Answer:
[tex]V=14m/s[/tex]
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
[tex]M_A=(60+5)*2[/tex]
[tex]M_A=130kgm/s[/tex]
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
[tex]130=(60*1)=(5*v)[/tex]
[tex]V=14m/s[/tex]
To avoid collision with B, A throws the bag with a speed of 26 m/sec.
Given-
The mass of the two astronauts A and B is 60 kg.
The mass of astronaut A is 65 kg with the box.
The speed of astronaut A is 2 m/sec.
The speed of astronaut B is 1 m/sec.
The mass of the bag is 5 kg.
The momentum of a body is the product of its mass and velocity. Thus the momentum of astronaut A is,
[tex]M_{A} =m_{a} \times v_{a}[/tex]
[tex]M_{A} =(60+5)\times2[/tex]
[tex]M_{A} =130[/tex]
To avoid the collision with B, A throws the bag with a speed v. For this the momentum after the trough must be equal to the momentum of A. Therefore,
[tex]M_{A} =5\times v_{r}[/tex]
[tex]130 =5\times v_{r}[/tex]
[tex]v_{r}= \dfrac{130}{5}[/tex]
[tex]v_{r}=26[/tex]
Hence, to avoid collision with B, A throws the bag with a speed of 26 m/sec.
For more about the velocity, find the link below-
https://brainly.com/question/862972
