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Two astronauts, A and B, both with mass of 60 kg, 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 and B catches it. Find the minimum value of v.
(1) 7.8 m/s
(2) 26.0 m/s
(3) 14.0 m/s
(4) 9.2 m/s

Respuesta :

okpalawalter8

Answer:

[tex]v = 14m/s[/tex]

Explanation:

From the question we are told that

Mass of both astronauts [tex]M= 60kg[/tex]

initial Speed of astronaut [tex]A=2m/s[/tex]

initial Speed of astronaut [tex]B=1m/s[/tex]

Mass of bag [tex]M_b =5kg[/tex]

Generally the initial momentum of astronaut  A+bag is mathematically given

as

[tex]A+bag =(60+5)*2[/tex]

[tex]A+bag =130kgm/s[/tex]

Generally  for collision to be avoided astronauts have to move with steady pace of 1m/s or less,Because astronauts A final speed should be equal or less to that of B's.

Therefore

[tex]130 -(60*1)=(5*v)[/tex]

[tex]v=\frac{70}{14}[/tex]

[tex]v = 14m/s[/tex]

[tex]v = 14m/s[/tex] is the 3rd option making option 3(14m/s)correct.

 

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