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

Answer:

$T=22.5sec$

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

 $T=\frac{d}{V}$

 $T=\frac{18}{0.8}$

 $T=22.5sec$