Answer
(a) The speed of the shuttle in the minus x direction is 0.0028 m/s.
(b) The shuttle exerts an average force of 53 N on the satellite.
Work Step by Step
(a) We can use conservation of momentum to find the speed $v_1$ of the shuttle (in the minus x direction) after ejecting the shuttle:
$m_1~v_1 = m_2~v_2$
$v_1 = \frac{m_2~v_2}{m_1} =\frac{(850~kg)(0.30~m/s)}{92,000~kg}$
$v_1 = 0.0028~m/s$
The speed of the shuttle in the minus x direction is 0.0028 m/s.
(b) $Impulse = \Delta p$
$F\cdot t = m \Delta v$
$F = \frac{m \Delta v}{t} = \frac{(850~kg)(0.30~m/s)}{4.8~s}$
$F = 53~N$
The shuttle exerts an average force of 53 N on the satellite.
