; overflow-wrap: normal; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">?1(?1?−?2?2,f?1)2+?2?22,f(?21?2−2?1?2??2,f+?22?22,f)+?1?2?22,f(?1?2+?22)?22,f+(−2?1?2?)?2,f+0=?1?2=?21?2=0$$\begin{array}{rl}{m}_1{\left(\frac{m_{1}v-m_2{v}_{2,\text{f}}}{m_1}\right)}^2+m_2{v}_{2,\text{f}}^2& =m_1{v}^2\\ (m_1^2{v}^2-2m_1{m}_{2}v{v}_{2,\text{f}}+m_2^2{v}_{2,\text{f}}^2)+m_1{m}_2{v}_{2,\text{f}}^2& =m_1^2{v}^2\\ (m_1{m}_2+m_2^2)v_{2,\text{f}}^2+(-2m_1{m}_{2}v)v_{2,\text{f}}+0& =0\end{array}$$

Using the quadratic formula,

v2,f=2m1m2v+4m12m22v2−4(m1m2+m22)(0)2(m1m2+m22)=2m1m2v+4m12m22v22m2(m1+m2)=4m1m2v2m2(m1+m2)=2m1vm1+m2" role="presentation" style="box-sizing: border-box; display: inline; line-height: normal; font-size: 18px; word-spacing: normal; overflow-wrap: normal; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">?2,f=2?1?2?+4?21?22?2−4(?1?2+?22)(0)⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√2(?1?2+?22)=2?1?2?+4?21?22?2⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√2?2(?1+?2)=4?1?2?2?2(?1+?2)=2?1??1+?2$$\begin{array}{rl}{v}_{2,\text{f}}& =\frac{2m_1{m}_{2}v+\sqrt{4m_1^2{m}_2^2{v}^2-4(m_1{m}_2+m_2^2)\left(0\right)}}{2(m_1{m}_2+m_2^2)}\\ & =\frac{2m_1{m}_{2}v+\sqrt{4m_1^2{m}_2^2{v}^2}}{2m_2(m_1+m_2)}\\ & =\frac{4m_1{m}_{2}v}{2m_2(m_1+m_2)}\\ & =\frac{2m_{1}v}{m_1+m_2}\end{array}$$

or

v2,f=2m1m2v−4m12m22v2−4(m1m2+m22)(0)2(m1m2+m22)=2m1m2v−4m12m22v22m2(m1+m2)=02m2(m1+m2)=0" role="presentation" style="box-sizing: border-box; display: inline; line-height: normal; font-size: 18px; word-spacing: normal; overflow-wrap: normal; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">?2,f=2?1?2?−4?21?22?2−4(?1?2+?22)(0)⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√2(?1?2+?22)=2?1?2?−4?21?22?2⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√2?2(?1+?2)=02?2(?1+?2)=0$$\begin{array}{rl}{v}_{\text{2,f}}& =\frac{2m_1{m}_{2}v-\sqrt{4m_1^2{m}_2^2{v}^2-4(m_1{m}_2+m_2^2)\left(0\right)}}{2(m_1{m}_2+m_2^2)}\\ & =\frac{2m_1{m}_{2}v-\sqrt{4m_1^2{m}_2^2{v}^2}}{2m_2(m_1+m_2)}\\ & =\frac{0}{2m_2(m_1+m_2)}\\ & =0\end{array}$$

Now substitute these possibilities for v2,f" role="presentation" style="box-sizing: border-box; display: inline; line-height: normal; word-spacing: normal; overflow-wrap: normal; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">?2,f$v_{\text{2,f}}$ into the expression for v1,f" role="presentation" style="box-sizing: border-box; display: inline; line-height: normal; word-spacing: normal; overflow-wrap: normal; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">?1,f$v_{\text{1,f}}$.

v1,f=m1v−m2(2m1vm1+m2)m1=(1−2m2(m1+m2))v=(m1+m2−2m2m1+m2)v=(m1−m2m1+m2)v" role="presentation" style="box-sizing: border-box; display: inline; line-height: normal; font-size: 18px; word-spacing: normal; overflow-wrap: normal; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">?1,f=?1?−?2(2?1??1+?2)?1=(1−2?2(?1+?2))?=(?1+?2−2?2?1+?2)?=(?1−?2?1+?2)?$$\begin{array}{rl}{v}_{1,\text{f}}& =\frac{m_{1}v-m_2\left(\frac{2m_{1}v}{m_1+m_2}\right)}{m_1}\\ & =(1-\frac{2m_2}{(m_1+m_2)})v\\ & =\left(\frac{m_1+m_2-2m_2}{m_1+m_2}\right)v\\ & =\left(\frac{m_1-m_2}{m_1+m_2}\right)v\end{array}$$

or

v1,f=m1v−m2(0)m1=v" role="presentation" style="box-sizing: border-box; display: inline; line-height: normal; font-size: 18px; word-spacing: normal; overflow-wrap: normal; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">?1,f=?1?−?2(0)?1=?$$\begin{array}{rl}{v}_{1,\text{f}}& =\frac{m_{1}v-m_2\left(0\right)}{m_1}\\ & =v\end{array}$$

The solution with v1,f=v" role="presentation" style="box-sizing: border-box; display: inline; line-height: normal; word-spacing: normal; overflow-wrap: normal; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">?1,f=?$v_{\text{1,f}}=v$ and v2,f=0" role="presentation" style="box-sizing: border-box; display: inline; line-height: normal; word-spacing: normal; overflow-wrap: normal; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">?2,f=0$v_{\text{2,f}}=0$ are the initial conditions, so ignore this solution because it would imply the two satellites do not interact at all. Use the first solution to find the final relative speed.

vfinal=v1,f−v2,f=(m1−m2m1+m2)v−2m1m1+m2v=(m1−m2−2m1m1+m2)v=((−1)(m1+m2)m1+m2)v=−v" role="presentation" style="box-sizing: border-box; display: inline; line-height: normal; font-size: 18px; word-spacing: normal; overflow-wrap: normal; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">?final=?1,f−?2,f=(?1−?2?1+?2)?−2?1?1+?2?=(?1−?2−2?1?1+?2)?=((−1)(?1+?2)?1+?2)?=−?$$\begin{array}{rl}{v}_{\text{final}}& =v_{1,\text{f}}-v_{2,\text{f}}\\ & =\left(\frac{m_1-m_2}{m_1+m_2}\right)v-\frac{2m_1}{m_1+m_2}v\\ & =\left(\frac{m_1-m_2-2m_1}{m_1+m_2}\right)v\\ & =\left(\frac{(-1)(m_1+m_2)}{m_1+m_2}\right)v\\ & =-v\end{array}$$

So, the final relative speed after the collision is just −0.250" role="presentation" style="box-sizing: border-box; display: inline; line-height: normal; word-spacing: normal; overflow-wrap: normal; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">0.250$0.250$ m/s, which means they are separating at the same relative speed at which they approached each other.

It may seem that you did a lot of algebra to arrive at a very simple answer. It is often the case in physics that the amount of math involved depends on how the problem was set up. You may want to think about whether there is a way of setting this problem up that would reduce the amount of work involved in finding the answer. For example, you arbitrarily picked one of the satellites to define the "rest frame" used for analyzing the problem. Would some other choice of rest frame simplify the analysis?