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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.

Sagot :

okpalawalter8

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

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