Answer:
<em>a) 0.22 m/s</em>
<em>b) 531.67 N</em>
<em>c) 508.81 J</em>
<em>d) 38.72 J</em>
Explanation:
the mass of the astronaut = 121 kg
astronaut's push of speed = 2.9 m/s
mass of the space capsule = 1600 kg
a) according to the conservation of momentum, the summation of the total momentum in a system must be equal to zero.
let us take the direction of the astronaut as positive.
Astronaut's momentum p = mv
where
m is the mass
v is the velocity
momentum p = 121 x 2.9 = 350.9 kg-m/s
The space capsules momentum = mv
==> 1600 x (-v) = -1600v this is because the space capsule moves in the opposite direction to the astronaut.
according to conservation of momentum
350.9 + (-1600v) = 0
350.9 = 1600v
v = 350.9/1600 =<em> 0.22 m/s</em>
b) magnitude of the force F is the rate of change of momentum.
The astronaut and the space capsule both change momentum from 0 to 350.9 kg-m/s. In 0.66 seconds, the force will be
F =
where
u is their initial velocity = 0 m/s
where v = 2.9
t = 0.66
substituting, we have
F = = 350.9/0.66 = <em>531.67 N this same force is experienced by the space capsule</em>
c) Kinetic energy of the astronaut =
m is the mass = 121 kg
v is the velocity = 2.9 m/s
KE = = <em>508.81 J</em>
d) Kinetic energy of the space capsule =
KE = = <em>38.72 J</em>