Question

A 105 kg astronaut carrying a 16 kg tool bag finds himself separated from his spaceship by 18 m and moving away from the spaceship at 0.1 m/s. To get back to the spaceship, he throws the tool bag away from the spaceship at 4.5 m/s (relative to the station). How long (in s) will he take to return to the spaceship

Answer 1

`T=22.5sec`

Step-by-step explanation:

From the question we are told that:

Mass of astronaut
`m_a=105kg`

Mass of tool
`m_t=16kg`

Distance
`d=18m`

Velocity of separation
`v_s= 0.1m/s`

Velocity of tool bag
`v_t=4.5m/s`

Generally the equation for momentum is mathematically given by

`P=mv`

Therefore

Initial Momentum before drop

`P_1=0.1(105+16)`

`P_1=12.1`

Initial Momentum after drop

`P_2=-16(4.5)+105V`

Therefore

Since
`P_1=P_2`

`-72+105V=12.1`

`V=0.8m/s`

Generally the equation for Time T is mathematically given by

`T=(d)/(V)`

`T=(18)/(0.8)`

`T=22.5sec`