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?

1 Answer
A08
Mar 11, 2017

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

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