A 62.9 kg spacewalking astronaut pushes off a 697.0 kg satellite, exerting a 117.0 N force for the 0.977 s it takes him to straighten his arms. How far apart are the astronaut and the satellite after 3.57 min?

Answer 1

m1=62.9 kg
m2=697 kg
F=117 N
t=0.977 s.
v1 = velocity of astronaut after contact
v2 = velocity of satellite after contact

By the conservation of momentum,
m1 v1 + m2 v2 =0
v2=-(m1 v1)/m2 ....(1)

By the conservation of energies,
F*t = (m1/2)v1² + (m2/2)v2² ....(2)
eliminate v2 by substitution of (1) in (2):
F*t = (m1/2)v1² + (m2/2)(m1*v1/m2)²
Solve for v1=1.8 m/s
substitute in (1) to get v2=0.2 m/s.

From v1 and v2 calculate distance after specified time.

The two velocities are in opposite directions, so relative velocity is the sum of absolute values.