There are other forces at work here nevertheless we will imagine it is just a conservation of momentum exercise. Also the given mass of the astronaut is light astronaut.
The solution for this problem is using the formula: m1V1=m2V2 but
we need to get V1:
V1= (m2/m1) V2
V1= (10/63) 12 = 1.9 m/s will be the final speed of the astronaut after
throwing the tank.
In outer space no other force like gravity and friction are nevertheless. The astronaut's final speed with respect to the shuttle after the tank is thrown is 1.9 m/s.
In outer space no other force like gravity and friction is nevertheless. Here only conversion of momentum is occurring.
The formula,
[tex]\rm \bold {m_1 V_1 = m_2 V_2}[/tex]
Where,
[tex]\rm \bold { m_1}[/tex] - mass first object = 63.0 kg
[tex]\rm \bold { m_2}[/tex] - mass ( second object) = 10.0 kg
[tex]\rm \bold { V_1}[/tex] - Velocity of the first object = ?
[tex]\rm \bold { V_2}[/tex] - Velocity of the second object = 12.0 m/s
Put the value in the formula,
[tex]\rm \bold { 63 \times V_1 = 12 \times 10}\\\\\rm \bold { V_2 = 1.9 m/s}[/tex]
Hence we can conclude that the astronaut's final speed with respect to the shuttle after the tank is thrown is 1.9 m/s.
To know more about relative speed, refer to the link:
https://brainly.com/question/13430965
