Question

Two astronauts, A and B, both with mass of 60Kg, are moving along a straight line in the same direction in a weightless spaceship. Relative to the spaceship the speed of A is 2 m/s and that of B is 1 m/s. A is carrying a bag of mass 5 Kg with him. To avoid collision with B A throws the bag with a speed v relative to the spaceship towards B catches it. Find the minimum value of v.

Answer 1

Answer:

$V=14m/s$

Explanation:

From the Question we are told that

Mass of A and B is 60kg

Speed of A=2m/s

Speed of B=1m/s

Mass of bag =5kg

Generally the momentum of the astronaut  A and bag is mathematically given as

  $M_A=(60+5)*2$

   $M_A=130kgm/s$

Generally to avoid collision the speed of astronaut be should be less than or equal to that of astronaut A with the bag

Therefore for minimum requirement speed of astronaut A should be given by astronaut B's speed which is equal to 1

Therefore

  $130=(60*1)=(5*v)$

   $V=14m/s$