Question

Two manned satellites approaching one another at a relative speed of 0.550 m/s intend to dock. The first has a mass of 2.50 ✕ 103 kg, and the second a mass of 7.50 ✕ 103kg. If the two satellites collide elastically rather than dock, what is their final relative velocity? Adopt the reference frame in which the second satellite is initially at rest and assume that the positive direction is directed from the second satellite towards the first satellite.


...............m/s

Answer 1

Answer: Their final relative velocity is -0.412 m/s.

Explanation:

According to the law of conservation,

      $m_{1}v_{1} + m_{2}v_{2} = (m_{1} + m_{2})v$

Putting the given values into the above formula as follows.

      $m_{1}v_{1} + m_{2}v_{2} = (m_{1} + m_{2})v$

     $2.50 \times 10^{3} kg \times 0 m/s + 7.50 \times 10^{3} kg \times -0.550 m/s = (2.50 \times 10^{3} kg + 7.50 \times 10^{3} kg)v$

           $-4.12 \times 10^{3} kg m/s = (10^{4} kg) v$

                   v = $\frac{-4.12 \times 10^{3} kg m/s}{10^{4} kg}$

                      = -0.412 m/s

Thus, we can conclude that their final relative velocity is -0.412 m/s.