Question

A space vehicle approaches a space station in orbit. The intent of the engineers is to have the vehicle slowly approach, reducing velocity, until a docking maneuver is completed and the vehicle is attached to the station. How does the total momentum of the docked vehicle and station compare to the momentum of each object before the docking maneuver?

Answer 1

Answer: The total momentum before the docking maneuver is $mV_{1}+MV_{2}$ and after the docking maneuver is $(m+M) U$

Explanation:

Linear momentum $p$ (generally just called momentum) is defined as mass in motion and is given by the following equation:  

$p=m.v$  

Where $m$ is the mass of the object and $v$ its velocity.

According to the conservation of momentum law:

"If two objects or bodies are in a closed system and both collide, the total momentum of these two objects before the collision $p_{i}$ will be the same as the total momentum of these same two objects after the collision $p_{f}$".

$p_{i}=p_{f}$

This means, that although the momentum of each object may change after the collision, the total momentum of the system does not change.

Now, the docking of a space vehicle with the space station is an inelastic collision, which means both objects remain together after the collision.

Hence, the initial momentum is:

$p_{i}=mV_{1}+MV_{2}$

Where:

$m$ is the mass of the vehicle

$V_{1}$ is the velocity of th vehicle

$M$ is the mass of the space station

$V_{2}$ is the velocity of the space station

And the final momentum is:

$p_{f}=(m+M)U$

Where:

$U$ is the velocity of the vehicle and space station docked

Answer 2

Answer: same

Explanation: