Question

What is the change in velocity of the international space station?

During a docking maneuver, a supply pod of 4000kg accidentally collides with the international space station of mass 100000kg with a relative velocity of `2ms^-1`. Assuming the pod attaches to the station, what is the change in velocity of the international space station?

Answer 1

`0.077ms^-1`, rounded to two decimal places.

Explanation:

Let `u` be the velocity of International space station before accident.

Therefore, velocity of supply pod `=(u+2) ms^-1`

Initial momentum of both bodies `=100000xxu+4000xx(u+2)`
`=10^5u+4xx10^3u+8xx10^3`

After the accident let `v` be velocity of the combined mass
Final momentum`=(10^5+4xx10^3)v`

From Law of conservation of momentum we have

`(10^5+4xx10^3)u+8xx10^3=(10^5+4xx10^3)v`
`=>(10^5+4xx10^3)v-(10^5+4xx10^3)u=8xx10^3`
`=>(10^5+4xx10^3)(v-u)=8xx10^3`
`=>(v-u)=(8xx10^3)/((10^5+4xx10^3))=8/104=1/13=0.077ms^-1`