Answer
After separation, the speed of the payload is $~~8200~m/s$
Work Step by Step
After separation, we can let $v$ be the speed of the rocket case. Then, since the payload capsule is at the front, the speed of the payload capsule must be $v+910.0~m/s$
We can use conservation of momentum to find the speed of the rocket case after separation:
$p_f = p_i$
$(290.0~kg)(v)+(150.0~kg)(v+910.0~m/s) = (290.0~kg+150.0~kg)(7600~m/s)$
$(290.0~kg)(v)+(150.0~kg)(v) = (440.0~kg)(7600~m/s)-(150.0~kg)(910.0~m/s)$
$v = \frac{(440.0~kg)(7600~m/s)-(150.0~kg)(910.0~m/s)}{290.0~kg+150.0~kg}$
$v = 7290~m/s$
After separation, the speed of the payload is $(7290~m/s+910.0~m/s)$ which is $~~8200~m/s$.
