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
Mar 11, 2017

Answer:

#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#